TITLE: Classifying a point on the unit ball of a normed space as extreme depending on the field
QUESTION [0 upvotes]: Sorry for the awkward title.
I'm looking for an example of a normed space $X$ and a point $x$ on $X$'s unit ball such that $x$ is an extreme point of the unit ball when considering $X$ as a vector space over the reals, but not when considering $X$ as a vector space over the complex field - or vice versa.
Or, if such a setting could not exist, some intuitive justification for that.
Thanks in advance.

REPLY [3 votes]: I suspect you haven't thought carefully enough about this. By definition, a point $x$ in some convex set $C \subseteq X$ is extreme if it cannot be written as $x = ty + (1-t)z$ for $y,z \in C$ and $t \in (0,1)$. In other words, the property depends only on the structure of $X$ as a real vector space, no complex numbers will ever enter the picture. So, at least for restriction of scalars it makes no difference. If, instead, you start with a real vector space $X$ and take its complexification $X^{\mathbb{C}} = X \otimes_{\mathbb{R}} \mathbb{C}$, we have $X^{\mathbb{C}} \cong X \oplus iX$ (as real vector spaces) and an embedding $i: X \hookrightarrow X^{\mathbb{C}}$ by $x \mapsto (x,0)$. In particular, scalar multiplication by real numbers on $X$ is unchanged (for that reason, it is called an extension of scalars), so again a point $x$ in $C$ is extreme if and only if $i(x)$ is extreme in $i(C)$. If you had something else in mind, please edit your question appropriately.


TITLE: A binomial sum is divisible by p^2
QUESTION [18 upvotes]: This is a question I have since longer time, but I have absolutely no idea how to proceed on it.
Let $p>3$ be a prime. Prove that $\displaystyle\sum\limits_{k=1}^{p-1}\frac{1}{k}\binom{2k}{k}\equiv 0\mod p^2$. Here, we work in $\mathbb{Z}_{\mathbb{Z}\setminus p\mathbb{Z}}$ (that is, $\mathbb{Z}$ localized at all numbers not divisible by $p$).
I know that it is $0\mod p$ (though I can't find the reference at the moment; it was some hard olympiad problem on MathLinks). The $0\mod p^2$ assertion is backed up by computation for all $p<100$. I am sorry if this is trivial or known. I would be delighted to see a combinatorial proof (= finding a binomial identity which reduces to the above when computed $\mod p^2$). Some number-theoretical arguments would be nice, too. However, I fear that if you use analytic number theory, I will not understand a single word.
EDIT: Epic fail at question title fixed.

REPLY [4 votes]: For the asymptotics: begin with the well-known generating function
$$ \sum_{k \ge 0} {2k \choose k} z^k = (1-4z)^{-1/2}. $$
We can remove the first term to get
$$ \sum_{k \ge 1} {2k \choose k} z^k = (1-4z)^{-1/2} - 1. $$
Dividing by $z$, and then integrating with respect to $z$, turns $z^k$ into $z^k/k$.
$$ \sum_{k \ge 1} {2k \choose k} {1 \over k} z^k = -2 \log (1 + \sqrt{1-4z}) + 2 \log 2. $$
(The $2 \log 2$ is a constant of integration.)
Finally, let $f(n) = \sum_{k=1}^n {1 \over k} {2k \choose k}$.  These are the partial sums of the previous sequence, so we have
$$ F(z) = \sum_{n \ge 1}  f(n) z^n = {-2 \over 1-z} \log( 1 + \sqrt{1-4z} ) $$
We can find the asymptotics of the coefficients of this generating function by singularity analysis.  The singularity closest to the origin is at $z = 1/4$.  Near $z = 1/4$,
$$ F(z) \sim {-8 \over 3} \sqrt{1-4z} $$
and by a transfer theorem (probably originally due to Flajolet and Odlyzko and in the Flajolet-Sedgewick book Analytic Combinatorics, although it's too late to look up the details right now) we get
$$ f(n) = [z^n] F(z) \sim {-8 \over 3} [z^n] \sqrt{1-4z} $$
Finally, $[z^n] \sqrt{1-4z} = -{2 \over n} {2n-2 \choose n-1}$, and so by Stirling's formula, as $n \to \infty$
$$ f(n) \sim {-8 \over 3} {-2 \over n} {4^{n-1} \over \sqrt{\pi n}} = {4 \over 3} {4^n \over \sqrt{\pi n^3}}. $$
Finally, because I was somewhat awkward in my indexing, we have to take $p = n-1$, and so
$$ f(p) \sim {1 \over 3} {4^p \over \sqrt{\pi p^3}} $$.
As has been observed by John Mangual and Mariano Suárez-Alvarez, this is about one-third of the $p$th Catalan number.  Examining the singularity near $z = 1/4$ more closely would lead to higher-order terms.


TITLE: What is the value of $\lim_{(x,y,z)\rightarrow(0,0,0)}\log(\frac{x}{yz})$?
QUESTION [2 upvotes]: What is the value of
  $\lim_{(x,y,z)\rightarrow(0,0,0)}\log(\frac{x}{yz})$?
(a)$0$
(b)$2$
(c)$4$
(d)does not exist

Let us take $y=mx$ and $z=mx$ where $m$ is an arbitrary parameter,then
$\lim_{(x,y,z)\rightarrow(0,0,0)}\log(\frac{x}{yz})=\lim_{m\rightarrow 0}\log(\frac{1}{m^2})=$does not exist
Is it correct??

REPLY [4 votes]: Note that the expression is well defined for $x/yz >0$, in this case

for $x=y=z=t$ with $t\to 0^+$

$$\log(\frac{x}{yz})=\log(\frac{1}{t})\to -\infty$$

for $x=t^2,\,y=z=t$ with $t\to 0^+$

$$\log(\frac{x}{yz})=\log(1)\to 0$$
thus the answer is correct since
$$\lim_{(x,y,z)\rightarrow(0,0,0)}\log(\frac{x}{yz})$$
doesn't exist.


TITLE: Find integrable $g$ such that $\left|\dfrac{\partial}{\partial t}e^{-tx^2}\right|\leqq g(x)$.
QUESTION [2 upvotes]: I have $\int_{-\infty}^{\infty} e^{-tx^2} dx=\sqrt{\dfrac{\pi}{t}}$ for $t>0$.
I want to differentiate this.
That is, I want to do
$$\dfrac{d}{dt}\int_{-\infty}^{\infty} e^{-tx^2} dx=\int_{-\infty}^{\infty} \dfrac{\partial}{\partial t}e^{-tx^2} dx=\dfrac{\partial}{\partial t}\sqrt{\dfrac{\pi}{t}}=-\dfrac{1}{2}\sqrt{\pi}t^{-\frac{3}{2}}.$$
But in order to do this, I have to find $g(x)$ s.t.
$g(x)$ is indipendent of $t$ and integrable on $(-\infty, \infty)$, and
$\bigg| \dfrac{\partial}{\partial t}e^{-tx^2}\bigg| \leqq g(x)$.
Of course, I know $\dfrac{\partial}{\partial t}e^{-tx^2}=-x^2e^{-tx^2}$ but I couldn't find such $g$.
What $g$ works ?

REPLY [3 votes]: "But in order to do this, I have to find $g(x)$ s.t. $g(x)$ is independent of $t$ and integrable on $(-\infty, \infty)$, and $\bigg| \dfrac{\partial}{\partial t}e^{-tx^2}\bigg| \leqq g(x)$." No, that's not right. Given some $t_0>0$ you only need that estimate in a small open interval centered at $t_0$ that stays away from $0.$


TITLE: Find all 4x4 A matrices so that $A^4=A^6$
QUESTION [5 upvotes]: Find all 4x4 A matrices so that $A^4=A^6$.
I think the method has to do something with eigenvalues, eigenvectors etc'...
Thanks in advance for any assistance!

REPLY [2 votes]: Hint:
We can write $A^6 - A^4 = 0$, which is to say that $A$ "satisfies" the polynomial $x^6 - x^4 = 0$.  We then know (by a theorem that is probably in your textbook) that the minimal polynomial of $A$ divides $x^6 - x^4 = x^4(x-1)(x+1)$.
What does this tell us about the Jordan-canonical form of $A$?
Connection between minimal polynomials and J-C form:

$q_A(x) = (x-\lambda_1)^{m_1}\cdots(x-\lambda_k)^{m_k}$, where $\lambda_1,\dots,\lambda_k$ are the eigenvalues of $A$, and $m_k$ is the length of the longest Jordan block associated with $\lambda_k$.
$m_k \geq 1$ for every eigenvalue of $A$. $A$ is diagonalizable if and only if $m_k=1$ for all $k$.


TITLE: Infinite series with binomial summation
QUESTION [2 upvotes]: If I have an infinite series with first term two and for any other term:
$$a_n=\sum_{k=1}^{n-1}\binom{4n-2}{4k-1}a_ka_{n-k}$$
What is the general term formula?

REPLY [2 votes]: Let $y=\sum_{n=1}^{\infty} \frac{a_n x^n}{(4n-1)!}$. Then by applying $\sum_{n=2}^{\infty}$ to the recurrence, we obtain
$$
\sum_{n=2}^{\infty} \frac{a_n x^n}{(4n-1)!}(4n-1)=\sum_{n=2}^{\infty}\sum_{k=1}^{n-1} \frac{a_k a_{n-k} x^k x^{n-k}}{(4k-1)!(4(n-k)-1)!}.
$$
The above can be written as
$$
4xy'-y - \frac{a_1 x}2 = y^2.
$$
Applying $a_1=2$, we obtain
$$
4xy'=x+y+y^2.
$$
This is Riccati equation according to Wolfram Alpha.
Following the steps of solving Ricatti equation, the substitution $\frac y{4x}=-\frac{u'}u$ yields
$$
u''+\frac3{4x} u'+\frac1{16x}u=0.
$$
According to this input in Wolfram Alpha, we obtain the general solution
$$
u=C_1 x^{\frac18}J_{-\frac14} (\frac{\sqrt x}2) + C_2 x^{\frac18} J_{\frac14}(\frac{\sqrt x}2).
$$
This output of Wolfram Alpha can easily be justified by suitable substitutions.
Due to our requirement that $\frac y{4x}$ needs to be analytic at $0$, the second part should not appear. Thus,
$$
u=C_1 x^{\frac18}J_{-\frac14} (\frac{\sqrt x}2).
$$
Hence, we recover $y$.
$$
y=4x\left( - \frac{ \left( x^{\frac18} J_{-\frac14} ( \frac{\sqrt x}2)\right)'}{x^{\frac18}J_{-\frac14}(\frac{\sqrt x}2)}\right).
$$
Since $J_v'(x)=\frac12 (J_{v-1} (x)- J_{v+1}(x))$, we obtain
$$
y=\frac{- \frac12 x^{\frac18}J_{-\frac14}(\frac{\sqrt x}2)- \frac12 x^{\frac58}J_{-\frac54}(\frac{\sqrt x}2)+\frac12 x^{\frac58}J_{\frac34}(\frac{\sqrt x}2)}{ x^{\frac18}J_{-\frac14}(\frac{\sqrt x}2)}.
$$
Then the power series expansion of the above solution will yield an expression for $a_n$.


TITLE: Question Regarding Linear ODE-Trouble Using Integrating Factor
QUESTION [3 upvotes]: Consider the equation $$\frac{dy}{dx} + 5y = e^{2x}$$
One method of attack as far as i know is to multiply both sides by $e^{5x}$.This gives $$e^{5x}\frac{dy}{dx} + y5e^{5x} = e^{2x}e^{5x} = e^{7x}$$
We now find that the LHS is,in fact,the derivative of $ye^{5x}$.
$$\therefore \frac{d}{dx}(ye^{5x}) = e^{7x}$$
Now what do i do?Integrate this way$$\int\frac{d}{dx}(ye^{5x}) = \int e^{7x} ?$$

REPLY [1 votes]: Yes, that's what you do.
Look at what you have: $${d\over dx} (ye^{5x}) = e^{7x}.$$
This is saying that $ye^{5x}$ is an antiderivative of $e^{7x}$. Thus, you can write
$$\tag{1}ye^{5x} =\int e^{7x}\, dx;$$
after all, the indefinite integral of a function is its general antiderivative.
At this point, you then evaluate the integral in (1) and solve for $y$.


TITLE: Why cannot $(2x^2 + x)^2$ be simplified to $2x^4 + x^2$?
QUESTION [4 upvotes]: So, I want to simplify an equation : $(2x^2 + x)^2$.
I thought this would simplify to $2x^4 + x^2$
But, if you input a value for $x$, the answers do not equal. For example, if you input $x = 3$, then:
$$(2x^2+x)^2
= 21^2
= 441$$
AND:
$$2x^4 + x^2
= 2(82) + 9
= 173$$
Can anyone explain why this is the case?

REPLY [1 votes]: Think of it like this.
$a^{2}$ means $a \times a$.
More generally, $a^{n}$ means $\underbrace{a \times a \times a \times \dots \times a}_{n \text{ times}}$.
So $2^{3}$ means $2 \times 2 \times 2$. That's $2 \times 2$, $3$ times.
Therefore $(2x^{2} + x)^{2}$ means $(2x^{2} + x) \times (2x^{2} + x)$.
Now, using your number, we see that:
$(2 \times 3^{2}) + 3 = 18 + 3 = 21$
And $21 \times 21$ is very different to $(2 \times 3^{4}) + 3^{2} = 162 + 9 = 171$.
What you have learned the hard way is that exponentiation is not distributive over addition. Which means that $(a + b)^{2} \neq a^{2} + b^{2}$. Why is this? Because $(a + b)^{2}$ means:
Multiply $(a + b)$ by $(a + b)$
Not multiply $a$ by $a$ and multiply $b$ by $b$ and add them together.
For a more simple example than yours, let's look at $(3 + 2)^{2}$
$(3 + 2)^{2} = (3 + 2) \times (3 + 2) = 5 \times 5 = 25$
But:
$(3 + 2)^{2} \neq 3 \times 3 + 2 \times 2 = 9 + 4 = 13$
N.B.: Here we are talking about the general case. There is certainly one specific example of where $(a + b)^{2} = a^{2} + b^{2}$ but this is basically just by coincidence.


TITLE: Internal Direct Sum Questiom
QUESTION [2 upvotes]: I'm posed with the following problem.  Given a vector space $\,V\,$ over a field (whose characteristic isn't $\,2$), we have a linear transformation from $\,V\,$ to itself.
We have subspaces
$$V_+=\{v\;:\; Tv=v\}\,\,,\,\, V_-=\{v\;:\; Tv=-v\}$$
I want to show that $\,V\,$ is the internal direct sum of these two subspaces.  I have shown that they are disjoint, but can't seem to write an arbitrary element in V as the sum of respective elements of the subspaces...
Edit:forgot something important!
$$\, T^2=I\,$$
Sorry!

REPLY [1 votes]: Another approach:
Lemma: In arbitrary characteristic, if $V$ is a $K$-vector space and $P:V\to V$ is an endomorphism with $P^2=\lambda P$, $\lambda\ne 0$, then $V=\ker P\oplus\ker (\lambda I-P)$.
Proof: If $v\in \ker P\cap\ker (\lambda I-P)$, then $\lambda v = (\lambda I-P)v+Pv = 0$, hence $v=0$. Thus $\ker P\cap\ker (\lambda I-P)=0$.
For $v\in V$ let $u=\lambda^{-1}Pv$ and $w=v-u$. Then clearly $v=u+w$ and $(\lambda I-P)u =Pv-\lambda^{-1}P^2v = 0$ and $Pw =Pv-Pu = Pv-\lambda^{-1}P^2v=0$, i.e. $u\in\ker(\lambda I-P)$ and $w\in \ker P$. We conclude that $V=\ker P\oplus\ker(\lambda I-P)$.
Corollary: If $\operatorname{char} K\ne 2$, $V$ is a $K$-vector space and $T:V\to V$ is an endomorphism with $T^2=I$, then $V=T_+\oplus T_-$ (with $T_\pm$ defined as in the OP).
Proof: Let $P=T+I$. Then $P^2 = T^2+2T+I^2=I+2T+I=2P$. Since $\lambda:=2\ne0$, the lemma applies. Here $\ker P=T_-$ and $\ker (2 I-P)=\ker(I-T)=T_+$.


text:
\begin{document}

\title[Toeplitz operators on Bergman space]
{On spectral properties of compact Toeplitz operators on Bergman space with logarithmically decaying symbol  and applications to banded matrices}

\author{M. Koïta}
\address{Institut Universitaire de Formation Professionnelle, Université de Ségou, B.P. 97, Ségou, Mali}
\email{kmahamet@yahoo.fr}

\author{S. Kupin}
\address{Institut de Mathématiques de Bordeaux UMR5251, CNRS, Université de Bordeaux, 351 ave. de la Libération, 33405 Talence Cedex, France}
\email{skupin@math.u-bordeaux.fr}

\author{S. Naboko}
\address{Physics Institute, St. Petersburg State University, Ulyanovskaya str. 1, St. Peterhof, St. Petersburg, 198-904 Russia}
\email{sergey.naboko@gmail.com}

\author{B. Touré}
\address{Institut Universitaire de Formation Professionnelle, Université de Ségou, B.P. 97, Ségou, Mali}
\address{Faculté des Sciences et  des Techniques, Université des Sciences, des Techniques et des Technologies de Bamako, Campus Universitaire de Badalabougou à Bamako, B.P. E-3206, Bamako, Mali}
\email{vbelco@yahoo.fr}

\subjclass[2010]{Primary: 47B35; Secondary: 30H20, 42C10}

\begin{abstract}
Let $L^2(\bd)$ be the space of measurable square-summable functions on the unit disk. Let $L^2_a(\bd)$ be the Bergman space, \ie the (closed) subspace of analytic functions in $L^2(\bd)$. $P_+$ stays for the orthogonal projection going from $L^2(\bd)$ to $L^2_a(\bd)$. For a function $\p\in L^\infty(\bd)$, the Toeplitz operator $T_\p: L^2_a(\bd)\to L^2_a(\bd)$ is defined as
$$
T_\p f=P_+\p f, \quad f\in L^2_a(\bd).
$$
The main result of this article are spectral asymptotics for singular (or eigen-) values of compact Toeplitz operators with logarithmically decaying symbols, that is
$$
\p(z)=\p_1(e^{i\th})\, (1+\log(1/(1-r)))^{-\g},\quad \g>0,
$$
where $z=re^{i\th}$ and $\p_1$ is a continuous (or piece-wise continuous) function on the unit circle. The result is applied to the spectral analysis of banded (including Jacobi) matrices.
\end{abstract}

\keywords{compact Toeplitz operators, Bergman spaces, spectral asymptotics, generalized Schatten-von Neumann classes, special classes of compact operators with logarithmically decaying singular values}

\maketitle

\vspace{-0.7cm}
\section*{Introduction}\label{s0}
The investigation of Toeplitz operators is an important topic of modern analysis. The theory of Toeplitz operators on Hardy spaces was developed extensively in 70-90's of the last century, see monographs by Nikolski \cite{nk1,nk2} and Böttcher-Silbermann \cite{bo1} for a detailed account on the topic. The study of Toeplitz operators on larger functional spaces (\ie Fock, Bergman spaces, etc.) is currently under progress.

In this article, we are interested in spectral asymptotics of singular values of a compact Toeplitz operator on the Bergman space with logarithmically decaying symbol. We need several definitions to give the formulations of our results.

Let $\bd:=\{z:|z|<1\}$ and $\bt:=\pt\bd=\{z : |z|=1\}$ be the unit disk and the unit circle, respectively. As usual, we denote by $L^2(\bd)=L^2(\bd, dA)$ the space of measurable square-summable functions on $\bd$ with respect to the normalized Lebesgue measure $dA(z):=dxdy/\pi, \ z\in\bd$. The Bergman space is defined as a (closed) subspace of analytic functions lying in $L^2(\bd)$, that is
\begin{equation}\label{e0011}
L^2_a(\bd):=L^2_a(\bd,dA)=\{f\in\mathcal{A}(\bd): ||f||^2_2=\int_\bd |f(z)|^2\, dA(z)<\infty\}.
\end{equation}
The Riesz orthogonal projection $P_+: L^2(\bd)\to L^2_a(\bd)$ is given by the integral operator
\begin{equation}\label{e0012}
(P_+f)(z):=\int_\bd \frac 1{(1-z\bar w)^2} f(w)\, dA(w),\quad f \in L^2(\bd).
\end{equation}
A detailed treatment of these and other objects pertaining to Bergman spaces can be found in Hedenmalm-Korenblum-Zhu \cite{hkz}, Zhu \cite[Chap. 4]{zh1}.

For a symbol $\p\in L^\infty(\bd)$, the corresponding Toeplitz operator is defined by the relation
\begin{equation}\label{e0013}
T_\p f:=P_+\p f, \quad f\in L^2_a(\bd).
\end{equation}
Many analytic  properties of these operators are very-well understood, see Zhu \cite[Chap. 7]{zh1} for a nice presentation of the subject.
For instance, we have trivailly
$$
||T_\p||\le ||\p||_\infty,
$$
so the Toeplitz operator corresponding to a bounded symbol is also bounded. The compactness of $T_\p$ is related to the behavior of the symbol $\p$ on the vicinity of the unit circle $\bt$. The following proposition is not difficult to prove.
\begin{proposition}[{\cite[Prop. 7.3]{zh1}}]\label{p001} Let $\p\in C(\bar \bd)$, the class of contionuous functions on the closure of $\bd$. Then, $T_\p$ is compact if and only if
$$
\lim_{|z|\to 1-}\p(z)=0, \quad z\in\bd.
$$
\end{proposition}
For a positive symbol $\p$, criteria for $T_\p$ to be compact or to belong to Schatten-von Neumann classes $\css_p, \ 0<p<\infty$, can be found in \cite[Sect. 7.3]{zh1}.

Subsequent results in this direction address the spectral asymptotics (\ie the asymptotics of eigen- and/or singular values) of Toeplitz operators with symbols from some special classes.

Consider a symbol $\p$ of the following form
\begin{equation}\label{e003}
\p(z):=\p_1(e^{i\th})\p_0(r)=\p_1(e^{i\th})\, (1-r)^{\g},
\end{equation}
where $\p_1\in L^{1/\g}(\bt), \ \g>0$, and $z:=re^{i\th}\in\bd$. By Proposition \ref{p001}, the Toeplitz operator $T_\p$ is obviously compact. Its singular values $\{s_n(T_\p)\}_n$ form a decreasing sequence and converge to zero.
An important result to come was obtained in Pushnitski \cite{pu1}.
\begin{theorem}[{\cite[Thm. 1.1]{pu1}}]\label{t001}
Consider the Toeplitz operator $T_\p$ with symbol defined in \eqref{e003}. Its singular values $\{s_n(T_\p)\}_n$ have the following asymptotics
\begin{eqnarray}\label{e004}
\lim_{n\to+\infty} n^\g s_n(T_\p)&=&\frac{\gga(\g+1)}{2^\g}||\p_1||_{L^\gi(\bt)}\\
&:=&\frac{\gga(\g+1)}{2^\g}\lp\int^{2\pi}_0 |\p_1(e^{i\th})|^\gi\, \frac{d\th}{2\pi}\rp^\g. \nn
\end{eqnarray}

Above, $\gga(.)$ is Euler gamma-function.
\end{theorem}
The above result was obtained in \cite{pu1} in slightly more general form. Defining the counting function $n(.,T_\p)$ for the sequence of singular numbers of $T_\p$ as
$$
n(s,T_\p):=\#\{n: s_n(T_\p)>s\}, \ s>0,
$$
we can rewrite \eqref{e004} in an equivalent manner
$$
\lim_{s\to0+} s^\gi n(s, T_\p)=\frac{\gga(\g+1)^\gi}2\, ||\p_1||^\gi_{L^\gi(\bt)}.
$$
The proof of the above theorem is purely operator-theoretic and it uses some basic facts on underlying Bergman space only. We also mention a recent article El-Fallah-El-Ibbaoui \cite{fa} on a closely related topic.

The main result of the present article is a counterpart of Theorem \ref{t001} in logarithmic scale.
Let $\p_1\in C(\bt)$, the space of continuous functions on the unit circle. For a fixed $\g>0$, consider
\begin{equation}\label{e0041}
\p(z):=\p_1(e^{i\th})\p_0(r),
\end{equation}
where
\begin{equation}\label{e0042}
\p_0(r):=\p_{0,\g}(r)=\frac 1{\lp1+\log\frac1{(1-r)}\rp^\g}, \qquad r\in [0,1).
\end{equation}
That is, the symbol $\p(z)$ logarithmically tends to zero when $z$ goes to the unit circle $\bt$.

\begin{theorem}\label{t1} Let $\p$ be the a symbol defined above, $\g>0$. The following asymptotics hold for the singular values of Toeplitz operator $T_\p$
\begin{equation}\label{e005}
\lim_{n\to+\infty} (\log(n+1))^{\g}s_n(T_\g)=||\p_1||_{L^\infty(\bt)}.
\end{equation}
\end{theorem}
Generalizations of the above theorem are given in Section \ref{s51}.

Recalling the definition of the counting function, we can rewrite \eqref{e005} as
\begin{equation}\label{e0051}
\lim_{s\to 0+} s^{1/\g}\log n(s,T_\p)=||\p_1||^{1/\g}_{L^\infty(\bt)},
\end{equation}
where $\p_1\not\equiv 0$ on $\bt$.
We stress that despite a partial similarity, certain important pieces of the proof of Theorem \ref{t1} seem to be more involved than those of Theorem \ref{t001}. Among other technical points, it relies essentially on the structure of Bergman space and it uses fine properties of the Riesz orthoprojector \eqref{e0012} on it.

The organization of the article is rather straightforward. A fast sample of the theory of compact operators with logarithmically decaying singular values is developed in Section \ref{s2}. This section also contains a result on asymptotic orthogonality of certain operators which will be the cornerstone for the proof of  Theorem \ref{t1}. It is proved in Section \ref{s3}. The asymptotic orthogonality for a specific family of Toeplitz operators is obtained in Section \ref{s4}. Section \ref{s5} gives slightly different versions the obtained results as well as their applications to the study of spectral properties of compact banded matrices with logarithmically decaying entries.

Throughout the article, ``generic'' constants change from one relation to another. Significant constants are sub-indexed like $C_1, C_2$, etc. Points of the unit disk $\bd$ are often written as $z=|z|e^{i\th}:=re^{i\th}, \
r\in[0,1), \th\in [0,2\pi)$. As usual, the unit circle $\bt=\{z=e^{i\th}: \th\in[0,2\pi)\}$ is identified with the interval $[0,2\pi]$.

\section{Some special classes of compact operators}\label{s2}

\subsection{Basic notions and Schatten-von Neumann classes of compact ope\-ra\-tors}\label{s21}
In the paper, we shall use a number of basic facts on compact operators, see Gohberg-Krein \cite[Chap. 2]{gk1} and Birman-Solomyak \cite[Chap. 11]{bs1}.

Let $A$ be a compact operator on a Hilbert space. The class of compact operators on the space is denoted by $\css_\infty$. It is well-known that the spectrum $\s(A_0)$ of a self-adjoint compact operator $A_0$ consists of  the closure of the set of eignevalues $\{\l_n(A_0)\}_\nno$ tending to zero. The singular values of a compact operator $A$ are defined as
$$
s_n(A):=\l_n(A^*A)^{1/2}, \quad n\ge 1.
$$
The sequence of singular values $\{s_n(A)\}_\nno$ is written taking into account the multiplicities; it is positive, decreasing and $s_n(A)$ goes to zero as $n\to +\infty$. It is a simple fact that for $A\in\css_\infty$ and a bounded operator $B$, one has
\begin{equation}\label{e02}
s_n(BA)\le ||B||s_n(A),
\end{equation}
see \cite[Chap. 11, Sect. 1]{bs1} or \cite[Chap. 2, Sect. 2]{gk1}.
Another important characteristic connected to the sequence
$\{s_n(A)\}_n$ is the counting function,
\begin{equation}\label{e01}
n(s,A):=\#\{n: s_n(A)>s\}, \quad s>0.
\end{equation}
For instance, for $A,B\in \css_\infty$ and $s>0$, one has
\begin{equation}\label{e61}
n(s,AB)\le n(s_1)+n(s_2, B),
\end{equation}
where $s=s_1s_2$ and $s_1,s_2>0$, see \cite[Chap. 11, Sect. 1]{bs1} once again.

The Schatten-von Neumann classes $\css_p, \ 0<p<\infty$, are given by
$$
\css_p:=\{A\in \css_\infty: ||A||^p_{\css_p}=\sum^\infty_{n=1} s_n(A)^p<\infty\}, \quad A\in \css_p \Leftrightarrow \int^\infty_0 s^{p-1} n(s,A)\, ds<\infty.
$$
The class $\css_2 \ (p=2)$ is called Hilbert-Schmidt class.

\subsection{Some specific classes of compact operators}\label{s22}
In this paper, we are mainly interested in classes of compact operators with logarithmically decaying singular values.
Below, we introduce the definitions and give the properties of operators from these classes. Up to certain technical aspects, the proofs of the assertions of this subsection follow Birman-Solomyak \cite[Chap. 11, Sect. 6]{bs1} and they are therefore omitted.

To start with, we consider a ``shifted'' counting function
\begin{equation}\label{e1}
\tn(s,A):=n(s, A)+2,
\end{equation}
compare with \eqref{e01}. We shall see that $\tn(.,A)$ is more convenient for our purposes from technical point of view. For $\g>0$, set
\begin{eqnarray}
\sg&:=&\lt\{A\in \css_\infty: s_n(A)=O\lp\frac 1{(\log(n+1))^\g}\rp\rt\} \label{e2}\\
&=&\lt\{A\in \css_\infty: \sup_{n\ge 1} \ (\log n)^\g s_n(A)<+\infty\rt\}, \nonumber \\
\sgo&:=&\lt\{A\in \css_\infty: s_n(A)=o\lp\frac 1{(\log(n+1))^\g}\rp\rt\} \label{e3}\\
&=&\lt\{A\in \css_\infty: \lim_{n\to+\infty} (\log n)^\g s_n(A)=0 \rt\}. \nonumber
\end{eqnarray}

For a compact operator $A$, the following relations are equivalent in a trivial way
\begin{equation}\label{e30}
\sup_{n\ge 1}\ (\log (n+1))^\g s_n(A)<\infty, \quad \sup_{s>0}\ s^\gi \log \tn(s,A)<\infty.
\end{equation}
Moreover, the equality
\begin{equation}\label{e301}
s_n(A)=\frac{C}{(\log (n+1))^\g}(1+o(1)), \ n\to+\infty,
\end{equation}
is equivalent to
\begin{equation}\label{e302}
\log\tn(s, A)=\frac{C^\gi}{s^\gi}(1+o(1)), \ s\to0+.
\end{equation}
Two remarks are in order. First,  one can work similarly with the usual and the ``shifted'' counting functions $n(.,A)$ and $\tn(.,A)$, respectively. Second, the classes $\sg$ and $\sgo$ can be similarly defined as
\begin{eqnarray*}
\sg&:=&\lt\{A\in \css_\infty: \log \tn(s,A)=O\lp\frac 1{s^\gi}\rp, \ s>0\rt\} \\
&=&\lt\{A\in \css_\infty: \sup_{s>0} s^\gi \log\tn(s,A)<+\infty\rt\},\\
\sgo&:=&\lt\{A\in \css_\infty: \log \tn(s,A)=o\lp\frac 1{s^\gi}\rp, \ s>0\rt\} \\
&=&\lt\{A\in \css_\infty: \lim_{s\to 0+} s^\gi \log \tn(s,A)=0 \rt\}.
\end{eqnarray*}

Once again, let $A\in \sg$. The next quantities will be useful in the sequel.
\begin{equation}\label{e4}
\ddg(A)=\limsup_{s\to 0+} s^\gi \log\tn(s,A),\qquad \dg(A)=\liminf_{s\to 0+} s^\gi \log\tn(s,A).
\end{equation}

\begin{proposition}\label{p1}
Let $A,B\in \sg$. Then
\begin{eqnarray}
&&\lpp\ddg(A+B)\rpp^\go\le \lpp\ddg(A)\rpp^\go+\lpp\ddg(B)\rpp^\go, \label{e51}\\
&&|\lpp\ddg(A)\rpp^\go-\lpp\ddg(B)\rpp^\go|\le \lpp\ddg(A-B)\rpp^\go, \label{e52}\\
&&|\lpp\dg(A)\rpp^\go-\lpp\dg(B)\rpp^\go|\le\lpp\ddg(A-B)\rpp^\go.\label{e53}
\end{eqnarray}
\end{proposition}
The proof of the above proposition follows \cite[Chap. 11, Sect. 6, Thm. 4, Cor. 5]{bs1}. The next proposition is its simple corollary.
\begin{proposition}[Ky-Fan-type lemma]\label{p2}
Let $A\in\sg$ and $B\in\sgo$. Then
\begin{equation}\label{e6}
\ddg(A+B)=\ddg(A),\qquad  \dg(A+B)=\dg(A).
\end{equation}
\end{proposition}

\begin{proposition}\label{p3} We have:
\begin{enumerate}
\item for any $p,\g>0$, $\css_p\subset \ss^0_\g$,
\item if $A,B\in \sg$, then $\dd_{2\g}(AB)\le\ddg(A)+\ddg(B)$,
and, consequently, $AB\in \ss_{2\g}$.
\item if $A\in \sg$ and $B\in\sgo$, then $\dd_{2\g}(AB)=0$, \ie $AB\in \ss^0_{2\g}$.
\end{enumerate}
\end{proposition}

\subsection{Asymptotical orthogonality of a set of operators}\label{s23}
Let $A$ be a compact operator and
$$
A=\sum^L_{k=1} A_k.
$$
Suppose that $A_k\in\css_\infty, \ k=1,\dots,L$, as well. The coming abstract proposition will prove to be a useful tool for our purposes. Up to some technical details, it is is similar to Pushnitski \cite[Theorem 2.2]{pu1}, and we give a fast sketch of its proof only.
\begin{proposition}\label{p4}
Suppose that $A, A_k, \ k=1,\dots, L$, are as above and the family $\{A_k\}_k$ is asymptotically orthogonal, that is
\begin{equation}\label{e08}
A^*_kA_j, A_kA^*_j\in \ss^0_{2\g}, \quad j\not=k,\ \  j,k=1,\dots,L.
\end{equation}
Then
\begin{equation}\label{e8}
\ddg(A)=\lsup s^\gi\log\lp\sum_{k=1}^L\tn(s,A_k)\rp,\quad \dg(A)=\linf s^\gi\log\lp\sum_{k=1}^L\tn(s,A_k)\rp.
\end{equation}
\end{proposition}

\begin{proof} Set $H_L=\oplus_{k=1}^L H$, and
$$
A_0=\diag\{A_1,\dots, A_L\}: H_L\to H_L,
$$
that is
$$A_0(f_1,\dots, f_L)=(A_1f_1,\dots A_Lf_L)
$$
for arbitrary $(f_1,\dots f_L)\in H_L$.
Consider also the embedding operator $J:H_L\to H$ given by
$$
J(f_1,\dots f_L)=f_1+\dots+f_L.
$$
We have $J^*f=(f,\dots,f):H\to H_L$. A straightforward computation shows that
\begin{eqnarray}
(JA_0)(JA_0)^*f&=&(A_1A_1^*+\dots+A_LA_L^*)f, \label{e91}\\
&&\nn \\
(JA_0)(JA_0)^*&=&
\begin{bmatrix}
A_1^*A_1&\dots &A^*_1A_L\\
\vdots& \ddots &\vdots \\
A^*_LA_1& \dots &A^*_LA_L
\end{bmatrix}, \label{e92}
\end{eqnarray}
where we used a natural block decomposition for the operator $(JA_0)(JA_0)^*: H_L\to H_L$.
Since the operator $A_0$ is block-diagonal on $H_L$, we see
$$
n(s,A_0)=\sum^L_{k=0}n(s, A_k), \ s>0.
$$
Now, by assumption \eqref{e08}, we have substracting the diagonal parts
$$
(JA_0)^*(JA_0)-A^*_0A_0\in \ss^0_{2\g}.
$$
Notice that the singular values of $T^*T$ and $TT^*$ coincide for any compact ope\-ra\-tor $T$, and, in particular,  $\dd_{2\g}((JA_0)^*(JA_0))=\dd_{2\g}((JA_0)(JA_0)^*)$. Hence, by Proposition \ref{p2},
\begin{eqnarray*}
\dd_{2\g}((JA_0)^*(JA_0))&=&\dd_{2\g}(A_0^*A_0)=\ddg(A_0)\\
&=&\lsup s^\gi \log\lp\sum^L_{k=1}n(s,A_k)+2\rp.
\end{eqnarray*}
Furthermore, since $AA^*=\sum^L_{k, j=1}A_kA_j^*$, relation \eqref{e08} and Proposition \ref{p2} yield once again
$$
AA^*-(JA_0)(JA_0)^*\in \ss^0_{2\g},
$$
and $\dd_{2\g}(AA^*)=\dd_{2\g}((JA_0)(JA_0)^*)$.
Putting together the above computations, we obtain
\begin{eqnarray}
\ddg(A)&=&\dd_{2\g}(AA^*)=\dd_{2\g}((JA_0)(JA_0)^*)=\dd_{2\g}((JA_0)^*(JA_0)) \nn\\
&=&\ddg(A_0)=\lsup s^\gi \log\lp\sum^L_{k=1}n(s,A_k)+2\rp. \label{e93}
\end{eqnarray}
Passing from counting functions $n(.,A_k)$ to $\tn(.,A_k)$ is also obvious, and so the first relation in \eqref{e8} is proved. The proof of the second relation in \eqref{e8} is the same (with $\limsup$ replaced by $\liminf$). The proposition is completed.
\end{proof}

\section{Proof of the main theorem}\label{s3}

\subsection{Some notation and starting remarks}
Recall the definition of function $\p$, see \eqref{e0041}, \eqref{e0042}. To begin with, we consider the simplest radial case $\p_1(e^{i\th})\equiv 1$, so that $\p(z):=\p_0(z)=\p_0(r)$.
\begin{lemma}\label{l02} We have
\begin{equation}\label{e102}
\lim_{n\to+\infty} (\log(n+1))^\g s_n(T_{\p_0})=||\p_1||_{L^\infty(\bt)}=1.
\end{equation}
\end{lemma}
\begin{proof}
Since the symbol $\p_0$ is radial,  the matrix of the positive compact ope\-ra\-tor $T_{\p_0}$ is diagonal when computed in the standard orthonormal basis of the Bergman space $\{e_n\}_{n=0,\dots},\ e_n(z)=\sqrt{n+1}\, z^n$. Passing to polar coordinates and using Lemma \ref{l01} (with $g(r)\equiv 1$ on $[0,1]$), we obtain
\begin{equation}\label{e002}
\begin{split}
s_n(T_\po)&=(T_\po e_n, e_n)=2(n+1)\int^1_0 r^{2n+1}\po(r)\,dr\\
&=2(n+1)\, \frac 1{(2n+1)(\log(2n+1))^\g}(1+o(1))\\
&=\frac1{(\log (n+1))^\g}\lp 1+o(1)\rp, \quad n\to+\infty.
\end{split}
\end{equation}
\end{proof}

Now, take a natural $L>0$ and let $I_j:=[2\pi (j-1)/L, 2\pi j/L), \ j=1,\dots, L$ be the partition of $[0,2\pi)$ into disjoint intervals of equal length. Set $\chi_{I_j}$ to be the characteristic functions of the intervals $I_j, \ j=1,\dots, L$, and, moreover
\begin{equation}\label{e101}
\tc_j:=\chi_{I_j}\p_0, \quad T_{\tc_j}:=P_+\tc_j=P_+(\chi_{I_j}\p_0),
\end{equation}
see \eqref{e0042}.
It is clear that the operators $T_{\tc_j}$ are unitarily equivalent to $T_{\tc_1}$ by rotation $z\mapsto e^{-i(2\pi (j-1)/L)}z, \ z\in\bd$, and so their singular values and counting functions coincide, $\tn(s,T_{\tc_j})=\tn(s,T_{\tc_1}), \ s>0, \ j=1,\ldots, L$.

\subsection{Proof of Theorem \ref{t1}}\label{s32}
The main tools of the proofs appearing in the present subsection are Proposition \ref{p4} and Theorem \ref{t20} saying that the above defined operators $T_{\tc_j}, \ j=1,\ldots, L$, form an asymptotically orthogonal family. The proof of Theorem \ref{t20} is given in Section \ref{s4}. Recall that $\p=\p_1\p_0$, see \eqref{e0041}, \eqref{e0042}.

\begin{lemma}\label{l1} The following claims hold true:
\begin{enumerate}
\item Let $\p:=\p_0$, so that $\p_1\equiv 1$ on $\bt$. Then
$$
\ddg(T_{\p_0})=\dg(T_{\p_0})=1.
$$
\item For $j=1,\ldots, L$,
\begin{equation*}
\ddg(T_{\tc_j})=\dg(T_{\tc_j})=1.
\end{equation*}
\end{enumerate}
\end{lemma}
\begin{proof}  Theorem \ref{t20} is crucial for the proof of the current lemma. Its first claim is a simple rewriting of relation \eqref{e102} with the help of observation on the equivalence of \eqref{e301} and \eqref{e302}.

Turning to the second claim of the lemma, we keep  $\p:=\p_0$ (or $\p_1\equiv 1$ on $\bt$).
Of course,
$$
\p_1:=\sum_{j=1}^L\, 1\cdot\chi_j, \quad T_\p:=T_{\p_0}=\sum_{j=1}^L\,1\cdot T_{\tc_j}.
$$
By Theorem \ref{t20}, the operators $T^*_{\tc_j} T_{\tc_k}, T_{\tc_j} T^*_{\tc_k}$ lie in $\ss^0_{2\g}$ for $j\not=k, \ j,k=1,\dots, L$, and so Proposition \ref{p4} shows that
\begin{eqnarray}\label{e10}
\ddg(T_{\p_0})&=&\lsup s^\gi\log\lp\sum^L_{j=1}\tn(s,T_{\tc_j})\rp\\
&=&\lsup s^\gi\log(L\tn(s,T_{\tc_1}))=\ddg(T_{\tc_1}). \nn
\end{eqnarray}
Similarly, one sees $\dg(T_{\p_0})=\dg(T_{\tc_1})$. It remains to recall that counting functions $\tn(., T_{\tc_j})$ and $\tn(., T_{\tc_1})$ coincide, and the lemma is proved.
\end{proof}

The next proposition says that the claim of Theorem \ref{t1} is rather simple to prove when $\p_1$ is a step (or a ``staircase'') function given by
$$
\p_1=\sum^L_{j=1} c_j\cdot \chi_j,
$$
where $c_j\in\bc, \ j=1,\ldots, L$.
\begin{proposition}\label{p41} For a step function $\p_1$ defined above, we have
$$
\ddg(T_\p)=\dg(T_\p)=||\p_1||^\gi_\lni.
$$
\end{proposition}
\begin{proof}
To ease the writing, we suppose that $c_1:=||\p_1||_\lni=\max_{j=1,\dots,L} |c_j|\ge 0$. The case $c_1=0$ being trivial, we assume that $|c_1|>0$. In particular,
$$
\tn(s, c_j T_{\tc_j})\le \tn(s, c_1 T_{\tc_1}),\ j=1,\dots, L,
$$
due to $|c_jT_{\tc_j}|\le |c_1|T_{\tc_j}$.
Applying Theorem \ref{t20} and Proposition \ref{p4} as in \eqref{e10}, we have
\begin{eqnarray*}
\ddg(T_\p)&=&\lsup s^\gi\log\lp\sum^L_{j=1}\tn(s,c_j T_{\tc_j})\rp\\
&=&\lsup s^\gi \log\lp \tn(s,c_1T_{\tc_1})+\sum^L_{j=2} \tn(s, c_jT_{\tc_j})\rp\\
&=&\lsup s^\gi \log \tn(s,c_1T_{\tc_1})\left\{1+\frac{\sum^L_{j=2} \tn(s, c_jT_{\tc_j})}{\tn(s,c_1T_{\tc_1})}\right\}.
\end{eqnarray*}
The term in the figure brackets on the RHS of this relation is greater or equal to one and bounded from above, so we continue as
\begin{eqnarray*}
\dots&=&\lsup s^\gi \log \tn(s,c_1T_{\tc_1})=c_1^\gi \lsup s^\gi\log\tn(s,T_1)\\
&=&c_1^\gi \ddg(T_1)=c^\gi_1.
\end{eqnarray*}
The computation for $\dg(T_\p)$ is completely analogous, and the proof is finished.
\end{proof}

\begin{lemma}\label{l2} For any $\p_1\in\lni$, we have
\begin{equation}\label{e401}
\ddg(T_\p)\le ||\p_1||_\lni^\gi.
\end{equation}
\end{lemma}
\begin{proof} Observe that inequality \eqref{e401} is homogeneous with respect to $||\p_1||_\lni^\gi$, so we can suppose $||\p_1||_\lni=1$ without loss of generality. Setting $F:=T_\po=P_+\po: L^2_a(\bd)\to L^2_a(\bd)$, we have
\begin{equation*}
\begin{split}
T_\p^*T_\p&=P_+\bar\p P_+\p P_+=P_+ \bar\po\bar\pon P_+\pon\po P_+\le P_+|\pon\po|^2 P_+\\
&\le ||\pon||^2_\lni P_+|\po|^2 P_+\le FF^*,
\end{split}
\end{equation*}
where we used repeatedly that $X^*YX\le ||Y||X^*X$ for bounded operators $X,Y$, and $Y=Y^*$. So we have that
$s^2_n(T_\p)=s_n(T_\p^*T_\p)\le s_n(FF^*)$, where $\{s_n(T_\p)\}_n$ and $\{s_n(FF^*)\}_n$ are singular values of operators $T_\p$ and $FF^*$, respectively. Consequently,
$$
\tn(s,T_\p)\le\tn(\sqrt s, FF^*),
$$
and we continue as
\begin{eqnarray*}
\ddg(T_\p)&=&\lsup s^\gi\log\tn(s,T_\p)\le \lsup s^\gi\log\tn(\sqrt s,FF^*)\\
&=&\lsup s^{1/(2\g)}\log \tn(s, FF^*)=\dd_{2\g}(FF^*)=\ddg(F)=\ddg(T_\po)=1
\end{eqnarray*}
by the first claim of Lemma \ref{l1}. The proof is finished. \end{proof}

\nt
{\it Proof of Theorem \ref{t1}.} \ Let $\p_1:\bt\to\bc$ be a complex-valued continuous function. For any given $\ep>0$, choose a step function $\ti\p_1$ with the property
$$
||\p_1-\ti\p_1||_\lni <\ep,
$$
and set $\ti\p:=\ti\p_1\p_0$.
Lemma \ref{l2} says that
$$
\ddg(T_\p-T_{\ti\p})=\ddg(T_{\p-\ti\p})\le||\p_1-\ti\p_1||_\lni^\gi\le \ep^\gi.
$$
On the other hand, Proposition \ref{p1} implies
\begin{eqnarray*}
|\lpp\ddg(T_\p)\rpp^{\g/(1+\g)}-\lpp\ddg(T_{\ti\p})\rpp^{\g/(1+\g)}|&\le&\lpp\ddg(T_\p-T_{\ti\p})\rpp^{\g/(1+\g)}<\ep^{1/(1+\g)},\\
|\lpp\dg(T_\p)\rpp^{\g/(1+\g)}-\lpp\dg(T_{\ti\p})\rpp^{\g/(1+\g)}|&\le&\lpp\ddg(T_\p-T_{\ti\p})\rpp^{\g/(1+\g)}<\ep^{1/(1+\g)}.
\end{eqnarray*}
We finish the proof passing to the limit with respect $\ep\to 0$.
\hfill $\Box$

\section{Proof of the asymptotic orthogonality of operators $T_{\tc_j}$ and $T_{\tc_k}, \ j\not =k$}\label{s4}
The purpose of this section is to prove the following theorem. Notice that in the case $\p_0(z)=(1-r)^\g, \ \g>0$, the proof of the similar result is rather simple.
\begin{theorem}\label{t20}
For $j\not=k, \ j,k=1,\dots,L$, the Toeplitz operators $T_{\tc_j}, T_{\tc_k}$, see \eqref{e101}, are asymptotically orthogonal, that is
$$
T_{\tc_k}^*T_{\tc_j}, T_{\tc_k}T_{\tc_j}^*\in \ss^0_{2\g}.
$$
\end{theorem}

By default, we assume that $j\not=k, \ j,k=1,\dots,L$ throughout this section. We present the argument for the operators $T_{\tc_k}^*T_{\tc_j}$, the reasoning for $T_{\tc_k}T_{\tc_j}^*$ is completely similar.

Since the proof of Theorem \ref{t20} in logarithmic case is rather involved, it is divided into a few steps.

\subsection{STEP 1. Toeplitz operators with symbols which are compactly supported in $\bd$}\label{s41}
Take a small $0<\d<1/2$ and define the characteristic function of $\{z: |z|\le 1-\d\}$,
\begin{equation*}
\chi_{00}(z)=
\left\{
\begin{array}{ll}
1,& |z|\le 1-\d\\
0,& |z|>1-\d
\end{array}
\right. ,
\quad \chi_{01}(z)=1-\chi_{00}(z),\quad z\in\bd.
\end{equation*}
Then, write $T_{\tc_j}$ as $T_{\tc_j}=T_{\cho\tc_j}+T_{\chn\tc_j}$. Since the $\mathrm{supp}\, (\cho\tc_j)$ is compact in $\bd$,  the singular values $\{s_n(T_{\cho\tc_j})\}_n$
decay exponentially, that is, there is a constant $C=C(\d)>0$ such that
$$
s_n(T_{\cho\tc_j})\le C (1-\d)^{2n},
$$
see Zhu \cite[Chap. 7]{zh1}. This bound is almost trivial and it follows from the estimate of the quadratic form of the operator $(T_{\cho\tc_j}$ with the help of explicit expression \eqref{e0012} for the projection $P_+$.

We have
$$
T_{\tc_k}^*T_{\tc_j}=T_{\cho\tc_k}^*T_{\cho\tc_j}+T_{\cho\tc_k}^*T_{\chn\tc_j}+T_{\chn\tc_k}^*T_{\cho\tc_j}+T_{\chn\tc_k}^*T_{\chn\tc_j}.
$$
All operators $T_{\tc_j}, T_{\cho\tc_j}, T_{\chn\tc_j}$ are bounded and so the singular values of the first three operator products on the RHS of the above relation decay exponentially as well. That is, proving that $T_{\tc_k}^*T_{\tc_j}\in \ss^0_{2\g}$ is equivalent to saying that $T_{\chn\tc_k}^*T_{\chn\tc_j}\in \ss^0_{2\g}$.

\subsection{STEP 2. Products of Toeplitz operators with smooth integral kernel}\label{s42}
Second, let $B_0=T_{\chn\tc_k}^*T_{\chn\tc_j}$. Recalling point (1) of Proposition \ref{p3}, we wish to prove $B_0\in\css_p\subset \ss^0_{2\g}$ for some $p>0$.

The above conclusion is rather simple to obtain in the case when $j<k-1$ or $j>k+1$, and, as always, $j,k=1,\dots, L$. This means that the intervals $I_j$ and $I_k$ do not touch.
\begin{lemma}\label{l3} Let $j<k-1$ or $j>k+1$. Then $T_{\chn\tc_k}^*T_{\chn\tc_j}\in\css_2$, the Hilbert-Schmidt class.
\end{lemma}

\begin{proof} The proof is well-known and quite simple, see Pushnitski \cite{pu1}. Set
$$
S_j:=\{z\in\bd: z=re^{i\th}, r\in (1-\d,1), \th\in I_j\},\quad j=1,\dots, L.
$$
By definition, we have $\chi_{S_j}=\chn\chi_j$ and $\ti\chi_{S_j}:=\chn\ti\chi_j$. Consequently,
$$
T_{\ti\chi_{S_k}}^* T_{\ti\chi_{S_j}}=P_+\ti\chi_{S_k} P_+\ti\chi_{S_j} P_+=P_+ (\ti\chi_{S_k} P_+\ti\chi_{S_j}) P_+.
$$
Since $||P_+||=1$, we prove that the operator $(\ti\chi_{S_k} P_+\ti\chi_{S_j})$ belongs to $\css_2$, and this will give the claim of the lemma by \eqref{e02} .
Indeed, the integral operator $(\ti\chi_{S_k} P_+\ti\chi_{S_j})$ can be written by \eqref{e0012} as
\begin{equation*}
\begin{split}
((\ti\chi_{S_k} P_+\ti\chi_{S_j})f)(w)&:=\int_\bd K(w,z) f(z)\, dA(z)\\
&:=\int_\bd \frac{\chi_{S_k}(w)\p_0(w)\chi_{S_j}(z)\p_0(z)}{(1-w\bar z)^2} f(z)\, dA(z),\quad f\in A^2(\bd).
\end{split}
\end{equation*}
Since the distance between regions $S_k$ and $S_j$ is strictly positive, the integral kernel $K(w,z)$ of the operator is a bounded function. Consequently, the kernel $K(.,.)$ of the operator lies in $L^2(\bd\times\bd; dA(w)\wedge dA(z)),$ and so the operator  $\ti\chi_{S_k} P_+\ti\chi_{S_j}$ is Hilbert-Schmidt, see \cite[Chap. 11, Sect. 3]{bs1}. The lemma is proved.
\end{proof}

\subsection{STEP 3. The case of $T_{\tc_k}^*T_{\tc_j}$ with neighboring intervals, $k=j\pm 1$; a bound on a kernel of an integral operator}\label{s43}
The arguments of the previous subsections show that it remains to prove that
$B_0:=T_{\chn\tc_k}^*T_{\chn\tc_j}\in \ss^0_{2\g}$ for $k=j\pm1,\ j,k=1,\dots,L$. By rotation, assume WLOG that $k=1$ and $j=L$. To simplify the notation, we set $\d=2\pi/L$ and
\begin{eqnarray*}
\o_1&:=&\chn\tc_1=\chi_{S_1}\p_0, \quad S_1:=\lt\{z=re^{i\th}: r\in (1-\d,1),\th\in \lp0,\frac{2\pi}L\rp\rt\},\\
\o_L&:=&\chn\tc_L=\chi_{S_L}\p_0,\quad S_L:=\lt\{z=re^{i\th}: r\in (1-\d,1),\th\in \lp\frac{2\pi(L-1)}{L},2\pi\rp\rt\}.
\end{eqnarray*}
Alternatively, one can write the sets $S_1$ and $S_L$ as
\begin{equation*}
\begin{split}
&S_1:=\{z=re^{i\th}: r\in (1-\d,1),\th\in (0,\d)\},\\
&S_L:=\{z=re^{i\th}: r\in (1-\d,1),\th\in (-\d,0)\},
\end{split}
\end{equation*}
so that $S_L=\bar S_1$ by complex conjugation. We put $B_0=T^*_{\o_1}T_{\o_L}$, and
\begin{eqnarray}
0&\le&D_0:=B_0^*B_0=(T_{\o_1}^*T_{\o_L})^*(T_{\o_1}^*T_{\o_L})=T^*_{\o_L}T_{\o_1}T^*_{\o_1}T_{\o_L}\nn \\
&=&P_+\o_LP_+\o_1P_+\o_1P_+\o_L P_+\label{e11}

\end{eqnarray}
Recalling inequality \eqref{e02}, we can omit two projections $P_+$ bordering the latter expression on the left and on the right. Consequently, the singular values (or eigenvalues) $s_n(D_0)$ are bounded from above by singular values (eigenvalues) $s_n(D)$ of the positive operator $D$ defined as
\begin{equation}\label{e12}
0\le\o_LP_+\o_1 P_+\o_1P_+\o_L\le D:=\o_LP_+\o_1^2P_+\o_L,
\end{equation}
where we used that $X^*YX\le ||Y||X^*X$ for two bounded operators $X$ and $Y, Y^*=Y$.

Now, we are interested in upper bounds on integral kernels for operators $D, D^m$ with a natural $m>0$. Let
\begin{eqnarray*}
(Df)(z_1)&:=&(D^1f)(z_1):=\int_{\bd, z_3} D_1(z_1,z_3)f(z_3)\, dA(z_3),\\
(D^m f)(z_1)&:=&\int_{\bd, z_{2m+1}} D_m(z_1,z_{2m+1})f(z_{2m+1})\, dA(z_{2m+1})\\
&=&\int_{\bd,z_{2m+1}}\lp\ldots\int_{\bd,z_3} \underbrace{D_1(z_1,z_3)\ldots D_1(z_{2m-1},z_{2m+1})}_{m}\rp
f(z_{2m+1})\\
&& \ \prod^m_{j=1}dA(z_{2j+1}),
\end{eqnarray*}
the notation for indices $z_j$ will be made clear a bit later. When needed, we shall indicate the variable of the integration as a sub-index of the integral as it is written above.

\begin{proposition}\label{p5} We have
\begin{equation}\label{e16}
\begin{split}
&D_1(z_1,z_3)=\int_{\bd, z_2}\frac{\tcsl(z_1)\tcso(z_2)^2\tcsl(z_3)}{(1-z_1\bar z_2)^2(1-z_2\bar z_3)^2}\, dA(z_2),\\
&D_m(z_1,z_{2m+1})=\underbrace{\int_{\bd,z_2}\dots\int_{\bd,z_{2m}}}_{2m-1}
\frac{\prod_{j=1}^m \tcsl(z_{2j-1})\tcso(z_{2j})^2\tcsl(z_{2j+1})}
{\prod_{j=1}^m (1-z_{2j-1}\bar z_{2j})^2(1-z_{2j}\bar z_{2j+1})^2}\, \prod_{j=2}^{2m}dA(z_j).
\end{split}
\end{equation}
\end{proposition}

\begin{proof}
Relations \eqref{e11}, \eqref{e12} and the form of the Riesz orthoprojector $P_+$ in the Bergman space \eqref{e0012} give that, for $f\in L^2_a(\bd)$,
\begin{eqnarray}
Df(z_1)&=&\int_{\bd, z_2}\frac{\tcsl(z_1)\tcso(z_2)}{(1-z_1\bar z_2)^2}\, \lp\int_{\bd, z_3}\frac{\tcso(z_2)\tcsl(z_3)}{(1-z_2\bar z_3)^2}\, f(z_3) dA(z_3)\rp\, dA(z_2)\nn \\
&=&\int_{\bd, z_3} D_1(z_1,z_3)f(z_3)\, dA(z_3), \label{e13}
\end{eqnarray}
where, by Fubini-Tonelli theorem,
\begin{eqnarray}
D_1(z_1,z_3)&=&\int_{\bd, z_2}\frac{\tcsl(z_1)\tcso(z_2)^2\tcsl(z_3)}
{(1-z_1\bar z_2)^2(1-z_2\bar z_3)^2}\, dA(z_2). \label{e14}
\end{eqnarray}
This is the first equality in \eqref{e16}; the second one is proved similarly by an elementary induction.
\end{proof}
We stress that $z_{2j}\in S_1, \ j=1,\dots, m$, and $z_{2j+1}\in S_L, \ j=0,\dots, m$, in the second relation \eqref{e16}.

Before going to the proof of a coming proposition,we introduce some notation.  First, set
\begin{equation}\label{e15}
\psi_0(r):=\frac 1{\lp 1+1/\log r\rp^\g}, \quad 1/2<r<1.
\end{equation}
Set $\dh=\sqrt 2\d$ and define slightly different domains
\begin{eqnarray}\label{e150}
\shl&:=&\{z: |z|<1, |z-1|<\dh, \ \im z<0\} \\
&=&\{z=1-re^{i\th}: 0\le r<\min\{\dh, 2\cos\th\}, \ \th\in (0,\pi/2)\}, \nn
\end{eqnarray}
see Figure \ref{fig1}. It is clear that $S_L\subset \shl$, and so $\chi_{S_L}\le \chi_\shl$. The region $\sho$ is defined for $S_1$ in the same manner.

Now, it is convenient to make the polar change of variables; notice that the new variables $(r_j,\th_j)$ are centered at the point $z_0=1$, and not the origin $z_0=0$. More precisely, we set
\begin{eqnarray}\label{e17}
&& z_{2j+1}:=1-r_{2j+1}e^{-i\th_{2j+1}}\in\sho, \ j=0,\dots,m,\\
&& z_{2j}:=1-r_{2j}e^{i\th_{2j}}\in\shl,  \ j=1,\dots,m.\nn
\end{eqnarray}
We have $z_{2j+1}\in\sho$ and $z_{2j}\in\shl$ for $r_k\in (0,\sqrt2\d), \th_k\in (0,\pi/2),\ k=1,\dots, 2m+1$.

\begin{figure}[tbp]

\includegraphics[width=14cm]{pic1.pdf}
\vspace{-2cm}
\caption{Domains $S_1, S_L$, and $\hat S_1, \hat S_L$.}
\label{fig1}
\end{figure}

\begin{proposition}\label{p6} We have the following bound on kernels $D_m$ \eqref{e16} in terms of the variables introduced in \eqref{e17}
\begin{equation}\label{e18}
\begin{split}
&|D_m(z_1,z_{2m+1})|\\
&\le
C\underbrace{\int^\dh_{0,r_2}\dots\int^\dh_{0,r_{2m}}}_{2m-1}
\frac{\chi_\shl(z_1)\psi_0(r_1)\, \lp\prod^{2m}_{j=2}\psi_0(r_j)^2\rp\, \chi_\shl(z_{2m+1})\psi_0(r_{2m+1})}
{\prod^{2m}_{j=1}(r_j+r_{j+1})}\, \prod^{2m}_{j=2}dr_j.
\end{split}
\end{equation}

\end{proposition}

\begin{proof}
First, we take the modulus under the integral in the second relation \eqref{e16}. Since $S_1\subset\sho$ and $S_L\subset\shl$, we see
$\tc_{S_1}\le \tc_\sho$ and $\tc_{S_L}\le \tc_\shl$, so we can replace $\tc_{S_1}, \tc_{S_L}$ with $\tc_\sho,\tc_\shl$
under the integral. Second, to uniformize the notation, we make the change of variables
$$
z_{2j}\in\sho \mapsto \bar z_{2j}\in\shl.
$$
In this way, we have that $z_k=1-r_ke^{i\th}\in\shl$ for all $k=2,\dots, 2m$.

Furthermore, we get with the help of the new variable $z_{2j}$
$$
1-z_{2j-1}z_{2j}=r_{2j-1}e^{i\th_{2j-1}}+r_{2j}e^{i\th_{2j}}-r_{2j-1}r_{2j}e^{i(\th_{2j-1}+\th_{2j})}.
$$
Consequently, there is a $C=C(\dh)>0$ such that
$$
|1-z_{2j-1}z_{2j}|\ge C(r_{2j-1}+r_{2j}).
$$
Remind that $\tc_\shl(z)=\chi_\shl(z)\p_0(|z|)$. Concerning these factors, we notice that $1-|z_k|\le |1-z_k|=r_k, \ k=2,\dots, 2m$, and so
$$
\p_0(z_k)\le \psi_0(r_k).
$$
Plugging all these bounds into the second relation \eqref{e16}, we obtain
\begin{equation*}
\begin{split}
&|D_m(z_1,z_{2m+1})|\le\underbrace{\int_{\bd,z_2}\dots\int_{\bd,z_{2m}}}_{2m-1}
\frac{\prod_{j=1}^m \tc_\shl(z_{2j-1})\tc_\shl(z_{2j})^2\tc_\shl(z_{2j+1})}
{\prod_{j=1}^m |1-z_{2j-1} z_{2j}|^2 |1-z_{2j}z_{2j+1}|^2}\, \prod_{j=2}^{2m}dA(z_j)\\
&\le C \underbrace{\int^{\pi/2}_{0,\th_2}\int^\dh_{0,r_2}r_2\, dr_2d\th_2\, \dots
\int^{\pi/2}_{0,\th_{2m}}\int^\dh_{0,r_{2m}}r_{2m}\, dr_{2m}d\th_{2m}}_{2m-1}
\frac{\tc_\shl(z_1)\, \lp\prod_{j=2}^{2m}\psi_0(r_j)^2\rp\, \tc_\shl(z_{2m+1}) }
{\prod_{j=1}^{2m} (r_j+r_{j+1})^2}\\
&\le C \lp\frac\pi 2\rp^{2m-1} \underbrace{\int^\dh_0 \dots\int^\dh_0}_{2m-1}
\frac{\tc_\shl(z_1)\, \lp\prod_{j=2}^{2m}\psi_0(r_j)^2\rp\, \tc_\shl(z_{2m+1}) }
{\prod_{j=1}^{2m} (r_j+r_{j+1})}\, \prod_{j=2}^{2m}dr_j,
\end{split}
\end{equation*}
where we used that $r_{j+1}/(r_j+r_{j+1})^2\le 1/(r_j+r_{j+1})$. The proposition is proved.
\end{proof}

\subsection{STEP 4. The case of $T_{\tc_k}^*T_{\tc_j}$ with $k=j\pm 1$; an auxiliary combinatorial lemma and the final computation}\label{s44}
The integral from the RHS of \eqref{e18} is computed on a cube $(r_2,\dots, r_{2m})\in \ccs=(0,\dh)^{2m-1}$. Roughly speaking, the main idea for the calculation of this subsection is to divide the cube in standard simplexes
$$
\ccs_{i_2,\dots,i_{2m}}=\{(r_2,\dots,r_{2m})\in\ccs: r_{i_2}\in (0,\dh),\ \dh>r_{i_2}>\dots>r_{i_{2m}}>0\},
$$
and to obtain an appropriate bound for \eqref{e18} from above integrating on every simplex. Here, $\{i_j\}_{j=2,3,\ldots, 2m}$ is a transposition of the set $\{2,3,\ldots, 2m\}$.

\begin{lemma}\label{l4} Let $r_j\ge 0, \ j=1,\dots,2l$, where $l>0$ is a natural number.  Then
\begin{equation}\label{e19}
\begin{split}
&(r_1+r_2)\lp \prod^{l-1}_{j=2}(r_j+r_{j+1})\rp\, (r_l+r_{l+1})\cdot
(r_1+r_{l+2})\lp \prod^{2l-1}_{j=l+2}(r_j+r_{j+1})\rp (r_{2l}+r_{l+1})\\
&\qquad \ge (\max_{i=1,\dots, 2l} r_i)^2\lp \prod^{2l}_{j=1}r_j\rp'',
\end{split}
\end{equation}
where $(\prod .)''$ means that we drop both the maximal and the minimal factors in the product.
\end{lemma}

Suppose that we have $r_{i_1}> r_{i_2}> \dots r_{i_{2l}}>0$. Then the lemma says that
\begin{equation*}
\begin{split}
&(r_1+r_2)\lp \prod^{l-1}_{j=2}(r_j+r_{j+1})\rp\, (r_l+r_{l+1})\cdot
(r_1+r_{l+2})\lp \prod^{2l-1}_{j=l+2}(r_j+r_{j+1})\rp\, (r_{2l}+r_{l+1})\\
&\qquad \ge r_{i_1}^2 \lp\prod^{2l-1}_{j=2}r_{i_j}\rp.
\end{split}
\end{equation*}

\begin{proof}
We can assume that $r_j>0, \ j=1,\dots, 2l$. Consider the decomposition of the set $\{r_1,\ldots,r_{2l}\}$ into two disjoint subsets $G_1, G_2$, where
$$
G_1=\{r_1,r_{l+1}\}, \quad G_2=\{(r_j)_{j=2,\dots, l}, \ (r_j)_{j=l+2,2l}\}.
$$
Using the symmetry of variables $r_j$ within each group, one can reduce the general situation to the analysis of four cases only: 1)
$\max_{j=1,\dots, 2l} r_j\in G_1,\ \min_{j=1,\dots, 2l} r_j\in G_1$; 2) $\max_{j=1,\dots, 2l} r_j\in G_1,\ \min_{j=1,\dots, 2l} r_j\in G_2$; 3) $\max_{j=1,\dots, 2l} r_j\in G_2,$ $\min_{j=1,\dots, 2l}$ $r_j\in G_1$; 4)
$\max_{j=1,\dots, 2l} r_j\in G_2, \min_{j=1,\dots, 2l} r_j\in G_2$.

\emph{Case 1:} Without loss of generality, suppose $r_1=\max_{j=1,\dots, 2l} r_j\in G_1,$ $r_{l+1}=$ $\min_{j=1,\dots, 2l} r_j\in G_1$. We have
\begin{eqnarray*}
&&(r_1+r_2)(r_2+r_3)\dots (r_l+r_{l+1})\cdot (r_1+r_{l+2})(r_{l+2}+r_{l+3})\dots (r_{2l}+r_l)\\
&&\ge r_1 r_2\dots r_l\cdot r_1 r_{l+2}\dots r_{2l}=r_1^2\, \prod^{2l}_{j=2, j\not=l+1} r_j.
\end{eqnarray*}

\emph{Case 2:} WLOG $r_1=\max_{j=1,\dots, 2l} r_j\in G_1,\ r_2=\min_{j=1,\dots, 2l} r_j\in G_2$.
Then
\begin{eqnarray*}
&&(r_1+r_2)(r_2+r_3)\dots (r_l+r_{l+1})\cdot (r_1+r_{l+2})(r_{l+2}+r_{l+3})\dots (r_{2l}+r_l)\\
&&\ge r_1 r_3\dots r_l r_{l+1}\cdot r_1 r_{l+2}\dots r_{2l}=r_1^2\, \prod^{2l}_{j=3} r_j.
\end{eqnarray*}

\emph{Case 3:} WLOG $r_2=\max_{j=1,\dots, 2l} r_j\in G_2, \ r_1=\min_{j=1,\dots, 2l} r_j\in G_1$. We have
\begin{eqnarray*}
&&(r_1+r_2)(r_2+r_3)\dots (r_l+r_{l+1})\cdot (r_1+r_{l+2})(r_{l+2}+r_{l+3})\dots (r_{2l}+r_l)\\
&&\ge r_2 r_2 r_3\dots r_l\cdot r_{l+2} r_{l+3}\dots r_{2l} r_{l+1}=r_2^2\, \prod^{2l}_{j=3} r_j.
\end{eqnarray*}

\emph{Case 4:} WLOG $r_2=\max_{j=1,\dots, 2l} r_j\in G_2,\ r_3=\min_{j=1,\dots, 2l} r_j\in G_2$. We get
\begin{eqnarray*}
&&(r_1+r_2)(r_2+r_3)\dots (r_l+r_{l+1})\cdot (r_1+r_{l+2})(r_{l+2}+r_{l+3})\dots (r_{2l}+r_l)\\
&&\ge r_2 r_2 r_4\dots r_{l+1}\cdot r_1 r_{l+2}\dots r_{2l}=r_2^2\, \prod^{2l}_{j=1, j\not=2,3} r_j.
\end{eqnarray*}
The proof is finished.
\end{proof}

\begin{proposition}\label{p7} The following assertions take place:
\begin{enumerate}
\item We have
\begin{equation}\label{e20}
\begin{split}
&||D^m||^2_{\css_2}\le C\underbrace{\int^\dh_0\dots\int^\dh_0}_{4m}
\frac{\prod^{4m}_{j=1} \psi_0(r_j)^2}{(r_1+r_2)^2 \lp\prod^{2m-1}_{j=2}(r_j+r_{j+1})\rp (r_{2m}+r_{2m+1})}\\
&\hspace{2,5cm} \frac 1{(r_1+r_{2m}) \lp\prod^{4m-1}_{j=2m+2}(r_j+r_{j+1})\rp (r_{4m}+r_{2m+1})}\, \prod^{4m}_{j=1}dr_j.
\end{split}
\end{equation}
\item In particular, $D^m\in \css_2$ (or, equivalently, $D\in \css_{2m}$) whenever $\g>1/(8m)$.
\end{enumerate}
\end{proposition}

\begin{proof}
The proof of the proposition is lengthy, but rather elementary. For the convenience of the reader, we first treat the case $m=1$, and then go to the general $m$.

\emph{First}, let $m=1$. Recall the notation introduced in Proposition \ref{p6}. Relation \eqref{e18} says that
\begin{equation}\label{e201}
|D(z_1,z_3)|\le C\int^\dh_0 \frac{\tc_\shl(z_1)\psi_0(r_2)^2\tc_\shl(z_3)}{(r_1+r_2)(r_2+r_3)}\, dr_2,\quad z_1,z_3\in\bd.
\end{equation}
 We continue as
\begin{eqnarray}
I_1&:=&||D||^2_{\css_2}=\int_{\bd, z_1}\int_{\bd, z_3} |D(z_1,z_3)|^2\, dA(z_1)dA(z_3) \nn\\
&=&\int_{\shl, z_1}\int_{\shl, z_3} |D(z_1,z_3)|^2\, dA(z_1)dA(z_3). \label{e21}
\end{eqnarray}
Now, we pass to the polar coordinates translated to point $z_0=1$
$$
z_1=1-r_1e^{i\th_1}, \ z_3=1-r_3e^{i\th_3}
$$
exactly as we did in Proposition \ref{p6}. Here, $r_j\in (0,\dh),\ \th_j\in (0,\pi/2), \ j=1,3$. We replace each factor $|D(z_1,z_3)|$ in the integal on RHS of \eqref{e21} by the integral bound given in \eqref{e18}. Hence, the quantity in the RHS of \eqref{e21} is written as a 4-tuple integral, and it is estimated above as
\begin{equation}\label{e22}
I_1 \le C\underbrace{\int^\dh_{0,r_1}\dots \int^\dh_{0,r_4}}_{4}
\frac{\prod_{j=1}^4 \psi_0(r_j)^2}{(r_1+r_2)(r_2+r_3)\cdot (r_1+r_4)(r_4+r_3)}\, \prod^4_{j=1}dr_j.
\end{equation}
This is point (1) of the proposition for $m=1$.

As already mentioned, we cut the cube $\ccs=\ccs(\dh)=(0,\dh)^4$ in simplexes
$$
\ccs_{i_1\dots i_4}=\{(r_1,r_2,r_3,r_4): \dh>r_{i_1}>\dots r_{i_4}>0\},
$$
where $i_k=1,\dots, 4, i_k\not=i_j, k\not=j, \ k,j=1,\dots,4$. Lemma \ref{l4} with $l=2$ says trivially that
$$
(r_1+r_2)(r_2+r_3)\cdot (r_1+r_4)(r_4+r_3)\ge r_{i_1}^2 r_{i_2}r_{i_3}.
$$
Consequently, the upper bound for integral \eqref{e22} on the simplex $\ccs_{i_1\dots i_4}$ reads as
\begin{eqnarray}\label{e23}
&&\int_{\ccs_{i_1\dots i_4}}\frac{\prod_{j=1}^4 \psi_0(r_j)^2}{(r_1+r_2)(r_2+r_3)\cdot (r_1+r_4)(r_4+r_3)}\, \prod^4_{j=1}dr_j \\
&\le& C \int^\dh_0 dr_{i_1}\, \frac{\psi_0(r_{i_1})^2}{r^2_{i_1}}\, \int^{r_{i_1}}_0 dr_{i_2}\,\frac{\psi_0(r_{i_2})^2}{r_{i_2}} \dots \int^{r_{i_3}}_0 dr_{i_4}\, \psi_0(r_{i_4})^2 \nn\\
&\le& C \int^\dh_0 dr_{i_1}\, \frac{\psi_0(r_{i_1})^2}{r^2_{i_1}}\, \int^{r_{i_1}}_0  dr_{i_2}\, \frac{\psi_0(r_{i_2})^2}{r_{i_2}}\, \int^{r_{i_2}}_0 dr_{i_3}\, \frac{\psi_0(r_{i_3})^4 r_{i_3}}{r_{i_3}} \nn\\
&\le& C \int^\dh_0 dr_{i_1}\, \frac{\psi_0(r_{i_1})^2}{r^2_{i_1}}\, \int^{r_{i_1}}_0 dr_{i_2}\, \frac{\psi_0(r_{i_2})^6 r_{i_2}}{r_{i_2}} \nn\\
&\le& C \int^\dh_0 dr_{i_1}\, \frac{\psi_0(r_{i_1})^8 r_{i_1}}{r^2_{i_1}}=C \int^\dh_0 dr_{i_1}\, \frac{\psi_0(r_{i_1})^8}{r_{i_1}} \nn
\end{eqnarray}
where we used that $\psi_0(r_{i+1})\le \psi_0(r_i)$ since $r_{i+1}\le r_i$ and the function $\psi_0$ is increasing.
The latter integral is convergent whenever $8\g>1$, which is point (2) of the current proposition with $m=1$.

\emph{Second}, we follow the lines of the proof for $m=1$ in the general case, but we use some combinatorics. We get
\begin{eqnarray*}
I_m&:=&||D^m||^2_{\css_2}=\int_{\bd, z_1}\int_{\bd, z_{2m+1}} |D_m(z_1,z_{2m+1})|^2\, dA(z_1)dA(z_{2m+1}) \nn\\
&=&\int_{\shl, z_1}\int_{\shl, z_{2m+1}} |D_m(z_1,z_{2m+1})|^2\, dA(z_1)dA(z_{2m+1}).
\end{eqnarray*}
Now, we bound $|D(z_1,z_{2m+1})|^2$ from above by the product of expressions from \eqref{e18}. So, we come to a $4m$-tuple integral
\begin{eqnarray*}
&&I_m\le C\underbrace{\int^\dh_0\dots\int^\dh_0}_{4m}
\frac{\prod^{4m}_{j=1} \psi_0(r_j)^2}{(r_1+r_2) \lp\prod^{2m-1}_{j=2}(r_j+r_{j+1})\rp (r_{2m}+r_{2m+1})}\\
&&\qquad\qquad \frac 1{(r_1+r_{2m}) \lp\prod^{4m-1}_{j=2m+2}(r_j+r_{j+1})\rp (r_{4m}+r_{2m+1})}\, \prod^{4m}_{j=1}dr_j,
\end{eqnarray*}
the computation being completely analogous to \eqref{e22}. Point (1) of the proposition is proved in the general case.

We now divide the cube $\ccs=\ccs(\dh)=(0,\dh)^{4m}$ in simplexes
$$
\ccs_{i_1\dots i_{4m}}=\{(r_1,\dots,r_{4m}): \dh>r_{i_1}>\dots >r_{i_{4m}}>0\},
$$
where $i_k=1,\dots, 4m, i_k\not=i_j, k\not=j, \ k,j=1,\dots, 4m$. Lemma \ref{l4} ($l=2m$) gives
\begin{equation*}
\begin{split}
&\lp \prod^{2m}_{j=1}(r_j+r_{j+1})\rp \cdot
(r_1+r_{2m+2})\lp \prod^{4m-1}_{j=2m+2}(r_j+r_{j+1})\rp (r_{4m}+r_{2m+1})\\
&\ge r^2_{i_1}\lp\prod^{4m-1}_{j=2} r_{i_j}\rp.
\end{split}
\end{equation*}
Consequently, the upper bound for the integral on $\ccs_{i_1\dots i_{4m}}$ is
\begin{eqnarray*}
&&\int_{\ccs_{i_1\dots i_{4m}}}\dots \le
C \int^\dh_0 dr_{i_1}\, \frac{\psi_0(r_{i_1})^2}{r^2_{i_1}}\, \int^{r_{i_1}}_0 dr_{i_2}\, \frac{\psi_0(r_{i_2})^2}{r_{i_2}}\, \dots \int^{r_{i_{4m-1}}}_0 dr_{i_{4m}}\, \psi_0(r_{i_{4m}})^2.
\end{eqnarray*}
As before, we have $\psi_0(r_{i+1})\le \psi_0(r_i)$ since $\psi_0$ is increasing and $0<r_{i+1}\le r_i$. We bound this integral by telescoping as in \eqref{e23} to obtain
\begin{eqnarray*}
&& \int_{\ccs_{i_1\dots i_{4m}}}\dots \le C \int^\dh_0 dr_{i_1}\, \frac{\psi_0(r_{i_1})^{8m} r_{i_1}}{r^2_{i_1}}=C \int^\dh_0 dr_{i_1}\, \frac{\psi_0(r_{i_1})^{8m}}{r_{i_1}}.
\end{eqnarray*}
The latter integral is convergent whenever $8m\g>1$, which is point (2) of the current proposition. The proof is finished.
\end{proof}

\nt
{\it Proof of Theorem \ref{t20}.}
Indeed, fix a natural $m>0$ such that $\g>1/(8m)$. Recall that
$$
(T^*_{\o_1}T_{\o_L})^*(T^*_{\o_1}T_{\o_L})\le D,
$$
and, by Proposition \ref{p7}, the compact positive operator $D$ lies in $\css_{2m}\subset \ss^0_{2\g}$. By monotonicity of singular values, the operator  $T^*_{\o_1}T_{\o_L}$ is in $\ss^0_{2\g}$ as well, and the theorem is proved. \hfill$\Box$

\section{Applications and concluding remarks}\label{s5}
The corollaries presented in this Section mainly follow Pushnitski \cite{pu1}. Once again, remind the notation for functions $\p_0,\p_1, \p$, \eqref{e0041}, \eqref{e0042}.

\subsection{Different verisons of the obtained results}\label{s51}

\begin{corollary}\label{c1}\hfill

\begin{enumerate}
\item Let $\p$ be a function continuous on $\bar\bd$ and having the property that
$$
\lim_{|z|\to1-0} |\p_{0,\g}(|z|)^{-1}\p(z)-\p_1(e^{i\th})|=0, \quad z=|z|e^{i\th}\in\bd,
$$
for some $\p_1\in C(\bt)$. Then the singular values of operator $T_\p$ have the asymptotics \eqref{e005}, \eqref{e0051}.
\item In particular, let
$$\p(z)=\p_1(e^{i\th})\p_{0,\g}(|z|)\, g(|z|),$$
where $g\in L^\infty[0,1)$ and  $g(1):=\lim_{r\to1-0} g(r)$. Then
$$
\lim_{n\to+\infty} \log(n+1)^\g s_n(T_\p)=\lp |g(1)|\,||\p_1||_{L^\infty(\bt)}\rp^\g,
$$
idem for the counterpart of formula \eqref{e0051}.
\end{enumerate}
\end{corollary}
The proof of the first claim of the corollary is similar to the reasoning of Section 3 and it is omitted. The second claim is an easy consequence of the first one.

The general operator-theoretic techniques developed in Pushnitski-Yafaev \cite{puy} permit one to treat the sequences of positive and negative eigen-values $\{\l^\pm_n(T_\p)\}_n$ of operator $T_\p$ with a real symbol $\p$. Recall that $\p^\pm_1=\max\{\pm\p_1,0\}$. For simplicity, we give the following corollary for a symbol $\p$ given in \eqref{e0041}, \eqref{e0042}.
\begin{corollary}\label{c2}
Let $\p$ be a real symbol as above. We have the following asymptotics for the sequences of positive and negative eigenvalues of operator $T_\p$, respectively
$$
\lim_{n\to+\infty} \log(n+1)^\g \l^\pm_n(T_\p)=||\p^\pm_1||_{L^\infty(\bt)}^\g.
$$
\end{corollary}

\begin{corollary}\label{c3}
Formulae \eqref{e005}, \eqref{e0051} hold for (finitely) piece-wise continuous functions $\p_1$ defined on $\bt$.
\end{corollary}
The last corollary follows from the fact that the piece-wise continuous finctions lie in the closure of step-functions on $\bt$ in $L^\infty(\bt)$-norm.

\subsection{An application to finite-banded matrices}\label{s52}
Clearly enough, the obtained results allow us to handle the singular values of finite-banded matrices with loga\-rithmically decaying entries. We give a couple of definitions to make this more precise.

Let $D:\ell^2(\bz_+)\to \ell^2(\bz_+)$ be a compact operator. Its matrix is denoted by
$$
D:=[d_{i,j}]_{i,j=0,\dots,\infty}
$$
in the standard basis of $\ell^2(\bz_+)$. For a fixed $N\in\bn$, we say that $D$ is a $(2N+1)$-banded matrix, if $d_{i,j}=0$ for $|i-j|>N+1$. Furthermore, let $\{b_{-N},\dots, b_{N}\}$ be a set of $2N+1$ complex coefficients. For a fixed $\g>0$, we say that a $(2N+1)$-banded matrix $D$ has logarithmically decaying entries, if
\begin{equation*}
d_{m,m+j}=\frac{b_j}{(\log m)^\g}(1+o(1)),\ \quad m\to +\infty, \ j=-N,\dots,N.
\end{equation*}
To give the next corollary, define a specific function $\p_1$ corresponding to coefficients $\{b_j\}_{j=-N,\dots,N}$ as
$$
\p_{1,b}(e^{i\th}):=\sum^N_{j=-N} b_je^{i j\th}.
$$
\begin{corollary}\label{c4} We have the following asymptotics for the singular values of the above $(2N+1)$-banded matrix $D$
\begin{equation}\label{e25}
\lim_{n\to+\infty} (\log n)^\g s_n(D)=||\p_{1,b}||_{L^\infty(\bt)}.
\end{equation}
\end{corollary}
The proof of the corollary follows at once from two facts. First, the singular values of Toeplits operator $T_\p$ with $\p=\p_{1,b}\p_0$ have asymptotics \eqref{e25}. Second, we write the operator $T_\p$ in the standard basis of the Bergman space $L^2_a(\bd)$ and we identify it with the obtained matrix on $\ell^2(\bz_+)$. It is easy to see now that $D-T_\p\in \ss^0_\g$, and the claim of the corollary follows from Proposition \ref{p2}.

\section{Appendix A: Asymptotics of a logarithmic integral}\label{s6}
The following lemma is an easy consequence of a Watson-type lemma for Laplace integrals proved in Kupin-Naboko \cite[Thm. 0.2, Cor. 2.4]{kn1}.

\begin{lemma}\label{l01} Let $g\in L^\infty[0,1)$, and $g(1):=\lim_{r\to 1-0}g(r)\not=0$. Then, for a  $\g>0$,
$$
\int^1_0 \frac {r^n}{(1+\log1/(1-r))^\g}g(r)\, dr=\frac{g(1)}{n(\log n)^\g}\lp 1+o(1)\rp.
$$
\end{lemma}

\bigskip
\nt{\bf Acknowledgments.}  The authors would like to thank Omar El-Fallah, Leonid Golinski, Karim Kellay and Alexander Pushnitski for heplful discussions.

The work is partially supported by the project ANR-18-CE40-0035.

S. Naboko kindly acknowledges the support by RScF-20-11-20032 grant and Knut and Alice Wallenberg Foundation grant. A part of this research was done during S. Naboko's visit to University of Bordeaux in October-November, 2019. He is grateful to the University for the hospitality.

B. Touré gratefully acknowledges the financial support coming from agreements between University of Bordeaux and the Embassy of France in Mali, 2018.

config: arxiv


TITLE: Why is a solution to $y'' + y = 0$ periodic
QUESTION [8 upvotes]: Solutions to $y'' + y = 0$ are 2$\pi$ periodic. Is this accidental or does this ODE have some symmetries associated with it that force the solutions to be periodic?

REPLY [3 votes]: Well, the equation $y'' + y = 0$ it conserves the quantity $y'^2/2+y^2/2$, so the trajectories are confined to this circle in phase space. Moreover, the equation is autonomous and has unique solutions, so the system can't wobble back and forth around the circle; it can only spiral around it. This is almost sufficient to show that any solution must be periodic; I think it only remains to show that the system continues moving at a rate which is bounded below by a positive number, rather than approaching a steady state.
(Answer based on a comment by Ian.)


TITLE: Pressure done by a bar/beam inserted in a continuous medium (wall)
QUESTION [1 upvotes]: Related to a DIY project, I'm facing the following question:

Assume a metal bar (beam) of square section,  $a \times a$, is inserted partially in a continuous medium (lets say, a wall) in a way that $h$ of the beam is inside the medium and $l$ is outside of the wall, and a force $F=200 N$ is done in the other extreme of the bar, perpendicular to it. The question is: how the pressure ( the force per unit area, $N/mm^2$ ) is distributed in the contact surfaces between the cantilever beam and the wall ?
Note: we could say that the beam has no weight, no friction forces and the system is stable, no displacements. An infinite wall in width, height and thickness. If easier, we can consider a beam of circular section instead of square.
Inside the wall the beam has 5 surfaces: right, left, top, bottom and back. In the ideal case of no friction, I think forces in right and left surfaces will be null. No idea if forces in back and top are also null. Force in bottom must allow to keep fixed all system, with an unknown distribution of pressure on it ( uniform? maximum at $h$ deep? ).
I've no knowledge to solve it as a continuous problem. A strong simplification that came to my mind is consider the system as equivalent to a lever with lengths $h$ and $l$, being the support axe the red line in the draw. That means to the force in the other extreme of the beam (green line in the draw) will be $F' = F \frac {l}{h}$. By example if h=5cm and l=50 cm, then $F'=200*50/5=2000N$.
Background: if we stand a TV of 20 kg using a $l=50 cm$ beam, how to estimate the forces over the wall ?

REPLY [1 votes]: This shows a combination of a steel rod and a concrete wall in 3D. The rod is embedded in the wall 1/6 of the length. Section of a rod 1x1. We take the effective force divided by the cross-sectional area as 1. The middle picture is the distribution of the vertical component of the deformations in the wall; in the right-hand picture, this is the distribution of the $\sigma_{zz}$ stress component. It can be seen that in such a situation, a stress increase of 30 times can be obtained. The calculations are performed using FEM and Mathematica 12.


TITLE: Is unsorted first-order logic equivalently expressive to infinitely-many-sorted first-order logic?
QUESTION [5 upvotes]: I know that unsorted FOL is equivalently expressive to finitely-many-sorted FOL. The only semantic criteria placed on the sorts in the many-sorted FOL are that every object in the domain is in one of the sorts, and the sorts are disjoint subsets of the domain. All the statements in unsorted FOL can be trivially mapped to statements in just one of the sorts of the many-sorted logic, in both the finite & infinite case, and both should have all the same models.
To translate from N-sorted FOL to unsorted FOL, we can add unary relation symbols $S_1, ...,  S_N$ and then we add axioms that state that they are disjoint and cover the whole set:

$\forall x, S_1(x) \vee ... \vee S_N(x)$

and for every pair of sort-symbols $S, S'$, the axiom:

$\neg(\exists x, S(x) \wedge S'(x))$

For every constant symbol c meant to be interpreted as being in sort S, we add the axiom:

$S(c)$

Then anywhere you do a quantification over a sort S in N-sorted logic like

$\forall x : S, P(x)$

we can translate this to an equivalent unsorted FOL statement in our new theory:

$\forall x, S(x) \implies P(x)$

To attempt to translate from infinitely-many-sorted FOL to unsorted FOL, we can add unary relation symbols $S_1, S_2,...$ for each sort, and then add the axioms

$\forall x, \phi(x)$

and for every pair of sort-symbols $S, S'$, the axiom

$\neg(\exists x, S(x) \wedge S'(x))$

and for every constant symbol c meant to be interpreted as being in sort S, the axiom:

$S(c)$

Where $\phi(x)$ stands for an FOL formula expressing that the object $x$ is in at least one of the sorts. We'll call this theory $T$. The formula $\phi(x)$ must be semantically equivalent to the following infinite disjunction of statements:

$S_1(x) \vee S_2(x) \vee ...$

The negation of $\forall x, \phi(x)$ must therefore be equivalent to the statement:

$\exists x, \neg S_1(x) \wedge \neg S_2(x) \wedge ...$

Skolemizing, we can add a fresh constant symbol $\sigma$ and create an equisatisfiable formula:

$\neg S_1(\sigma) \wedge \neg S_2(\sigma) \wedge ... $

This is an infinite conjunction of finite first-order statements, and thus represents a first-order theory, call it $\neg T$. Because we assumed $T$ was a first-order theory, then $T \cup \neg T$ is also a first-order theory.
$T \cup \neg T$ is not satisfiable, because $\neg T$ semantically contradicts $\forall x, \phi(x)$, in the case where $x = \sigma$. By the compactness theorem, there must exist some finite subset of $T \cup \neg T$ which is not satisfiable.
However, it's straightforward to see that every finite subset is satisfiable: choose a structure where the objects are the constant symbols $c$ such that the axiom $S(c)$ is included in your subset, for some sort symbol $S$, and map every sort symbol $S$ to the unary relation that holds for an object $c$ iff the axiom $S(c)$ is included in your subset; choose your structure such that for the interpretations of any two sort symbols $S, S'$ are disjoint sets of the domain; if for every axiom $S(c)$ included in your subset, the axiom $\neg S(\sigma)$ is also included in your subset, then choose another sort symbol $S'$ such that the axiom $\neg S'(\sigma)$ is not included in your subset, add a new object in the domain to represent $\sigma$ and map $S'$ to a unary relation on the domain that only contains $\sigma$; otherwise, there exists an axiom $S(c)$ in your subset such that the axiom $\neg S(\sigma)$ is not included in your subset, in which case interpret $\sigma$ as the same object that $c$ is interpreted as. Whatever finite subset of the theory is chosen, there should exist a structure defined in this fashion which models that subset.
This contradicts the compactness theorem. Therefore, $T \cup \neg T$ could not actually be a first-order theory. But since $\phi(x)$ is the only thing in the theory which hasn't been expressed directly with a first-order sentence, it follows that $\phi(x)$ must not be expressible in unsorted FOL.
Is my reasoning correct here? If so does this imply that unsorted FOL is not capable of simulating infinitely-many-sorted FOL, as a consequence of the compactness theorem, and that the latter is therefore more expressive?

REPLY [5 votes]: If we had infinitely many sorts $\tau_1, \tau_2, \ldots$, why not add an infinite collection of unary predicates $S_1, S_2, \ldots$, add axioms for all $i \not = j$:
$$
(\exists x) S_i(x) \\
S_i(x)  \to \lnot S_j(x)
$$
along with the axioms for constants of particular sorts, and replace each quantifier
$(\exists x^{\tau_i}) \ldots$ with $(\exists x)[S_i(x) \land \ldots]$ and similarly for $(\forall x^{\tau_i})$.  For notation, say that this interpretation takes a sentence $\phi$ in infinitely many sorts to $\phi^*$ in one sort with new unary relations.
This seems to interpret any model with infinitely many sorts into a model of the expanded theory with only one sort. Conversely, a model in the new language and expanded theory might have some elements that do not satisfy $S_i$ for any $i$, but if we discard those then the resulting model immediately leads to an equivalent model in infinitely many sorts. The elements that are not in any sort are irrelevant to sentences of the form $\phi^*$, because the quantifiers in $\phi^*$ are explicitly tied to particular sorts.
Letting $T^*$ be the interpretation and expansion of a theory $T$, in this way we still have that $T$ has a model with infinitely many sorts if and only if $T^*$ has a model with one sort.


TITLE: Fourier optimization problem related to the Prime Number Theorem
QUESTION [6 upvotes]: Let $\kappa>0$ be given. What is the function
$f:\mathbb{R}\to [0,\infty)$ with $\int_\mathbb{R} f(x) dx = 1$ such that
$$\int_\mathbb{R} |x| f(x) dx + \kappa \int_{|t|\geq T}\left| \frac{\widehat{f}(t)}{t} \right| dt$$
is minimal?

Commentary.
The motivation of this question resides in its applications to bounds in analytic number theory -- for example, explicit versions of the prime number theorem. The question is essentially equivalent to the one I have asked in Best smoothing for the Prime Number Theorem? and Optimizing a smoothing function with the Prime Number Theorem in mind . The main difference is that now I have a plausible candidate: we could work with $$f(x) = \frac{1/\sigma}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left(\frac{x}{\sigma}\right)^2}.$$
Since then
$$\int_{\mathbb{R}} |x| f(x) dx = 2 \int_0^\infty \frac{x/\sigma}{\sqrt{2\pi}} e^{-\frac{1}{2} \left(\frac{x}{\sigma}\right)^2} dx = - \frac{2\sigma}{\sqrt{2\pi}} \int_0^\infty \left(e^{-\frac{1}{2} \left(\frac{x}{\sigma}\right)^2}\right)' dx = \sqrt{\frac{2}{\pi}} \cdot\sigma$$
and, by $\widehat{f}(t) = e^{-2 \pi^2 \sigma^2 t^2}$,
$$\frac{\partial}{\partial \sigma}\int_{|t|\geq T} \left|\frac{\widehat{f}(t)}{t}\right| dt = 2\frac{\partial}{\partial \sigma} \int_{t\geq T} \frac{e^{-2\pi^2 \sigma^2 t^2}}{t} dt
=  - 2 \int_{t\geq T} 4 \pi^2 \sigma t e^{-2\pi^2\sigma^2 t^2} dt = - 2\frac{e^{-2\pi^2 \sigma^2 T^2}}{\sigma}
,$$
we see that the optimal value of $\sigma$ is the one for which $\sigma = \sqrt{2\pi} \kappa \cdot e^{-2 \pi^2 \sigma^2 T^2}$. We write $\sigma = \frac{\sqrt{2 \log T + \eta}}{2 \pi T}$, and see we must solve for $\frac{\sqrt{2 \log T + \eta}}{2\pi} = \sqrt{2\pi} \kappa e^{-\eta/2}$, so $\eta = -2 \log \frac{\sqrt{\log T}}{2\pi^{3/2}\kappa} + O\left(\frac{1}{\log T}\right) = - \log \log T + 2 \log(2 \pi^{3/2} \kappa) + O\left(\frac{1}{\log T}\right)$.
Then
$$\begin{aligned}\int_\mathbb{R} |x| f(x) dx + \kappa \int_{|t|\geq T}\left| \frac{\widehat{f}(t)}{t} \right| dt &= \sqrt{\frac{2}{\pi}} \sigma + 2 \kappa \int_{t\geq T} \frac{e^{-2\pi^2 \sigma^2 t^2}}{t} dt\\ &=
\sqrt{\frac{2}{\pi}} \sigma - \kappa \textrm{Ei}(-2\pi^2 \sigma^2 T^2)
= \sqrt{\frac{2}{\pi}} \sigma- \kappa \textrm{Ei}(- \log T - \eta/2) \\
&\sim \frac{\sqrt{\log(2\pi^{3/2} \kappa T) - \frac{1}{2} \log \log T}}{\pi^2 T} +
\kappa \sqrt{\frac{2}{\pi}} \frac{e^{-\eta/2}}{T (\log T + \frac{\eta}{2})}\\
&= \frac{1}{\pi^2 T} \left(\sqrt{\log(2\pi^{3/2} \kappa T)} - \frac{\log \log T}{4 \sqrt{\log 2\pi^{3/2} \kappa T}} + \frac{1}{\sqrt{2\log T}} + \dotsc\right).
\end{aligned}$$
Can one do better?

REPLY [9 votes]: If you care about the correct order of magnitude only (i.e., don't mind a bounded constant factor separated from both $0$ and infinity), the problem is fairly straightforward.
Scaling By playing with $uf(ux)$ instead of $f$ and choosing $u=1/\kappa$, one can immediately see that the answer for the minimum is $\kappa F(\kappa T)$ where $F$ is the answer with $\kappa=1$, so I'll assume $\kappa=1$ from now on.
The case $T\gtrsim 1$. Take any smooth non-negative even $\psi$ supported on $[-2,2]\setminus[-1,1]$ and such that $\int_{\mathbb R}\psi=1$ and put $\psi_T(t)=T^{-1}\psi(t/T)$. Then the inverse Fourier transform $\Psi_T$ of $\psi_T$ satisfies $\Psi_T(x)\ge 1-2T|x|$. Thus, integrating
$$
(2T)^{-1}=(2T)^{-1}\int_{\mathbb R}f \le \int_{\mathbb R}[|x|+(2T)^{-1}\Psi_T]f =
\\
\int_{\mathbb R}|x|f+(2T)^{-1}\int_{\mathbb R}\psi_T \widehat f
\lesssim \Phi(f)
$$
where $\Phi(f)$ is your functional to be minimized if $T\gtrsim 1$, so in this regime $\Phi(T)\gtrsim T^{-1}$. To get a matching upper bound, just take any positive $f_0$ with integral $1$, Fourier transform supported on $[-1,1]$ and some minimally decent decay like $|f_0(x)|\lesssim \frac 1{1+|x|^3}$ and put $f(x)=Tf_0(Tx)$, so that the Fourier transform of $f$ vanishes outside $[-T,T]$ and we are left with $\int_{\mathbb R}|x|f=T^{-1}\int_{\mathbb R}|x|f_0\approx T^{-1}$.
Now we want to investigate what happens as $T\to 0$.
Note that if $\int_{|x|>T^{-1/2}}|f|\ge T^{1/4}$, say, then $\Phi(f)\ge T^{-1/4}$ already. Otherwise
$$
|\widehat f(t)|\ge 1-T^{-1/2}t-T^{1/4}>1/2
$$
when $t\in (T,T^{3/4})$, say, so the Fourier part of $\Phi(f)$ is $\gtrsim \log(1/T)$.
Hence $\Phi(f)\gtrsim \log(1/T)$ in this case. To have a matching upper bound just take any fixed smooth fast decaying function.
The final answer is that, up to a constant factor, the minimum is $\frac 1T$ if $\kappa T\ge 1$ and $\kappa\log\frac e{\kappa T}$ if $\kappa T<1$.


TITLE: Finding the angle between two points in $l^2$ metric space
QUESTION [0 upvotes]: let $p = (1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...)$ and $q=(\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},...)$
I am asked to find angle between these two in the metric space $l^2$
I can apply the formula $$\frac{<p,q>}{||p|||q||}=cos \theta$$
Where the numerator is the scalar product.
$$<p,q>  = {\sum_1^\infty a_kb_k}$$ in this metric. and $$||p|| = \sqrt{ \sum_1^\infty |a_k|^2}$$
Calculating the scalar product we end up with the series
$$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}.... = 1$$
However I couldn't manage to find $||p||||q||$ any hints?

REPLY [0 votes]: The well known series in this case is $$\sum_{n=1}^\infty \frac{1}{n^2}  =\frac{\pi^2}{6}$$ This equals $||p||^2$ while $||q||^2$ misses the 1 term so equals $\frac{\pi^2}{6} - 1$


TITLE: Complex Log Function
QUESTION [0 upvotes]: This problem is giving me quite a bit of grief. Heres my issue I do the first problem step by step and get the correct solution. I then try the second problem step by step and get the wrong answer. Specifically I get them to be equal again. Can someone explain why the process taken in question 1 does not work in 2. Thanks for the help.     textbook solution

REPLY [0 votes]: The expanation for the difference between Part (i) and Part (ii) is that in the complex numbers $e^u=e^v$ does $not$ imply $e^{u/5}=e^{v/5}$.
Note that $(\sqrt 3\; -i)^{1/5}$ is not a number but a set of $5$ numbers.
In Part (i), "$z$ is a log of $(\sqrt 3\;-i)^{1/5}\;$" iff $(e^z)^5=\sqrt 3 -i$..... while "$z$ is $1/5$ of a log of $\sqrt 3\;-i $" iff "$5z$ is a log of $\sqrt 3\;-i\;$" iff $e^{5z}=\sqrt 3\;-i.$ (... & observe that $(e^z)^5=e^{5z}.)$
In Part (ii), "$z$ is a log of $(\sqrt 3\;-i)^5\;$" iff $e^z=(\sqrt 3\;-i)^5,$ but $e^z=(\sqrt 3\;-i)^5$  does $not$ imply that $e^{z/5}=\sqrt 3\;-i$.... while "$z$ is $5$ times a log of $\sqrt 3\; -i\;$" iff  $e^{z/5}=\sqrt 3\;-i.$


TITLE: Microsoft Paint Spraypaint Tool
QUESTION [7 upvotes]: The following question is motivated by the spraypaint tool in Microsoft Paint.  In a sense, I'm asking how many clicks on average are necessary to fill each pixel within the range of the tool if we leave the cursor fixed.  The following is a more precise statement of the question:

Consider $n$ dice, each with $k$ sides.  A roll consists of rolling all $n$ dice.  Determine $E(n,k)$, the expected number of rolls needed so that all numbers $1,\ldots,k$ have appeared on some roll.

The translation to spraypaint can be seen by assuming the range of the spraypaint tool is $k$ pixels, and each click paints $n$ random pixels.
I remember looking at this problem years ago, and only partially solving it.  I believe I worked on the case $n=1$:
$$E(1,k)=kH_k=k\sum_{i=1}^k\frac{1}{i}$$
Here $H_k$ is the $k$th harmonic number.  I'm not entirely convinced of this result.  For $n=2$ I could only find a closed form for asymptotic behavior as $k$ got large.  My memory of the solution is a bit fuzzy though.

REPLY [1 votes]: For $n \gt 1$ you can imagine the die rolls in a batch of $n$ are sequential.  Except for the last roll, the situation is the same, so $E(n,k) \approx \frac 1nE(1,k)$.  It will actually be a little higher than this, as you will (in effect) keep rolling up to the next multiple of $n$, when with $n=1$ you would quit.  The heuristic answer would be that on average you would waste half the rolls, so $E(n,k) \approx \frac 1nE(1,k)+\frac 12$.  To do better, I think you would have to assess the probability of completion at each multiple of $n$.


TITLE: Minimization of a function including quadratic and exponential terms
QUESTION [4 upvotes]: Apologies for the simple question. Is it possible to analytically minimize a function such as
$$f(x) = a^x + c x (x-1) \qquad x > 0, \quad 0 < a,c < 1$$
Nothing else can be said about the constants $a$ or $c$.

REPLY [6 votes]: Alright, here's my expansion of my comment.
Let's rearrange the equation in VFG's answer like so:
$$\frac{\ln\;a}{2c}=\left(\frac12-x\right)a^{-x}$$
and multiply both sides by $\sqrt{a}$:
$$\frac{\sqrt{a}\ln\;a}{2c}=\left(\frac12-x\right)a^{\frac12-x}$$
We're close to a form where we can now use the Lambert function, but let's change bases first:
$$\frac{\sqrt{a}\ln\;a}{2c}=\left(\frac12-x\right)\exp\left(\frac{\ln\;a}{2}-x\ln\;a\right)$$
We're much closer now, so we multiply both sides by a factor of $\ln\;a$
$$\frac{\sqrt{a}(\ln\;a)^2}{2c}=\left(\frac{\ln\;a}{2}-x\ln\;a\right)\exp\left(\frac{\ln\;a}{2}-x\ln\;a\right)$$
and then perform the inversion:
$$W\left(\frac{\sqrt{a}(\ln\;a)^2}{2c}\right)=\frac{\ln\;a}{2}-x\ln\;a$$
which is now easily solved for $x$:
$$x=\frac12-\frac1{\ln\;a}W\left(\frac{\sqrt{a}(\ln\;a)^2}{2c}\right)$$
Now, the Lambert function has two real branches $W(z)$ (principal branch) and $W_{-1}(z)$; one is real for $z\geq-e^{-1}$, and the other is real only in the interval $[-e^{-1},0)$. If both $a$ and $c$ are in the interval $(0,1)$, the argument of the Lambert function would be much greater than 1, and thus we can dispose of the $W_{-1}$ branch, as well as consider $W(z)$ only for $z\geq 0$, which is true for $a$ and $c$ in the interval $(0,1)$. Thus, the Lambert solution using the principal branch is valid.