text:
\begin{document}
\maketitle

\selectlanguage{english}

\begin{abstract}
A classical result due to Deshouillers, Dress and Tenenbaum asserts that on average the distribution of the divisors of the integers follows  the arcsine law.
In this paper, we investigate the distribution of  smooth divisors of the integers,  that is, those  divisors  which are free of large prime factors.
We show that on average these divisors are distributed according to a probability law that we will describe.
\end{abstract}
\section{Introduction}
Let $n\geqslant1$ be an interger. We denote by $P(n)$  the largest prime divisor of $n\geqslant 2$ and we set $P(1)=1$. Let  $y\in ]1,+\infty[$ be a real number. Consider the set of   $y$-smooth divisors of $n$, that is, those  divisors of  $n$ which are  free of prime factors  exeeding $y$.
 $$\mathcal{D}_{n,y}:=\big\{d| n : P(d)\leqslant y\big\},$$
 and denote by $\tau(n,y)$ its cardinality. For each integer $n\geqslant1$ and for each real number $y>1$, we define the random variable $$X_{n,y}:\mathcal{D}_{n,y}\longrightarrow [0,1],$$
 which takes the values $\log d / \log n$ with uniform probability  $1/ \tau(n,y)$ and  for  $v\in[0,1]$, its distribution function
 $$F_{n,y}(v):=\mathbb{P}(X_{n,y}\leqslant v)=\frac{1}{\tau(n,y)}\,\sum_{d|n, d\leqslant n^{v},P(n)\leqslant y} 1.$$
It is easy to see that the sequence $(F_{n,y})_{n\geqslant1}$ does not converge pointwise  in $[0,1]$. We consider its mean in the interval $[0,1]$
 \begin{eqnarray}
 \frac{1}{x}\sum_{n\leqslant x}F_{n,y}(v)=\frac{1}{x}\sum_{n\leqslant x}\mathbb{P}(X_{n,y}\leqslant v)\cdot
 \end{eqnarray}
 The aim of this  paper is to show that this mean  converges to a distribution function which will be  described.
Deshouillers, Dress  and Tenenbaum   [4]  studied of the analogue of this mean by considering  the set of all   divisors of  $n$, that is without  constraint on their prime factors. By denoting here by  $X_{n}$ the analogue of the random variable $X_{n,y}$ defined on the set of  all divisors of $n$, they showed that
 \begin{eqnarray}\frac{1}{x}\sum_{n\leqslant x}\mathbb{P}(X_{n}\leqslant v) =\frac{2}{\pi}\,\arcsin(\sqrt{v})+O\left( \frac{1}{\sqrt{\log x}}\right),
 \end{eqnarray}
 uniformely for $v\in[0,1]$. This arcsin law is a Dirichlet  law in one dimension  with parameters equal $(1/2,1/2)$. Studying Couples of divisors, the authors of the present paper [7]  showed that they are distributed according  a two dimensional Dirichet Law . The method works in higher  dimension but becomes very  technique. de la Bret\`eche et Tenenbaum [2] studied  also  couples of divisors by using a probabilistic model that preserves the equiprobability of first marginal law and allows them to deduce the second marginal law. They also obtained a  Dirichlet  law.\\
Recently, Basquin [1]  studied an analogue  of this question of law of divisors by considering the set of divisors of smooth integers $n$. This question is naturally to the  de Bruijn function:
$$\Psi(x,y):= \sharp\{n\leqslant x: P(n)\leqslant y\}\cdot$$
The asymptotic behavior of de Bruijn's function is  known in a large range of  $xy$-plane. It is conneted to Dickman's function  $\rho$, which is the continuous solution in  $]0,+\infty[$  to the differential-difference  equation with inital an condition:
$$
\left
\lbrace
\begin{array}{lcl}
 u\rho^{\prime}(w)+\rho(w-1)=0,\quad (w>1)\\
 \rho(w)=1,\qquad \qquad (0\leqslant w \leqslant 1)\\
 \rho(w)=0, \quad (w<0),
\end{array}
\right.
$$
which the asymptotic behavior is well known. For example, we have
$$\log \rho (w)=-(1+0(1))w\log w, \quad (w\rightarrow +\infty).$$
Before quoting Basquin's result  and  formulate the behavior of  $\Psi(x,y)$, let us  introduce some notations that will be maintained  throughout the rest this paper. For $1< y \leqslant x$, we set
$$
u:=\frac{\log x}{\log y},
$$
and we denote by $(H_{\epsilon})$ the subset of $\mathbb{R}^{2}$ defined by the condition
$$x\geqslant x_{0}(\epsilon), \quad \exp\left((\log\log x)^{\frac{5}{3} +\epsilon}\right)\leqslant y\leqslant x,$$
where $ x_{0}(\epsilon)>0$ is a suffisantely large constant depending of  $\epsilon>0$.
Here it is  sufficient to quote the following  asymptotic formula for $\Psi(x,y)$ due to Hildebrand [6] and valid in  the range $(H_{\epsilon})$
$$\Psi(x,y)= x \rho(u)\left( 1+O\left(\frac{\log (u+1)}{\log y}\right) \right)\cdot$$
Let us introduce  functions  $\rho_{k}$  for  $k\in ]0,+\infty[$. Each function  $\rho_{k}$  is the continuous solution to the differential-difference equation with initial condition :
$$\left
\lbrace
\begin{array}{lcl}
w\rho'_{k}(w)+(1-k)\rho_{k}(w)+k\rho_{k}(w-1)=0, \quad  (w>1)\\
\rho_{k}(w)=\frac{1}{\Gamma(k)}\,w^{k-1},\quad (0<w \leqslant 1)\\
\rho_{k}(w)=0,\quad (w\leqslant 0).
\end{array}
\right.
$$
In particular, we have  $\rho_{1}=\rho$. Function  $\rho_{k}$ is the k-th fractionnal convolution  power of $\rho$ -- see Hensley's work [5].
Its asymptotic behavior  of $\rho_{k}$ is well known -- see in particular Smida's papers  [8],[9],  where we find  properties of these functions and their connection to the asymptotic behavior  of Dickman's function $\rho$. In particular, we have  the formula [8],
$$\rho_{k}(u)=k^{u(1+O(\frac{1}{\log u})}\rho(u),\quad (u\rightarrow +\infty).$$
Basquin showed that
$$\frac{1}{\Psi(x,y)}\,\sum_{n\leqslant x,P(n)\leqslant y} \mathbb{P}(X_{n}\leqslant v)=\frac{1}{\rho(u)}\,\int_{0}^{uv} \rho_{\frac{1}{2}}(s)\rho_{\frac{1}{2}}(u-s)\,ds +O\left(   \frac{\log (u+1)}{\log y}+\frac{1}{\sqrt{\log y}}       \right),$$
uniformely for  $v\in[0,1]$ and  $(x,y)\in (H_{\epsilon})$,
and he deduced  that as   $u\longrightarrow +\infty$, the distribution function converges to  the  normal distribution. More precisely, he showed that
$$\frac{1}{\rho(u)}\,\int_{0}^{uv} \rho_{\frac{1}{2}}(s)\rho_{\frac{1}{2}}(u-s)\,ds=\Phi\left(u\sqrt{2\xi'(u)}(v-\frac{1}{2})\right)+O\left(\frac{1}{u}\right),$$
where  $\xi'(u)\sim 1/u $, as $u\rightarrow +\infty$ and
$$\Phi(w)=\frac{1}{\sqrt{\pi}} \int_{-\infty}^{w} e^{-t^{2}}\,dt,$$
is the normal distribution function.
\section{Statements of results}
To state our results, let us introduce the Buchstab function  $\omega$. This function was discovered by Bushstab  and comes from the study of the uncancelled elements in the sieve of Eratosthenes -- see de Bruijn's  beautiful paper [3]. It is the unique continuous solution for  $v>1$ to the  differential-difference equation, with initial condition:
$$
\left \lbrace
\begin{array}{lcl}
v\omega'(v)+\omega(v) -\omega(v-1)=0,\quad (v>2)\\
\omega(v)=\frac{1}{v},\quad (1\leqslant v\leqslant 2)\\
\omega(v)=0,\quad v<1.
\end{array}
\right.
$$
Its asymptotic behavior is known -- see de Bruijn [3] and Tenenbaum's book  [10 ] chap. III.6). In particular for $v\geqslant0$, we have
\begin{eqnarray}
\omega(v)=e^{-\gamma}+O\left(\rho(v)e^{\frac{-c\,v}{\log^{2}(v+2)}}\right),
\end{eqnarray}
where  $\gamma$ is the Euler function and $c$ a positive constant.
In the first theorem below we show the convergence  of the mean of distribution functions  (1) to a distribution function,
in the second one we describe the limit law as $u\longrightarrow  +\infty$ and in the third one, we give as an example  expressions of  the limit law for
$1\leqslant u \leqslant  2$.
\begin{Theo}
Uniformly for $v\in[0,1]$ and
 $(x, y)$ in $(H_{\epsilon})$, we have
 $$
\begin{array}{lcl}
\frac{1}{x}\sum_{n\leqslant x}\mathbb{P}(X_{n,y}\leqslant v)&=&\int_{0}^{uv} \left(\int_{0}^{u-s-1}\rho_{\frac{1}{2}}(z)\omega(u-s-z )\,dz\right)\rho_{\frac{1}{2}}(s)\,ds \\[0.5cm]
&+&\int_{0}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds\\[0.5cm]
&+&O\left( \frac{\log (u+1)}{\log y}+\frac{1}{\sqrt{\log y}}\right)\cdot
\end{array}
$$
\end{Theo}
We notice that for $y=x$, that is to say $u=1$, the  formula of Theorem 2.1  is reduced to formula (2) obtained  in [4]. Indeed,
the first integral vanishes  because  $\rho_{\frac{1}{2}}(z)=0$ for $z\leqslant 0$ and
$$\int_{0}^{v} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(1-s)\,ds=\frac{2}{\pi}\int_{0}^{\sqrt{v}}\,\frac{dt}{\sqrt{1-t^{2}}}=\frac{2}{\pi}\arcsin(\sqrt{v}).$$
Let us denote
$$F(u,v)=\int_{0}^{uv} \left(\int_{0}^{u-s-1}\rho_{\frac{1}{2}}(z)\omega(u-s-z )\,dz\right)\rho_{\frac{1}{2}}(s)\,ds+\int_{0}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds.$$
We have
\begin{Theo}
 For  $v\in[0,1]$ and  as $u\rightarrow +\infty$, we uniformly have
 $$F(u,v)=\frac{1}{\sqrt{e^{\gamma}}}  \int_{0}^{uv}\rho_{\frac{1}{2}}(s)ds+O\left(\rho_{2}(u) \right),$$
 where $\gamma$ is the  d'Euler constant.
\end{Theo}
\begin{Theo}
\begin{enumerate}
\item
For $v\in[0,\frac{u-1}{u}]$ and $1<u\leqslant 2$, we have
$$
 F(u,v)=\frac{2}{\pi}\arcsin(\sqrt{v})+\frac{1}{\pi}\left(\;\log(u)+\log(1-v)\;\right)\arcsin\left(  \sqrt{\frac{uv}{u-1}} \,  \right)-\frac{1}{2}\log(1-v).
 $$
 \item
 For $v\in[\frac{u-1}{u},\frac{1}{u}]$ and $1<u\leqslant 2$, we have
 $$F(u,v)=\frac{2}{\pi}\arcsin(\sqrt{v})+\frac{1}{2}\log u.$$
 \item
 For $v\in[\frac{1}{u},1]$ and $1<u\leqslant 2$, we have
 $$F(u,v)=\frac{2}{\pi}\arcsin(\sqrt{v})+\frac{1}{\pi}\left(\;\log(u)+\log(v)\;\right)\arcsin\left(  \sqrt{\frac{u(1-v)}{u-1}} \,  \right)-\frac{1}{2}\log(v)\cdot$$
 \end{enumerate}
\end{Theo}
Let us set
$$\begin{array}{lcl}
S(x,y,v)&:=&\sum_{n\leqslant x}\mathbb{P}(X_{n,y}\leqslant v)=\sum_{n\leqslant x}\frac{1}{\tau(n,y)}\,\sum_{d|n, d\leqslant n^{v},P(d)\leqslant y} 1\\[0.5cm]
&=& S_{1}(x,y,v)-S_{2}(x,y,v),
\end{array}
$$
with
$$S_{1}(x,y,v):= \sum_{n\leqslant x}\frac{1}{\tau(n,y)}\,\sum_{\stackrel{d|n, d\leqslant x^{v}}{P(d)\leqslant y}}1=\sum_{d\leqslant x^{v}, P(d)\leqslant y}\,\sum_{m\leqslant x/d}\frac{1}{\tau(dm,y)}, $$
and
$$S_{2}(x,y,v):= \sum_{n\leqslant x}\frac{1}{\tau(n,y)}\,\sum_{\stackrel{d|n, n^{v}<d\leqslant x^{v}}{P(d)\leqslant y}}1.$$
We will show that the main contribution to Theorem 2.1 comes from the estimation of  $S_{1}(x,y,v)$. The proof rests on the estimation
$$\sum_{n\leqslant x}\frac{1}{\tau(dn,y)}\; ,$$
for $(x,y)$ in $(H_{\epsilon})$ and $d\geqslant 1$, $y$-smooth. To study  this quantity, we used the factorisation  $n=ab$ with  $a$ is $y$-smooth and  $b$  has all its prime factors greater than $y$. This allowed us to use results on this topic  available in the literature.
\section{Proof of  Theorem 2.1 }
\subsection{Preparatory Lemmas}
Let us introduce some notations which will be used in the sequel. For each fixed integer $d\geqslant 1$ we define a multiplicative function
$$\gamma_{d}(n)=\frac{\tau(d)}{\tau(dn)},$$
where $\tau$ is the divisor function.
For each prime number $p$, we denote  by $v_{p}(d)$ the p-adic valuation of $d$. We have
 $$\gamma_{d}(p^{\alpha})=\frac{v_{p}(d)+1}{v_{p}(d)+\alpha+1}\cdot$$
 we consider  the Dirichlet series of $\gamma_{d}(n)$
$$F_{d}(s):=\sum_{n\geqslant1}\frac{\gamma_{d}(n)}{n^{s}}, \quad ( \Re (s)>1).$$ We have
  $F_{d}(s)=\zeta^{\frac{1}{2}}(s)G_{d}(s)$ the half-plane  $\Re (s)>1$,
with $\zeta$ is the Riemann zeta function and
 $$
 \begin{array}{lcl}
 G_{d}(s)&=&\prod_{p}\left( 1-\frac{1}{p^{s}}\right)^{\frac{1}{2}}\left( \sum_{\alpha\geqslant 0}\frac{v_{p}(d)+1}{(v_{p}(d)+\alpha+1)p^{\alpha s}}\right) \\[0.5cm]
 &=& B(s) K_{d}(s),
 \end{array}
 $$
 with
  $$B(s):=\prod_{p}\left(1-\frac{1}{p^{s}}\right)^{\frac{1}{2}}\left(1+\sum_{\alpha \geqslant 1}\frac{1}{(\alpha+1)p^{\alpha s}}\right),$$
 and
 $$K_{d}(s):=\prod_{p^{\beta}\| d}\left(1+ \sum_{\alpha\geqslant 1}\frac{\beta+1}{(\beta+\alpha+1)p^{\alpha s}}\right)\left(1+\sum_{\alpha\geqslant 1}\frac{1}{(\alpha+1)p^{\alpha s}}\right)^{-1}\cdot$$
 For each fixed integer $d\geqslant1$, we define Dirichlet series
$$K_{d}(s):=\sum_{n\geqslant 1}\frac{\delta_{d}(n)}{n^{s}}; \quad B(s)=\sum_{n\geqslant 1}\frac{b(n)}{n^{s}};\quad G_{d}(s)=\sum_{n\geqslant 1}\frac{h_{d}(n)}{n^{s}}\cdot$$
Then we have
\begin{eqnarray}
F_{d}(s)=\sum_{n\geqslant1}\frac{\gamma_{d}(n)}{n^{s}}=\zeta^{\frac{1}{2}}(s)\sum_{n\geqslant 1}\frac{h_{d}(n)}{n^{s}}
\end{eqnarray}
and
\begin{eqnarray}
\sum_{n\geqslant 1}\frac{h_{d}(n)}{n^{s}}=\left( \sum_{n\geqslant 1}\frac{b(n)}{n^{s}}\right)\left(\sum_{n\geqslant 1}\frac{\delta_{d}(n)}{n^{s}}\right)\cdot
\end{eqnarray}
\begin{Lem}
Let $d\geqslant1$ be a fixed integer. It exists a real number  $0<\eta< 1/3$ such that the series
$$\sum_{n\geqslant 1}\frac{h_{d}(n)}{n^{s}}$$
is absolutely  convergent in the half-plane $\Re(s)=\sigma\geqslant1-\eta$ and for $d\geqslant1$  we uniformly have
$$\sum_{n\geqslant 1}\frac{h_{d}(n)}{n^{\sigma}}\ll_{\eta} \prod_{p|d}(1+\frac{2}{p^{\sigma}})\cdot$$
\end{Lem}
The  result of this lemma can be deduced from from a general study developped in dans [4],  page 275. Indeed, we apply  lemme 1 of [4], page 276, we get that the series
$$\sum_{n\geqslant 1}\frac{\delta_{d}(n)}{n^{s}}=K_{d}(s)$$is absolutely convergent in the half-plane   $\Re(s)=\sigma\geqslant1-\eta$  and we have
\begin{eqnarray}\sum_{n\geqslant 1}\frac{|\delta_{d}(n)|}{n^{\sigma}}\ll_{\eta} \prod_{p|d}(1+\frac{2}{p^{\sigma}})\cdot
\end{eqnarray}
The  lemme 2 of  [4], page 278, applies to the series
$$\sum_{n\geqslant 1}\frac{b(n)}{n^{s}}=B(s)$$
with  the exponent  $\alpha=1/2$ and $\psi(n)=1/\tau(n)$. We obain
 \begin{eqnarray}\sum_{n\geqslant 1}\frac{|b(n)|}{n^{\sigma}}\leqslant \prod_{p}( 1+ O(\frac{1}{p^{2-2\eta}}))\ll_{\eta}1.
 \end{eqnarray}
 in the half-plane $\Re(s)=\sigma\geqslant1-\eta$. Lemma 3.1 follows from  $(5)$, $(6)$ and $(7)$.
 \hfill{
$\square$}\\[0.5cm]
We set
 $$M_{\eta,d}:=\prod_{p|d}(1+\frac{2}{p^{1-\eta}})\cdot$$
From the formula $(4)$ we define a mutiplicative function  $h_{d}$ by the convolution identity $\gamma_{d}=\tau_{\frac{1}{2}}\ast h_{d}$.
From lemma 3.1 we get that its Dirichlet  series satisfies the conditions (1.18) of  Th\'eor\`eme 3 of  [9], page 25,  since we have
$$\sum_{\stackrel{n>t}{P(d)\leqslant y}}\frac{|h_{d}(n)|}{n}\leqslant\frac{1}{t^{\eta/2}}\sum_{\stackrel{n>t}{P(d)\leqslant y}}\frac{|h_{d}(n)|}{n^{1-(\eta)/2})}\leqslant\frac{1}{t^{\eta/2}}\sum_{n\geqslant1}\frac{|h_{d}(n)|}{n^{1-(\eta/2)}}\ll_{\eta} \frac{M_{\eta,d}}{t^{\eta/2}}\cdot$$
The proof of Théorème 3 of [9]  paragraph 6 page 45  works  and we obtain the first result of  lemma 3.2 below. The second result is a consequence of   Théorème T of [4]  applied to the function  $1/\tau(n)$, and by noting that for $0<u\leqslant 1$
 \begin{eqnarray}
 \rho_{\frac{1}{2}}(u)=\frac{1}{\sqrt{\pi} \sqrt{u}}= \frac{\sqrt{\log y}}{\sqrt{\pi}\,\sqrt{\log x}}\cdot
 \end{eqnarray}
  \begin{Lem}
 \begin{enumerate}
 \item
  let be $\eta\in]0,\frac{1}{2}[$ and let  $\epsilon>0$  be fixed . For $d\geqslant1$ and $(x,y) \in (H_{\epsilon})$ we uniformly have
$$\sum_{\stackrel{n\leqslant x}{P(n)\leqslant y}}\gamma_{d}(n)=\frac{x}{\sqrt{\log y} }\;\rho_{\frac{1}{2}}(u)\left( G_{d}(1)+O\left(M_{\eta,d}\,\left(\frac{\log (u+1)}{\log y}+\frac{1}{\sqrt{ \log y}}\right)\right) \right)\cdot$$

\item
For  $1<x\leqslant y$ and $d\geqslant1$, we uniformly have
$$\sum_{\stackrel{n\leqslant x}{P(n)\leqslant y}}\gamma_{d}(n)=\frac{x}{\sqrt{\log y} }\;\rho_{\frac{1}{2}}(u)\left( G_{d}(1)+O\left(\frac{M_{\eta,d}}{\log x}\right) \right)\cdot$$
\end{enumerate}
\end{Lem}
  Basquin [2] obtained the first result of this lemma by using another  convolution identity and by applying a general result of  Tenenbaum et Wu [11]. We set
$$g(d):=\frac{K_{d}(1)}{\tau(d)}=\prod_{p^{\beta}\| d}\left(\sum_{\alpha\geqslant 0}\frac{1}{(\beta+\alpha+1)p^{\alpha }}\right)\left(\sum_{\alpha\geqslant 0}\frac{1}{(\alpha+1)p^{\alpha }}\right)^{-1}\cdot$$
$g$ is a multiplicative function and for  $\Re(s)>1$, we have
$$\sum_{n\geqslant 1}\frac{g(n)}{n^{s}}=\zeta^{\frac{1}{2}}(s)\,\sum_{n\geqslant 1}\frac{\beta(n)}{n^{s}},$$
where $\beta$ is a multiplicative function satisfying $g=\tau_{\frac{1}{2}}\ast \beta$. We have
$$H(s):=\sum_{n\geqslant 1}\frac{\beta(n)}{n^{s}}=\prod_{p}\left(1-\frac{1}{p^{s}}\right)^{\frac{1}{2}}\left( \sum_{\alpha\geqslant0} \frac{g(p^{\alpha})}{p^{\alpha s}} \right)\cdot$$
Let us note for later use that
\begin{eqnarray}
G_{d}(1)=K_{d}(1)B(1)=\tau(d) g(d) B(1).
\end{eqnarray}
 Lemme 2 of  [4], page 278, applies to the series  $H(s)$
 with  exponent  $\alpha=1/2$ and the function  $\psi(n)=g(n)$. It follows that the series $H(s)$  is absolutely convegent  in the half-plane  $\Re(s)=\sigma\geqslant1-\eta$ and we have
$$\sum_{n\geqslant 1}\frac{|\beta(n)|}{n^{\sigma}} \ll_{\eta}1.$$
 The conditions (1.18) of application of
Th\'eor\`eme 3 of  [9], page 25, are satisfied and we obtain the first result of the following lemma 3.3. The second result is an immediate consequence of lemme 3 page 282 of [4] and  relation (8) above.
\begin{Lem}
\begin{enumerate}
\item
 Uniformly in the range $(H_{\epsilon})$. We have
$$\sum_{\stackrel{n\leqslant x}{P(n)\leqslant y}}g(n)=H(1)\,\frac{x}{\sqrt{\log y} }\;\rho_{\frac{1}{2}}(u)\left( 1+O\left(\frac{\log (u+1)}{\log y}+\frac{1}{\sqrt{ \log y}}\right) \right)\cdot$$

\item
For $1<x\leqslant y$, we have
$$\sum_{n\leqslant x}g(n)=H(1)\,\frac{x}{\sqrt{\log y} }\;\rho_{\frac{1}{2}}(u)\left( 1+O\left(\frac{1}{\log x } \right)\right)\cdot$$
\end{enumerate}
\end{Lem}
 \begin{Rem}
 We set
$$N(d):=\frac{M_{\eta,d}}{\tau(d)}\cdot$$
The function $N$ is multiplicative and positive and satisfies
$N(p)=1+O(\frac{1}{\sqrt{p}})$ and $N(p^{\alpha})\ll 1$.
So,  from Lemme 3.1 of  Basquin [1],  and partial summation we get
$$\sum_{\stackrel{d\leqslant x}{P(d)\leqslant y}}\frac{N(d)}{d}\ll_{\eta} \sqrt{\log y}\,,$$
uniformly in  $(H_{\epsilon})$.
\end{Rem}
\begin{Lem}
We have $B(1)H(1)=1$.
\end{Lem}
\textbf{Proof.}
We have
$$B(1)=\prod_{p}\left(1-\frac{1}{p}\right)^{\frac{1}{2}}\left(\sum_{j \geqslant 0}\frac{1}{(j+1)p^{j}}\right),\quad H(1)=\prod_{p}\left(1-\frac{1}{p}\right)^{\frac{1}{2}}\left(\sum_{\alpha\geqslant0}\frac{g(p^{\alpha})}{p^{\alpha}}\right),$$
and
$$ \sum_{\alpha\geqslant0}\frac{g(p^{\alpha})}{p^{\alpha}}= \sum_{\alpha\geqslant0}\left(  \sum_{j \geqslant 0}\frac{1}{(\alpha +j+1)p^{\alpha+j}} \right)\left(  \sum_{j \geqslant 0}\frac{1}{(j+1)p^{j}}  \right)^{-1},$$
then
$$B(1)H(1)=\prod_{p}\left(1-\frac{1}{p}\right)\sum_{\alpha\geqslant0}\left(  \sum_{j \geqslant 0}\frac{1}{(\alpha +j+1)p^{j+\alpha}} \right)\cdot$$
By noticing that for $0<|x|<1$,
$$\frac{x^{\alpha +j}}{\alpha +j+1 }=\frac{1}{x}\int_{0}^{x}t^{\alpha +j} dt,$$
 we get
$$\sum_{\alpha\geqslant0}\left(  \sum_{j \geqslant 0}\frac{x^{\alpha +j}}{\alpha +j+1} \right)=  \frac{1}{x}\int_{0}^{x}\frac{dt}{(1-t)^{2}}=\frac{1}{1-x}, $$
and then  $B(1)H(1)=1$.
\hfill{
$\square$}\\[0.5cm]
Let us  denote by $P_{-}(n)$ the smallest prime factors of $n>1$ and we set  $P_{-}(1)=+\infty$.
 The function
$$ \Phi(x,y):= \sum_{\stackrel{n\leqslant x}{P_{-}(n)>y}}1$$
 has been studied by  de Bruijn [3]. In the lemma below we quote an asymptotic formula which is sufficient for our purpose.

\begin{Lem}
Uniformly in the range $(H_{\epsilon})$ we have
$$\Phi(x,y)=\frac{x\,\omega(u)}{\log y} -\frac{y}{\log y}+O\left(\frac{x}{(\log y)^{2}}\right).$$

\end{Lem}
\textbf{Proof.} Lemma follows from  Corollaire 6.14 of [10] chap. III.6 page 547 and Mertens Formula. \hfill{
$\square$}
\begin{Lem}
 Uniformly for each integer $d$ such that  $P(d)\leqslant y $ and $ (x,y)$  in $(H_{\epsilon})$ we have
$$
\begin{array}{lcl}
\sum_{n\leqslant x}\frac{1}{\tau(dn,y)}&=&\frac{x}{\sqrt{\log y}}\,\frac{G_{d}(1)}{\tau(d)}\left(\int_{0}^{u-1}\rho_{\frac{1}{2}}(z)\omega(u-z)\,dz +\rho_{\frac{1}{2}}(u)\right)\\[0.5cm]
&+& O\left( \frac{M_{\eta,d}}{\tau(d)} \left(\frac{x\,\log (u_1)}{(\log y)^{\frac{3}{2}}}+\frac{x}{\log y}\right)\right)\cdot$$
\end{array}$$
\end{Lem}
\textbf{Proof.} We write $n=ab$ with $P(a)\leqslant y $ and ($b=1$ or  if $p| b\Rightarrow p>y $).   Then for $d$ $y$-smooth, we have
$$\tau(dn,y)=\tau(dab,y)=\tau(da).$$
and then
$$T_{d}(x,y):=\sum_{n\leqslant x}\frac{1}{\tau(dn,y)} =\sum_{a\leqslant x, P(a)\leqslant y}\frac{1}{\tau(da)}\, \Phi(\frac{x}{a},y),$$
First we consider the range $$\exp\left((\log\log x)^{\frac{5}{3} +\epsilon}\right)\leqslant y\leqslant \frac{x}{a}\; ,\quad  x\geqslant x_{0}(\epsilon).$$
Write
$$\sum_{n\leqslant x}\frac{1}{\tau(dn,y)} = \sum_{a\leqslant x/y, P(a)\leqslant y}\frac{1}{\tau(da)}\, \Phi(\frac{x}{a},y)+\sum_{x/y<a\leqslant x, P(a)\leqslant y}\frac{1}{\tau(da)} :=T_{d}(x,y)+\overline{T}_{d}(x,y).$$
Lemma 3.2 gives
$$\overline{T}_{d}(x,y)=\frac{G_{1}(d)}{\tau(d)}\;\frac{x}{\sqrt{\log y}}\,\rho_{\frac{1}{2}}(u)+O\left( \frac{M_{\eta,d}}{\tau(d)} \left(\frac{x\,\log (u+1)}{(\log y)^{\frac{3}{2}}}+\frac{x}{\log y}\right)\right)\cdot$$
To study  $T_{d}(x,y)$ we apply Lemma 3.5. We obtain
$$
\begin{array}{lcl}
T_{d}(x,y)&=&\frac{x}{\log y}\sum_{a\leqslant x/y, P(a)\leqslant y}\frac{1}{a\tau(da)}\omega(u-\frac{\log a}{\log y})+ O\left( \frac{x}{\log^{2} y}\sum_{a\leqslant x/y, P(a)\leqslant y}\frac{1}{a\tau(da)} \right)\\
&=:& \frac{x}{\log y}\; L_{1}+O\left( \frac{x}{\log^{2} y} \;L_{2}   \right)\cdot
\end{array}$$
Now let us study $L_{1}$.  Partial summation and Lemma 3.2 give
$$\begin{array}{lcl}
L_{1}
&=&\frac{1}{\tau(d)}\int_{1^{-}}^{x/y}\frac{1}{t}\omega(u-\frac{\log t}{\log y})\,d\left(\sum_{a\leqslant t, P(a)\leqslant y}\gamma_{d}(a)\right)\\[0.5cm]
&=&\frac{1}{\tau(d)}\int_{1}^{x/y}\left(\sum_{a\leqslant t, P(a)\leqslant y}\gamma_{d}(a)\right) \,\left(\frac{1}{t^{2}}\omega(u-\frac{\log t}{\log y})\,+\omega'(u-\frac{\log t}{\log y})\frac{1}{t^{2} \log y}\right)\,dt\\[0.5cm]
&+&O\left( \frac{M_{\eta,d}}{\tau(d)} \frac{1}{\sqrt{\log y}}\right)\\[0.5cm]
&=&\frac{1}{\sqrt{\log y}}\frac{G_{d}(1)}{\tau(d)}\int_{1}^{x/y}\left(\frac{1}{t}\omega(u-\frac{\log t}{\log y})\,+\omega'(u-\frac{\log t}{\log y})\frac{1}{t \log y}\right)\,\rho_{\frac{1}{2}}(\frac{\log t}{\log y})\,dt\\[0.5cm]
&+& O\left(\frac{M_{\eta,d}}{\tau(d)} \int_{1}^{x/y}\left(\frac{\log (\frac{\log t}{\log y}+1)}{(\log y)^{\frac{3}{2}}}+\frac{1}{ \log y} \right) \omega(u-\frac{\log t}{\log y})\rho_{\frac{1}{2}}(\frac{\log t}{\log y})\frac{dt}{t}\right)\\[0.5cm]
&+&O\left(\frac{M_{\eta,d}}{\tau(d)} \int_{1}^{x/y}\left(\frac{\log (\frac{\log t}{\log y}+1)}{(\log y)^{\frac{3}{2}}}+\frac{1}{ \log y} \right) |\omega'(u-\frac{\log t}{\log y})|\rho_{\frac{1}{2}}(\frac{\log t}{\log y})\frac{dt}{t \log y}\right)\\[0.5cm]
&+&O\left( \frac{M_{\eta,d}}{\tau(d)} \frac{1}{\sqrt{\log y}}\right)\cdot
\end{array} $$
 By  change of variable $z=\frac{\log t}{\log y}$. We obtain
$$
\begin{array}{lcl}
L_{1}
 &=&\sqrt{\log y}\,\frac{G_{d}(1)}{\tau(d)}\int_{0}^{u-1}\rho_{\frac{1}{2}}(z)\omega(u-z)\,dz + O\left( \frac{M_{\eta,d}}{\tau(d)} \left(1+\frac{\log u}{(\log y)^{\frac{1}{2}}}\right)\right),
\end{array}
$$
 We estimate $L_{2}$ in the same way. Lemma 3.2 and By partial summation give
$$L_{2}\ll\sqrt{\log y}\;\frac{M_{\eta,d}}{\tau(d)} \int_{0}^{u}\rho_{\frac{1}{2}}(z)dz\ll\sqrt{\log y}\;\frac{M_{\eta,d}}{\tau(d)}\cdot$$
We get our result in the considered range  from these different estimates. In the range  $\frac{x}{a} <y\leqslant x$  we have
$\Phi(x/a,y)=1$. The result follows immediately from Lemma 3.2.
\hfill{
$\square$}
\begin{Lem}
  Set  $\epsilon_{x}=\frac{\log 2 }{\log x}$. Uniformly for $(x,y)\in (H_{\epsilon})$ \\and $0\leqslant v < \epsilon_{x}$ we have
 $$ S(x,y,v)=O\left(\frac{x}{\sqrt{\log y}}\right)\cdot$$
 \end{Lem}
\textbf{Proof} The condition on  $v$ implies  $d=1$. By   Lemme 3.6 with  $d=1$ we get
$$S(x,y,v)=\sum_{n\leqslant x}\frac{1}{\tau(n,y)}\,\sum_{d|n, d\leqslant n^{v},P(d)\leqslant y} 1 \leqslant \sum_{n\leqslant x}\frac{1}{\tau(n,y)}\ll \frac{x}{\sqrt{\log y }}\cdot$$
\hfill{
$\square$}
\begin{Lem} Let be $\epsilon_{x}=\frac{\log 2 }{\log x}$. Uniformly for  $(x,y)\in (H_{\epsilon})$ and $ \epsilon_{x}\leqslant v\leqslant1$, we have
$$S_{2}(x,y,v)=O\left(\frac{x}{\log y }  \right)\cdot$$
\end{Lem}
\textbf{Proof} We have
$$S_{2}(x,y,v)=\sum_{n\leqslant x}\frac{1}{\tau(n,y)}\,\sum_{d|n, n^{v}<d\leqslant x^{v},P(d)\leqslant y} 1\leqslant \sum_{d\leqslant x^{v},P(d)\leqslant y}\,\sum_{m\leqslant x^{1-v}}\frac{1}{\tau(dm,y)}\cdot$$

If $1-\epsilon_{x}\leqslant v\leqslant1$, then $x^{1-v}<2$. In this case, we have

$$S_{2}(x,y,v)\leqslant \sum_{d\leqslant x,P(d)\leqslant y}\,\frac{1}{\tau(d)}\ll\frac{x}{\sqrt{\log y}}\,\rho_{\frac{1}{2}}(u)\ll\frac{x}{\sqrt{\log y}}\cdot$$
Suppose that  $\epsilon_{x}\leqslant v\leqslant 1-\epsilon_{x}$.
If   $y\leqslant \min\{ x^{v},x^{1-v}\}$.  We apply Lemmas 3.6 and 3.3,  we get
$$
\begin{array}{lcl}
S_{2}(x,y,v)&\ll& \sum_{d\leqslant x^{v},P(d)\leqslant y}\frac{x^{1-v}}{\sqrt{\log y}}\,\frac{G_{d}(1)}{\tau(d)}\left(\int_{0}^{(1-v)u-1}\rho_{\frac{1}{2}}(z)\omega((1-v)u-z)\,dz+\rho_{\frac{1}{2}}((1-v)u)\right) \\[0.8cm]
 &\ll& \frac{x^{1-v}}{\sqrt{\log y}}\,\sum_{d\leqslant x^{v},P(d)\leqslant y} g(d)\\ [0.8cm]
 &\ll&\frac{x}{\log y} \,\rho_{\frac{1}{2}}(uv)\ll \frac{x}{\log y},
\end{array}
$$
since $uv\geqslant1$ et $(1-v)u\geqslant1$.
 We proceed in the same way in other cases. For  $x^{1-v}< y \leqslant x^{v}$, the inner sum is
$$\sum_{m\leqslant x^{1-v}}\frac{1}{\tau(dm,y)}=\sum_{m\leqslant x^{1-v}}\frac{1}{\tau(dm)}\ll \frac{x^{1-v}}{\sqrt{(1-v)\log x}} \ll x^{1-v}$$
from Lemma  3.2.  We apply Lemme 3.3 to the outer sum we get
$$S_{2}(x,y,v)\ll \frac{x}{\sqrt{\log y}}\rho_{\frac{1}{2}}(uv)\ll \frac{x}{\sqrt{\log y}}\cdot$$
The case $x^{v}< y \leqslant x^{1-v}$ is similar. Last  if $y>\max\{x^{1-v},x^{v}\}$, we set
$\epsilon'_{x}=\frac{\log 2}{\sqrt{\log x}}$ and consider  three cases  $\epsilon_{x}\leqslant v\leqslant \epsilon'_{x}$ , $ \epsilon'_{x}< v\leqslant 1-\epsilon'_{x}$ et $1-\epsilon'_{x}< v \leqslant 1-\epsilon_{x}$.
In each situation we have $\frac{1}{\sqrt{v(1-v)}}\ll\sqrt{\log x}$.
 By applying Lemmas 3.2 and 3.3 we get
$$ S_{2}(x,y,v)\ll \frac{x}{\log x} \frac{1}{\sqrt{v(1-v)}}\ll \frac{x}{\sqrt{\log x}}\cdot$$
 \hfill{
$\square$}
\subsection{Proof of Theorem 2.1 }
Taking into account Lemmas 3.7 and  3.8,  it remains to estimate $S_{1}(x,y,v)$ for
$\epsilon_{x}\leqslant v\leqslant1$.
We consider two following situations : $\epsilon_{x}\leqslant v\leqslant1-\epsilon_{x}$ et $1-\epsilon_{x}<v\leqslant1$.
 First suppose that $1-\epsilon_{x}< v\leqslant1$. Dans ce cas $x/2 <x^{v}$. We write
 $$
 \begin{array}{lcl}
 S_{1}(x,y,v)&=&\sum_{d\leqslant x/2, P(d)\leqslant y}\,\sum_{m\leqslant x/d}\frac{1}{\tau(dm,y)}+ \sum_{x/2<d\leqslant x^{v}, P(d)\leqslant y}\,\sum_{m\leqslant x/d}\frac{1}{\tau(dm,y)}\\[0.8cm]
 &=&:\widehat{S}_{1} +\widehat{S}_{2}.
 \end{array}$$
 Let us study $ \widehat{S}_{2}$. $ x/2< d \iff x/d<2$ thus $m=1$. We have
 $$ \widehat{S}_{2}=\sum_{x/2<d\leqslant x^{v}, P(d)\leqslant y}\frac{1}{\tau(d)}\ll\sum_{d\leqslant x, P(d)\leqslant y}\frac{1}{\tau(d)}\ll\frac{x}{\sqrt{\log y}}\rho_{\frac{1}{2}}(u)\ll\frac{x}{\sqrt{\log y}}\cdot$$
The evalation of $\widehat{S}_{1}$ is similar to the envaluation of  $S_{1}(x,y,v)$ under  complementary condition $\epsilon_{x}\leqslant v\leqslant 1-\epsilon_{x}$, sinc  $ x/2=x^{1-\epsilon_{x}}$.
Let us study  $S_{1}(x,y,v)$ under the condition $\epsilon_{x}\leqslant v\leqslant 1-\epsilon_{x}$. First, consider the range $$\exp\left((\log\log x)^{\frac{5}{3} +\epsilon}\right)\leqslant y\leqslant x/d\; ,\quad  x\geqslant x_{0}(\epsilon).$$
We write
 $$
 \begin{array}{lcl}
 S_{1}(x,y,v)&:=&\sum_{n\leqslant x}\frac{1}{\tau(n,y)}\,\sum_{\stackrel{d|n, d\leqslant x^{v}}{P(d)\leqslant y}}1=\sum_{d\leqslant x^{v}, P(d)\leqslant y}\,\sum_{m\leqslant x/d}\frac{1}{\tau(dm,y)}\\[0.5cm]
&=&\sum_{d\leqslant x^{v}, P(d)\leqslant y}\,T_{d}(\frac{x}{d},y)\cdot
\end{array}
$$
We apply Lemma 3.6.  We obtain
$S_{1}(x,y,v)=S_{1,1}+O(S_{1,2})$ avec, en posant  $$u_{d}:=u-\frac{\log d}{\log y},$$
$$S_{1,1}:=\frac{x}{\sqrt{\log y}}\,\sum_{d\leqslant x^{v}, P(d)\leqslant y}\frac{G_{d}(1)}{d\tau(d)}\left(\int_{0}^{u_{d}-1}\rho_{\frac{1}{2}}(z)\omega(u_{d}-z)\,dz+\rho_{\frac{1}{2}}(u_{d})\,\right),$$
 and
$$S_{1,2}:=x\,\left(\frac{1}{\log y}+\frac{\log(u+1)}{(\log y)^{\frac{3}{2}}}\right)\,\sum_{d\leqslant x^{v}, P(d)\leqslant y}\frac{M_{\eta,d}}{d\tau(d)}\ll x\,\left(\frac{1}{\sqrt{\log y}}+\frac{\log(u+1)}{\log y}\right), $$
by using Remark 3.1.
It remains to estimate  $S_{1,1}$. From (9), we have  $\frac{G_{d}(1)}{\tau(d)}=B(1) g(d)$.
 By partial summation we get
 $$
\begin{array}{lcl}
 S_{1,1}&=&\frac{x}{\sqrt{\log y}}\,B(1)\sum_{ d\leqslant x^{v}, P(d)\leqslant y}\frac{g(d)}{d}\left(\int_{0}^{u_{d}-1}\rho_{\frac{1}{2}}(z)\omega(u_{d}-z)\,dz+\rho_{\frac{1}{2}}(u_{d})\right)\\
 &=&\frac{B(1)x^{1-v}}{\sqrt{\log y}}\left(\int_{0}^{(1-v)u-1} \rho_{\frac{1}{2}}(z)\omega((1-v)u-z )\, dz+\rho_{\frac{1}{2}}((1-v)u) \right)\left(\sum_{n\leqslant x^{v}, P(n)\leqslant y}g(n)\right)\\
 &-&\frac{x}{\sqrt{\log y}}\,B(1) \;\int_{1}^{x^{v}}
\left( \sum_{n\leqslant t, P(n)\leqslant y}g(n)\right)\;d\left(\frac{1}{t}\left(\int_{0}^{u_{t}-1} \rho_{\frac{1}{2}}(z)\omega(u_{t}-z )\, dz+\rho_{\frac{1}{2}}(u_{t})\right)\right)\\
&:=& R_{1}-R_{2}.
 \end{array}
 $$
From  Lemmas 3.3 and  3.4
 $$
R_{1}\ll\frac{x}{\log y}\, \rho_{\frac{1}{2}} (uv)\left(1+ \rho_{\frac{1}{2}}((1-v)u)\right)
\ll \frac{x}{\sqrt{\log y}}\cdot
$$
the last upper bound follows by using the same way as in proof of Lemma 3.8.
Now consider  $R_{2}$.  We have
 $$d\left( \frac{1}{t}  \left(\int_{0}^{u_{t}-1}\rho_{\frac{1}{2}}(z)\,\omega(u_{t}-z )\, dz+\rho_{\frac{1}{2}}(u_{t}) \right)\right) =D_{1}+D_{2},$$
 with
 $$D_{1}:=-\left(\int_{0}^{u_{t}-1}\,\rho_{\frac{1}{2}}(z)\omega(u_{t}-z )\,dz +\rho_{\frac{1}{2}}(u_{t})\right)\,\frac{dt}{t^{2}},$$
and
$$D_{2}:=\frac{1}{t}d\left( \int_{0}^{u_{t}-1}\rho_{\frac{1}{2}}(z)\,\omega(u_{t}-z )\, dz +\rho_{\frac{1}{2}}(u_{t})\right).$$
 Write $ R_{2}=R_{2,1}+R_{2,2}$
with
$$R_{2,1}:=\frac{x}{\sqrt{\log y}}\,B(1) \;\int_{1}^{x^{v}}  \left(\sum_{n\leqslant t, P(n)\leqslant y}g(n)\right) D_{1},$$
and
$$R_{2,2}:=\frac{x}{\sqrt{\log y}}\,B(1) \;\int_{1}^{x^{v}}  \left(\sum_{n\leqslant t, P(n)\leqslant y}g(n)\right) D_{2}.$$
To evaluate  $R_{2,1}$ we apply Lemmas 3.3 and  3.4. We get
$$
\begin{array}{lcl}
R_{2,1}&=&
-\frac{x}{\log y}\int_{1}^{x^{v}}\rho_{\frac{1}{2}}(\frac{\log t}{\log y})\,\left(\int_{0}^{u_{t}-1}\,\rho_{\frac{1}{2}}(z)\omega(u_{t}-z )\,dz+\rho_{\frac{1}{2}}(u_{t})\right)\; \times\\[0.5cm]
&&\left(1+O\left(\frac{\log(\frac{\log t}{\log y}+1)}{\log y}+\frac{1}{\sqrt{\log y}}\right)\;\right)\;\frac{dt}{t}\cdot
\end{array}$$
A change of variable  $s=\frac{\log t}{\log y}$ gives

$$
\begin{array}{lcl}
R_{2,1}&=&-x \int_{0}^{uv}  \rho_{\frac{1}{2}}(s)\left(\int_{0}^{u-s-1}\rho_{\frac{1}{2}}(z)\omega(u-s-z )dz+\rho_{\frac{1}{2}}(u-s)\right)\; \times\\[0.5cm]
&&\left(1+O\left(\frac{\log (s+1)}{\log y}+\frac{1}{\sqrt{\log y}}\right)\right)ds\\[0.5cm]
&=&- x \,\int_{0}^{uv}  \rho_{\frac{1}{2}}(s)\,\left(\int_{0}^{u-s-1}\,\rho_{\frac{1}{2}}(z)\omega(u-s-z )\,dz\right)\,ds-x\,\,\int_{0}^{uv}  \rho_{\frac{1}{2}}(s)\rho_{\frac{1}{2}}(u-s) ds\\[0.5cm]
& +&O\left(\frac{x\,\log(u+1)}{\log y}+\frac{x}{\sqrt{\log y}}\right)\\[0.5cm]
\end{array}
$$
Now consider $R_{2,2}$. an easy calculation gives

$$D_{2}\ll \frac{1}{t^{2} (u_{t}-1)\log y}$$

 $$
 \begin{array}{lcl}
 D_{2}&=&-\frac{1}{t^{2} (u_{t}-1)\log y}\left(\int_{0}^{u_{t}-1}\rho_{\frac{1}{2}}(z)\,\omega(u_{t}-z )\, dz\right)\;dt\\[0.5cm]
 &-&\frac{1}{t^{2}(u_{t}-1) \log y}\left(\int_{0}^{u_{t}-1}z\rho'_{\frac{1}{2}}(z)\,\omega(u_{t}-z )\, dz\right)\;dt\\[0.5cm]
 &-&\frac{1}{t^{2} (u_{t}-1)\log y}\left(\int_{0}^{u_{t}-1}(u_{t}-z)\rho_{\frac{1}{2}}(z)\,\omega'(u_{t}-z )\, dz\right)\;dt\\[0.5cm]
 &-&\frac{1}{t^{2} \log y}\,\rho'_{\frac{1}{2}}(u_{t})\\[0.5cm]
 &:=& D_{2,1}+D_{2,2}+D_{2,3}+D_{2,4}.
  \end{array}$$
We have
$$R_{2,2}=\frac{x}{\sqrt{\log y}}\,B(1) \;\int_{1}^{x^{v}}  \left(\sum_{n\leqslant t, P(n)\leqslant y}g(n)\right) \left(D_{2,1}+D_{2,2}+D_{2,3}+D_{2,4}\right)\;dt=\widetilde{D}_{1}+\widetilde{D}_{2}+\widetilde{D}_{3}+\widetilde{D}_{4}$$
where
$$\widetilde{D}_{i}=\frac{x}{\sqrt{\log y}}\,B(1) \;\int_{1}^{x^{v}}  \left(\sum_{n\leqslant t, P(n)\leqslant y}g(n)\right) D_{2,i}$$
 for $i\in \{1,2,3,4\}$.
By using the same method as previously, we get
$$
\widetilde{D}_{1}\ll\frac{x}{\log y}\int_{0}^{uv}\frac{\rho_{\frac{1}{2}}(s)}{u-s-1}\,\left(\int_{0}^{u-s-1}\,\rho_{\frac{1}{2}}(z)\omega(u-s-z )\,dz\right)\,ds.$$
For  $s\geqslant u-2$ we have   $\omega (u-s-z)=0$ since  $0 \leqslant z \leqslant u-s-1\leqslant1$ and for  $s\leqslant u-2$, we have $u-s-1\geqslant1$. It follows  that
$$\widetilde{D}_{1}\ll \frac{x}{\log y}\int_{0}^{u-2}\rho_{\frac{1}{2}}(s)ds\,\int_{0}^{u-1}\,\rho_{\frac{1}{2}}(z)dz
\ll\frac{x}{\log y}\cdot$$
The study of  $\widetilde{D}_{2}$ , $\widetilde{D}_{3}$ and $\widetilde{D}_{4}$ is similar by using propreties of  $\rho'_{\frac{1}{2 }}$ and $\omega'$. We get the same result. We omit details.
Assembling these estimates we get our result in the considered range.
In the complementary range  $x/d< y\leqslant x$, proof is similar, we omit it.
 \hfill{
$\square$}
\section{Proof of  Theorem 2.2}
\subsection{Lemmas}
We need a weak form of the following lemma
\begin{Lem}
For each fixed integer $N\geqslant1$ and uniformly for $w>1$ we have
$$\int_{w}^{\infty } \rho_{\frac{1}{2}}(t) dt=2\sum_{k=1}^{N}(w+k)\rho_{\frac{1}{2}}(w+N)+ O\left(\frac{\rho_{\frac{1}{2}}(w)}{w^{N}}\right).$$
\end{Lem}
\textbf{Proof} Set
$$I(w):=\int_{w}^{\infty } \rho_{\frac{1}{2}}(t) dt.$$
By change of variable  $ z=t+1$ and by using  the differerential equation satisfied by  $\rho_{\frac{1}{2}}$ we obtain
\begin{eqnarray}
I(w)=\int_{w+1}^{\infty } \rho_{\frac{1}{2}}(z-1) dz= -2\int_{w+1}^{\infty } z\rho'_{\frac{1}{2}}(z) dz-\int_{w+1}^{\infty } \rho_{\frac{1}{2}}(z) dz  = -2 J(w)-I(w+1).
\end{eqnarray}
with  $$J(w):=\int_{w+1}^{\infty } z\rho'_{\frac{1}{2}}(z) dz.$$
Integration by parts gives
$$J(w)=-(w+1)\rho_{\frac{1}{2}}(w+1)-\int_{w+1}^{\infty } \rho_{\frac{1}{2}}(z)dz=-(w+1)\rho_{\frac{1}{2}}(w+1)-I(w+1).$$
Substituting in  (11) we obtain
\begin{eqnarray}
I(w)-I(w+1)=2(w+1)\rho_{\frac{1}{2}}(w+1).
\end{eqnarray}
And by iteration we obtain for every integer  $k\geqslant1$,
$$I(w+k-1)-I(w+k)=2(w+k)\rho_{\frac{1}{2}}(w+k),$$
Summing these inequalities, we obtain
$$I(w)=2(w+1)\rho_{\frac{1}{2}}(w+1)+2\sum_{k=2}^{\infty}(w+k)\rho_{\frac{1}{2}}(w+k),$$

The lemma follows from this formula and the result
$$\rho_{\frac{1}{2}}(w+k)=O\left(\frac{\rho_{\frac{1}{2}}(w)}{w^{k}}\right),$$
 uniformly for $w>1$ and  $k>0$
which is a consequence of the asymptotic formula for  $\rho_{\frac{1}{2}}(w)$ (see  [8]).
 \hfill{
$\square$}
\begin{Lem}
For $v\in[0,1]$ and as  $u\rightarrow \infty$ we uniformly have
$$H(u,v):= \frac{1}{e^{\gamma}}  \int_{0}^{uv}\left( \int_{0}^{u-s-1}\rho_{\frac{1}{2}}(w)dw\right)\,\rho_{\frac{1}{2}}(s)ds= \frac{1}{\sqrt{e^{\gamma}}}  \int_{0}^{uv}\rho_{\frac{1}{2}}(w)dw+O\left( u\rho(u)\right),$$
where  $\gamma$ is Euler's constant.
\end{Lem}
\textbf{Proof }
 We have
$$
\begin{array}{lcl}
H(u,v)&=& \frac{1}{e^{\gamma}}  \int_{0}^{uv}\left( \int_{0}^{+\infty} \rho_{\frac{1}{2}}(w)dw -\int_{u-s-1}^{\infty}\rho_{\frac{1}{2}}(w)dw\right)\,\rho_{\frac{1}{2}}(s)ds\\[0.5cm]
&=& \frac{1}{\sqrt{e^{\gamma}}}  \int_{0}^{uv} \rho_{\frac{1}{2}}(s)ds - \frac{1}{e^{\gamma}}  \int_{0}^{uv}\left(\int_{u-s-1}^{\infty}\rho_{\frac{1}{2}}(w)dw\right)\,\rho_{\frac{1}{2}}(s)ds,$$
\end{array}$$
since by using Laplace transform, we have  $$\int_{0}^{+\infty} \rho_{\frac{1}{2}}(w)dw=\widehat{\rho_{\frac{1}{2}}}(0)=(\widehat{\rho}(0))^{1/2}=\sqrt{e^{\gamma}}.$$
We apply Lemma 4.1 in a weak version we obtain
$$
\begin{array}{lcl}
H(u,v)&=& \frac{1}{\sqrt{e^{\gamma}}}  \int_{0}^{uv} \rho(s)ds+O\left( u \int_{0}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds \right)\\
&=&\frac{1}{\sqrt{e^{\gamma}}}  \int_{0}^{uv} \rho(s)ds+O(u\rho(u)),
\end{array}
$$
since
$$ \int_{0}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds\leqslant  \int_{0}^{u} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds =(\rho_{\frac{1}{2}}\ast \rho_{\frac{1}{2}})(u)\ll\rho(u).$$
\hfill{
$\square$}\\[0.5cm]
\subsection{Proof of Theorem 2.2}
We have
$$F(u,v)=\int_{0}^{uv} \left(\int_{0}^{u-s-1}\rho_{\frac{1}{2}}(z)\omega(u-s-z )\,dz\right)\rho_{\frac{1}{2}}(s)\,ds+O(\rho(u)).$$
 And
 $$
 \begin{array}{lcl}
\int_{0}^{uv} \left(\int_{0}^{u-s-1}\rho_{\frac{1}{2}}(z)\omega(u-s-z )\,dz\right)\rho_{\frac{1}{2}}(s)\,ds= \frac{1}{\sqrt{e^{\gamma}}}  \int_{0}^{uv}\rho_{\frac{1}{2}}(s)ds-\frac{1}{\sqrt{e^{\gamma}}}  \int_{0}^{uv}\rho_{\frac{1}{2}}(s)\left(\int_{u-s-1}^{\infty}\rho_{\frac{1}{2}}(z)dz\right) ds+
  O\left(  \int_{0}^{uv} \rho_{\frac{1}{2}}(s) \left(\int_{0}^{u-s-1}\rho_{\frac{1}{2}}(z)\rho(u-s-z)dz\right)\,ds\   \right)
   \end{array}
   $$
 By using  formula  (3) and Lemma  4.2, we get
 $$
 \begin{array}{lcl}
&& \int_{0}^{uv} \left(\int_{0}^{u-s-1}\rho_{\frac{1}{2}}(z)\omega(u-s-z )\,dz\right)\rho_{\frac{1}{2}}(s)\,ds=
\frac{1}{e^{\gamma}}\int_{0}^{uv} \rho_{\frac{1}{2}}(s) \left(\int_{0}^{u-s-1}\rho_{\frac{1}{2}}(z)dz\right)\,ds\, \\[0.5cm]
 &+& O\left(  \int_{0}^{uv} \rho_{\frac{1}{2}}(s) \left(\int_{0}^{u-s-1}\rho_{\frac{1}{2}}(z)\rho(u-s-z)dz\right)\,ds\   \right)\\[0.5cm]
&=&\frac{1}{\sqrt{e^{\gamma}}}  \int_{0}^{uv}\rho_{\frac{1}{2}}(s)ds+O(u\rho(u))+O\left( \rho_{2}(u)\right)\\[0.5cm]
&=&\frac{1}{\sqrt{e^{\gamma}}}  \int_{0}^{uv}\rho_{\frac{1}{2}}(s)ds+O\left( \rho_{2}(u)\right)
\end{array}
$$
since
$$
\begin{array}{lcl}
\int_{0}^{uv} \rho_{\frac{1}{2}}(s) \left(\int_{0}^{u-s-1}\rho_{\frac{1}{2}}(z)\rho(u-s-z)dz\right)\,ds& \leqslant &\int_{0}^{uv} \rho_{\frac{1}{2}}(s)(\rho_{\frac{1}{2}}\ast \rho)(u-s)\,ds\\
&\leqslant&(\rho_{\frac{1}{2}}\ast \rho_{\frac{1}{2}}\ast \rho)(u)=(\rho\ast \rho)(u)=\rho_{2}(u),
\end{array}
$$
and   $u\rho(u)\ll \rho_{2}(u).$
\hfill{
$\square$}
\section{Proof of Theorem 2.3}
\subsection{Lemmas}
\begin{Lem}  For $0\leqslant \xi <1$ we have
$$R(\xi):=\frac{1}{\pi} \int_{0}^{\xi} \int_{0}^{-s+\xi} \frac{ds'\,dz'} {\sqrt{s'}\sqrt{z'}(1-s'-z')}=
-\log (1-\xi)\cdot$$
\end{Lem}
\textbf{Proof}
We use change of variables $(s',z')\mapsto (w, r)=(\frac{s'}{s'+z'},s'+z')$. Then we have  a $dsdz=rdwdr$.
We obtain
 $$
R(\xi) =\left(\frac{1}{\pi}\int_{0}^{1}w ^{-\frac{1}{2}} (1-w)^{-\frac{1}{2}}\;dw\right)\left(  \int_{0}^{\xi}(1-r)^{-1} \;dr \right)=-\log (1-\xi),
$$
since
$$\frac{1}{\pi}\int_{0}^{1}w ^{-\frac{1}{2}} (1-w)^{-\frac{1}{2}}\;dw=\frac{B(\frac{1}{2},\frac{1}{2})}{\pi}=\frac{\Gamma^{2}(\frac{1}{2})}{\pi}=1.$$
\hfill{
$\square$}
\begin{Lem} For $\beta \leqslant 1-\frac{1}{u}$ we have
$$
\begin{array}{lcl}
S(\beta)&:=&\frac{1}{\pi}\int_{0}^{\beta} \left(\int_{0}^{-s+1-\frac{1}{u}}\frac{ds \,dz}{\sqrt{s}\,\sqrt{z}\,(1-s-z)}\right)\\[0.5cm]
&=&\frac{2}{\pi}\left(\;\log(u)+\log(1-\beta)\;\right)\arcsin\left(  \sqrt{\frac{u\beta}{u-1}} \,  \right)-\log(1-\beta)\cdot
\end{array}
$$
\end{Lem}
\textbf{Proof} We write $S(\beta)=I_{1}-I_{2}$ with
$$I_{1}=\frac{1}{\pi}\int_{0}^{1-\frac{1}{u}} \left(\int_{0}^{-s+1-\frac{1}{u}}\frac{ds \,dz}{\sqrt{s}\,\sqrt{z}\,(1-s-z)}\right),\,I_{2}=\frac{1}{\pi}\int_{\beta}^{1-\frac{1}{u}} \left(\int_{0}^{-s+1-\frac{1}{u}}\frac{ds \,dz}{\sqrt{s}\,\sqrt{z}\,(1-s-z)}\right)\cdot$$
From Lemma 5.1, $I_{1}=R(1-\frac{1}{u})=\log(u)$. Let us study  $I_{2}$. By change of  variables $(s',z')\mapsto(s,z)=(s'+\beta,z')$ we obtain
$$I_{2}=\frac{1}{\pi}\int_{0}^{1-\frac{1}{u} -\beta} \left(\int_{0}^{-s'+1-\frac{1}{u} -\beta}\frac{ds' \,dz'}{\sqrt{s'+\beta}\,\sqrt{z'}\,(1-s'-\beta-z')}\right)\cdot$$
Now we put the change of variables
$(s'+\beta)=rw$ et $s'+\beta+z'=r$ then we have     $ds'dz' =rdwdr$ and
$$I_{2}=\frac{1}{\pi}\int_{\frac{u\beta}{u-1}}^{1}  \frac{dw}{\sqrt{w} \sqrt{1-w} }\int_{\beta}^{1-\frac{1}{u}}\frac{dr}{1-r}\cdot$$
Since
$$\int_{\beta}^{1-\frac{1}{u}}\frac{dr}{1-r}=\log(u)+\log(1-\beta),$$
then by change of variable $t=\sqrt{w}$, we get
$$\frac{1}{\pi}\int_{\frac{u\beta}{u-1}}^{1}  \frac{dw}{\sqrt{w} \sqrt{1-w}} =\frac{2}{\pi}\int_{\sqrt{\frac{u\beta}{u-1}}}^{1} \frac{dt}{ \sqrt{1-t^{2}} }=1-\frac{2}{\pi}\arcsin\left(  \sqrt{\frac{u\beta}{u-1}} \,  \right)\cdot$$
Finally,
$$I_{2}=\left(\;\log(u) +\log(1-\beta)\;\right)\left(1-\frac{2}{\pi}\arcsin \left(  \sqrt{\frac{u\beta}{u-1}} \,\right) \right)\cdot$$
It follows
$$S(\beta)=\frac{2}{\pi}( \log(u)+\log(1-\beta))\arcsin \left(  \sqrt{\frac{u\beta}{u-1}}\right) -\log(1-\beta).$$
\hfill{
$\square$}
\begin{Lem}
For $0<w\leqslant1$ we have  $$\rho_{\frac{1}{2}}(w)=\frac{1}{\sqrt{\pi} \sqrt{w}},$$
and for  $1\leqslant w\leqslant2$, we have
$$\rho_{\frac{1}{2}}(w)=\frac{1}{\sqrt{\pi} \sqrt{w}}
- \frac{\log(\sqrt{w}+\sqrt{w-1})}{\sqrt{\pi} \sqrt{w}}.$$
\end{Lem}
\textbf{Proof} The first formula is the definition of $\rho_{\frac{1}{2}}$  for $0< w \leqslant1$ and the second one follows  from differential equation
satisfied by  $\rho_{\frac{1}{2}}$ pour  $1\leqslant w \leqslant2 $. \hfill{
$\square$}\\[0.2cm]
Let us consider the integral
$$I:=\int_{0}^{uv} \left(\int_{0}^{u-s-1}\rho_{\frac{1}{2}}(z)\omega(u-s-z )\,dz\right)\rho_{\frac{1}{2}}(s)\,ds,$$
 for  $1\leqslant u\leqslant 2$. We notice  that on one hand
 for  $s>u-1$
$$\int_{0}^{u-s-1}\rho_{\frac{1}{2}}(z)\omega(u-s-z )\,dz=0,$$
since $\rho_{\frac{1}{2}}(z)=0$ pour $z\leqslant0$ and then we can restrict the study of the l'integral on  $s$ at the interval $[0,M]$ where $M=\min\{u-1,uv\}\leqslant 1$.
And on the other hand, $0\leqslant z\leqslant u-s-1$ then $1\leqslant u-s-z\leqslant u-s\leqslant u\leqslant 2$  and   $\omega (u-s-z)=1/(u-s-z)$.  It follows
 \begin{eqnarray}
I=\frac{1}{\pi}\int_{0}^{M} \left(\int_{0}^{-s+u-1}\frac{dz}{\sqrt{z}\,(u-s-z)}\right)\frac{ds}{\sqrt{s}}\cdot
\end{eqnarray}
 We will give two expessions of  $I$.
By the change of variable $t=\sqrt{z}$ in the inner integral $(13)$ we get
\begin{eqnarray}
I=\frac{2}{\pi}\,\int_{0}^{M}\frac{\log(\sqrt{u-s}+\sqrt{u-s-1})}{\sqrt{s} \sqrt{u-s}}\,ds,
\end{eqnarray}
and by the change of variables $(s',z')\mapsto (s,z)=(us',uz')$ and by puting  $M'=\frac{M}{u}=\min \{ 1-\frac{1}{u},v \}$
in  (13), we obtain
\begin{eqnarray}
 I=\frac{1}{\pi}\int_{0}^{M'}\int_{0}^{-s'+1-\frac{1}{u}}\frac{ds'\,dz'}{\sqrt{s'}  \sqrt{z'}(1-s'-z')}\cdot
\end{eqnarray}
By using notations of  lemma 5.2 and the expressions  (13) et (14) of $I$, we obtain
\begin{eqnarray}
I=
\left \lbrace
\begin{array}{lcl}
S(1-\frac{1}{u})\quad \text{si}\quad M=u-1,\\[0.2cm]
S(v)\qquad \quad \, \text{si}\quad M=uv.
\end{array}
\right.
\end{eqnarray}
Now we will study the integral
$$\int_{0}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds,\quad (1\leqslant u \leqslant 2).$$

\begin{Lem}
For $v\in[0,\frac{u-1}{u}]$ with  $1\leqslant u \leqslant2$ we have
$$\int_{0}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds=\frac{2}{\pi}\arcsin(\sqrt{v})-\frac{1}{2} S(v).$$
\end{Lem}
\textbf{Proof}
For $v\in[0,\frac{u-1}{u}]$ we have  $uv\leqslant u-1$ then $0\leqslant s \leqslant uv\leqslant u-1\leqslant1$ and
$1\leqslant u-s \leqslant 2$. By using Lemma 5.3 we get
$$\int_{0}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds=\frac{1}{\pi}\int_{0}^{uv} \frac{ds}{\sqrt{s}\sqrt{u-s}}-\frac{1}{\pi}\int_{0}^{uv} \frac{\log(\sqrt{u-s}+\sqrt{u-s-1})}{\sqrt{s}\sqrt{u-s}}\,ds:=J_{1}-J_{2}.$$
By the change of variable $t=\sqrt{s}$ we have
$$J_{1}=\frac{2}{\pi}\int_{0}^{\sqrt{v}} \frac{dt}{\sqrt{1-t^{2}}}=\frac{2}{\pi}\arcsin(\sqrt{v}).$$
From (16) we have $J_{2}=\frac{1}{2}S(v)$. This completes the proof of Lemma 5.4.
\hfill{
$\square$}
\begin{Lem}
For $v\in[\frac{u-1}{u},\frac{1}{u}]$ with $1\leqslant u\leqslant 2$ we have
$$\int_{0}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds=\frac{2}{\pi}\arcsin(\sqrt{v})-\frac{1}{2}S(1-\frac{1}{u}).$$
\end{Lem}
\textbf{Proof}  For  $v\in[\frac{u-1}{u},\frac{1}{u}]$we have  $u-1 \leqslant uv \leqslant1$. We write
$$\int_{0}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds=\int_{0}^{u-1} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds+\int_{u-1}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds=J_{1}+J_{2}\cdot$$
Consider $J_{1}$. As in previous Lemma we have
$$
\begin{array}{lcl}
J_{1}&=&\frac{1}{\pi}\int_{0}^{u-1} \frac{ds}{\sqrt{s}\sqrt{u-s}}-\frac{1}{\pi}\int_{0}^{u-1} \frac{\log(\sqrt{u-s}+\sqrt{u-s-1})}{\sqrt{s}\sqrt{u-s}}\,ds\\[0.5cm]
&=& \frac{2}{\pi} \arcsin(\sqrt{\frac{u-1}{u}})-\frac{1}{2}S(1-\frac{1}{u}).
\end{array}
$$
Now consider  $J_{2}$. We have  $u-1\leqslant s\leqslant uv\leqslant1$ et $u(1-v)\leqslant u-s\leqslant 1$. It follows
$$J_{2}=\frac{1}{\pi}\int_{u-1}^{uv} \frac{ds}{\sqrt{s}\sqrt{u-s}}=\frac{2}{\pi}\arcsin(\sqrt{v})-\frac{2}{\pi}\arcsin(\sqrt{\frac{u-1}{u}}).$$
This completes the proof of Lemma 5.5.
\hfill{
$\square$}
\begin{Lem}
For $v\in[\frac{1}{u}, 1]$ with $1\leqslant u\leqslant 2$ we have
$$\int_{0}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds=-\frac{3}{2}S(1-\frac{1}{u})+\frac{2}{\pi}\arcsin(\sqrt{v})+S(1-v).$$
\end{Lem}
 \textbf{Proof}
As $u\leqslant 2$ we have $\frac{u-1}{u}\leqslant \frac{1}{u}\leqslant v\leqslant1$ and then  $u-1\leqslant uv\leqslant u$. We write
$$ \int_{0}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds=\int_{0}^{u-1} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds+\int_{u-1}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds=J_{1}+J_{2}\cdot$$
$J_{1}$ has been studied in lemma 5.5. It remains to calculate $J_{2}$. As  $\frac{1}{u} \leqslant v$ then $uv\geqslant1$. We write
$$J_{2}=\int_{u-1}^{1} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds+\int_{1}^{uv} \rho_{\frac{1}{2}}(s)\,\rho_{\frac{1}{2}}(u-s)\,ds := J_{2,1}+J_{2,2}.$$
We have
$$J_{2,1}=\frac{ 1}{\pi}\int_{u-1}^{1}\frac{ds}{\sqrt{s}\sqrt{u-s}}=\frac{2}{\pi}\arcsin(\frac{1}{\sqrt{u}})-\frac{2}{\pi}\arcsin(\sqrt{\frac{u-1}{u}}).$$
 We write
$$J_{2,2}=\frac{1}{\pi}\int_{1}^{uv}\frac{ds}{\sqrt{s}\sqrt{u-1}}-\frac{1}{\pi}\int_{1}^{uv}\frac{\log(\sqrt{s}+\sqrt{s-1})}{\sqrt{s}\sqrt{u-s}}\,ds:=\widehat{J_{1}}-\widehat{J_{2}}.$$
We have
$$\widehat{J_{1}}=\frac{2}{\pi}\arcsin(\sqrt{v})-\frac{2}{\pi}\arcsin(\frac{1}{\sqrt{u}}).$$
Now it remains to study $\widehat{J_{2}}$. We put the change of variable  $s'=u-s$. We get
$$\widehat{J_{2}}=\frac{1}{\pi}\int_{u(1-v)}^{u-1}\frac{\log(\sqrt{u-s'}+\sqrt{u-s'-1})}{\sqrt{s'}\sqrt{u-s'}}\,ds'$$
As $v\geqslant \frac{1}{u}$ we have $u(1-v)\leqslant u-1\leqslant 1$. Therefore using notations in (14) (15) and (16) and Lemmas 5.1 et 5.2 we obtain
$$\widehat{J_{2}}=S(1-\frac{1}{u})-S(1-v).$$
We complete the proof by grouping  different  above estimates.
\hfill{
$\square$}\\[0.5cm]
\subsection{Proof of Theorem 2.3}
Theorem 2.3 follows  from  (15) and different Lemmas  of section 5.   \hfill{
$\square$}\\[0.5cm]

\selectlanguage{english}

config: arxiv


TITLE: Generalization of the Beltrami identity to two functions of two variables
QUESTION [3 upvotes]: Let $I$ be the functional defined by
$$I[u]=\int_{a}^b L(x,u(x),u(x)')\,dx$$
where $a$ and $b$ are constants and $u'=du/dx$. The Beltrami identity states that  if $\partial L/\partial x=0$, then the Euler-Lagrange equation
$$ \frac{\partial L}{\partial u}-\frac{d}{dx}\left(\frac{\partial L}{\partial u'}\right)=0$$
can be integrated once, giving rise to
$$
L-u'\frac{\partial L}{\partial u'}=\text{constant}.$$
Now assume that $I$ is a functional of two functions that depends on two variables, that is
$$
I[u,v]=\int_a^b\int_c^d L(x,y,u(x,y),v(x,y),u_{,x}(x,y),u_{,y}(x,y),v_{,x}(x,y),v_{,y}(x,y))\,dxdy,$$
where $a$, $b$, $c$ and $d$ are constants and $u_{,x}=\partial u/\partial x$, etc. If $\partial L/\partial x=0$ (or $\partial L/\partial y=0$). Is there any identity analogous to the Beltrami identity for the Euler-Lagrange equation?

REPLY [0 votes]: Well, for a Lagrangian density ${\cal L}(x,u(x),\partial u(x))$ in  $n$ variables $\{x^1,\ldots,x^n\}$ with no explicit dependence
$$\frac{\partial {\cal L}}{\partial x^k}~=~0$$
along the $k$'th coordinate, there is a corresponding continuity equation
$$ \sum_{j=1}^n\frac{d T^{j}{}_k}{dx^j}~=~0,$$
where the canonical stress-energy-momentum tensor is
$$  T^{j}{}_{\ell}~:=~\frac{\partial {\cal L}}{\partial u_{,j}} u_{,\ell}-\delta^j_{\ell} {\cal L}. $$
One can in principle create a conserved quantity by integrating $T^{j}{}_k$ over (an appropriate region of) the hyperplane $\{x^j={\rm const}\}$, cf. Noether's theorem. Note that the choice of hyperplane is not unique.


text:
\begin{document}

\maketitle

\begin{abstract}
We consider whether algorithmic choices in over-parameterized linear matrix factorization introduce implicit regularization.
We focus on noiseless matrix sensing over rank-$r$ positive semi-definite (PSD) matrices in $\mathbb{R}^{n \times n}$, with a sensing mechanism that satisfies restricted isometry properties (RIP).
The algorithm we study is \emph{factored gradient descent}, where we model the low-rankness and PSD constraints with the factorization $UU^\top$, for $U \in \mathbb{R}^{n \times r}$.
Surprisingly, recent work argues that the choice of $r \leq n$ is not pivotal: even setting $U \in \mathbb{R}^{n \times n}$ is sufficient for factored gradient descent to find the rank-$r$ solution, which suggests that operating over the factors leads to an implicit regularization.
In this contribution, we provide a different perspective to the problem of implicit regularization. We show that under certain conditions, the PSD constraint by itself is sufficient to lead to a unique rank-$r$ matrix recovery, without implicit or explicit low-rank regularization.
\emph{I.e.}, under assumptions, the set of PSD matrices, that are consistent with the observed data, is a singleton, regardless of the algorithm used.
\end{abstract}

\section{Introduction}

In this work, we study how \emph{over-parameterization} relates to regularization \citep{zhang2016understanding}.
By over-parameterization, we mean that the number of parameters to estimate is larger than the available data, thus leading to an under-determined system.\footnote{It helps picturing over-parameterization via a simple linear system of equations: when the number of parameters is more than the number of equations, then there is an infinite number of solutions, and which is the one we choose depends on additional regularization bias.}
E.g., deep neural networks are usually designed over-parameterized, with ever growing number of layers, and, eventually, a larger number of parameters \citep{telgarsky2016benefits}.
What is surprising though is the lack of \emph{overfitting} in such networks: while there could be many different parameter realizations that lead to zero training error, the algorithms select models that also generalize well to unseen data, despite over-parameterization \citep{keskar2016large, poggio2017theory, soltanolkotabi2017theoretical, dinh2017sharp, cooper2018loss}.

The authors of \citep{li2017algorithmic} show that the success of over-parameterization can be theoretically fleshed out in the context of shallow, linear neural networks.
They consider the case of low-rank and positive semi-definite (PSD) factorization in matrix sensing \citep{recht2010guaranteed}:
given measurements $y = \mathcal{A}(X^\star) \in \mathbb{R}^m$ ---where $X^\star \in \mathbb{R}^{n \times n}$ has rank $r \ll n$ and is PSD, and $\mathcal{A} : \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^m$ satisfies the restricted isometry property--- they prove that a \emph{square and full-rank} factorized gradient descent algorithm over $U \in \mathbb{R}^{n \times n}$, where $X = UU^\top$, converges to $X^\star$.
\emph{I.e.}, whereas the algorithm has the expressive power to find any matrix $X$ that is consistent with the noiseless data (and due to over-parametrization there are infinitely many such $X$'s), in contrast, it automatically converges to the minimum rank solution.
This argument was previously conjectured in \citep{gunasekar2017implicit}.

This could be seen as a first step towards understanding over-parameterization in general non-linear models, whose objectives are more involved and complex.
Such network simplifications have been followed in other recent works in machine learning and theoretical computer science, such as in convolutional neural networks \citep{du2017convolutional},

and landscape characterization of generic objectives \citep{baldi1989neural, boob2017theoretical, safran2017spurious}.

In this work, we provide a different perspective on the interpretation of over-parameterization in matrix sensing.
We show that, in the noiseless case, \emph{the PSD constraint by itself could be sufficient to lead to a unique matrix recovery from observations, without the use of implicit or explicit low-rankness.}
In other words, the set of PSD matrices that satisfy the measurements is a singleton,

irrespective of the algorithm used.

\textbf{Notation.}
Vectors are denoted with plain lower case letters; matrices are denoted with capital letters; and mappings, from one Euclidean space to another, are denoted with capital calligraphic letters.
Given $x \in \mathbb{R}^n$, its $\ell_1$-norm is defined as $\|x\|_1 = \sum_{i = 1}^n |x_i|$, where $x_i$ denotes its $i$-th entry; similarly, we define the $\ell_2$-norm as $\|x\|_2 = \sqrt{\sum_{i = 1}^n x_i^2}$.
The $\ell_0$-pseudonorm, $\|x\|_0$, is defined as the number non-zero entries in $x$.
Given $x$, $\texttt{diag}(x) \in \mathbb{R}^{n \times n}$ is the diagonal matrix with diagonal entries the vector $x$.
For two matrices $X, Y$ with appropriate dimensions, we define their inner product as $\left\langle X, Y \right\rangle = \text{Tr}(X^\top Y)$, where $\text{Tr}(\cdot)$ is the trace operator.
Given $X \in \mathbb{R}^{n \times n}$, the nuclear norm is defined as $\|X\|_* = \sum_{i = 1}^n \sigma_i(X)$, where $\sigma_i(X)$ is the $i$-th singular value.
The spectral norm is denoted as $\|X\| = \sigma_{\max}(X)$, where $\sigma_{\max}(\cdot)$ is the maximum singular value.

\subsection{Related work}
\textbf{Implicit regularization in matrix sensing.}
This area was initiated by the conjecture in \citep{gunasekar2017implicit}: The authors suggest that non-convex gradient descent on a full-dimensional factorization $UU^\top$, where $U \in \mathbb{R}^{n \times n}$, converges to the minimum nuclear norm solution.
\citep{li2017algorithmic} sheds light on this conjecture: they theoretically explain the regularization inserted by algorithms, even beyond learning matrix factorization models, such as one-hidden-layer neural nets with quadratic activation; see also \citep{du2018power}.

\textbf{Implicit regularization beyond matrix sensing.}
For the general linear regression setting, \citep{wilson2017marginal} shows that, under specific assumptions, adaptive gradient methods, like AdaGrad and Adam, converge to a different solution than the simple (stochastic) gradient descent (SGD); see also \citep{gunasekar2018characterizing}.
SGD has been shown to converge to the so-called \emph{minimum norm solution};
see also \citep{soudry2017implicit} for the case of logistic regression.
This behavior is also demonstrated using DNNs in \citep{wilson2017marginal}, where simple gradient descent generalizes as well as the adaptive methods.

\textbf{No spurious local minima.} There is a recent line of work, focusing on non-convex problems, that state conditions under which problem formulations actually have no-spurious local minima, when we transform the problem from its convex formulation to a non-convex one.
Characteristic examples include that factored gradient descent does not introduce spurious local minima in matrix completion \cite{ge2016matrix} and matrix sensing \cite{bhojanapalli2016global, park2017non}, and all local minima are global in some tensor decompositions \cite{ge2015escaping} and dictionary learning \cite{sun2016complete};
see \cite{ge2017no, sun2015nonconvex} for a complete overview of these results.
Further, there is literature that characterizes the landscape of factorization problems, using the strict saddle property, to indicate that we can escape easily any saddle point \cite{ge2015escaping, zhu2018global}; in this work, we take a different path showing that by construction the set of solutions that satisfy our observations is a singleton, demystifying the behavior of factored gradient descent in over-parameterized matrix sensing.

\section{Nonnegativity and sparsity: the vector analog of PSD and low rankness}\label{sec:sparsity}
We briefly describe the work of \citep{bruckstein2008uniqueness}, as we borrow ideas from that paper.
Consider the problem of finding a non-negative, sparse solution to an over-parameterized linear system of equations: $A x^\star = b$.
Here, the \textit{sensing matrix} $A \in \mathbb{R}^{m \times n}$, where $m < n$, the unknown $x^\star \in \mathbb{R}^n$ satisfies $x^\star \geq 0$ (entrywise) and is sufficiently sparse $\|x^\star\|_0 \leq k$, and the measurements are $b \in \mathbb{R}^{m}$.

This scenario suggests the following optimization problem as a solution:
\begin{equation}
\begin{aligned}
& \underset{x \in \mathbb{R}^n}{\text{min}}
& & f(x)
& \text{subject to}
& & b = Ax \text{ and } x \geq 0.
\end{aligned} \label{eq:f1}
\end{equation}
Here, $f$ is a function metric that measures the quality of the candidate solutions.
Examples are $f(x) = \|x\|_2^2$ (\emph{i.e.},  the minimum norm solution that satisfies the constraints), $f(x) = \|x\|_1$ (\emph{i.e.}, the solution that has small $\ell_1$-norm, and promotes sparsity), and $f(x) = \|x\|_0$ (\emph{i.e.}, the solution with the smallest number of non-zeros).
These tasks have been encountered in statistics, computer vision and signal processing applications \citep{zass2007nonnegative, shashua2005non, hazan2005sparse}, and they are popular in the compressed sensing literature \citep{donoho2006compressed, foucart2013mathematical}, when $x^\star$ is assumed sparse.

Let us disregard for the moment the positivity constraints on $x$.
By definition, an over-parameterized linear inverse problem has infinite number of solutions.
Unless we use the information that $x^\star$ is sparse, its reconstruction using only $b$ and $A$ is an ill-posed problem, and there is no hope in finding the true vector without ambiguity.

Therefore, to reconstruct $x^\star$ in an over-parametrized setting, prior knowledge should be exploited by the optimization solver.
Compressed sensing is an example where additional constraints restrict the feasible set to a singleton: under proper assumptions on the sensing matrix $A$ --such as the restricted isometry property \citep{candes2008restricted}, or the coherence property \citep{bruckstein2008uniqueness}-- and assuming sufficient number of measurements $m < n$, one can show that the feasible set $\{x : Ax = b \text{ and } \|x\|_0 \leq k\}$ contains only one element, for sufficiently small $k$.

Re-inserting the positivity constraints in our discussion, \citep{bruckstein2008uniqueness} show that, when a sufficiently sparse solution $x^\star$ generates $b = Ax^\star$, and assuming the row-span of $A$ intersects with the positive orthant, then \emph{the non-negative constraint by itself is sufficient to identify the sparse $x^\star$, and reduce the cardinality of the feasible solutions $\{x : Ax = b\}$ to singleton.}

In other words, the inclusion of a sparsity inducing $f$ in \eqref{eq:f1} is not needed,

even if we know a priori that $x^\star$ is sparse; non-negativity is sufficient to find a unique solution to the feasibility problem:
\begin{equation*}
\begin{aligned}
& \text{find}
& & x
& \text{such that}
& & b = Ax \text{ and } x \geq 0,
\end{aligned}
\end{equation*}
that matches $x^\star$.
This way, we can still use convex optimization solvers --linear programming in this particular case-- and avoid hard non-convex problem instances.

\section{The matrix sensing problem for PSD matrices}\label{sec:lowrank}
Let us now describe the \emph{matrix sensing} problem, draw the connections with the vector case, and study the over-parametrization $X = UU^\top$, for $U \in \mathbb{R}^{n \times n}$. Following \citep{li2017algorithmic}, we consider the PSD-constrained case, where the optimum solution is both low-rank and PSD.

A rough description is as \emph{a problem of linear system of equations over matrices}.
It is derived by the generative model $b = \mathcal{A}\left(X^\star\right)$, where $X^\star \in \mathbb{R}^{n \times n}$ is the low-rank, PSD ground truth.
Let the true rank of $X^\star$ be $r \ll n$.
The mapping $\mathcal{A}:\mathbb{R}^{n \times n} \rightarrow \mathbb{R}^m$ is such that the $i$-th entry of $\mathcal{A}(X)$ is given by $\left(\mathcal{A}(X)\right)_i = \left\langle A_i, X \right\rangle$, for $A_i \in \mathbb{R}^{n \times n}$ independently drawn symmetric measurement matrices.

We study the PSD-constrained formulation, where we aim to find $X^\star$ via:
\begin{equation}
\begin{aligned}
&\underset{X \in \mathbb{R}^{n \times n}}{\text{min}}
& & f(X)
& \text{subject to}
& & b = \mathcal{A}(X),~X \succeq 0.
\end{aligned} \label{eq:MS}
\end{equation}
$f(X)$ again represents a function metric that promotes low-rankness; standard choices include the nuclear norm $f(X) = \|X\|_*$ (which imposes ``sparsity" on the set of singular values and hence low-rankness), and the non-convex $f(X) = \text{rank}(X)$ metric.

Practical methods for this scenario include
$(i)$ the PSD-constrained basis pursuit algorithm for matrices \citep{chen2001atomic, goldstein2010high} that solve \eqref{eq:MS} for $f(X) := \|X\|_*$ using interior-point methods \citep{liu2009interior};
and $(ii)$ projected gradient descent algorithms, that solve an equivalent form of \eqref{eq:MS} for wisely chosen $\lambda> 0$:
\begin{equation}
\begin{aligned}
&\underset{X \in \mathbb{R}^{n \times n}}{\text{min}}
& & g(X) := \tfrac{1}{2} \|b - \mathcal{A}(X)\|_2^2 \\
& \text{subject to}
& & X \succeq 0, f(X) \leq \lambda.
\end{aligned} \quad \quad \text{via} \quad X_{i+1} = \Pi_{\mathcal{C}} \left( X_i - \eta \nabla g(X_i) \right), \label{eq:MS1}
\end{equation}

for $\Pi_{\mathcal{C}}(Y) := \arg\min_{X \in \mathcal{C}} \tfrac{1}{2} \|X - Y\|_F^2$, and $\mathcal{C} := \{X : X \succeq 0, f(X) \leq \lambda\}$ \citep{kyrillidis2011recipes, kyrillidis2014matrix, khanna2017iht}. In the latter, the objective $f$ appears in the constraint set as $f(X) := \text{rank}(X)$ or $f(X) := \|X\|_*$ .

Recently, we have witnessed a series of works \citep{zhao2015nonconvex, park2016provable, park2016non, sun2016guaranteed,
bhojanapalli2016global, park2016finding, kyrillidis2017provable, ge2017no, hsieh2017non}, that operate directly on the factorization $X = UU^\top$, and do not include any PSD and rank constraints.
This is based on the observation that, for any rank-$r$ and PSD $X$, the factorization $UU^\top$, for $U \in \mathbb{R}^{n \times r}$, guarantees that $X (= UU^\top)$ is at the same time PSD and at most rank-$r$.
This re-parameterizes \eqref{eq:MS} as:
\begin{equation*}
\begin{aligned}
& \text{find}
& & U \in \mathbb{R}^{n \times r}
& \text{subject to}
& & b = \mathcal{A}(UU^\top),
\end{aligned}
\end{equation*}
and \eqref{eq:MS1} as:
\begin{equation*}
\begin{aligned}
&\underset{U \in \mathbb{R}^{n \times r}}{\text{min}}
& & g(UU^\top) := \tfrac{1}{2} \|b - \mathcal{A}(UU^\top)\|_2^2
\end{aligned}
\end{equation*}
Observe that in both cases, there are no metrics that explicitly favor low-rankness or any PSD constraints; these are implicitly encoded by the factorization $UU^\top$.
Algorithmic solutions for the above criteria include the factorized gradient descent \citep{bhojanapalli2016dropping, park2016finding} that obeys the following recursion:

\begin{align}
U_{i+1} = U_i - \eta \nabla g(U_i U_i^\top) \cdot U_i. \label{eq:FGD}
\end{align}

Current theory \citep{bhojanapalli2016dropping, park2016finding} assumes that $r$ is known a priori, in order to set the dimensions of the factor $U \in \mathbb{R}^{n \times r}$, accordingly.
The only work that deviates from this perspective is the recent work in \citep{li2017algorithmic}, where the authors prove that even square $U \in \mathbb{R}^{n \times n}$ in \eqref{eq:FGD} still converges to the low-rank ground truth $X^\star$, with proper initialization and step size selection.
The result relies on restricted isometry assumptions of $\mathcal{A}$.
In a manner, this suggests that \emph{operating on the factorized space, the algorithm implicitly favors low-rank solutions, even if there is expressive power to select a full rank-$n$ $\widehat{X} = \widehat{U} \widehat{U}^\top$ as a solution.}
The following subsection provides a different perspective on the matter: \emph{the implicit PSD constraint in $UU^\top$ could be sufficient to reduce the feasibility set to singleton, no matter what algorithm is used for solution.}

\textbf{When positivity constraints are sufficient for unique recovery under RIP.}
We note that the \textit{Restricted Isometry Property} (RIP) assumption is made in \citep{zhao2015nonconvex, park2016provable, park2016non, sun2016guaranteed, bhojanapalli2016global, park2016finding, kyrillidis2017provable, ge2017no, hsieh2017non, li2017algorithmic}.
There are various versions of RIP, with the most well-known being the RIP-$\ell_2/\ell_2$ \citep{chen2015exact}.
Due to the construction of our sensing matrices as outer products of Gaussians---and thus the connection with rank-one measurements in matrix sensing---for our theory, we will also use a variant of RIP, RIP-$\ell_2 / \ell_1$ \citep{chen2015exact}.
The equivalence or superiority of one RIP definition over the other is not known, to the best of our knowledge.
The RIP of linear maps on low rank matrices is key in our disucssion \citep{candes2011tight, liu2011universal}:

\begin{definition}[RIP in $\ell_2 / \ell_1$ \cite{chen2015exact}]\label{def:RIP}
A linear map $\mathcal{F} : \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^m$ satisfies the $r$-RIP-$\ell_2 / \ell_1$ with constant $\delta_r$, if

$(1 - \delta_r)\|X\|_F \leq \|\mathcal{F}(X)\|_1 \leq (1 + \delta_r)\|X\|_F$,

is satisfied for all matrices $X \in \mathbb{R}^{n \times n}$ such that $\text{rank}(X) \leq r$. Here, $\|\cdot\|_1$ denotes the $\ell_1$-norm over matrices.
\end{definition}

\begin{corollary}[\citep{needell2009cosamp}]\label{eq:RIPrelation}
Let $\gamma$ and $r$ be positive integers. Then, $\delta_{\gamma r} \leq \gamma \cdot \delta_{2r}$.
\end{corollary}

We extend the results in the previous section, and \emph{prove that, under appropriate conditions, the set of solutions $\{X \in \mathbb{R}^{n \times n} : b = \mathcal{A}(X), X \succeq 0\}$ is a singleton.}
To generalize, in our theoretical developments we will consider the case where $\mathcal{A}$ is generated through a Gaussian process.

\textbf{Sensing mappings comprised of Wishart matrices and their properties.}
In particular, consider the sensing map $\left(\mathcal{A}(X)\right)_i = \langle A_i,  X \rangle$, where $A_i$ are non-singular Wishart matrices for $i = 1, \dots, m$, and $m$ is the total number of measurements.

\begin{definition}[Wishart Distribution, \cite{muirhead2009aspects}]\label{def:wishart}
Suppose that $Z \in \mathbb{R}^{p \times n}$ where each column $z_1, \ldots, z_p \sim \mathcal{N}_n(0, \Sigma)$ (multivariate normal with zero mean).
Define $n \times n$ matrix $A$ by $A = \sum_{i=1}^{p} z_i^T z_i = Z^T Z$.
We say that $A$ follows a Wishart distribution with $p$ degrees of freedom and covariance matrix $\Sigma \succeq 0$, which we denote by $A \sim W_n(p, \Sigma)$.
\end{definition}

Wishart matrices are commonly used to estimate covariance in high dimensional statistics \citep{vershynin2012close, chen2015exact}, and they come with nice properties that we will exploit in our theory.
To generate $\mathcal{A}$, we generate Wishart matrices $A_i$ as defined above, for $\Sigma = \sigma^2 I_n \succ 0$, and $I_n$ is the $n \times n$ identity matrix.

The parameter $p$ is user-defined and set to $p > n + 1$.
By assumption of $\Sigma$, all $A_i$'s are non-singular Wishart matrices \citep{eaton}.
This ensures that, $\forall A_i$, our theory holds by the properties of non-singular Wishart matrices: $i)$ the density function of $A_i$ exists, $ii)$ $A_i^{-1}$ exists, and $iii)$ $A_i$ is positive definite.

By definition of $A_i$ as positive definite matrices, this results to the following observation:

\begin{equation}
	\exists~\varphi = [\varphi_1,~\varphi_2,~\ldots,~\varphi_m]^\top \quad \text{such that} \quad B = \sum_{i=1}^m \varphi_i A_i \in \mathbb{R}^{n \times n} \quad \text{and} \quad B \succ 0. \label{eq:span}
\end{equation}
I.e., there exists at least one vector $\varphi$ such that the weighted sum of $A_i$'s is a positive definite matrix; this can be easily derived from the fact that by construction all $A_i$'s are positive definite; this also relates to the Farkas' Lemma for semidefinite programs \citep{lovasz2003semidefinite}.

To proceed, we require the following definitions of Wishart matrices: \vspace{-0.2cm}
\begin{itemize}[leftmargin=0.5cm]
    \item By \citep{muirhead2009aspects}, we know that $B$ is a non-singular wishart matrix satisfying: $B \sim W_n(m \cdot p, \Sigma)$. I.e., the weight sum of Wishart matrices satisfies the Wishart distribution.
    \item As a non-singular matrix, $B^{-1}$ exists and follows an inverse Wishart distribution.
	In particular, $B^{-1} \sim W_n^{-1}(mp + n + 1, \Sigma^{-1})$ \citep{supp}. \vspace{-0.2cm}
\end{itemize}
Further, since $B \succ 0$, there exists a unique $V \in \mathbb{R}^{n \times n}$ such that $B = VV^\top$. \vspace{-0.2cm}
\begin{itemize}[leftmargin=0.5cm]
	\item Regarding the decomposition $B = VV^\top$, we can extract information about $V$'s by Bartlett's Decomposition \citep{eaton}.
	In particular, the matrix $V$ is a lower triangular matrix, where the random variables $V_{kj}|k \geq j$ are mutually independent: for $k > j$, $V_{kj}$ follow a normal distribution as $V_{kj} \sim \mathcal{N}(0, \sigma^2)$, and diagonal elements of $V$ follow a chi-squared distribution as $V_{jj}^2 \sim \sigma^2 \cdot \mathcal{X}_{m-j+1}^2$, for all $j = 1, \ldots, n$.
\end{itemize}

\textbf{Equivalent reformulation of matrix sensing with fixed trace.}
Given the above set up, we will make the following connections, starting with the following change of variables.
Given the full rank $V$ such that $B = VV^\top$, and for each $A_i$, we define a new mapping $\mathcal{M}: \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^m$, such that:
\begin{align*}
\left( \mathcal{M}(X) \right)_i = \left \langle M_i,  X \right \rangle = \left \langle V^{-1} A_i \left(V^{-1}\right)^\top, X \right \rangle, \quad \text{for all} ~~i, \quad \text{where $M_i := V^{-1} A_i \left(V^{-1}\right)^\top$.}
\end{align*}

Given $X \in \mathbb{R}^{n \times n}$ and $X \succeq 0$, define the auxiliary variable $Y = V^\top X V \in \mathbb{R}^{n \times n}$; observe that, for full rank $V$, $Y \succeq 0$.
Then, for any $X \succeq 0$, we have:
\begin{align*}
b_i = \left(\mathcal{A}(X) \right)_i &= \langle A_i, X \rangle = \langle A_i, X VV^{-1} \rangle = \langle A_i \left(V^{-1}\right)^\top , X V \rangle \\
					  &= \langle A_i \left(V^{-1}\right)^{\top}, (V^{\top})^{-1} V^{\top} X V \rangle  \\
					  &= \langle V^{-1}A_i \left(V^{-1}\right)^{\top}, V^{\top} X V \rangle \\
					  &= \langle M_i, Y \rangle = \left(\mathcal{M}(Y) \right)_i,
\end{align*}
where the last equality is due to the definitions of $\left( \mathcal{M}(\cdot) \right)_i$ and $Y$.
For the rest of the discussion, we assume that $b = \mathcal{A}(X^\star)$, for rank-$r$ $X^\star$.

The above indicates the one-to-one correspondence between the original feasibility set and the corresponding set after the change of variables:
\begin{align}
\{X \in \mathbb{R}^{n \times n} : b = \mathcal{A}(X), X \succeq 0\} \quad \text{and} \quad \{Y \in \mathbb{R}^{n \times n} : b = \mathcal{M}(Y), Y \succeq 0\}.\label{eq:equivsets}
\end{align}
Further, the rank of the solutions, $X^\star$ and $Y^\star$, are the same.
After the change of variables to $\mathcal{M}$, for $X$ and $Y$ that belong to the above sets, we observe:
\begin{align*}
\text{Tr}(Y) \stackrel{(i)}{=} \text{Tr}(V^\top X V) &= \text{Tr}(X VV^\top) = \text{Tr}(X B) \stackrel{(ii)}{=} \text{Tr}\left(X \sum_{i=1}^m \varphi_i A_i\right) \\
								         &= \sum_{i = 1}^m \varphi_i \cdot \langle A_i, X \rangle \stackrel{(iii)}{=} \sum_{i = 1}^m \varphi_i \cdot b_i := c, \quad \text{for constant} ~~c.
\end{align*}
Here, $i)$ is due to the definition of $Y = V^\top X V$, $ii)$ is due to the assumption that the span of $\mathcal{A}$ is strictly positive and equals $B$, according to \eqref{eq:span}, and $iii)$ is due to $b_i = \langle A_i, X\rangle$, for $X$ being in the feasibility set.
\emph{This dictates that the trace of matrices in the set $\{Y \in \mathbb{R}^{n \times n} : b = \mathcal{M}(Y), Y \succeq 0\}$ is constant and does not depend on $X$ directly}; it only depends on the measurement vector $b$ and the vector $\varphi$ defined above.

Let us focus on the set $\{Y \in \mathbb{R}^{n \times n} : b = \mathcal{M}(Y), Y \succeq 0\}$.
By definition, $b = \mathcal{M}(Y^\star)$, where $Y^\star$ is rank-$r$ and relates to $X^\star$ in $Y^\star = V^\top X^\star V$.
Assume that $\mathcal{M}: \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^m$, is a linear map that satisfies the RIP in Definition \ref{def:RIP}.
Consider the convex optimization criterion with estimate $\widehat{Y}$:
\begin{align}
\widehat{Y} = \underset{Y \in \mathbb{R}^{n \times n}}{\text{argmin}} ~\|Y\|_*  \quad \text{subject to} \quad b = \mathcal{M}(Y). \label{eq:cvxsolver}
\end{align}

The following result is from \citep{chen2015exact}.
\begin{theorem}[Theorem 1 in \citep{chen2015exact} (Informal)]\label{thm:recht}
Assume $\mathcal{M}(\cdot)$ satisfies the RIP-$\ell_2/\ell_1$ for some $\delta_{\gamma r} < 1$, and for some integer $r \geq 1$.
Then, \eqref{eq:cvxsolver}, in the absence of
noise, allows perfect recovery of the unique $Y^\star$ that satisfies the measurements, with exponentially high probability, provided we have enough samples $O(\gamma nr)$.
\end{theorem}

Let us interpret and use this theorem.
Assume that $\text{rank}(Y^\star) = r$, and  $\delta_{\gamma r} < 1$.
Under these assumptions, the minimizer $\widehat{Y}$ of \eqref{eq:cvxsolver} is identical to the unique, rank-$r$ matrix $Y^\star$ that satisfies the the set of observations $b = \mathcal{M}(Y^\star)$.
Taking into account the PSD nature of $Y$, we have

$\|Y\|_* = \text{Tr}(Y) = c$.
Note that we do not include the constraint $\text{Tr}(Y) = c$, because any feasible solution should satisfy this condition (see above).
Also, we do not include the PSD constraint; the problem \eqref{eq:cvxsolver} is sufficient to guarantee uniqueness.

By the above theorem, any other PSD solution, $Y^\sharp$, that satisfies the measurements $b$ must have a nuclear norm larger than $\|\widehat{Y}\|_*$.
Being PSD, this also means $\text{Tr}(Y^\sharp) > c$, which implies that any other PSD solution is not in the feasible set $\{Y \in \mathbb{R}^{n \times n} : b = \mathcal{M}(Y), Y \succeq 0\}$.
Hence, this set contains only one element, by contradiction.

Due to the one-to-one correspondence between the sets in \eqref{eq:equivsets}

then, we infer that the first set is also a singleton.
This further implies that \emph{the inclusion of any metric $f$ that favors low-rankness in \eqref{eq:MS}-\eqref{eq:MS1} or restricting $U$ to be a tall matrix with wisely chosen $r$ in \eqref{eq:FGD} makes no difference, as there is only one matrix that fits measurements $b$.}

\textbf{RIP-$\ell_2/\ell_1$ for the new sensing mapping $\mathcal{M}(\cdot)$.}
Key assumption is that the RIP-$\ell_2/\ell_1$ holds for the transformed sensing map $\mathcal{M}(\cdot)$---and not the original sensing map $\mathcal{A}$.
Thus, in general, it is required to find such transformation between $\mathcal{A}$ and $\mathcal{M}$.

We know that $M_i := V^{-1} A_i \left(V^{-1}\right)^\top$, where $B = \sum_{i=1}^m \varphi_i A_i = VV^\top$.
By construction through the Wishart distribution, we know that $M_i$, by Theorem 3.2.4 from \citep{muirhead2009aspects}, satisfies:
$$M_i := V^{-1} A_i (V^{-1})^T \sim W_n(p, \Sigma_M),$$
where $\Sigma_M = V^{-1} \Sigma (V^{-1})^T = \sigma^2(VV^T)^{-1} = \sigma^2 B^{-1}$.

Using the following definition of sub-exponential random variables, we show Lemma \ref{wishart-sub-exp}.
\begin{definition}[Sum of Sub-Exponential Random Variables, \citep{sum-se}] \label{sum-se}
Suppose that $X_1, \ldots, X_n$ are independent $(\tau_i^2, b_i)$ sub-Exponential random variables.
Then, the sum $\sum_{i=1}^{n} X_i$ is $(\sum_{i=1}^{n} \tau_i^2, b*)$ sub-Exponential, where $b* = \max_i(b_i)$.
\end{definition}
\begin{lemma}\label{wishart-sub-exp}
Non-singular Wishart matrices are sub-exponential matrices.
\end{lemma}

\textit{Proof.}
Let $A$ be a non-singular Wishart random matrix.
According to our constructions so far, $A$ may be characterized by $A \sim W_n(p, \Sigma)$, $p > n$, and $\Sigma = \sigma^2 I_n \succ 0$.
Recall that $A := Z^T Z$, where $Z \in \mathbb{R}^{p \times n}$, where each row of $Z$ is generated from a multivariate normal distribution with zero mean.
It is well known that each row $Z_i$, $1 \leq i \leq p$, is a sub-Gaussian random vector \citep{vershynin2017four}, and that all elements of these vectors are sub-Gaussian random variables.
From \cite{ar-ast}, we know that both the square of a sub-Gaussian random variable, and the product of independent sub-Gaussian random variables, are sub-Exponential; see also Lemma 7 in \cite{ahmed2014compressive}.

We then can use the following result from \cite{foucart2019iterative}:
\begin{theorem}[\citep{foucart2019iterative}] \label{foucart}
Given sub-exponential sensing matrices $M_i$ for the matrix sensing setting over rank-$r$ matrices, the RIP-$\ell_2/\ell_1$ requirement in Definition \ref{def:RIP} is satisfied with probability:
\begin{align*}
    \mathbb{P}\left( \left|\|\mathcal{M}(X)\|_1 - \|X\|_F \right | > \delta \|X\|_F \right) \leq 2 e^{-\kappa \delta^2 m};
\end{align*}
for $\kappa$ constant related to the  subexponential distribution, and $m \geq c(\delta) n r$, for constant $c(\delta)$.
\end{theorem}

This completes the proof: In plain words, using Gaussian-generated sensing matrices, we can generate Wishart sensing maps $\mathcal{A}$, such that the solution cardinality that satisfies the observations is a singleton.

\begin{remark}{\label{rem:00}}
The above show that the specific RIP-$\ell_2/\ell_1$ assumption on $\mathcal{M}$ is a sufficient, \emph{but not a necessary}, condition to guarantee that the feasibility set  $\{X \in \mathbb{R}^{n \times n} : b = \mathcal{A}(X), X \succeq 0\}$ is a singleton, $X^\star$.
It remains an open question to find necessary conditions and possibly different sufficient conditions  --such as the incoherence condition in \citep{bruckstein2008uniqueness} for matrices, or the RIP-$\ell_2/\ell_2$-- that also lead to a singleton set. In~\citep{baldwin2015informational} the authors find particular instances of sensing maps that, while not satisfy RIP condition, lead to a singleton set.
It is interesting to study the sensing construction in the current paper for RIP-$\ell_2/\ell_2$ conditions using:
\begin{theorem}[\citep{chen2015exact}, Theorem 5] \label{chen-rip}
Let $\mathcal{P}(\cdot): \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^m$ be a sensing map, with individual matrices $P_i \in \mathbb{R}^{n \times n}$. Suppose that for all $1 \leq i \leq m$,

$\|P_i\|_2 \leq K$,

$\left\|\mathbb{E}\left[\mathcal{P}_i^* \mathcal{P}_i\right] - \mathcal{I}\right\|_2 \leq \frac{c_5}{n},$

hold for some quantity $K \leq n^2$.

For any small constant $\delta > 0$, if $m > c_0 r K^2 \log^7n$, then with probability at least $1 - 1/n^2$, one has

    $\mathcal{P}$ satisfies RIP-$\ell_2/\ell_2$ w.r.t. all matrices of rank at most $r$ and obeys $\delta_r \leq \delta$.

\end{theorem}
\end{remark}

\section{Experiments}\label{sec:experiments}
The aim of the following experiments is twofold: in Section 4.1, to show that the theory applies in practice; in Section 4.2, to show that a sensing map $\mathcal{A}$ beyond the Wishart distribution can be sufficient to lead to a good approximation of $X^\star$, without the use of explicit regularization for low-rankness.

\subsection{Using Wishart matrices in simulated matrix sensing problems}

\begin{wrapfigure}{r}{0.6\textwidth}
    \vspace{-0.4cm}
    \centering
    \includegraphics[width=0.9\linewidth]{plot2-crop.pdf}
	\caption{Results for simulation using Wishart sensing matrices $A_i$.}\label{f1}
\end{wrapfigure}
The following example shows some preliminary results, shown in Figure \ref{f1}.
Here, we compare $i)$ least squares in $X$ with no constraints (CVX - second order method), $ii)$ least squares in $X$ with PSD constraints (CVX - second order method), and $iii)$ least squares in $X$ with PSD constraints (using projected gradient descent).
The plot assumes small $n = 15$ for proof of concept, where $X \in \mathbb{R}^{n \times n}$, due to the computational restrictions the second order method poses; same behavior is observed for any value we tested.
The degrees of freedom are $n(n+1)/2$.
While, criterion $i)$ finds a good solution after observing $n(n+1)/2$ samples, as expected,
criteria $ii)-iii)$ find a relatively good solution well before that, which implies that the PSD constraint alone is sufficient to find the solution, only from a limited set of random measurements.

We generate $X^\star$ as a rank-$1$ PSD matrix, where $X^\star = \lambda v v^\top$ for random scalar $\lambda$ and vector $v$.
The measurement matrices $A_i$ are generated from a Gaussian distribution: $b \overset{i.i.d.}{\sim} \mathcal{N}_n(0, I_n)$ and $A_i = \frac{1}{2 \sqrt{n}}
b b^\top$.
Therefore the $A_i$ are symmetric and PSD, according to the construction described in the main text.
The measurements $y \in \mathbb{R}^m$ are generated by $y_i = \text{tr}(A_i^T X^\star)$ for $i = 1, \ldots, m$.

\subsection{Beyond Wishart matrices: quantum state tomography}
In this subsection, we consider the setting of quantum state tomography (QST):
We generate measurements according to $b_i = \left \langle A_i, X^\star \right \rangle ,~i = 1, \dots, m$, where
$A_i = \left(I \pm \otimes_{j=1}^q s_j\right)/2$ and the Pauli observable $\otimes_{j=1}^q s_j$ is randomly generated.
In all settings, for simplicity, we assume $X^\star \in \mathbb{C}^{2^q \times 2^q}$ is rank-1, PSD and normalized $\text{Tr}(X^\star) = 1$, to satisfy the QST setting.

Given $b$ and $\mathcal{A}$, we consider:
\begin{equation}
\begin{aligned}
&\underset{X \in \mathbb{C}^{n \times n}}{\text{min}}
& & \|X\|_* \\
& \text{subject to}
& & b = \mathcal{A}(X), \\
& & & X \succeq 0.
\end{aligned}
\quad ~~\Bigg |~~ \quad
\begin{aligned}
&\underset{X \in \mathbb{C}^{n \times n}}{\text{min}}
& & \|X\|_F \\
& \text{subject to}
& & b = \mathcal{A}(X), \\
& & & X \succeq 0.
\end{aligned}
\quad ~~\Bigg |~~ \quad
\begin{aligned}
&\underset{X \in \mathbb{C}^{n \times n}}{\text{min}}
& & \tfrac{1}{2} \|b - \mathcal{A}(X)\|_2^2 \\
& \text{subject to}
& & X \succeq 0.
\end{aligned} \label{eq:ls}
\end{equation}
\emph{I.e.}, $i)$ the left criterion is the \emph{nuclear-norm minimization} problem, with explicit regularization towards low-rank solutions \citep{recht2010guaranteed};
$ii)$ the middle criterion is the \emph{minimum-norm solution} problem, where the objective regularizes towards $X$ with the minimum Frobenius norm;
$iii)$ the right criterion is the \emph{PSD constrained, least-squares} problem, where the task is to fit the data subject to PSD constraints.
In the two latter settings, there is no explicit regularization towards low-rank solutions.

We use the CVX Matlab implementation, in its low-precision setting, to solve all problems in \eqref{eq:ls} \citep{gb08, cvx}.
The results are presented in Table \ref{table:1}: $\text{dist}(\widehat{X}, X^\star)$ denotes the entrywise distance $\|\widehat{X} - X^\star\|_F$.
Since the estimates $\widehat{X}$ in all criteria in \eqref{eq:ls} are only approximately low-rank\footnote{Due to numerical precision limits, non of the solutions are $X^\star$ nor rank-1 in the strict sense.}, we also report the entrywise distance between $X^\star$ and the best rank-1 approximation of $\widehat{X}$, denoted as $\widehat{X}_1$.
We consider four different settings for $(n^2, m)$ parameters; our experiments are restricted to small values of $q$ in $n^2 = (2^q)^2$, due to the high computational complexity of the CVX solvers (by default we use the \texttt{SDPT3} solver \citep{toh1999sdpt3}).
Note that this is a second-order algorithm.

\begin{table*}[!ht]
\centering
	\begin{scriptsize}
	\rowcolors{2}{white}{black!05!white}
	\begin{tabular}{c c c c c c c c c}
		\toprule
		& \multicolumn{2}{c}{$\min \|X\|_*$} & & \multicolumn{2}{c}{$\min \|X\|_F^2$} & & \multicolumn{2}{c}{$\min \tfrac{1}{2} \|b - \mathcal{A}(X)\|_2^2$} \\
		\cmidrule{2-3} \cmidrule{5-6} \cmidrule{8-9}
		$(n^2, m)$ & $\text{dist}(\widehat{X}, X^\star)$ & $\text{dist}(\widehat{X}_1, X^\star)$ & & $\text{dist}(\widehat{X}, X^\star)$ & $\text{dist}(\widehat{X}_1, X^\star)$  & & $\text{dist}(\widehat{X}, X^\star)$ & $\text{dist}(\widehat{X}_1, X^\star)$  \\
		$(256, 128)$ &  $3.58 \cdot 10^{-5}$ & $ 3.45 \cdot 10^{-5}$ & &  $4.66 \cdot 10^{-3}$ & $4.49 \cdot 10^{-3}$ & &  $1.54 \cdot 10^{-4}$ & $1.43 \cdot 10^{-4}$ \\
		$(1024, 288)$ &  $1.65 \cdot 10^{-5}$ & $1.63 \cdot 10^{-5}$ & &  $1.68 \cdot 10^{-3}$ & $1.64 \cdot 10^{-3}$ & &  $8.08 \cdot 10^{-5}$ & $7.61 \cdot 10^{-5}$ \\
		$(4096, 640)$ &  $1.84 \cdot 10^{-5}$ & $1.82 \cdot 10^{-5}$ & &  $1.51 \cdot 10^{-3}$ & $1.48 \cdot 10^{-3}$ & &  $1.04 \cdot 10^{-4}$ & $9.81 \cdot 10^{-5}$ \\
		$(16384, 1536)$ &  $1.28 \cdot 10^{-5}$ & $1.27 \cdot 10^{-5}$ & & $1.00 \cdot 10^{-3}$ & $9.98 \cdot 10^{-3}$ & & $5.45 \cdot 10^{-5}$ & $5.20 \cdot 10^{-5}$ \\
		\midrule
	\end{tabular}
	\end{scriptsize} \vspace{-0.3cm}
	\caption{Experimental results for \eqref{eq:ls}. $\text{dist}(\widehat{X}, X^\star)$ defines the entrywise distance $\|\widehat{X} - X^\star\|_F$.} \label{table:1}
\end{table*}

Table \ref{table:1} support our claim: \emph{All three criteria, and for all cases, lead to the same solution, while they all use different ``regularization'' in optimization}.
Any small differences can be assumed due to numerical precision, and not equivalent initial conditions.
We observe consistently that, using the nuclear-norm bias, we obtain a better approximation of $X^\star$.
Thus, \emph{using explicit regularization helps}.

\subsection{Behavior of first-order, non-convex solvers on $UU^\top$ parameterization}
In view of the previous results, here we study the behavior of first-order, non-convex solvers, that utilize the re-parameterization of $X$ as $UU^\top$.
We borrow the iteration in \citep{bhojanapalli2016dropping, park2016finding}, where:

$U_{i+1} = U_i - \eta \nabla g(U_i U_i^\top) \cdot U_i$,

for $g(UU^\top) := \tfrac{1}{2} \|b - \mathcal{A}(UU^\top)\|_2^2$.
We consider two cases: $i)$ $U \in \mathbb{C}^{n \times r}$ where $r$ is the rank of $X^\star$, and is assumed known \emph{a priori}; this is the case in \citep{bhojanapalli2016dropping, park2016finding} and has explicit regularization, as the algorithm operates only on the space of rank-$r$ matrices. $ii)$ $U \in \mathbb{C}^{n \times n}$ where we can operate over the whole space $\mathbb{C}^{n \times n}$; this is the case studied in \citep{li2017algorithmic}.

In both cases, the initialization $U_0$ and step size $\eta$ follow the prescriptions in \citep{park2016finding}, and they are computed using the same procedures for both cases.
Table \ref{table:2} reports our findings.
To ease comparison, we repeat the results of the least-squares objective in \eqref{eq:ls}.
We observe that all algorithms converge close to $X^\star$: obviously, using the \emph{a priori} information that $X^\star$ is rank-1 biases towards a low-rank estimate, where faster convergence rates are observed.
In the contrary, using $U \in \mathbb{C}^{n \times n}$ shows slower convergence towards the vicinity of $X^\star$;
nevertheless, the reported results suggests that still one can achieve a small distance to $X^\star$ ($\|\widehat{X} - X^\star\|_F \lesssim 10^{-2}$).
Finally, while $\widehat{X}$ could be full-rank, most of the energy is contained in a small number of principal components, indicating that all algorithms favor (approximately) low-rank solutions.

\begin{table*}[!ht]
\centering
	\begin{scriptsize}
	\rowcolors{2}{white}{black!05!white}
	\begin{tabular}{c c c c c c c c c}
		\toprule
		& \multicolumn{2}{c}{$\min \tfrac{1}{2} \|b - \mathcal{A}(X)\|_2^2$} & & \multicolumn{2}{c}{$U \in \mathbb{C}^{n \times r}$} & & \multicolumn{2}{c}{$U \in \mathbb{C}^{n \times n}$} \\
		\cmidrule{2-3} \cmidrule{5-6} \cmidrule{8-9}
		$(n^2, m)$ & $\text{dist}(\widehat{X}, X^\star)$ & $\text{dist}(\widehat{X}_1, X^\star)$ & & $\text{dist}(\widehat{X}, X^\star)$ & $\text{dist}(\widehat{X}_1, X^\star)$  & & $\text{dist}(\widehat{X}, X^\star)$ & $\text{dist}(\widehat{X}_1, X^\star)$  \\
		$(256, 128)$ &  $1.54 \cdot 10^{-4}$ & $1.43 \cdot 10^{-4}$ & &  $9.52 \cdot 10^{-5}$ & - & &  $3.12 \cdot 10^{-2}$ & $2.82 \cdot 10^{-2}$ \\
		$(1024, 288)$ &  $8.08 \cdot 10^{-5}$ & $7.61 \cdot 10^{-5}$ & &  $4.47 \cdot 10^{-5}$ & - & &  $1.87 \cdot 10^{-2}$ & $1.76 \cdot 10^{-2}$ \\
		$(4096, 640)$ &  $1.04 \cdot 10^{-4}$ & $9.81 \cdot 10^{-5}$ & &  $4.07 \cdot 10^{-5}$ & - & &  $2.51 \cdot 10^{-2}$ & $2.37 \cdot 10^{-2}$ \\
		$(16384, 1536)$ &  $5.45 \cdot 10^{-5}$ & $5.20 \cdot 10^{-5}$ & & $2.47 \cdot 10^{-5}$ & - & & $1.41 \cdot 10^{-2}$ & $1.35 \cdot 10^{-2}$ \\
		\midrule
	\end{tabular}
	\end{scriptsize} \vspace{-0.3cm}
	\caption{Results for $UU^\top$ parameterization. $\text{dist}(\widehat{X}, X^\star)$ defines the entrywise distance $\|\widehat{X} - X^\star\|_F$.} \label{table:2}
\end{table*}

\section{Conclusion}
In this manuscript we provide theoretical and practical evidence that in PSD, low-rank matrix sensing, the solution set is a singleton, under RIP assumptions and appropriate transformations on the sensing map $\mathcal{A}$.
In these cases, the PSD constraint itself provides guarantees for unique matrix recovery.
The question whether the above can be generalized to less restrictive linear sensing mappings $\mathcal{A}$ remains open.
Note that RIP is a sufficient but not a necessary condition; we believe that generalizing our work to more broad settings and assumptions is an interesting research direction.

\bibliographystyle{unsrt}
\bibliography{biblio}

\end{document}


TITLE: Bound on complex triples satisfying some simple equations.
QUESTION [1 upvotes]: Let $a,b,c \in \mathbb{C}$ be three complex numbers satifying the following equations :
$$\left\{ \begin{array}{rl} a+b+c &= 3 \\ a^2+b^2+c^2 &=9 \end{array} \right.$$
I want to prove that $\max(|a|, |b|, |c|)\geq 2$.
I can formulate this as an optimisation problem for a multiple real variable function and use Lagrange multipliers, but it would probably be extremely long. The problem seems simple enough that there should be a direct and elegant proof.
So, any help would be very appreciated. Thanks.

REPLY [1 votes]: Plotting the Regions
Since $ab+bc+ca=\frac12\left((a+b+c)^2-\left(a^2+b^2+c^2\right)\right)=0$, Newton's identities say that $a$, $b$, and $c$ are roots of
$$\newcommand{\Re}{\operatorname{Re}}
z^3-3z^2-abc=0\tag1
$$
Suppose that $|a|\le|b|\le|c|\lt2$
$$
\begin{align}
|c^3-3c^2|
&=|c|^2|c-3|\tag{2a}\\
&=|abc|\tag{2b}\\
&\le|c|^3\tag{2c}
\end{align}
$$
That is, $|c-3|\le|c|$, which means that $c$ is closer to $3$ than it is to $0$. The dividing line between "closer to $0$" and "closer to $3$" is $\Re(z)=\frac32$. Therefore, in addition to $|c|\lt2$, we know that $\Re(c)\ge\frac32$.
Given $c$, we can use $(1)$ and Newton's identities to show that $a+b=3-c$ and $ab=c^2-3c$. Then we can then compute
$$
\begin{align}
a&=\frac{3-c-\sqrt{9+6c-3c^2}}2\tag{3a}\\
b&=\frac{3-c+\sqrt{9+6c-3c^2}}2\tag{3b}
\end{align}
$$
These could be switched, but it will be seen that, given $(3)$ and the possible locus of $c$, $\Re(b)\ge2$.
Here is a diagram of the regions where $c$, $a+b$, $ab$, $b$, and $a$ lie.

The yellow moon-shaped region, where $b$ lies, belies the assumption that $|a|\le|b|\le|c|\lt2$. Let us now prove that $\Re(b)\ge2$.

Getting Real
Since
$$
\begin{align}
\cos(\theta)
&=2\cos^2(\theta/2)-1\tag{4a}\\
&\le\cos^2(\theta/2)\tag{4b}
\end{align}
$$
if we use the square root with the branch cut along the negative real axis, where $\cos(\theta/2)\gt0$, we have
$$
\overbrace{\Re\left(\sqrt{z}\right)}^{\sqrt{r}\cos(\theta/2)}\ge\overbrace{\sqrt{\Re(z)_+}}^{\sqrt{r}\sqrt{\cos(\theta)_+}}\tag5
$$
where $x_+=\max(x,0)$ for $x\in\mathbb{R}$.
With $c=x+iy$,
$$
\Re(9+6c-3c^2)=9+6x-3x^2+3y^2\tag6
$$
Therefore, for $\frac32\le x\lt2$
$$
\begin{align}
\Re(b)
&=\Re\left(\frac{3-c+\sqrt{9+6c-3c^2}}2\right)\tag{7a}\\
&\ge\frac{3-x+\sqrt{9+6x-3x^2}}2\tag{7b}\\[12pt]
&\ge2\tag{7c}
\end{align}
$$
Explanation:
$\text{(7a)}$: derived in $(3b)$
$\text{(7b)}$: apply $(5)$ and $(6)$
$\text{(7c)}$: the function is decreasing on $[1,3]$ and equals $2$ at $x=2$
$\text{(7c)}$ is a contradiction since we assumed $|a|\le|b|\le|c|\lt2$.


TITLE: Let $A$ is a real symmetric matrix, $B$ is a real antisymmetric matrix, $A^2 = B^2$, prove $A = B = 0$.
QUESTION [0 upvotes]: Let $\boldsymbol{A}$ is a real symmetric matrix, $\boldsymbol{B}$ is a real antisymmetric matrix, $\boldsymbol{A}^2 = \boldsymbol{B}^2$, prove $\boldsymbol{A} = \boldsymbol{B} = \boldsymbol{0}$.

I tried the second-order matrix.
Let $\boldsymbol{A} = \begin{bmatrix} a_{11} & a_{12} \\ a_{12} & a_{22} \end{bmatrix}$,
$\boldsymbol{B} = \begin{bmatrix} b_{11} & b_{12} \\ {-b_{12}} & b_{22} \end{bmatrix}$, $a_{ij}, b_{ij} \in \mathbb{R}$.
\begin{align}
\boldsymbol{A}^2 &= \begin{bmatrix} {a_{11}^2 + a_{12}^2} & {a_{11}a_{12} + a_{12}a_{22}} \\ {a_{11}a_{12} + a_{12}a_{22}} & {a_{12}^2 + a_{22}^2} \end{bmatrix} \\
\boldsymbol{B}^2 &= \begin{bmatrix} {b_{11}^2 - b_{12}^2} & {b_{11}b_{12} + b_{12}b_{22}} \\ {-b_{11}b_{12} - b_{12}b_{22}} & {-b_{12}^2 + b_{22}^2} \end{bmatrix}
\end{align}
Because $\boldsymbol{A}^2 = \boldsymbol{B}^2$, I got
\begin{align}
a_{11}^2 + a_{12}^2 &= b_{11}^2 - b_{12}^2 \\
a_{11}a_{12} + a_{12}a_{22} &= b_{11}b_{12} + b_{12}b_{22} \\
a_{11}a_{12} + a_{12}a_{22} &= -b_{11}b_{12} - b_{12}b_{22} \\
a_{12}^2 + b_{22}^2 &= -b_{12}^2 + b_{22}^2
\end{align}
Hence,
\begin{align}
a_{11}a_{12}+a_{12}a_{22}=b_{11}b_{12}+b_{12}b_{22}=0
\end{align}
I lost my momentum here.

REPLY [1 votes]: note the orthogonality
$-1\cdot \text{trace}\big(BA\big)= \text{trace}\big(B^TA\big)= \text{trace}\big(B^TA\big)^T= \text{trace}\big(A^TB\big)= \text{trace}\big(AB\big)= \text{trace}\big(BA\big)$
$\implies  \text{trace}\big(BA\big)= 0$
using this we can compute the squared Frobenius norm of A+B
$\Big \Vert A+B\Big \Vert_F^2= \text{trace}\big(A^TA\big) +\text{trace}\big(B^TA\big)+\text{trace}\big(A^TB\big)+\text{trace}\big(B^TB\big) $
$= \text{trace}\big(A^2\big) +0+0 -\text{trace}\big(B^2\big) $
$=0 $
$\implies A+B = \mathbf 0\implies A = -B $
so $A$ is both symmetric and skew symmetric, i.e. $A=A^T=-A^T \implies A=\mathbf 0=-B = B$


TITLE: Why doesn't a ray of light have enough momentum to make us fall?
QUESTION [4 upvotes]: Why can't light be so powerful that it has enough momentum to make us fall?

REPLY [2 votes]: As I indicated in my comment on Rod's answer, some very powerful lasers do exist - and while their photons don't have much momentum, they do "pack a mean punch".
In fact, laser ablation (where a laser beam produces significant local heating and material is ejected at high speed) may just produce the phenomenon needed.
Let's take the example of the LLNL laser at the National Ignition Facility. In a few nano seconds, it dumps a few MJ. Rod's answer assumed "perfect reflection". I would like to create an absorbing layer, and see what kind of punch that can give.
For example, if the ablated material had a velocity of 1000 m/s, you could get a 50 Ns pulse with just 50 gram of material. Let's assume that $\frac12 kT$ of the energy of the ablated atoms is available to provide thrust in the right direction, then for carbon atoms and a velocity of 1000 m/s you would need a temperature of about 700 K. This is quite possible, as demonstrated in this paper
If you wanted to ablate less material, you would need a correspondingly higher temperature.
My point is - if you shot a real laser at a person who was wearing an appropriate body armor, it would not be terribly hard (and certainly within the range of available technology) to "zap" a bit of their armor off and push them over.
The key difference between this answer and Rod's is in the mass used: the photons have a really, really terribly relationship between energy and momentum - the moment you can bring some "real world" mass into play, it becomes possible.


TITLE: Sum of Completion Times
QUESTION [0 upvotes]: I am trying to understand the solution of sum of completion times. I found the following stuff in the link:
Prompt Scheduling for Selfish Agents
Paper says:

Consider the case of a single queue, a selfish agent will simply join
  the queue immediately upon arrival, there is no reason to delay. Thus,
  jobs will be processed in first-in-first-out (FIFO) order. However,
  this may be quite bad in terms of the sum of completion times. Imagine
  a job with processing time L, arriving at time zero, followed by
  sqrt(L) jobs of length 1, all of which arrive immediately after the
  first. As the first job will only be done at time L, the sum of
  completion times for these 1 +sqrt(L) jobs is about L^(3/2)

I can’t understand how this value of L^(3/2) is obtained.
Is it simply:

L + L^(1/2)

Somebody please guide me.
Zulfi.

REPLY [1 votes]: The length-$L$ job obviously has a turnaround time (enqueue-to-done-processing) of $L$. The first length-$1$ job has a turnaround time of $L+1$, the second $L+2$ and so on, up to $L+\sqrt L$ for the last job. The aforementioned sum of completion times is
$$L+(L+1)+(L+2)+\dots+(L+\sqrt L)\sim L\sqrt L+\frac{\sqrt L(\sqrt L+1)}2\sim L^{3/2}$$


TITLE: Newton's Universal Law of Gravitation doubt
QUESTION [2 upvotes]: The Universal Law of Gravitation states that the module of the force, $F$ is
$$F = \frac{GmM}{r^2},$$
where $m$ and $M$ are the mass of the two objects and $r$ is the distance between the two objects. From this, it is also derived that the gravitational potential energy is equal to
$$ E = -\frac{GmM}{r}$$

So does this mean that when the two objects are together, i.e. $r=0$, the force and the gravitational energy is infinite? How is it possible that energy and force be infinite?

Please can you explain my doubt?

REPLY [3 votes]: At $r=0$, the force is $+\infty$, and the potential energy is $-\infty$. This arises if you consider point masses, which we model mathematically using Dirac-delta distributions. Physically, we cannot have two point masses overlapping ($r=0$) since this would require an infinite amount of energy. Why? Well, fix one mass, and bring the other mass from $r=\infty$ towards $r=0$. Then, as the masses get closer, more and more energy is required to bridge the gap. So there is no contradiction, because you can never construct any physical system with overlapping point masses. Furthermore, in any real system, the objects are not point masses, but have some finite volume. Even if you consider an electron as a point mass, the electromagnetic force would far outweigh the gravitational force.

REPLY [1 votes]: The thing is, that, as every equation, these are based on model views. The model in this case includes the representation of the mass object as a point mass. Hence we are able to attribute to it some coordinate and measure distances between objects, which, in fact, consist of several gravitating "sub"objects. One of the properties of the point mass is that it can't be overlapped with another object (including another point mass). That is, the equation, which represents the model, includes a restriction on r = 0. Thus, in frames of this model (or equation), it is improper to consider r = 0 - as, obviously, model wouldn't work in this case.