TITLE: Existence of bounded linear functional following from inequality $\left| \langle x^*, x \rangle \right| \leq C \lVert Ax \rVert$
QUESTION [0 upvotes]: Given that $X, Y$ are real normed spaces, $A:X\to Y$ is a bounded linear operator and $x^*\in X^*$, the following holds:
If there is some $c\geq 0$ such that
$$ \tag{1} \left| \langle x^*, x \rangle \right| \leq C \lVert Ax \rVert $$
for all $x\in X$, then $ x^*\in \operatorname{im}A^*$
The proof I am reading starts with this observation: Assuming (1), there is a bounded linear functional $\psi: \operatorname{im}A \to \mathbb{R}$ such that $\psi\circ A = x^* $ (later it is extended to an operator $y^*\in Y^*$).
This observation is not obvious to me, why does $\psi$ exist? Any attempt to construct such an operator failed...

REPLY [2 votes]: Notice that for every $x \in X$ with $x \in \ker A$ we have $|\langle x^*, x \rangle| \leq C \|Ax\| = 0$ and thus $x \in \ker x^*$. Because $\ker A \subseteq \ker x^*$ there exists a unique linear map $\psi \colon \operatorname{im} A \to \mathbb{R}$ with $x^* = \psi \circ A$ by the universal property of the quotient vector space $X/\ker A \cong \operatorname{im} A$.
As Daniel Fischer has pointed out in the comments we have
$$
 |\langle \psi, Ax \rangle| = |\langle x^*, x \rangle| \leq C\|Ax\|
 \quad\text{for all $x \in X$}
$$
and thus $|\psi(y)| \leq C\|y\|$ for all $y \in \operatorname{im} A$. So $\psi$ is bounded with $\|\psi\| \leq C$.


TITLE: A morphism of sheaves is surjective if it is surjective at the level of stalk for every point.
QUESTION [0 upvotes]: I would like to show that if $\phi:\mathscr{F} \rightarrow \mathscr{G}$ be a morphism of sheaves (on a topological space $X$) such that $\phi_P:\mathscr{F}_{P} \rightarrow \mathscr{G}_{P}$ is surjective for all $P \in X$ then $\phi$ is also surjective.
My effort: I am using the fact $\text{Im}(\phi)_P=\text{Im}(\phi_P)$ for all $P \in X$. Then we have $\text{Im}(\phi)_P=\mathscr{G}_P$ implying that $(\mathscr{G}/\text{Im}(\phi))_P=\mathscr{G}_P/\text{Im}(\phi)_P=0$. Now since $\mathscr{G}/\text{Im}(\phi)$ is a sheaf we've $\mathscr{G}/\text{Im}(\phi)=0$. From here how do I conclude that $\text{Im}(\phi)=\mathscr{G}$.
I need some help. Thanks.

REPLY [1 votes]: By $im(\phi) = \mathscr G$ I will assume you mean that the canonical map $im(\phi) \longrightarrow \mathscr G$ is an isomorphism. This argument holds in any abelian category.
As you have proven, $\mathscr G/im(\phi) = 0$. This is precisely the cokernel of $\phi$, and the image of a map in an abelian category is defined to be the kernel of its cokernel. We have the following universal property:

In symbols, $\pi \circ \psi = 0$ implies the existence and uniqueness of the dashed map making the diagram commute.
Now, as we said, $im(\phi) = ker(cok(\phi))=ker(\pi)$. So by "the canonical map $im(\phi) \longrightarrow \mathscr G$", I mean $\iota$. We will proceed to show that it is an isomorphism.
Since $\mathscr G/im(\phi)=0$, we necessarily have $\pi = 0$. We will therefore show that $id: \mathscr G \longrightarrow \mathscr G$ satisfies the universal property of $ker(\pi)$. Indeed, let $\psi: \mathscr H \longrightarrow \mathscr G$ satisfy $\pi \circ \psi = 0 \circ \psi$. Thus, we have the commutative diagram:

and certainly, only $\psi$ can fill in for the dashed map and allow the diagram to commute. Hence, this satisfies the same universal property as $ker(\pi)$. Hence, they are isomorphic. This arises by applying the universal property in both directions. In fact, this yields the following commutative diagram.

Hence, we have shown that the canonical map $im(\phi)=ker(\pi)\longrightarrow \mathscr G$ is an isomorphism. This is the closest to equality we can get in an abelian category.


TITLE: Construction of Angular Momentum eigenvectors
QUESTION [0 upvotes]: I have a problem that asks (verbatim)

Carryout the construction of the eigenvectors of total angular momentum and its z component for $j_1$=3/2 and $j_2$=1/2

I am not completely sure where to start out with this.  Our professor has not given the best explanation.  The equations I have been trying to use are the following:
$$\textbf{L}^2|l,m\rangle =\hbar^2l(l+1)|l,m\rangle$$
$$\textbf{L}_z|l,m\rangle=\hbar m|l,m\rangle$$
I figure I probably want to use the same equations, but for $\bf J$ instead of $\bf L$.  I am also confused what the differences between $j_1$ and $j_2$ would represent.  I would greatly appreciate any help!

REPLY [2 votes]: I'm gonna guess the problem involves the angular momentum states with $\ell=1$ and the spin-1/2 states.  I'm making this guess because, when $j$ is half-integer, there is no "stand-alone" representation of the states in terms of regular (spherical or otherwise) coordinates, i.e. there is nothing like the spherical harmonics $Y_{\ell m}(\theta,\phi)$ which depend on the physical polar and azimuthal angles $\theta$ and $\phi$.  Moreover, in putting together the states of $\ell=1$ and $s=1/2$, one does obtain $j=1/2$ and $j=3/2$.
There are two ways to proceed.  The first is the "cheap" way which is to use pre-tabulated Clesbch-Gordan coefficients.  Thus, in general
$$
\vert j m\rangle = \sum_{\alpha\beta}
\vert 1;\alpha\rangle \vert \textstyle \frac{1}{2} ;\beta\rangle
C^{jm}_{1\alpha;\frac{1}{2}\beta}\, ,\qquad
\alpha \in \{1,0,-1\}\, ,\quad \beta\in\{\frac{1}{2},-\frac{1}{2}\}
$$
where $C^{jm}_{1\alpha;\frac{1}{2}\beta}$ is a Clebsch-Gordan coefficient, with $j=1/2$ or $3/2$.
It's just a matter of looking up the appropriate coefficient.
A more explicitly way is to start with
$$
\vert 11\rangle \vert \textstyle\frac{1}{2}\frac{1}{2}\rangle
$$
and note that this state is killed by
$L_+=L^{(1)}_++L^{(2)}_+$ where $L^{(1)}_+$ acts on the first factor alone, and $L^{(2)}_+$ acts on the second factor alone, i.e.
$$
L^{(1)}_+\vert 1\alpha\rangle \vert \textstyle\frac{1}{2}\beta\rangle = \sqrt{(1-\alpha)(1+\alpha+1)}
\vert 1,\alpha+1\rangle \vert \textstyle\frac{1}{2}\beta\rangle
$$
etc.  Thus, one easily verifies that
\begin{align}
L_+\vert 11\rangle \vert \textstyle\frac{1}{2}\frac{1}{2}\rangle&=0\,  ,\\
L_z\vert 11\rangle \vert \textstyle\frac{1}{2}\frac{1}{2}\rangle&=\textstyle\frac{3}{2}
\vert 11\rangle \vert \textstyle\frac{1}{2}\frac{1}{2}\rangle\, .
\end{align}
Since the only way to kill an angular momentum state is when its $L_z$-eigenvalue is equal to $j$, this identifies
$$
\vert \textstyle\frac{3}{2}\frac{3}{2}\rangle
= \vert 11\rangle \vert \textstyle\frac{1}{2}\frac{1}{2}\rangle\, .
$$
To get the other states in the sequence, use the lowering operator $L_-$.  For instance
$$
L_-\vert \textstyle\frac{3}{2}\frac{3}{2}\rangle
=\sqrt{3}\vert \textstyle\frac{3}{2}\frac{1}{2}\rangle
=\sqrt{2}\vert 10\rangle\vert \frac{1}{2}\frac{1}{2}\rangle+\vert 11\rangle \vert \frac{1}{2},-\frac{1}{2}\rangle
$$
and thus
$$
\vert \textstyle\frac{3}{2}\frac{1}{2}\rangle
=\sqrt{\frac{2}{3}}\vert 10\rangle\vert \frac{1}{2}\frac{1}{2}\rangle+\sqrt{\frac{1}{3}}\vert 11\rangle \vert \frac{1}{2},-\frac{1}{2}\rangle  \, . \tag{1}
$$
One then notes there are two possible eigenstates of $L_z$ with eigenvalue $1/2$.  They are $\sqrt{2}\vert 10\rangle\vert \frac{1}{2}\frac{1}{2}\rangle$ and $\vert 11\rangle \vert \frac{1}{2},-\frac{1}{2}\rangle$.  Clearly, (1) is a linear combination of these two. Consider now the combination orthogonal to (1):
$$
\vert\chi\rangle=\sqrt{\textstyle\frac{1}{3}}\vert 10\rangle\vert \textstyle\frac{1}{2}\frac{1}{2}\rangle-\sqrt{\frac{2}{3}}\vert 11\rangle \vert \frac{1}{2},-\frac{1}{2}\rangle  \, . \tag{2}
$$
One readily verifies that
$$
L_+\vert\chi\rangle=
\sqrt{\textstyle\frac{2}{3}}\vert 11\rangle\vert \textstyle\frac{1}{2}\frac{1}{2}\rangle-\sqrt{\frac{2}{3}}\vert 11\rangle \vert \frac{1}{2},\frac{1}{2}\rangle=0 \tag{3}
$$
which makes $\vert\chi\rangle$ the $\vert\frac{1}{2}\frac{1}{2}\rangle$ since it is killed by $L_+$ and has $L_z$-eigenvalue $1/2$.
We can use the lowering operator $L_-$ on (1) to obtain the $\vert \frac{3}{2},-\frac{1}{2}\rangle$ state and likewise on (3) to obtain the $\vert\frac{1}{2},-\frac{1}{2}\rangle$ states, and finally complete the calculation by using $L_-$ on $\vert \frac{3}{2},-\frac{1}{2}\rangle$.


TITLE: Questions on the integral of pseudo Gaussian kernel and its derivative on $(0,\infty)$
QUESTION [1 upvotes]: Consider pseudo Gaussian densities for $0<s<t$ and $x,y\in\mathbb R$
$$f(s,x,t,y):=\frac{1}{\sqrt{2\pi A(s,t,y)}}\exp\left(-\frac{(y-x)^2}{2A(s,t,y)}\right)\quad\mbox{and} \quad g(s,x,t,y):=\frac{1}{\sqrt{2\pi B(s,t,y)}}\exp\left(-\frac{(y-x)^2}{2B(s,t,y)}\right),$$
where $A(s,t,y):=\int_s^t k(u,y)/(1+a(u))du$ and $B(s,t,y):=\int_s^tk(u,y)/(1+b(u))du$. I'm interested in the dependence of the difference $f-g$ on the parameters $a,b$. Assume  $a, b: \mathbb R_+ \to [0,1]$ are continuous, $k: \mathbb R_+\times\mathbb R \to [1,2]$ is $1-$Lipschitz. Does there exist $C>0$ depending only on $T>0$ s.t.
\begin{eqnarray}
\left|\int_0^\infty f(0,0,t,y)dy - \int_0^\infty g(0,0,t,y)dy\right | &\le& Ct^{1/2}\|a-b \|_t,\quad \forall 0<t\le T\quad (\ast)  \\
\int_0^t \left|\int_0^\infty \partial_sf(s,0,t,y)dy - \int_0^\infty \partial_s g(s,0,t,y)dy\right |ds &\le& C(t-s)^{1/2}\|a-b \|_t,\quad \forall 0<s<t\le T\quad (\star),
\end{eqnarray}
where $\|a-b \|_t:=\max_{0\le u\le t}|a(u)-b(u)|$.
PS : $(\star)$ can be shown if
\begin{eqnarray}
\left|\int_0^\infty \partial_sf(s,0,t,y)dy - \int_0^\infty \partial_s g(s,0,t,y)dy\right | \le C(t-s)^{-1/2}\|a-b \|_t.
\end{eqnarray}
Iosif Pinelis has shown in How does the integral of pseudo Gaussian kernel on $(0,\infty)$ depend on its variance? that
$$\left|\int_0^\infty p(x)dx \int_0^\infty f(s,x,t,y)dy - \int_0^\infty p(x)dx\int_0^\infty g(s,x,t,y)dy\right |\le C(t-s)^{1/2}\|a-b \|_t,$$
where $p$ is some probability density on $(0,\infty)$.

REPLY [1 votes]: $\newcommand{\De}{\Delta}\newcommand{\vpi}{\varphi}$Let
\begin{equation*}
    A:=A(y):=A(0,t,y),\quad B:=B(y):=B(0,t,y),
\end{equation*}
so that
\begin{equation*}
    f(0,0,t,y)=\frac1{\sqrt{2\pi}}\,\vpi_{A(y)}(y), \quad g(0,0,t,y)=\frac1{\sqrt{2\pi}}\,\vpi_{B(y)}(y),
\end{equation*}
where
\begin{equation*}
    \vpi_a(u):=\frac1{\sqrt a}\,\exp\Big(-\frac{u^2}{2a}\Big).
\end{equation*}
Note that
\begin{equation*}
    \frac{\partial}{\partial a}\vpi_a(u)
    =\frac12\Big(\frac{u^2}{a^{5/2}}-\frac1{a^{3/2}}\Big)\exp\Big(-\frac{u^2}{2a}\Big).
\end{equation*}
So,
\begin{equation*}
    2\sqrt{2\pi}\,\int_0^\infty dy\,[f(0,0,t,y)-g(0,0,t,y)]  \tag{1}\label{1}
    =\int_0^1 dv\,I(v),
\end{equation*}
where
\begin{equation*}
    I(v):=I_1(v)-I_2(v), \tag{2}\label{2}
\end{equation*}
\begin{equation*}
    I_1(v):=\int_0^\infty dy\,H(y)
\frac{y^2}{c_v(y)^{5/2}}\exp\Big(-\frac{y^2}{2c_v(y)}\Big),
\end{equation*}
\begin{equation*}
    I_2(v):=\int_0^\infty dy\,H(y)
\frac1{c_v(y)^{3/2}}\exp\Big(-\frac{y^2}{2c_v(y)}\Big),
\end{equation*}
\begin{equation*}
    H:=B-A,
\end{equation*}
\begin{equation*}
    c_v:=A+v(B-A).
\end{equation*}
Next is the crucial step:
\begin{equation*}
    I_1(v)=I_{11}(v)+I_{12}(v), \tag{3}\label{3}
\end{equation*}
where
\begin{equation*}
    I_{11}(v):=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\Big(\frac{y}{c(y)}-\frac{y^2c'(y)}{c(y)^2}\Big)
\frac{y}{c(y)^{3/2}}H(y),
\end{equation*}
\begin{equation*}
    I_{12}(v):=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\frac{y^3c'(y)}{c(y)^{7/2}} H(y),
\end{equation*}
\begin{equation*}
    c(y):=c_v(y),
\end{equation*}
and $c'$ is the derivative of the Lipschitz function $c=c_v$; this derivative exists almost everywhere (a.e.), since $k$ is $1$-Lipschitz. Moreover,
\begin{equation*}
    |c'|\le t
\end{equation*}
a.e.
Note also that for $y\ge0$ and $t>0$
\begin{equation*}
t/2\le c(y)\le2t,\quad  |H(y)|\ll t\,\De a,\quad    |H'(y)|\ll t\,\De a,
\end{equation*}
where
\begin{equation*}
    \De a:=\|a-b\|_t
\end{equation*}
and $E\ll F$ means that $|E|\le CF$ for some universal real constant $C$.
So,
\begin{equation*}
    |I_{12}(v)|\ll\int_0^\infty dy\,\exp\Big(-\frac{y^2}{4t}\Big)
\frac{y^3\,t}{t^{7/2}}\, t\,\De a \asymp\sqrt t\,\De a. \tag{4}\label{4}
\end{equation*}
Integrating by parts, we have
\begin{equation*}
\begin{aligned}
    I_{11}(v)&=-\int_0^\infty dy\,\Big[\frac d{dy}\exp\Big(-\frac{y^2}{2c_v(y)}\Big)\Big]
\frac{y}{c(y)^{3/2}}H(y) \\
&=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\frac d{dy}\Big[\frac{y}{c(y)^{3/2}}H(y)\Big] \\
&=I_2(v)+I_{111}(v)-\frac32\,I_{112}(v),
\end{aligned}
\tag{5}\label{5}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
    I_{111}(v)&:=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\frac{y}{c(y)^{3/2}}H'(y),
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
    I_{112}(v)&:=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big)
\frac{yc'(y)}{c(y)^{5/2}}H(y).
\end{aligned}
\end{equation*}
Similarly to \eqref{4}, we get
\begin{equation}
    |I_{111}(v)|+|I_{112}(v)|\ll \sqrt t\,\De a. \tag{6}\label{6}
\end{equation}
Collecting \eqref{1}, \eqref{2}, \eqref{3}, \eqref{4}, \eqref{5}, and \eqref{6}, we conclude that
\begin{equation*}
    \Big|\int_0^\infty dy\,[f(0,0,t,y)-g(0,0,t,y)]\Big|\ll \sqrt t\,\De a,
\end{equation*}
which proves the first inequality of the two ones in question.
The other inequality can apparently be proved similarly. Since this answer is already rather long and complicated, I will leave the other inequality as an exercise (or to be posted elsewhere).


TITLE: Surjectivity of an operator on $L^2(R)$ defined as identity minus convolution
QUESTION [4 upvotes]: I'm having trouble with the following exercise. Define $T:L^2(\mathbb{R})\to L^2(\mathbb{R})$ by means of the formula
$$(Tf)(x)=\int_{0}^{1}f(x+y)dy.$$
I proved that, for all $f\in L^2(\mathbb{R})$, $Tf$ is a continuous function and $||Tf||_2\le||f||_2$, with equality iff $f(x)=0$ almost everywhere.
Then there is the following question, on which I'm stuck: is $S:=I-T$ surjective? Here, $I$ is the identity operator. This should be doable at the level of "Papa Rudin".
Edit:
Thanks to the last paragraph in @FTP's great answer, I think I have an idea for a more "elementary" (but more complicated) solution.
We know that $X:=L^1(\mathbb{R})\cap L^2(\mathbb{R})$ is dense in $L^2(\mathbb{R})$. Suppose $S$ is surjective. Since $S$ is injective and continuous, by the Open Mapping Theorem, it follows that $S$ is invertible.
Now, let $g\in X$ and $f=S^{-1}g$. By density, there is $\{f_n\}\subset X$ such that $f_n\to f$ in $L^2(\mathbb{R})$. Write $g_n=Sf_n$. Since each $f_n$ is integrable, it follows by Fubini-Tonelli that each $g_n$ is integrable, and we actually have
$$\int_{\mathbb{R}}g_n(x)dx\int_{\mathbb{R}}(Sf_n)(x)dx=0.\quad (*)$$
Also, by the continuity of $S$, we obtain that $g_n\to g$ in $L^2(\mathbb{R})$.
If we choose $g$ very nice with positive integral, can we obtain contradiction with equation (*)?

REPLY [5 votes]: The operator $T$ is the convolution of $f$ with function $g=\chi_{[-1, 0]}$. Therefore, on the Fourier transform side we have $\widehat{Tf} = \hat f\hat g$. Here
$$
\hat g(\xi) = \int_{-1}^0 e^{-2\pi i \xi x}\,dx = \frac{e^{2\pi i \xi}-1}{2\pi \xi i}
\tag1 $$
Notably, $\hat g$ is smooth and $\hat g(0)=1$. We don't really need (1) to see this, the obvious properties that $g$ has compact support and $\int g=1$ are enough.
So, $I-T$ is $\hat f\mapsto (1-\hat g)\hat f$ on the Fourier side. This is clearly injective, as $1-\hat g \ne 0$ almost everywhere. If this map was  surjective, it'd be invertible (open mapping theorem), and the inverse can be nothing else but  $\hat f\mapsto (1-\hat g)^{-1} \hat f$. But   the latter is not a bounded operator on $L^2$  since $(1-\hat g)^{-1} \notin L^\infty$.
Without OMT
One can avoid the open mapping theorem and simply present a function $\phi\in L^2$ such that $\phi/(1-\hat g)\notin L^2$. For example,
$$
\phi(\xi) = \max(0, |\xi|^{-1/3}-1)
$$
would do. Let $\check \phi$ be the inverse Fourier transform of $\phi$; then $\check \phi\notin \operatorname{ran}(I-T)$. Indeed, if $\check \phi = (I-T)f$, then $\hat f(\xi) = \phi(\xi)/(1-\hat g(\xi))$ which is not in $L^2$.
Without Fourier transform
Informally, the above is a long-winded justification of "since $Tf$ and $f$ have the same integral, the integral of $f-Tf$ must be zero". The preceding sentence doesn't make a proof because $f$ and $Tf$ need not be in $L^1(\mathbb{R})$. But one can consider the regularization
$$
\lim_{c\to 0^+} \int_\mathbb{R} e^{-cx^2} (f(x)-Tf(x))\,dx
\tag 2$$
and show (with some help from Fubini) that (2) is $0$ for every $f\in L^2$. Then it becomes obvious that, for example, any positive function cannot be in $\operatorname{ran}(I- T)$.


TITLE: On the automorphism group of the dihedral group $\text{Aut} (D_3)$
QUESTION [3 upvotes]: I am trying to calculate $\text{Aut} (D_3)$, the automorphism group of the group of symmetries of the triangle. But I got stuck and now I have two questions about this.
Let me share my thoughts first:
First I noted that an isomorphism maps elements to elements of same order. Therefore, any $\varphi : D_3 \to D_3$ maps reflections to reflections and rotations to rotations. As candidates I got:
(1) the identity map $\varphi_1 (x) = x$
(2) the map $\varphi_2 (x) = x$ except $R_{120}\mapsto R_{240}$ and $R_{240}\mapsto R_{120}$
(3) the map $\varphi_3 (x) = x$ except $F_1 \mapsto F_2$ and $F_2 \mapsto F_1$
(4)  the map $\varphi_4 (x) = x$ except $F_2 \mapsto F_3$ and $F_3 \mapsto F_2$
(5)  the map $\varphi_5 (x) = x$ except $F_1 \mapsto F_3$ and $F_3 \mapsto F_1$
Then since $F_1 F_2 = R_{120}$, $\varphi_3 (R_{120}) = R_{240}$ and $\varphi_3 (R_{240}) = R_{120}$. Now I wanted to show that also $\varphi_2 (F_1) = F_2 $ and $\varphi_2(F_2) = F_1$ but unfortunately this doesn't have to be true since $\varphi_2$ could map $F_1$ to $F_3$ for example.
Now my question is:

Can one say something like we relabel the reflections and therefore
  $\varphi_2 (F_1) = F_2$?

If yes then all these are equal and there are only two automorphisms --
That I have found anyway because my next question is:

How can I be sure that I found all automorphisms of a given group? Are
  there upper and lower bounds on the number?

REPLY [1 votes]: Let's characterize $D_3$ as the group generated by $r,t$ with
    $$r^3 = t^2 = (rt)^2 = e$$
    $\hspace{10pt}$Suppose that $\phi:D_3 \to D_3$ is an automorphism.
Then, $\phi(t)$ must be of order two $\Rightarrow$ $\phi(t) \in \{t,rt, r^2 t\}$.
Similarly we can see that $\phi(r)$ must be of order three $\Rightarrow$ $\phi(r) \in \{r,r^2\}$.
Since these two choices determine $\phi$ (and any two are compatible), we have $2 \times 3 = 6$ possible choices for $\phi$.


TITLE: If $f$ is continuous and $f(x_0) = \alpha > 0$, why is there some interval on which $f(x) > \frac{\alpha}{2}$?
QUESTION [0 upvotes]: I do not understand the second line of the proof. How do we know that $f(x)>\frac{\alpha}{2}$?

REPLY [3 votes]: As $f$ is continuous at $x_0$, then for every $\epsilon > 0$, there is $\delta > 0$ such that $|f(x) - f(x_0)| < \epsilon$ whenever $|x - x_0| < \delta$. Setting $\varepsilon = \frac{\alpha}{2}$, we see that there is a $\delta > 0$ such that if $|x - x_0| < \delta$, then $|f(x) - f(x_0)| < \frac{\alpha}{2}$. Now note that
$$|f(x) - f(x_0)| < \frac{\alpha}{2} \Longleftrightarrow -\frac{\alpha}{2} < f(x) - f(x_0) < \frac{\alpha}{2} \Longleftrightarrow f(x_0) - \frac{\alpha}{2} < f(x) < f(x_0) + \frac{\alpha}{2}.$$
In particular, if $x \in (x_0 -\delta, x_0 + \delta)$, then $|x - x_0| < \delta$, so $f(x) > f(x_0) - \frac{\alpha}{2} > \alpha - \frac{\alpha}{2} = \frac{\alpha}{2}$.

Note, the same proof can be used to show that for any $\beta < \alpha$, we could find an interval $I$ centred at $x_0$ with $f(x) > \beta$ on $I$. In that case, choose $\epsilon = \alpha - \beta$.


text:
\begin{document}

\title{\bf On the number of $k$-compositions of $n$ satisfying certain coprimality conditions}
\author{L\'aszl\'o T\'oth  \\
Department of Mathematics \\
University of P\'ecs \\
Ifj\'us\'ag \'utja 6, 7624 P\'ecs, Hungary \\
E-mail: {\tt ltoth@gamma.ttk.pte.hu}}
\date{}
\maketitle

\centerline{Acta Math. Hungar. {\bf 164} (1) (2021), 135--156}

\begin{abstract} We generalize the asymptotic estimates by Bubboloni, Luca and Spiga (2012) on the number of $k$-compositions of $n$
satisfying some coprimality conditions. We substantially refine the error term concerning the number of $k$-compositions of $n$ with pairwise
relatively prime summands. We use a different approach, based on properties of multiplicative arithmetic functions of $k$ variables and on an
asymptotic formula for the restricted partition function.
\end{abstract}

{\sl Key Words and Phrases}: $k$-compositions, coprime summands, $t$-wise relatively prime summands, multiplicative arithmetic function of several variables, restricted partition function.

{\sl 2010 Mathematics Subject Classification}: 05A17, 11N37, 11P81.

\medskip
{The research was financed by NKFIH in Hungary, within the framework of the 2020-4.1.1-TKP2020 3rd thematic programme of the University of P\'ecs.}

\section{Motivation}

Let $n,k\in \N:=\{1,2,\ldots \}$. For $n\ge k$, the $k$-compositions of $n$ are the representations $x_1+\cdots+x_k=n$, where
$(x_1,\ldots,x_k)\in \N^k$. It is well known that the number of all $k$-compositions of $n$ is $\binom{n-1}{k-1}$. Gould \cite{Gou1964} considered the
number $R_k(n)$ of $k$-compositions of $n$ such that $\gcd(x_1,\ldots,x_k)=1$, obtained that its Lambert series is
\begin{equation*}
\sum_{n=1}^{\infty} R_k(n) \frac{x^n}{1-x^n} = \frac{x^k}{(1-x)^k},
\end{equation*}
and then deduced that
\begin{equation} \label{R_n_k}
R_k(n) = \sum_{d\divides n} \binom{d-1}{k-1}\mu(n/d),
\end{equation}
where $\mu$ is the M\"{o}bius function. A short direct proof of identity \eqref{R_n_k} is the following.
By using that $\sum_{d\mid n} \mu(d)=1$ or $0$, according to $n=1$ or $n>1$, we obtain that
\begin{equation} \label{proof_R_n_k}
R_k(n) = \sum_{\substack{x_1,\ldots,x_k\ge 1\\x_1+\cdots+ x_k=n}} \sum_{d\divides \gcd(x_1,\ldots,x_k)} \mu(d)
= \sum_{d\divides n} \mu(d) \sum_{\substack{y_1,\ldots,y_k\ge 1\\y_1+\cdots+ y_k=n/d}} 1
= \sum_{d\divides n} \mu(d) \binom{n/d-1}{k-1},
\end{equation}
giving \eqref{R_n_k}.

Let $J_t(n)=n^t \prod_{p\divides n} (1-1/p^t)$ denote the Jordan function of order $t$, where $J_1(n)=\varphi(n)$ is Euler's totient function.
If $k=2$, then it follows that $R_2(n)=\varphi(n)$ for every $n\ge 2$. Identity \eqref{R_n_k} shows that for $k\ge 3$,
\begin{equation} \label{R_k_n_asympt}
R_k(n) = \frac1{(k-1)!}J_{k-1}(n) + O(n^{k-2}),
\end{equation}
as $n\to \infty$. In fact, \eqref{R_n_k} implies that for every $k$, $R_k(n)$ is a linear combination of the Jordan functions $J_t(n)$ ($1\le t\le k-1$).
More exactly,
for every $n\ge k\ge  2$,
\begin{equation*}
R_k(n) = \frac1{(k-1)!} \sum_{t=1}^{k-1} a_{k,t} J_t(n) = \frac1{(k-1)!}\left(J_{k-1}(n) -\frac{k(k-1)}{2} J_{k-2}(n)+\cdots +a_{k,1} \varphi(n) \right),
\end{equation*}
where
\begin{equation*}
a_{k,t}(n) = \sum_{j=t}^{k-1} (-1)^{j-t} s(k-1,j) \binom{j}{t},
\end{equation*}
$s(n,k)$ denoting the signed Stirling numbers of the first kind.

Bubboloni, Luca and Spiga \cite{BLS2012} investigated the number of $k$-compositions of $n$ with some different coprimality constraints.
Namely, let
\begin{equation*}
A_k(n):= \# \{(x_1,\ldots,x_k)\in \N^k: x_1+\cdots +x_k=n, \gcd(x_1,x_2\cdots x_k)=1\}
\end{equation*}
and
\begin{equation*}
B_k(n):= \# \{(x_1,\ldots,x_k)\in \N^k: x_1+\cdots +x_k=n, \gcd(x_i,x_j)=1, 1\le i< j\le k\}.
\end{equation*}

Note that, if $k=2$, then $A_2(n)=B_2(n)=R_2(n)=\varphi(n)$ for every $n\ge 2$. They proved in \cite{BLS2012} that for any $k\ge 3$,
\begin{equation} \label{A_k}
A_k(n) = C_k f_k(n)\frac{n^{k-1}}{(k-1)!} + O\left(n^{k-2} (\log n)^{k-1} \right)
\end{equation}
and
\begin{equation} \label{B_k}
B_k(n) = D_k g_k(n)\frac{n^{k-1}}{(k-1)!} + O\left(n^{k-1} (\log n)^{-1} \right),
\end{equation}
as $n\to \infty$, where $C_k$ and $D_k$ are explicit constants, depending only on $k$, while $f_k(n)$ and $g_k(n)$ are depending on $k$ and on the prime
factorization of $n$. Actually, it was also proved in \cite{BLS2012} that $2/3<f_k(n)<2$ and $1/(2k)<g_k(n)<2k$ for every $n\in \N$,
and explicit constants for the above $O$-terms were given. Note also, that in \cite{BLS2012} the sum $A_k(n)$ was denoted by $A_{k-1}(n)$, and
the corresponding results were formulated for $k+1$ instead of $k$. As described in \cite{BLS2012}, these results have applications in group
theory and Galois theory.

In a recent preprint, Thomas \cite{Tho2020} considered the case $k=3$ and obtained for the sum $B_3(n)$ the error term $O(n^{3/2+\varepsilon})$, which
refines \eqref{B_k}.

In this paper we define for $1\le s \le k-1$,
\begin{equation*}
A_{k,s}(n):= \# \{(x_1,\ldots,x_k)\in \N^k: x_1+\cdots +x_k=n, \gcd(x_1\cdots x_s, x_{s+1}\cdots x_k)=1\},
\end{equation*}
representing the number of $k$-compositions of $n$ such that the first $s$ summands are coprime to the last $k-s$ summands.
We remark the symmetry property $A_{k,s}(n)= A_{k,k-s}(n)$, valid for every $1\le s\le k-1$.
Also, for $2\le t \le k$ we define
\begin{equation*}
B_{k,t}(n):= \# \{(x_1,\ldots,x_k)\in \N^k: x_1+\cdots +x_k=n, \gcd(x_{i_1},\ldots,x_{i_t})=1, 1\le i_1<\ldots < i_t\le k\},
\end{equation*}
representing the number of $k$-compositions of $n$ such that the summands are $t$-wise relatively prime. If $s=1$ and $t=2$, then
$A_{k,1}(n)=A_k(n)$ and $B_{k,2}(n)=B_k(n)$, defined above. Also, if $t=k$, then $A_{k,k}(n)=R_k(n)$.

We obtain asymptotic formulas with error terms for $A_{k,s}(n)$ and $B_{k,t}(n)$, which generalize formulas \eqref{R_k_n_asympt}, \eqref{A_k}
and \eqref{B_k}. In the case $t=2$ our error term substantially refines the error term of formula \eqref{B_k}, obtained
in \cite{BLS2012,Tho2020}. We use a different approach, inspired by \eqref{proof_R_n_k}, based on properties of multiplicative arithmetic
functions of $k$ variables and the restricted partition function
\begin{equation} \label{def_N}
N(n;a_1,\ldots,a_k):= \# \{(x_1,\ldots,x_k)\in \N^k: a_1x_1+\cdots +a_kx_k=n\}.
\end{equation}

The treatments for the sums $A_{k,s}(n)$ and $B_{k,t}(n)$ are quite similar. For the proofs we use the method and some results from our
earlier work \cite{Tot2016}, concerning the asymptotic density of the $k$-tuples of positive integers
with $t$-wise relatively prime components, with some other notation. The main results of the present paper are given in Section
\ref{Section_Main_results}, and their proofs are included in Section \ref{Section_Proofs}. Some preliminaries, needed for the proofs, are presented in Section \ref{Section_Preliminaries}.

We will use the notation $n=\prod_p p^{\nu_p(n)}$ for the prime power factorization of $n\in \N$, the product being over the primes $p$,
where all but a finite number of the exponents $\nu_p(n)$ are zero. The number of distinct prime factors of $n$ will be denoted by $\omega(n)$.

\section{Main results} \label{Section_Main_results}

For $k\ge 3$ and $1\le s\le k-1$ let
\begin{equation} \label{notation_F}
F_{k,s}(x): = \frac1{x} \left(x^k+ (x-1)^k- x^s(x-1)^{k-s}- x^{k-s}(x-1)^s + (-1)^{k-1}\right),
\end{equation}
which is a polynomial of degree $k-3$. Also, for $k\ge 3$ and $2\le t\le k$ let
\begin{align}  \label{notation_G}
G_{k,t}(x): = & \frac1{x} \left(\sum_{j=0}^{t-1} \binom{k}{j}(x-1)^{k-j}+(-1)^{k-t} \binom{k-1}{t-1} \right)
\nonumber
\\ = & \frac1{x} \left(x^k- \sum_{j=t}^{k-1}(-1)^{j-t} \binom{k}{j} \binom{j-1}{t-1} x^{k-j} \right),
\end{align}
which is a polynomial of degree $k-1$.

We prove the following results.

\begin{theorem} \label{Theorem_A} Let $k\ge 3$ and let $1\le s \le k-1$. Then
\begin{equation*}
A_{k,s}(n) = C_{k,s} f_{k,s}(n)\frac{n^{k-1}}{(k-1)!} + \begin{cases} O\left(n^{k-2}(\log n)^{\max(s,k-s)} \right), & \text{if $k\ge 4$,} \\
O\left(n(\log n)^2 \log \log n \right), & \text{if $k=3$,}
\end{cases}
\end{equation*}
where
\begin{equation*}
C_{k,s} = \prod_p \left(1- \frac{F_{k,s}(p)}{p^{k-1}} \right),
\end{equation*}
\begin{equation*}
f_{k,s}(n)= \prod_{p\divides n} \left(1 + \frac{(-1)^{k-1}}{p^{k-1}-F_{k,s}(p)} \right).
\end{equation*}
\end{theorem}

\begin{theorem} \label{Theorem_B} Let $k\ge 3$ and let $2\le t \le k$. Then
\begin{equation*}
B_{k,t}(n) = D_{k,t} g_{k,t}(n) \frac{n^{k-1}}{(k-1)!} + \begin{cases} O\left(n^{k-2} \right), & \text{if $k\ge 4$, $t\ge 3$, or $k=t=3$,} \\
O\left(n^{k-2} (\log n)^{k-1} \right), & \text{if $k\ge 4$, $t=2$,} \\
O\left(n (\log n)^2 (\log \log n)^2 \right), & \text{if $k=3$, $t=2$,}
\end{cases}
\end{equation*}
where
\begin{equation*}
D_{k,t} = \prod_p \frac{G_{k,t}(p)}{p^{k-1}},
\end{equation*}
\begin{equation*}
g_{k,t}(n)= \prod_{p\divides n} \left(1 + \frac{(-1)^{k-t+1}\binom{k-1}{t-1}}{G_{k,t}(p)}\right).
\end{equation*}
\end{theorem}

We note that $f_{k,s}(n)$ and $g_{k,t}(n)$ are positive bounded functions of $n$, which specify the asymptotic behavior of $A_{k,s}(n)$ and
$B_{k,t}(n)$. In the case $s=1$ we recover from Theorem \ref{Theorem_A} formula \eqref{A_k} with the same error term for $k\ge 4$, and with a
slightly weaker error term for $k=3$. In the case $t=2$ we recover from Theorem \ref{Theorem_B} formula \eqref{B_k} with a substantially better error term for every $k\ge 3$.

\section{Preliminaries to the proofs} \label{Section_Preliminaries}

\subsection{Arithmetic functions of several variables}

Let $f:\N^k\to \C$ be an arithmetic function of $k$ variables. Its Dirichlet series is given by
\begin{equation*}
D(f;z_1,\ldots,z_k)= \sum_{n_1,\ldots,n_k=1}^{\infty}
\frac{f(n_1,\ldots,n_k)}{n_1^{z_1}\cdots n_k^{z_k}}.
\end{equation*}

Similar to the one variable case, if $D(f;z_1,\ldots,z_k)$ is
absolutely convergent for $(z_1,\ldots,z_k)\in \C^k$, then it is
absolutely convergent for every $(s_1,\ldots,s_k)\in \C^k$ with $\Re
s_j \ge \Re z_j$ ($1\le j\le k$).

The Dirichlet convolution of the functions $f,g:\N^k\to \C$ is defined by
\begin{equation*}
(f*g)(n_1,\ldots,n_k)= \sum_{d_1\mid n_1, \ldots, d_k\mid n_k}
f(d_1,\ldots,d_k) g(n_1/d_1, \ldots, n_k/d_k).
\end{equation*}

If $D(f;z_1,\ldots,z_k)$ and $D(g;z_1,\ldots,z_k)$ are
absolutely convergent, then
$D(f*g;z_1,\ldots,z_k)$ is also absolutely convergent and
\begin{equation*}
D(f*g;z_1,\ldots,z_k) = D(f;z_1,\ldots,z_k) D(g;z_1,\ldots,z_k).
\end{equation*}

We also recall that a nonzero arithmetic function of $k$ variables $f:\N^k\to \C$ is said to be multiplicative if
\begin{equation*}
f(m_1n_1,\ldots,m_kn_k)= f(m_1,\ldots,m_k) f(n_1,\ldots,n_k)
\end{equation*}
holds for every $m_1,\ldots,m_k,n_1,\ldots,n_k\in \N$ such that $\gcd(m_1\cdots m_k,n_1\cdots n_k)=1$.
If $f$ is multiplicative, then it is determined by the values
$f(p^{\nu_1},\ldots,p^{\nu_k})$, where $p$ is prime and
$\nu_1,\ldots,\nu_k\in \N_0:=\N \cup \{0\}$. More exactly, $f(1,\ldots,1)=1$ and
for every $n_1,\ldots,n_k\in \N$,
\begin{equation*}
f(n_1,\ldots,n_k)= \prod_p f(p^{\nu_p(n_1)}, \ldots,p^{\nu_p(n_k)}),
\end{equation*}
the product being over the primes $p$. If $f,g:\N^k \to \C$ are multiplicative, then their Dirichlet convolution $f*g$ is also multiplicative.
If $k=1$, i.e., in the case of functions of a
single variable we recover the familiar notion of multiplicativity.

If $f$ is multiplicative, then its Dirichlet series can be expanded into a (formal)
Euler product, that is,
\begin{equation} \label{Euler_product}
D(f;z_1,\ldots,z_k)=  \prod_p \sum_{\nu_1,\ldots,\nu_k=0}^{\infty}
\frac{f(p^{\nu_1},\ldots, p^{\nu_k})}{p^{\nu_1z_1+\cdots +\nu_k
z_k}},
\end{equation}
the product being over the primes $p$. More exactly, if $f$ is multiplicative, then the series
$D(f;z_1,\ldots,z_k)$ is absolutely convergent if and only if
\begin{equation*}
\sum_p \sum_{\substack{\nu_1,\ldots,\nu_k=0\\ \nu_1+\cdots +\nu_k \ge 1}}^{\infty}
\frac{|f(p^{\nu_1},\ldots, p^{\nu_k})|}{p^{\nu_1 \Re z_1+\cdots +\nu_k
\Re z_k}} < \infty
\end{equation*}
and in this case equality \eqref{Euler_product} holds. See Delange \cite{Del1969} and the survey by the author \cite{Tot2014}.

\subsection{The restricted partition function}

Let $\N_0:=\{0,1,2,\ldots\}$ and consider the restricted partition function
\begin{equation} \label{def_P}
P(n;a_1,\ldots,a_k):= \# \{(x_1,\ldots,x_k)\in \N_0^k: a_1x_1+\cdots +a_kx_k=n\},
\end{equation}
where $a_1,\ldots,a_k$ are given positive integers. It is known that in the case $\gcd(a_1,\ldots,a_k)=1$ the function $P(n;a_1,\ldots,a_k)$
can be represented as a $(k-1)$-degree polynomial in $n$ plus a periodic sequence of $n$. More exactly, one has
\begin{equation} \label{form_P}
P(n;a_1,\ldots,a_k)= c_{k-1}n^{k-1}+ c_{k-2} n^{k-2}+\cdots + c_1n+c_0 + s(n;a_1,\ldots,a_k),
\end{equation}
where
\begin{equation} \label{coeff_P}
c_{k-1} = \frac1{(k-1)!a_1\cdots a_k}, \quad  c_{k-2}= \frac{a_1+\cdots +a_k}{2(k-2)!a_1\cdots a_k},
\end{equation}
and $s(n;a_1,\ldots,a_k)$ is a periodic sequence with period $\lcm(a_1,\ldots,a_k)$. See Comtet \cite[Sect.\ 2.6, Th.\ C]{Com1974}.
There is a rich bibliography on properties of the restricted partition function, in particular on its polynomial part. See, e.g., the recent
papers \cite{AC2005, CF2018, DV2017} and their references.

For our purposes the variables $x_1,\ldots,x_k$ are positive. If $(x_1,\ldots,x_k)\in \N^k$ is a solution of the equation $a_1x_1+\cdots + a_kx_k=n$,
then $(y_1,\ldots,y_k)=(x_1-1,\ldots,x_k-1)\in \N_0^k$ is a solution for $a_1y_1+\cdots + a_ky_k=n-a_1-\cdots -a_k$, and conversely. Hence the connection
between the functions \eqref{def_N} and \eqref{def_P} is given by
\begin{equation} \label{connect_N_P}
N(n;a_1,\ldots,a_k) = P(n-a_1-\cdots-a_k;a_1,\ldots,a_k).
\end{equation}

We need the following result.

\begin{lemma} \label{Lemma_Partition} Let $k\in \N$, $k\ge 2$ be fixed and let $a_1,\ldots,a_k\in \N$ such that $\gcd(a_1,\ldots,a_k)=1$. Then
\begin{equation*}
N(n;a_1,\ldots,a_k) = \frac{n^{k-1}}{(k-1)!a_1\cdots a_k}+ O\left(\frac{n^{k-2}(a_1+\cdots +a_k)}{a_1\cdots a_k} \right),
\end{equation*}
as $n\to \infty$, uniformly for $n,a_1,\ldots, a_k$.
\end{lemma}

\begin{proof} According to \eqref{form_P} and \eqref{coeff_P} we have
\begin{equation*}
P(n;a_1,\ldots,a_k) = \frac{n^{k-1}}{(k-1)!a_1\cdots a_k}+ O\left(\frac{n^{k-2}(a_1+\cdots +a_k)}{a_1\cdots a_k} \right).
\end{equation*}

By using \eqref{connect_N_P} we conclude that the same asymptotics holds for $N(n;a_1,\ldots,a_k)$ as well.
\end{proof}

\section{Proofs} \label{Section_Proofs}

\subsection{The sum $A_{k,s}(n)$}

We define the function
\begin{equation*}
\vartheta_{k,s}(n_1,\ldots,n_k)= \begin{cases} 1, & \text{ if $\gcd(n_1\cdots n_s, n_{s+1} \cdots n_k)=1$}, \\ 0, &
\text{ otherwise,}
\end{cases}
\end{equation*}
which is symmetric in the variables $x_1,\ldots,x_s$ and in $x_{s+1},\ldots,x_k$, but not symmetric in all variables. However, $\vartheta_{k,s}$
is a  multiplicative function of $k$ variables. Hence we have
\begin{equation*}
\vartheta_{k,s}(n_1,\ldots,n_k)= \prod_p \vartheta_{k,s} (p^{\nu_p(n_1)}, \ldots,p^{\nu_p(n_k)})
\end{equation*}
for every $n_1,\ldots,n_k\in \N$. Also, for every prime $p$ and every $\nu_1,\ldots,\nu_k\in \N_0$,
\begin{equation} \label{vartheta_val}
\vartheta_{k,s}(p^{\nu_1}, \ldots,p^{\nu_k})= \begin{cases}
1, & \text{ if $\nu_1=\cdots =\nu_s=0$, $\nu_{s+1},\ldots,\nu_k\in \N_0$}, \\
\ & \text{ or $\nu_1,\ldots,\nu_s\in \N_0$, $\nu_{s+1} =\cdots =\nu_k=0$}, \\
0, & \text{ otherwise.}
\end{cases}
\end{equation}

For the multiple Dirichlet series of the function $\vartheta_{k,s}$ we
have the next result.

\begin{lemma} \label{Lemma_Dir_series} If $1 \le s \le k-1$, then
\begin{equation} \label{product_repr}
\sum_{n_1,\ldots,n_k=1}^{\infty}
\frac{\vartheta_{k,s}(n_1,\ldots,n_k)}{n_1^{z_1}\cdots n_k^{z_k}}=
\zeta(z_1)\cdots \zeta(z_k) H_{k,s}(z_1,\ldots,z_k),
\end{equation}
where $\zeta$ is the Riemann zeta function, and
\begin{equation} \label{H_k_s}
H_{k,s}(z_1,\ldots,z_k)= \prod_p \left(\prod_{1\le j\le s} \left(1-\frac1{p^{z_j}}\right) + \prod_{s+1\le j\le k} \left(1-\frac1{p^{z_j}} \right)
- \prod_{1\le j\le k} \left(1-\frac1{p^{z_j}}\right) \right),
\end{equation}
absolutely convergent if $\Re z_j \ge 0$ \textup{($1\le j\le k$)} and $\Re (z_j+z_{\ell}) >1$ \textup{($1\le j\le s$, $s+1 \le \ell \le k$)}.
In particular, for $z_1=\cdots =z_k=1$,
\begin{equation*}
H_{k,s}(1,\ldots,1)= \prod_p  \left(1- \frac{p F_{k,s}(p)+(-1)^k}{p^k}\right)
\end{equation*}
is absolutely convergent, where $F_{k,s}(p)$ is defined by \eqref{notation_F}.
\end{lemma}

\begin{proof} The function $\vartheta_{k,s}(n_1,\ldots,n_k)$ is multiplicative, hence its Dirichlet series can be expanded into an
Euler product. Using \eqref{vartheta_val} we deduce
\begin{equation*}
\sum_{n_1,\ldots,n_k=1}^{\infty} \frac{\vartheta_{k,s}(n_1,\ldots,n_k)}{n_1^{z_1}\cdots n_k^{z_k}}=
\prod_p \sum_{\nu_1,\ldots,\nu_k=0}^{\infty}
\frac{\vartheta_{k,s}(p^{\nu_1},\ldots,p^{\nu_k})}{p^{\nu_1 z_1+\cdots +\nu_k z_k}}
\end{equation*}
\begin{equation*}
= \prod_p \left( \sum_{\nu_1,\ldots,\nu_s=0}^{\infty} \frac1{p^{\nu_1 z_1+\cdots +\nu_s z_s}} +
\sum_{\nu_{s+1},\ldots,\nu_k=0}^{\infty} \frac1{p^{\nu_{s+1} z_{s+1} +\cdots +\nu_k z_k}} -1 \right)
\end{equation*}
\begin{equation*}
= \prod_p \left(\prod_{1\le j\le s} \left(1-\frac1{p^{z_j}}\right)^{-1} + \prod_{s+1\le j\le k} \left(1-\frac1{p^{z_j}} \right)^{-1}
- 1 \right),
\end{equation*}
which gives the desired formula by factoring out $\prod_{1\le j\le k} \prod_p \left(1-\frac1{p^{z_j}} \right)^{-1} =\zeta(z_1)\cdots \zeta(z_k)$.
Observe that, performing the multiplications in \eqref{H_k_s}, all terms with $p^e$
cancel out, where $e=z_{i_1}+\cdots +z_{i_j}$ such that $1\le i_1<\cdots <i_j\le s$ or $s+1\le i_1<\cdots <i_j\le k$. This gives the
result on the absolute convergence. For example, if $k=3$ and $s=2$, then
\begin{equation*}
H_{3,2}(z_1,z_2,z_3)=  \prod_p \left( \left(1-\frac1{p^{z_1}}\right) \left(1-\frac1{p^{z_2}}\right)+ \left(1-\frac1{p^{z_3}}\right) - \left(1-\frac1{p^{z_1}}\right)\left(1-\frac1{p^{z_2}}\right)\left(1-\frac1{p^{z_3}}\right)\right)
\end{equation*}
\begin{equation*}
=  \prod_p \left(1- \frac1{p^{z_1+z_3}} -\frac1{p^{z_2+z_3}} + \frac1{p^{z_1+z_2+z_3}}\right),
\end{equation*}
absolutely convergent if $\Re z_j \ge 0$ \textup{($1\le j\le 3$)} and $\Re (z_1+z_3) >1$, $\Re (z_2+z_3) >1$.

If $z_1=\cdots =z_k=1$, then
\begin{equation*}
H_{k,s}(1,\ldots,1)= \prod_p  \left( \left(1-\frac1{p}\right)^s + \left(1-\frac1{p} \right)^{k-s} - \left(1-\frac1{p}\right)^k \right)
\end{equation*}
\begin{equation*}
= \prod_p  \left(1- \frac{p F_{k,s}(p)+(-1)^k}{p^k}\right),
\end{equation*}
by a direct computation.
\end{proof}

\begin{lemma} \label{Lemma_convo} Let $1\le s \le k-1$. For every $n_1,\ldots,n_k\in \N$,
\begin{equation} \label{lalambda}
\vartheta_{k,s}(n_1,\ldots,n_k) = \sum_{d_1\divides n_1, \ldots, d_k\divides
n_k} \lambda_{k,s}(d_1,\ldots,d_k),
\end{equation}
where the function $\lambda_{k,s}$ is multiplicative, and for any prime powers $p^{\nu_1},\ldots,p^{\nu_k}$
\textup{($\nu_1,\ldots,\nu_k\in \N_0$)}
\begin{equation} \label{lambda_val}
\lambda_{k,s}(p^{\nu_1},\ldots,p^{\nu_k})=
\begin{cases} 1, & \nu_1=\cdots=\nu_k=0,\\
(-1)^{\nu_1+\cdots+\nu_k-1}, & \nu_1,\ldots,\nu_k\in \{0,1\}, \\\
\ & \nu_1+\cdots+\nu_s\ge 1, \nu_{s+1}+\cdots +\nu_k\ge 1, \\
0, & \text{otherwise}.
\end{cases}
\end{equation}
\end{lemma}

\begin{proof} To deduce \eqref{lalambda}, observe that according to \eqref{product_repr}, the multiplicative function $\vartheta_{k,s}$ is the Dirichlet convolution of the
constant $1$ function and the function $\lambda_{k,s}$, where
\begin{equation*}
H_{k,s}(z_1,\ldots,z_k)= \sum_{n_1,\ldots,n_k=1}^{\infty} \frac{\lambda_{k,s}(n_1,\ldots,n_k)}{n_1^{z_1}\cdots n_k^{z_k}}.
\end{equation*}

The function $\vartheta_{k,s}$ is multiplicative, hence $\lambda_{k,s}$ is also multiplicative,
and
\begin{equation} \label{H_product}
H_{k,s}(z_1,\ldots,z_k) =\prod_p \sum_{\nu_1,\ldots,\nu_k=0}^{\infty}
\frac{\lambda_{k,s}(p^{\nu_1},\ldots,p^{\nu_k})}{p^{\nu_1 z_1+\cdots +\nu_k z_k}}.
\end{equation}

Identifying the terms of the sums under the products over the primes $p$ in \eqref{H_product} and \eqref{H_k_s}, respectively,
we conclude \eqref{lambda_val}. Also see the example given in the proof of
Lemma \ref{Lemma_Dir_series}.
\end{proof}

We will need the following properties of the function $\lambda_{k,s}$.

\begin{lemma} \label{Lemma_prop_lambda}
a) Let $\nu_1,\ldots,\nu_k\ge 0$. Then $\lambda_{k,s}(p^{\nu_1},\ldots,p^{\nu_k})=0$ provided that (i) $\nu_i\ge 2$ for at least one $i$ such that $1\le i\le k$, or (ii) $\nu_i=1$ for
at least one $i$ such that $1\le i\le s$ and $\nu_{s+1}=\cdots=\nu_k=0$, or (iii) $\nu_i=1$ for at least one $i$ with $s+1\le i\le k$ and $\nu_1=\cdots =\nu_s=0$.

b) $\lambda_{k,s}(p,\ldots,p)=(-1)^{k-1}$ for every prime $p$.

c) Let $j_1,\ldots,j_k,\delta\in \N$. If $\gcd(j_1\cdots j_k,\delta)\ne 1$, then  $\lambda_{k,s}(j_1\delta,\ldots,j_k\delta)=0$. If $\gcd(j_1\cdots j_k,\delta) = 1$, then $\lambda_{k,s}(j_1\delta,\ldots,j_k\delta) = \lambda_{k,s}(j_1,\ldots,j_k) \lambda_{k,s}(\delta,\ldots,\delta)$.

d) Let $\delta\in \N$. Then $\lambda_{k,s}(\delta,\ldots,\delta) =(-1)^{(k-1)\omega(\delta)}\mu^2(\delta)$.
\end{lemma}

\begin{proof} a), b) Follow from \eqref{lambda_val}.

c) If $\gcd(j_1\cdots j_k,\delta)\ne 1$, then at least one $j_i\delta$ is not squarefree and $\lambda_{k,s}(j_1\delta,\ldots,j_k\delta)=0$ by property a). If $\gcd(j_1\cdots j_k,\delta) = 1$, then $\lambda_{k,s}(j_1\delta,\ldots,j_k\delta) = \lambda_{k,s}(j_1,\ldots,j_k) \lambda_{k,s}(\delta,\ldots,\delta)$ by the multiplicativity of the function $\lambda_{k,s}$.

d) If $\delta$ is not squarefree, then $\lambda_{k,s}(\delta,\ldots,\delta)=0$ by a). If $\delta=p_1\cdots p_r$ is squarefree, then using multiplicativity and b) we have $\lambda_{k,s}(\delta,\ldots,\delta) = \lambda_{k,s}(p_1,\ldots,p_1) \cdots \lambda_{k,s}(p_r,\ldots,p_r)= (-1)^{(k-1)\omega(\delta)}$.
\end{proof}

\begin{lemma} \label{Lemma_Repr_A_N} Let $1 \le s \le k-1$. For every $n\ge k$ we have
\begin{equation} \label{form_A_k_t}
A_{k,s}(n) = \sum_{\delta \divides n} \mu^2(\delta) (-1)^{(k-1)\omega(\delta)} \sum_{\substack{1\le j_1,\ldots,j_k\le n/\delta \\ \gcd(j_1,\ldots,j_k)=1\\
\gcd(j_1\cdots j_k, \delta)=1 }}  \lambda_{k,s}(j_1,\ldots,j_k) N(n/\delta;j_1,\ldots,j_k).
\end{equation}
\end{lemma}

\begin{proof} By the definitions of $A_{k,s}(n)$ and $\vartheta_{k,s}(n_1,\ldots,n_k)$, and by Lemma \ref{Lemma_convo} we have
\begin{equation*}
A_{k,s}(n)= \sum_{\substack{x_1,\ldots,x_k\ge 1\\x_1+\cdots+ x_k=n}} \vartheta_{k,s}(x_1,\ldots,x_k)
= \sum_{\substack{x_1,\ldots,x_k\ge 1\\ x_1+\cdots+ x_k=n}} \sum_{d_1\divides x_1, \ldots, d_k\divides
x_k} \lambda_{k,s}(d_1,\ldots,d_k)
\end{equation*}
\begin{equation*}
= \sum_{1\le d_1,\ldots,d_k\le n} \lambda_{k,s}(d_1,\ldots,d_k)
\sum_{\substack{a_1,\ldots,a_k\ge 1\\ d_1a_1+\cdots+ d_ka_k=n}} 1
\end{equation*}
\begin{equation*}
= \sum_{1\le d_1,\ldots,d_k\le n}  \lambda_{k,s}(d_1,\ldots,d_k) N(n;d_1,\ldots,d_k).
\end{equation*}

Let $\delta=\gcd(d_1,\ldots,d_k)$. If $\delta \notdivides n$, then the equation $d_1a_1+\cdots+ d_ka_k=n$ has no solutions in $a_1,\ldots,a_k$. If
$\delta \divides n$, then $N(n;d_1,\ldots,d_k)= N(n/\delta;d_1/\delta,\ldots,d_k/\delta)$. We deduce, by grouping the terms according to the values
of $\delta$,
\begin{equation*}
A_{k,s}(n) = \sum_{\delta \divides n} \sum_{\substack{1\le d_1,\ldots,d_k\le n\\ \gcd(d_1,\ldots,d_k)=\delta}}  \lambda_{k,s}(d_1,\ldots,d_k)
N(n/\delta;d_1/\delta,\ldots,d_k/\delta)
\end{equation*}
\begin{equation} \label{sum_for_A}
= \sum_{\delta \divides n} \sum_{\substack{1\le j_1,\ldots,j_k\le n/\delta \\ \gcd(j_1,\ldots,j_k)=1}}  \lambda_{k,s}(j_1\delta,\ldots,j_k\delta)
N(n/\delta;j_1,\ldots,j_k).
\end{equation}

Now we use properties c) and d) of Lemma \ref{Lemma_prop_lambda}. If $\gcd(j_1\cdots j_k,\delta)\ne 1$, then $\lambda_{k,s}(j_1\delta,\ldots,j_k\delta)=0$.
If $\gcd(j_1\cdots j_k,\delta) = 1$, then $\lambda_{k,s}(j_1\delta,\ldots,j_k\delta) = \lambda_{k,s}(j_1,\ldots,j_k) \lambda_{k,s}(\delta,\ldots,\delta)$, where $\lambda_{k,s}(\delta,\ldots,\delta)=(-1)^{(k-1)\omega(\delta)}\mu^2(\delta)$.

Inserting these into \eqref{sum_for_A} gives formula \eqref{form_A_k_t}.
\end{proof}

\begin{lemma} \label{Lemma_T_delta} If $1\le s \le k-1$, then for every $\delta \in \N$,
\begin{equation*}
T_{\delta}:= \sum_{\substack{j_1,\ldots,j_k=1 \\ \gcd(j_1,\ldots,j_k)=1\\ \gcd(j_1\cdots j_k,\delta)=1 }}^{\infty}
\frac{\lambda_{k,s}(j_1,\ldots,j_k)}{j_1\cdots j_k}
=  \prod_{p\,\centernot\divides \, \delta} \left(1 - \frac{F_{k,s}(p)}{p^{k-1}} \right).
\end{equation*}
\end{lemma}

\begin{proof} We have
\begin{equation*}
T_{\delta}= \sum_{\substack{j_1,\ldots,j_k=1 \\ \gcd(j_1\cdots j_k,\delta)=1 }}^{\infty}
\frac{\lambda_{k,s}(j_1,\ldots,j_k)}{j_1\cdots j_k} \sum_{c \divides \gcd(j_1,\ldots,j_k)} \mu(c)
\end{equation*}
\begin{equation*}
 = \sum_{\substack{c,v_1,\ldots,v_k=1 \\ \gcd(v_1\cdots v_k,\delta)=1\\ \gcd(c,\delta)=1}}^{\infty}
\frac{\mu(c) \lambda_{k,s}(cv_1,\ldots,cv_k)}{cv_1\cdots cv_k}.
\end{equation*}

Here we can assume that $c$ is squarefree and we use property c) of Lemma \ref{Lemma_prop_lambda} (with $c$ instead of $\delta$ and with $v_1,\ldots,v_k$ instead of $j_1,\ldots,j_k$). We obtain that

\begin{equation*}
T_{\delta}= \sum_{\substack{c=1\\ (c,\delta)=1}}^{\infty} \frac{\mu(c)\lambda_{k,s}(c,\ldots,c)}{c^k}
\sum_{\substack{v_1,\ldots,v_k=1 \\ \gcd(v_1\cdots v_k,c\delta)=1}}^{\infty}
\frac{\lambda_{k,s}(v_1,\ldots,v_k)}{v_1\cdots v_k}.
\end{equation*}

Now from Lemma \ref{Lemma_Dir_series} we have
\begin{equation*}
T_{\delta} = \sum_{\substack{c=1\\ (c,\delta)=1}}^{\infty} \frac{\mu(c)\lambda_{k,s}(c,\ldots,c)}{c^k} \prod_{p\, \centernot\divides \, c\delta}
\left(1- \frac{p F_{k,s}(p)+(-1)^k}{p^k} \right)
\end{equation*}
\begin{equation*}
= \prod_{p\, \centernot\divides \, \delta} \left(1- \frac{p F_{k,s}(p)+(-1)^k}{p^k} \right)  \sum_{\substack{c=1\\ (c,\delta)=1}}^{\infty} \frac{\mu(c)\lambda_{k,s}(c,\ldots,c)}{c^k} \prod_{p \divides c} \left(1- \frac{p F_{k,s}(p)+(-1)^k}{p^k} \right)^{-1}
\end{equation*}
\begin{equation*}
= \prod_{p\, \centernot\divides \, \delta} \left(1- \frac{p F_{k,s}(p)+(-1)^k}{p^k} \right) \prod_{p\, \centernot\divides \, \delta}
\left(1- \frac{\lambda_{k,s}(p,\ldots,p)}{p^k \left(1- \frac{p F_{k,s}(p)+(-1)^k}{p^k}  \right) }\right)
= \prod_{p\, \centernot\divides \, \delta}  \left(1 - \frac{F_{k,s}(p)}{p^{k-1}} \right),
\end{equation*}
using that $\lambda_{k,s}(p,\ldots,p)=(-1)^{k-1}$ (see property b) of Lemma \ref{Lemma_prop_lambda}).
\end{proof}

\begin{proof}[Proof of Theorem {\rm \ref{Theorem_A}}]

By applying Lemma \ref{Lemma_Partition} we have
\begin{equation*}
N(n/\delta;j_1,\ldots,j_k)= \frac{(n/\delta)^{k-1}}{(k-1)!j_1\cdots j_k}+  O\left(\frac{(n/\delta)^{k-2}(j_1+\cdots +j_k)}{j_1\cdots j_k} \right),
\end{equation*}
and inserting it into \eqref{form_A_k_t} gives
\begin{equation} \label{A_to_go}
A_{k,s}(n) = \frac{n^{k-1}}{(k-1)!}  \sum_{\delta \divides n} \frac{\mu^2(\delta)}{\delta^{k-1}} (-1)^{(k-1)\omega(\delta)} \sum_{\substack{j_1,\ldots,j_k\le n/\delta \\ \gcd(j_1,\ldots,j_k)=1\\ \gcd(j_1\cdots j_k,\delta)=1 }}
\frac{\lambda_{k,s}(j_1,\ldots,j_k)}{j_1\cdots j_k}
\end{equation}
\begin{equation} \label{A_error}
+ O\left(n^{k-2} \sum_{\delta \divides n} \frac{\mu^2(\delta)}{\delta^{k-2}} \sum_{j_1,\ldots,j_k\le n/\delta}
\frac{(j_1+\cdots +j_k)|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_1\cdots j_k} \right).
\end{equation}

The inner sum $I(n,\delta)$ in \eqref{A_to_go} can be written as
\begin{equation}  \label{def_I_R}
I(n,\delta): = \sum_{\substack{j_1,\ldots,j_k\le n/\delta \\ \gcd(j_1,\ldots,j_k)=1\\ \gcd(j_1\cdots j_k,\delta)=1 }}
\frac{\lambda_{k,s}(j_1,\ldots,j_k)}{j_1\cdots j_k}  = \sum_{\substack{j_1,\ldots,j_k =1 \\ \gcd(j_1,\ldots,j_k)=1\\
\gcd(j_1\cdots j_k,\delta)=1 }}^{\infty} \frac{\lambda_{k,s}(j_1,\ldots,j_k)}{j_1\cdots j_k} + R(n,\delta),
\end{equation}
where for the remainder term $R$ one has
\begin{equation} \label{first_est_R}
R(n,\delta) \ll  \sideset{}{'}\sum_{j_1,\ldots,j_k} \frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_1\cdots j_k}
\end{equation}
with $\sum^{'}$ meaning that $j_1,\ldots, j_k\in \N$ and $j_1,\ldots,j_k\le n/\delta$ does not hold, that is, there exists at least one $i$
($1\le i\le k$) such that $j_i>n/\delta$ (the gcd conditions can be omitted).

Hence, according to \eqref{A_to_go}, \eqref{def_I_R} and using Lemma \ref{Lemma_T_delta} the main term for $A_{k,s}(n)$ is
\begin{equation*}
\frac{n^{k-1}}{(k-1)!} \prod_p  \left(1 - \frac{F_{k,s}(p)}{p^{k-1}} \right) \sum_{\delta \divides n} \frac{\mu^2(\delta)}{\delta^{k-1}}
(-1)^{(k-1)\omega(\delta)} \prod_{p\divides \delta}  \left(1 - \frac{F_{k,s}(p)}{p^{k-1}} \right)^{-1}
\end{equation*}
\begin{equation*}
= \frac{n^{k-1}}{(k-1)!}  \prod_p  \left(1 - \frac{F_{k,s}(p)}{p^{k-1}} \right)
\prod_{p\divides n} \left(1 + \frac{(-1)^{k-1}}{p^{k-1}}  \left(1 - \frac{F_{k,s}(p)}{p^{k-1}} \right)^{-1} \right)
\end{equation*}
\begin{equation*}
= \frac{n^{k-1}}{(k-1)!}  \prod_p  \left(1 - \frac{F_{k,s}(p)}{p^{k-1}} \right)
\prod_{p\divides n} \left(1 + \frac{(-1)^{k-1}}{p^{k-1}- F_{k,s}(p)} \right)
\end{equation*}
\begin{equation*}
=\frac{n^{k-1}}{(k-1)!} C_{k,s} f_{k,s}(n).
\end{equation*}

To estimate the error under \eqref{A_error} note that for any $i$ ($1\le i\le k$),
\begin{equation*}
S:= \sum_{j_1,\ldots,j_k\le x} \frac{j_i|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_1\cdots j_k} =
\sum_{j_1,\ldots,j_k\le x} \frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_1\cdots j_{i-1} j_{i+1}\cdots j_k}
\end{equation*}
\begin{equation*}
\le \prod_{p\le x} \sum_{\nu_1,\ldots,\nu_k=0}^{\infty} \frac{|\lambda_{k,s}(p^{\nu_1},\ldots,p^{\nu_k})|}{p^{\nu_1+\cdots
+\nu_{i-1}+\nu_{i+1}+\cdots + \nu_k}}
\end{equation*}
\begin{equation*}
= \prod_{p\le x} \left(1+\frac{c_1}{p}+ \frac{c_2}{p^2}+\cdots +\frac{c_{k-1}}{p^{k-1}}\right),
\end{equation*}
by \eqref{lambda_val}, where $c_1,\ldots,c_{k-1}$ are certain positive
integers. We need the precise value of $c_1$. If $1\le i\le s$, then we have $p$ in the denominator if and only
if $\nu_i=1$ and exactly one of $\nu_{s+1},\ldots,\nu_k$ is $1$, all other values $\nu_j$ being $0$, which occurs $k-s$ times.
That is, $c_1=k-s$. In the case $s+1\le i\le k$, the denominator is $p$ if and only
if $\nu_i=1$ and exactly one of $\nu_1,\ldots,\nu_s$ is $1$, all other
values $\nu_j$ being $0$, which occurs $s$ times. So, in this case $c_1=s$.

We deduce that
\begin{equation} \label{estimate_S_i}
S\ll  \prod_{p\le x} \left(1+\frac1{p}\right)^{\max(s,k-s)} \ll (\log x)^{\max(s,k-s)},
\end{equation}
by Mertens' theorem. For $x=n/\delta$ we obtain that the error in \eqref{A_error} is
\begin{equation} \label{result_R_2}
 \ll n^{k-2} \sum_{\delta \divides n} \frac{\mu^2(\delta)}{\delta^{k-2}} (\log n/\delta)^{\max(s,k-s)} \le n^{k-2} (\log n)^{\max(s,k-s)}
\prod_{p\divides n} \left(1+\frac1{p^{k-2}} \right),
\end{equation}
where $P_k:= \prod_{p\divides n} \left(1+\frac1{p^{k-2}}\right) \ll 1$ if $k\ge 4$ and $P_3=\prod_{p\divides n} \left(1+\frac1{p} \right)=
\frac{n}{\varphi(n)} \prod_{p\divides n} \left(1-\frac1{p^2} \right) \ll \log\log n$ if $k=3$.

Now we estimate the error $R(n,\delta)$ from \eqref{first_est_R}. Consider the sums
\begin{equation*}
\sideset{}{'}\sum_{j_1,\ldots,j_k} \frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_1\cdots j_k}
\end{equation*}
with $n/\delta=x$, and assume that $j_i>x$  for $i\in I\subseteq \{1,\ldots,k\}$ and $j_{\ell}\le x$ for $\ell \notin I$. That is, we need to
estimate the sum
\begin{equation*}
U:= \sum_{\substack{j_i>x, \, i\in I\\ j_{\ell}\le x, \, \ell \notin I}} \frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_1\cdots j_k}.
\end{equation*}

Let $m=\# I$. We distinguish the following cases:

Case i) $m\ge 3$: if $0<\varepsilon <1/2$, then
\begin{equation*}
U = \sum_{\substack{j_i>x, \, i\in I\\ j_{\ell}\le x, \, \ell \notin I}} \frac{|\lambda_{k,s}(j_1,\ldots,j_k)|
\prod_{i\in I} j_i^{\varepsilon-1/2}}{\prod_{i\in I} j_i^{\varepsilon+1/2} \prod_{\ell \notin I} j_{\ell}}
\end{equation*}
\begin{equation*}
\le x^{m(\varepsilon-1/2)} \sum_{j_1,\ldots,j_k=1}^{\infty} \frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}
{\prod_{i\in I} j_i^{\varepsilon+1/2} \prod_{\ell \notin I} j_{\ell}}\ll x^{m(\varepsilon-1/2)},
\end{equation*}
since the latter series is convergent by Lemma \ref{Lemma_Dir_series}. Using that
$m(\varepsilon -1/2)< -1$ for $0<\varepsilon< (m-2)/(2m)$, here we
need $m\ge 3$, we obtain that $U\ll \frac1{x}$.

Case ii) $m=1$: Assume that $I=\{a\}$, $j_a>x$ and $j_{\ell} \le x$ for all $\ell \ne a$. Consider a prime $p$.
If $p\divides j_{\ell}$ for some $\ell \ne a$, then $p\le x$. If $p\divides j_a$ and $p>x$, then $p\notdivides j_{\ell}$ for every
$\ell \ne a$ and $\lambda_{k,s}(j_1,\ldots,j_k)=0$ by its definition \eqref{lambda_val}. Hence it is enough to select the primes
$p\le x$. We deduce
\begin{equation*}
U < \frac1{x} \sum_{\substack{j_a > x\\ j_{\ell} \le x, \, \ell \ne a}}
\frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_1\cdots j_{a-1}j_{a+1}\cdots j_k}
\end{equation*}
\begin{equation*}
\le \frac1{x} \prod_{p\le x} \sum_{\nu_1,\ldots, \nu_k=0}^{\infty}
\frac{|\lambda_{k,s}(p^{\nu_1},\ldots,p^{\nu_k})|}{p^{\nu_1+\cdots +\nu_{a-1}+\nu_{a+1}+ \cdots + \nu_k}} \ll \frac1{x}(\log x)^{\max(s,k-s)},
\end{equation*}
similar to the estimate \eqref{estimate_S_i}.

Case iii) $m=2$: Assume that $I=\{a,b\}$, that is, $j_a,j_b>x$ and $j_{\ell} \le x$ for all $\ell \ne a,b$. Furthermore, suppose that
$1\le a,b\le s$ or $s+1\le a,b\le k$. Let $p$ be a prime. If $p\divides j_{\ell}$ for some $\ell \ne a,b$, then $p\le x$. If $p\divides j_a$
or $p\divides j_b$, and $p>x$, then $p\notdivides j_{\ell}$ for every $\ell \ne a,b$ and $\lambda_{k,s}(j_1,\ldots,j_k)=0$ by \eqref{lambda_val}.
So, it is enough to consider the primes $p\le x$. Similar to the case ii) we deduce that
\begin{equation*}
U < \frac1{x} \sum_{\substack{j_a, j_b > x\\ j_{\ell} \le x, \, \ell \ne a,b}}
\frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_1\cdots j_{a-1}j_{a+1}\cdots j_k}
\end{equation*}
\begin{equation*}
\le \frac1{x} \prod_{p\le x} \sum_{\nu_1,\ldots, \nu_k=0}^{\infty}
\frac{|\lambda_{k,s}(p^{\nu_1},\ldots,p^{\nu_k})|}{p^{\nu_1+\cdots +\nu_{a-1}+\nu_{a+1}+ \cdots + \nu_k}} \ll \frac1{x}(\log x)^{\max(s,k-s)}.
\end{equation*}

Hence, we need to handle one more case, namely when $j_a,j_b>x$, where $1\le a\le s$, $s+1\le b\le k$ and $j_{\ell} \le x$ for all $\ell \ne a,b$.
We split the sum $U$ into two sums, namely
\begin{equation*}
U = \sum_{\substack{j_a, j_b>x \\ j_{\ell} \le x, \, \ell \ne a,b}}
\frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_1\cdots j_k}
\end{equation*}
\begin{equation*}
= \sum_{\substack{j_a> x^{3/2}, \, j_b>x\\ j_{\ell} \le x, \, \ell \ne a,b}}
\frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_1\cdots j_k} +
\sum_{\substack{x^{3/2}\ge j_a> x, \, j_b>x\\ j_{\ell} \le x, \, \ell \ne a,b }}
\frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_1\cdots j_k} =: U_1+U_2,
\end{equation*}
say, where
\begin{equation*}
U_1= \sum_{\substack{j_a> x^{3/2}, \, j_b >x\\ j_{\ell} \le x, \, \ell \ne a,b }}
\frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_a^{1/3} \prod_{\ell \ne a} j_{\ell}}
\frac1{j_a^{2/3}}
\end{equation*}
\begin{equation*}
\le \frac1{x} \sum_{\substack{j_1,\ldots,j_k=1}}^{\infty}
\frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}{j_a^{1/3} \prod_{\ell \ne a} j_{\ell}}
\ll \frac1{x},
\end{equation*}
since the series is convergent.

At the same time,
\begin{equation*}
U_2 \le \frac1{x} \sum_{\substack{x^{3/2}\ge j_a, \, j_b>x \\ j_{\ell} \le x, \, \ell \ne a,b }}
\frac{|\lambda_{k,s}(j_1,\ldots,j_k)|}{\prod_{1\le \ell \le k, \ell \ne b} j_{\ell}}
\end{equation*}
where $j_a\le x^{3/2}$, $j_b>x$, $j_{\ell} \le x$ ($\ell \ne a, b$).
Let $p$ be a  prime. If $p\divides j_{\ell}$ for an $\ell \in \{1,\ldots,k\}\setminus \{b\} $,
then $p\le x^{3/2}$. If $p\divides j_b$ and $p>x^{3/2}$, then $p\notdivides
j_{\ell}$ for every $\ell \ne b$ and $\lambda_{k,s}(j_1,\ldots,j_k)=0$ by its definition. Hence it is enough
to take the primes $p\le x^{3/2}$. We deduce, as above,
\begin{equation*}
U_2 \le \frac1{x} \prod_{p\le x^{3/2}} \sum_{\nu_1,\ldots, \nu_k=0}^{\infty}
\frac{|\lambda_{k,s}(p^{\nu_1},\ldots,p^{\nu_k})|}{p^{\nu_1+\cdots +\nu_{b-1}+\nu_{b+1} +\cdots
+\nu_k}}
\end{equation*}
\begin{equation*}
\ll \frac1{x}(\log x^{3/2})^{\max(s,k-s)} \ll \frac1{x}(\log
x)^{\max(s,k-s)}.
\end{equation*}

Hence for every $m\ge 1$,
\begin{equation*}
U \ll \frac1{x}(\log x)^{\max(s,k-s)},
\end{equation*}
which applied for $x=n/\delta$ gives by \eqref{A_to_go}, \eqref{def_I_R} and \eqref{first_est_R} that the error coming from $R(n,\delta)$ is
\begin{equation*}
\ll n^{k-1} \sum_{\delta \mid n} \frac{\mu^2(\delta)}{\delta^{k-1}} R(n,\delta) \ll n^{k-2} \sum_{\delta \divides n} \frac{\mu^2(\delta)}{\delta^{k-2}} (\log n/\delta)^{\max(s,k-s)},
\end{equation*}
leading to the same estimate as that for the error \eqref{A_error} obtained in \eqref{result_R_2}.

This finishes the proof of Theorem \ref{Theorem_A}.
\end{proof}

\subsection{The sum $B_{k,t}(n)$}

Similar to the proof of Theorem \ref{Theorem_A}, consider the characteristic function of the $k$-tuples of positive integers with
$t$-wise relatively prime components, i.e., let
\begin{equation*}
\varrho_{k,t}(n_1,\ldots,n_k)= \begin{cases} 1, & \text{ if $\gcd(n_{i_1},\ldots n_{i_t})=1$ for $1\le i_1<\ldots <i_t\le k$},
\\ 0, &
\text{ otherwise.}
\end{cases}
\end{equation*}

The function $\varrho_{k,t}(n_1,\ldots,n_k)$ is symmetric in all variables and it is a multiplicative function of $k$ variables.
For every prime $p$ and every $\nu_1,\ldots,\nu_r\in \N_0$ one has
\begin{equation*}
\varrho_{k,t}(p^{\nu_1}, \ldots,p^{\nu_k})= \begin{cases} 1, &
\text{if there are at most $t-1$ values $\nu_i\ge 1$}, \\ 0, &
\text{otherwise.}
\end{cases}
\end{equation*}

The following result on the multiple Dirichlet series of the function $\varrho_{k,t}$ was
proved by the author \cite{Tot2016}. Let $e_j(x_1,\dots,x_k)=\sum_{1\le i_1<\ldots<i_j\le k}
x_{i_1}\cdots x_{i_j}$ denote the elementary symmetric polynomials
in $x_1,\ldots,x_k$ of degree $j$.

\begin{lemma}[{\cite[Th.\ 2.1]{Tot2016}}] \label{Lemma_Dir_series_rho} Let $2\le t\le k$. Then
\begin{equation*}
\sum_{n_1,\ldots,n_k=1}^{\infty} \frac{\varrho_{k,t}(n_1,\ldots,n_k)}{n_1^{z_1}\cdots n_r^{z_k}}=
\zeta(z_1)\cdots \zeta(z_k) L_{k,t}(z_1,\ldots,z_k),
\end{equation*}
where
\begin{equation*}
L_{k,t}(z_1,\ldots,z_k)= \prod_p \left( 1- \sum_{j=t}^k (-1)^{j-t} \binom{j-1}{t-1} e_j(p^{-z_1},\ldots,p^{-z_k})\right)
\end{equation*}
is absolutely convergent if $\Re (z_{i_1}+\cdots+ z_{i_j})>1$ for every $1\le i_1< \ldots < i_j\le k$ with $t\le j\le
k$.

In particular, for $z_1=\cdots =z_k=1$,
\begin{equation*}
L_{k,t}(1,\ldots,1)= \prod_p  \frac{pG_{k,t}(p)+(-1)^{k-t+1}\binom{k-1}{t-1}}{p^k}
\end{equation*}
is absolutely convergent, where $G_{k,t}(p)$ is defined by \eqref{notation_G}.
\end{lemma}

\begin{lemma} \label{Lemma_convo_rho} Let $2\le t\le k$. Then for every $n_1,\ldots,n_k\in \N$,
\begin{equation*}
\varrho_{k,t}(n_1,\ldots,n_k) = \sum_{d_1\divides n_1, \ldots, d_k\divides
n_k} \psi_{k,t}(d_1,\ldots,d_k),
\end{equation*}
where the function $\psi_{k,t}$ is multiplicative, and for any prime powers $p^{\nu_1},\ldots,p^{\nu_k}$ \textup{($\nu_1,\ldots,\nu_k\in \N_0$)},
\begin{equation} \label{psi}
\psi_{k,t}(p^{\nu_1},\ldots,p^{\nu_k})=
\begin{cases} 1, & \nu_1=\ldots=\nu_k=0,\\
(-1)^{j-t+1}\binom{j-1}{t-1}, & \nu_1,\ldots,\nu_k\in \{0,1\}, \ j:=\nu_1+\ldots+\nu_k \ge t,\\
0, & \text{otherwise}.
\end{cases}
\end{equation}
\end{lemma}

\begin{proof} This follows by Lemma \ref{Lemma_Dir_series_rho}, where
\begin{equation*}
L_{k,t}(z_1,\ldots,z_k)= \sum_{n_1,\ldots,n_r=1}^{\infty} \frac{\psi_{k,t}(n_1,\ldots,n_k)}{n_1^{z_1}\cdots n_k^{z_k}}.
\end{equation*}

See the proof of the similar results included in Lemma \ref{Lemma_convo}.
\end{proof}

We will need the following properties the function $\psi_{k,t}$, given by \eqref{psi}, which are similar to the function $\lambda_{k,s}$. Cf. Lemma \ref{Lemma_prop_lambda}.

\begin{lemma} \label{Lemma_prop_psi} Let $j_1,\ldots,j_k,\delta\in \N$.

a) If $\gcd(j_1\cdots j_k,\delta)\ne 1$, then  $\psi_{k,t}(j_1\delta,\ldots,j_k\delta)=0$. If $\gcd(j_1\cdots j_k,\delta) = 1$, then
\begin{equation*}
\psi_{k,t}(j_1\delta,\ldots,j_k\delta) = \psi_{k,t}(j_1,\ldots,j_k) \psi_{k,t}(\delta,\ldots,\delta).
\end{equation*}

b) $\psi_{k,t}(\delta,\ldots,\delta) =\left((-1)^{k-t+1}\binom{k-1}{t-1}\right)^{\omega(\delta)}\mu^2(\delta)$.
\end{lemma}

\begin{lemma} \label{Lemma_Repr_B_N} Let $2 \le t \le k$. For every $n\ge k$ we have
\begin{equation} \label{form_B_k_t}
B_{k,t}(n) = \sum_{\delta \divides n} \mu^2(\delta) \left((-1)^{k-t+1} \binom{k-1}{t-1}\right)^{\omega(\delta)}
\sum_{\substack{1\le j_1,\ldots,j_k\le n/\delta \\ \gcd(j_1,\ldots,j_k)=1\\ \gcd(j_1\cdots j_k, \delta)=1 }}  \psi_{k,t}(j_1,\ldots,j_k)
N(n/\delta;j_1,\ldots,j_k).
\end{equation}
\end{lemma}

\begin{proof} We follow the same steps as in the proof of Lemma \ref{Lemma_Repr_A_N}. First, by the definitions of $B_{k,t}(n)$ and
$\varrho_{k,t}(n_1,\ldots,n_k)$,
and by Lemma \ref{Lemma_convo_rho} we have
\begin{equation*}
B_{k,t}(n)= \sum_{\substack{x_1,\ldots,x_k\ge 1\\x_1+\cdots+ x_k=n}} \varrho_{k,t}(x_1,\ldots,x_k)
= \sum_{\substack{x_1,\ldots,x_k\ge 1\\ x_1+\cdots+ x_k=n}} \sum_{d_1\divides x_1, \ldots, d_k\divides x_k}
\psi_{k,t}(d_1,\ldots,d_k)
\end{equation*}
\begin{equation*}
= \sum_{1\le d_1,\ldots,d_k\le n} \psi_{k,t}(d_1,\ldots,d_k)
\sum_{\substack{a_1,\ldots,a_k\ge 1\\ d_1a_1+\cdots+ d_ka_k=n}} 1
\end{equation*}
\begin{equation*}
= \sum_{1\le d_1,\ldots,d_k\le n}  \psi_{k,t}(d_1,\ldots,d_k) N(n;d_1,\ldots,d_k).
\end{equation*}

By grouping the terms according to the values of $\delta=\gcd(a_1,\ldots,a_k)$,
\begin{equation} \label{sum_for_B}
B_{k,t}(n) = \sum_{\delta \divides n} \sum_{\substack{1\le j_1,\ldots,j_k\le n/\delta \\ \gcd(j_1,\ldots,j_k)=1}}
\psi_{k,t}(j_1\delta,\ldots,j_k\delta)
N(n/\delta;j_1,\ldots,j_k).
\end{equation}

By using Lemma \ref{Lemma_prop_psi}  we obtain identity \eqref{form_B_k_t}.
\end{proof}

\begin{lemma} \label{Lemma_V_delta} If $2 \le t \le k$, then for every $\delta \in \N$,
\begin{equation*}
V_{\delta}: =\sum_{\substack{j_1,\ldots,j_k=1 \\ \gcd(j_1,\ldots,j_k)=1\\ \gcd(j_1\cdots j_k,\delta)=1 }}^{\infty}
\frac{\psi_{k,t}(j_1,\ldots,j_k)}{j_1\cdots j_k}
=  \prod_{p\, \centernot\divides \, \delta} \frac{G_{k,t}(p)}{p^{k-1}},
\end{equation*}
\end{lemma}

\begin{proof} We have by using the property of the M\"{o}bius function and the properties of the function $\psi_{k,t}$ in  Lemma \ref{Lemma_prop_psi},
\begin{equation*}
V_{\delta}= \sum_{\substack{j_1,\ldots,j_k=1 \\ \gcd(j_1\cdots j_k,\delta)=1 }}^{\infty}
\frac{\psi_{k,t}(j_1,\ldots,j_k)}{j_1\cdots j_k} \sum_{c\divides \gcd(j_1,\ldots,j_k)} \mu(c)
\end{equation*}
\begin{equation*}
 = \sum_{\substack{c,v_1,\ldots,v_k=1 \\ \gcd(cv_1\cdots v_k,\delta)=1 }}^{\infty}
\frac{\mu(c) \psi_{k,t}(cv_1,\ldots,cv_k)}{cv_1\cdots cv_k} =
\sum_{\substack{c=1\\ (c,\delta)=1}}^{\infty} \frac{\mu(c)\psi_{k,t}(c,\ldots,c)}{c^k}
\sum_{\substack{v_1,\ldots,v_k=1 \\ \gcd(v_1\cdots v_k,c\delta)=1}}^{\infty}
\frac{\psi_{k,t}(v_1,\ldots,v_k)}{v_1\cdots v_k}.
\end{equation*}

We deduce from Lemma
\ref{Lemma_Dir_series_rho} that
\begin{equation*}
V_{\delta} = \sum_{\substack{c=1\\ (c,\delta)=1}}^{\infty} \frac{\mu(c)\psi_{k,t}(c,\ldots,c)}{c^k} \prod_{p\, \centernot\divides \, c\delta}
\frac{pG_{k,t}(p)+(-1)^{k-t+1}\binom{k-1}{t-1}}{p^k}
\end{equation*}
\begin{equation*}
= \prod_{p\, \centernot\divides \, \delta} \frac{pG_{k,t}(p)+(-1)^{k-t+1}\binom{k-1}{t-1}}{p^k}  \sum_{\substack{c=1\\ (c,\delta)=1}}^{\infty}
\frac{\mu(c)\psi_{k,t}(c,\ldots,c)}{c^k} \prod_{p\divides c}  \left(\frac{pG_{k,t}(p)+(-1)^{k-t+1}\binom{k-1}{t-1}}{p^k}\right)^{-1}
\end{equation*}
\begin{equation*}
= \prod_{p\, \centernot\divides \, \delta} \frac{pG_{k,t}(p)+(-1)^{k-t+1}\binom{k-1}{t-1}}{p^k} \prod_{p\, \centernot\divides \, \delta}
\left(1- \frac{\psi_{k,t}(p,\ldots,p)}{p^k}\cdot  \frac{p^k}{pG_{k,t}(p)+(-1)^{k-t+1}\binom{k-1}{t-1}}\right)
\end{equation*}
\begin{equation*}
= \prod_{p\, \centernot\divides \, \delta} \frac{G_{k,t}(p)}{p^{k-1}},
\end{equation*}
using that $\psi_{k,t}(p,\ldots,p)=(-)^{k-t+1}\binom{k-1}{t-1}$. Cf. property b) of Lemma \ref{Lemma_prop_psi}.
\end{proof}

We also need the following estimates proved in \cite[Sect.\ 4]{Tot2016}, by similar arguments as given in the proof of Theorem \ref{Theorem_A}.

\begin{lemma} \label{Lemma_error_psi} Let $2\le t\le k$. Then for any $i$ \textup{($1\le i\le k$)},
\begin{equation} \label{estimate_1}
\sum_{j_1,\ldots,j_k\le x} \frac{|\psi_{k,t}(j_1,\ldots,j_k)|}{j_1\cdots j_{i-1}j_{i+1}\cdots j_k} \ll
\begin{cases} 1, & \text{ if $t\ge 3$}, \\ (\log x)^{k-1}, & \text{ if $t=2$}. \end{cases}
\end{equation}

Furthermore,
\begin{equation} \label{estimate_2}
\sideset{}{'}\sum_{j_1,\ldots,j_k} \frac{|\psi_{k,t}(j_1,\ldots,j_k)|}{j_1\cdots j_k} \ll
\begin{cases} x^{-1}, & \text{ if $t\ge 3$}, \\ x^{-1} (\log x)^{k-1}, & \text{ if $t=2$}, \end{cases}
\end{equation}
with $\sum^{'}$ meaning that there is at least one $i$ \textup{($1\le i\le k$)} such that $j_i>x$.
\end{lemma}

\begin{proof}[Proof of Theorem {\rm \ref{Theorem_B}}]
Identity \eqref{form_B_k_t} and Lemma \ref{Lemma_Partition} lead to
\begin{equation} \label{B_to_go}
B_{k,t}(n) = \frac{n^{k-1}}{(k-1)!}  \sum_{\delta \divides n} \frac{\mu^2(\delta)}{\delta^{k-1}} \left((-1)^{k-t+1} \binom{k-1}{t-1}\right)^{\omega(\delta)}
\sum_{\substack{j_1,\ldots,j_k\le n/\delta
\\ \gcd(j_1,\ldots,j_k)=1\\ \gcd(j_1\cdots j_k,\delta)=1 }} \frac{\psi_{k,t}(j_1,\ldots,j_k)}{j_1\cdots j_k}
\end{equation}
\begin{equation} \label{Berror_ezaz}
+ O\left(n^{k-2} \sum_{\delta \divides n} \frac{\mu^2(\delta)}{\delta^{k-2}} \binom{k-1}{t-1}^{\omega(\delta)}
\sum_{j_1,\ldots,j_k\le n/\delta} \frac{(j_1+\cdots+j_k)|\psi_{k,t}(j_1,\ldots,j_k)|}{j_1\cdots j_k}\right).
\end{equation}

The inner sum $J$ in \eqref{B_to_go} can be written as
\begin{equation}  \label{def_J_R_tilde}
J:= \sum_{\substack{j_1,\ldots,j_k\le n/\delta \\ \gcd(j_1,\ldots,j_k)=1\\ \gcd(j_1\cdots j_k,\delta)=1 }}
\frac{\psi_{k,t}(j_1,\ldots,j_k)}{j_1\cdots j_k}  = \sum_{\substack{j_1,\ldots,j_k =1 \\ \gcd(j_1,\ldots,j_k)=1\\
\gcd(j_1\cdots j_k,\delta)=1 }}^{\infty} \frac{\psi_{k,t}(j_1,\ldots,j_k)}{j_1\cdots j_k} + \widetilde{R}(n,\delta),
\end{equation}
where for the remainder term $\widetilde{R}(n,\delta)$ one has
\begin{equation} \label{first_est_R_tilde}
\widetilde{R}(n,\delta) \ll  \sideset{}{'}\sum_{j_1,\ldots,j_k} \frac{|\psi_{k,t}(j_1,\ldots,j_k)|}{j_1\cdots j_k}
\end{equation}
with $\sum^{'}$ meaning that there exists at least one $i$
($1\le i\le k$) such that $j_i>n/\delta$.

Hence, according to \eqref{B_to_go}, \eqref{def_J_R_tilde} and using Lemma \ref{Lemma_V_delta} the main term for $B_{k,t}(n)$ is
\begin{equation*}
\frac{n^{k-1}}{(k-1)!}  \prod_p \frac{G_{k,t}(p)}{p^{k-1}} \sum_{\delta \divides n} \frac{\mu^2(\delta)}{\delta^{k-1}}
\left((-1)^{k-t+1} \binom{k-1}{t-1}\right)^{\omega(\delta)}  \prod_{p\divides \delta} \left(\frac{G_{k,t}(p)}{p^{k-1}}\right)^{-1}
\end{equation*}
\begin{equation*}
= \frac{n^{k-1}}{(k-1)!}  \prod_p \frac{G_{k,t}(p)}{p^{k-1}}
\prod_{p\divides n} \left(1 + \frac{(-1)^{k-t+1}\binom{k-1}{t-1}}{G_{k,t}(p)} \right)
= \frac{n^{k-1}}{(k-1)!} D_{k,t} g_{k,t}(n).
\end{equation*}

Now we estimate the error $\widetilde{R}_1$ coming from \eqref{first_est_R_tilde}. By \eqref{B_to_go}, \eqref{def_J_R_tilde}, \eqref{first_est_R_tilde}
and applying \eqref{estimate_2} in the case $x=n/\delta$ we obtain that the error $\widetilde{R}_1$ is as follows. If $k\ge t\ge 3$, then
\begin{equation*}
\widetilde{R}_1 \ll n^{k-2} \sum_{\delta \divides n} \frac{\mu^2(\delta)}{\delta^{k-2}} \binom{k-1}{t-1}^{\omega(\delta)} =
n^{k-2} \prod_{p\divides n} \left(1+ \frac1{p^{k-2}} \binom{k-1}{t-1}\right),
\end{equation*}
which is $\ll n^{k-2}$ if $k\ge 4$. If $k=t=3$, then $\widetilde{R}_1 \ll n \prod_{p\divides n} \left(1+ \frac1{p} \right) \ll n\log \log n$.
However, formula \eqref{R_k_n_asympt} shows that in the latter case the final error term is $O(n)$.

If $k=3$, $t=2$, then
\begin{equation*}
\widetilde{R}_1 \ll n \sum_{\delta \divides n} \frac{\mu^2(\delta)}{\delta} 2^{\omega(\delta)} \left(\log n/\delta \right)^2 \le
n(\log n)^2 \prod_{p\divides n} \left(1+ \frac{2}{p}\right)
\end{equation*}
\begin{equation*}
\le n (\log n)^2 \prod_{p\divides n} \left(1+ \frac1{p}\right)^2
\ll n (\log n)^2 (\log \log n)^2.
\end{equation*}

To estimate the error $\widetilde{R}_2$ under \eqref{Berror_ezaz} we apply \eqref{estimate_1}. We deduce that if $t\ge 3$, then
\begin{equation*}
\widetilde{R}_2 \ll n^{k-2} \sum_{\delta \divides n} \frac{\mu^2(\delta)}{\delta^{k-2}} \binom{k-1}{t-1}^{\omega(\delta)} =
n^{k-2} \prod_{p\divides n} \left(1+ \frac1{p^{k-2}} \binom{k-1}{t-1}\right),
\end{equation*}
which is $\ll n^{k-2}$ if $k\ge 4$ and is $ \ll n \prod_{p\divides n} \left(1+ \frac1{p} \right) \ll n\log \log n$ if $k=t=3$.
However, in the latter case the final error term is $O(n)$ by \eqref{R_k_n_asympt}, already mentioned above.

If $t=2$, $k\ge 3$, then we have
\begin{equation*}
\widetilde{R}_2 \ll n^{k-2} \sum_{\delta \divides n} \frac{\mu^2(\delta)}{\delta^{k-2}} (k-1)^{\omega(\delta)} \left( \log n/\delta \right)^{k-1}\le
n^{k-2} (\log n)^{k-1} \prod_{p\divides n} \left(1+ \frac{k-1}{p^{k-2}}\right)
\end{equation*}
which is $\ll n^{k-2}(\log n)^{k-1}$ if $k\ge 4$, and is $\ll n (\log n)^2 (\log \log n)^2$ if $k=3$.

This completes the proof of Theorem \ref{Theorem_B}.
\end{proof}

\section{Acknowledgement} The author thanks the referee for useful comments and suggestions concerning the presentation of the proofs of the paper.

config: arxiv


TITLE: Prove that $\lim_{h \to 0}\frac{1}{h}\int_0^h{\cos{\frac{1}{t}}dt} = 0$
QUESTION [12 upvotes]: I'm trying to prove that $$\lim_{h \to 0}\frac{1}{h}\int_0^h{f(t)dt} = 0$$ where $$f(t) = \begin{cases}\cos{\frac{1}{t}} &\text{ if } t \neq 0\\ 0&\text{otherwise}\end{cases}.$$ Can someone give me a hint where to start? Darboux sums somehow seem to lead me nowhere.
NOTE: I cannot assume that $f$ has an antiderivative $F$.

REPLY [6 votes]: Let $g=\frac1h$ and $s=\frac1t$. Then integration by parts gives
$$
\begin{align}
\lim_{h\to0}\frac1h\int_0^h{\cos\!\left(\frac1t\right)\mathrm{d}t}
&=\lim_{g\to\infty}g\int_g^\infty{\frac{\cos(s)}{s^2}\,\mathrm{d}s}\\
&=\lim_{g\to\infty}g\left[-\frac{\sin(g)}{g^2}+2\int_g^\infty\frac{\sin(s)}{s^3}\,\mathrm{d}s\right]\\
&=\lim_{g\to\infty}O\left(\frac1g\right)\\[6pt]
&=0
\end{align}
$$
Since $\sin(s)=O(1)$.


TITLE: Has anyone considered axioms to the effect that: "The axiom of constructibility fails very very badly?"
QUESTION [4 upvotes]: If I'm not mistaken, the axiom of constructibility basically says that the universe has no (non-trivial) inner models. Has anyone considered axioms of the opposite flavour, basically asserting that the universe has many inner models?
Apologies if I've misunderstood the axiom of constructibility, it is very confusing to me.

REPLY [3 votes]: The axiom of constructibility is indeed equivalent to "there are no inner models", but this is not what it "essentially says". This would be sort of a backwards way of understanding the axiom.
Instead the axiom says that we can find a hierarchy where every set appears in some stage, and it is first-order definable over the previous stage in the hierarchy. In contrast, when we consider the von Neumann we take all the subsets of the previous step, definable or not.
It turns out that two models with the same ordinals have the same constructible universe. So in turn it follows that the axiom $V=L$ is equivalent to saying that there is no inner model (note, however, that "there are no inner models" is not a first-order statement).
If anything is the opposite of $V=L$ it would be the existence of $0^\#$. This axiom implies that not only $V\neq L$, but it is in fact very far away from $L$. What does that mean? If we start with $L$ and we add using forcing a real number, or even a proper class of new sets, then the resulting model will be somewhat close to $L$. In what sense do we mean close? If $X$ is any uncountable set of ordinals, then there is some $Y\in L$ which is a set of ordinals, $X\subseteq Y$ and $|X|=|Y|$. This is known as the covering lemma.
The existence of $0^\#$ is equivalent to saying that the covering lemma fails, and it turns out that it fails in many many places, when it fails. For example the set $\{\aleph_n\mid n\in\omega\}$ cannot be covered by any set of size $<\aleph_\omega$ which is in $L$. The reason is that $\aleph_\omega$ is regular in $L$, so any unbounded subset must have size $\aleph_\omega$.
If you want to extend this further, there is the notion of relative constructibility which says that there is some set $A$ from which the universe is constructible. And so we can define the notion of $A^\#$ in a similar fashion.
So the "ultimate anti-$L$ axiom" would be something of the form $\forall A(V\neq L[A])$. This axiom is a consequence of large cardinals such as strongly compact cardinals (and in fact much much weaker than that). And these axioms are used on a regular basis in set theory.