text:
\begin{document}
\title{Maximum weight spectrum codes with reduced length}
\author{G\'{e}rard D.\ Cohen and Ludo Tolhuizen \thanks{G.D.\ Cohen (deceased) was with T\'{e}l\'{e}com Paris, CRNS LTCI;
L.\ Tolhuizen  (ludo.tolhuizen@philips.com) is with Philips Research, Eindhoven}}
\date{}
\maketitle
\begin{abstract}
A $q$-ary linear  code of dimension $k$  is called a maximum weight spectrum (MWS) code if it has the
maximum possible number ({\em viz.} $(q^k-1)/(q-1)$) of  different non-zero weights.
We construct MWS codes from quasi-minimal codes, thus obtaining of much shorter length than hitherto known.
 By an averaging argument,
we show the existence of MWS codes of even shorter length.
\end{abstract}

\section{Introduction}

In a recent preprint \cite{Shi} Shi {\em et al.} introduced and studied $L(k,q)$, the maximum number of non-zero weights that a linear code
of dimension $k$ over $\mathbb{F}_q$ can have.
 As two non-zero linearly dependent words have equal weight, it is clear that \cite[Proposition 2]{Shi}
\begin{equation}\label{eq:basic}
L(k,q) \leq \frac{q^k-1}{q-1}.
\end{equation}
Shi {\em et al.} provide constructions showing that equality holds in (\ref{eq:basic}) for $q=2$ and all $k$, and for $k=2$ and all prime powers $q$.
They conjectured that equality holds in (\ref{eq:basic}) holds for all $k$ and all prime powers $q$.
Shortly afterwards, Alderson and Neri showed this conjecture to be true \cite{Ald}, and  called $q$-ary $k$-dimensional codes with exactly
$(q^k-1)/(q-1)$ non-zero words Maximum Weight Spectrum codes, or MWS codes for short. Alderson and Neri provide two code constructions, and
lower bounds on the length of MWS codes.
Another proof of the conjecture of Shi {\em et al} was provided in \cite{Men}.

The present note continues the study of MWS codes. More particularly, we construct  MWS codes from quasi-minimal codes \cite{Coh},
and thus obtain MWS  codes of much shorter length than hitherto known. Finally, by an averaging argument, we show the existence of MWS codes of even shorter length.

The second author dedicates this manuscript to the memory of G\'{e}rard Cohen, who sadly passed away during
its preparation.
\section{Notation and definitions}
In this section, we establish notation and provide some defintions.

For prime powers $q$, and positive integers $n$ and $k$, a $k$-dimensional code over $\mathbb{F}_q$ of length $n$ and minimum Hamming distance
 $d$ is denoted as an $[n,k,d]_q$ code.
Similarly, a  $k$-dimensional code over $\mathbb{F}_q$ of length $n$ is denoted as an $[n,k]_q$ code.
For $x\in\mathbb{F}_q^n$, the support of $x$, denoted by supp$(x)$, is the set of indices of the non-zero entries of $x$. That is,
 supp($x$) =  supp$(x_1,\ldots, x_n)  = \{ i\mid x_i\neq 0\}$.

\begin{Def}\label{QM}
A  code $C$ is called {\em quasi-minimal} (QM) if any pair of linearly indepedent codewords have different supports.
That is, if $c,c'\in C$ are linearly independent, then supp$(c) \neq $supp$(c')$,
\end{Def}
Quasi-minimal codes were introduced and studied in \cite{Coh}. Obviously, the $[n,n]_2$ code is quasi-minimal.
\begin{Def}\label{MWS}
A code $C$ is called {\em Maximum Weight Spectrum} (MWS) if any pair of linearly independent codewords have different weights.
That is, if $c,c'\in C$ are linearly independent, then wt$(c) )\neq \mbox{wt}(c').$
\end{Def}
\begin{Prop}
Any MWS code is a QM code.
\end{Prop}
{\bf Proof} Obvious.
\begin{Def}
For any property $P$, we define $n_{P,k,q}$ as the shortest length of a $q$-ary code of dimension $k$ that has property $P$.
That is,
\begin{equation}
n_{P,k,q} = \min\{ n\in \mathbb{N} \mid \mbox{ there exists an } [n,k]_q \mbox{ code  with property } P \}
\end{equation}
\end{Def}
The constructions of \cite[Sec.\ 3] {Ald} and \cite[Sec.\ 4,5]{Ald} imply that
\begin{equation}
 n_{MWS,k,q} \leq 2^{\frac{q^k-1}{q-1}-1} \mbox { and } n_{MWS,k,q} \leq {\cal O}(q^{k(k+1)-4}) .
\end{equation}
Moreover, as shown in \cite[Lemma 5.1]{Ald},
\begin{equation}\label{eq:lowbound}
 n_{MWS,k,q} \geq \frac{q}{2}\cdot \frac{q^k-1}{q-1}.
\end{equation}
By combining  \cite[Thm.1]{Shi} and
(\ref{eq:lowbound}), it follows that
\begin{equation}
n_{MWS,k,2} = 2^k-1.
\end{equation}
As shown in \cite{Ald},  we have that
\begin{equation}
n_{MWS,2,q} = \frac{1}{2}q(q+1).
\end{equation}

\section{Explicit construction of MWS codes from QM codes}\label{sec:geometric}
For each $n\geq 1$ and each prime power $q$, we define the embedding function $f:\mathbb{F}_q^n \rightarrow \mathbb{F}_q^{2^n-1}$ as
\begin{equation}
f(c_0,c_1,\ldots, c_{n-1}) = (c_0,c_1,c_1,c_2,c_2,c_2,c_2,\ldots, \overbrace{c_i,c_i,\ldots, c_i}^{2^i \mbox{\small{times}}}, \ldots, c_{n-1},..,c_{n-1}).
\end{equation}
\begin{Prop}\label{Prop:QMtoMWS}
If $C$ is an $[n,k]_q$ QM code, then $f(C)$ is an $[2^n-1,k]_q$ MWS code.
\end{Prop}
{\bf Proof}
It is clear that for each $c\in\mathbb{F}_q^n$, we have that
\begin{equation}\label{eq:weight} \mbox{wt}\left(f(c)\right) = \sum_{i\in\mbox{\small{supp}}(c)} 2^i.
\end{equation}
Equation~\ref{eq:weight} implies that there is a one-to-one relation between the support of a code word $c$ and the weight of $f(c).
\;\;\;\Box$ \\ \\
We remark that $f(C)$ is a a special case of what is called a "generalized $r$-repetition of $C$" in \cite{Ald}.
\begin{Prop}\label{prop:lowbound}
For all $k$ and prime powers $q$, we have that $n_{MWS,k,q}\leq 2^{n_{QM,k,q}}$.
\end{Prop}
{\bf Proof} Direct consequence of Proposition~\ref{Prop:QMtoMWS}. $\;\;\;\Box$ \\ \\

We can take for $C$  an $[n,k_q]$ code for which all non-zero words have equal weight, for example,
an $[n_k=(q^{k}-1)/(q-1),k,q^{k-1}]_q$ code, the  dual of the $[n_k,n_k-k,3]_q$ Hamming code. This choice is equivalent to
what is done in \cite[Sec.\ 3]{Ald}, and results in a  $[2^{(q^k-1)/(q-1)-1},k]_q$  MWS code. \\
If we take for $C$ an $[k,k]_2$ code, we obtain a $[2^k-1,k]_2$ MWS code. This is in fact the explicit construction from \cite[Thm. 1]{Shi}.

We now apply results from \cite{Coh} concerning QM codes.
It is shown in \cite{Coh} that an $[n,k,d]_q$ code is quasi-minimal whenever $\frac{d}{n}> \frac{q-2}{q-1}$. Combining this with
the Gilbert-Varshamov bound, it follows
\cite{Coh} that there exists a quasi-minimal  $[n,k]_q$ code whenever $\frac{k}{n} \leq 1-h_q((q-2)/(q-1))$, where
$h_q$ is the $q$-ary
entropy function,
\begin{equation} h_q(x) = -x\log_q(x) -(1-x)\log_q(1-x) + x\log_q(q-1).
\end{equation}
In other words, we have that
\begin{equation}
n_{QM,k,q} \leq k\cdot \lambda_q \mbox{ where } \lambda_q = (1-h_q(\frac{q-2}{q-1}))^{-1}.
\end{equation}
 We conclude that for all inegers $k$ and prime powers $q$,
\begin{equation} n_{MWS,q,k} \leq 2^{k\lambda_q}.
\end{equation}

By direct computation,  it can be found that

$\lambda_q =  2q^3\log q(1+{\cal O}(q^{-1}))$, where $\log$ denotes the natural logarithm.
In the next section, we provide a construction of MWS codes of smaller length.
\\ \\ \\
{\bf Remark} We can obtain sharper upper bounds on $n_{MWS,q,k}$ by using sharper bounds for $n_{QM,k,q}$. \\
First, it can be shown\cite{AsBa} that an $[n,k]_q$ code with minimum distance $d$ and maximum weight $D$ is QM whenever $\frac{d}{D} > \frac{q-2}{q-1}$.
Combining this with a generalization of GV we obtain a QM code of improved rate. Unfortunately, we  do not have a closed form expression for the rate,
it involves is a root of an algebraic equation involving the $q$-ary entropy function. \\

Alternatively, by a small adaptation of the proof of \cite[Thm. 2]{Coh}, it can be shown that
\[ n_{QM,k,q} \leq  k \mu_q \mbox{ with } \mu_q=2 / \log_q(\frac{q^2}{q^2-2q+2}). \]
This seems to yield the shortest length for QM codes but is "totally" non-constructive.
Note that  $\mu_q= q\log q(1+{\cal O}(q^{-1}))$, so  for large $q$, $\mu_q$ is approximately $2q^2$ as small as $\lambda_q$.

\section{A non-constructive bound upper bound to $n_{MWS,k,q}$ }\label{sec:randomcode}

\begin{Thm}\label{Thm:4kexistence}
For each prime power $q$, there exists an integer $D_q$ such that $n_{MWS,k,q}\leq D_q\cdot q^{4k}$ for all $k\geq 1$.
\end{Thm}
In order to prove this theorem, we use  several lemmas.
Throughout this section,  we denote the number of words of weight $w$ of a code $C$ by $A_w(C)$.
\begin{Lem}\label{lem:suff} A $q$-ary code $C$  is an MWS code if and only if
\begin{equation}\label{eq:wtdis}  \sum_{w=1}^n A_w(C) (A_w(C)-(q-1)) \;\; < 2 (q-1)^2.
\end{equation}
\end{Lem}
{\bf Proof} If $C$ is an MWS code, then $A_w(C)\in\{0,q-1\}$ for each $w\geq 1$, so (\ref{eq:wtdis}) is satisfied.
If $C$ is not an MWS code, then $A_w(C)\geq 2(q-1)$ for at least one $w. \;\;\;\Box$
\begin{Lem}
Let $n,k$ be integers and let $q$ be a power of a prime.
If
\begin{equation}\label{eq:bound}
 q^{2k-2n} \sum_{w=0}^n {n\choose w}^2 (q-1)^{2w} < 2 (q-1)^2,
\end{equation}
then there is a $q$-ary $[n,k]$ MWS code.
\end{Lem}
{\bf Proof}  We denote by ${\cal C}$ the set of all $q$-ary $[n,k]$ codes.
Each pair of linearly independent vectors in $\mathbb{F}_q^n$  clearly is in equally many, say $\lambda$, codes in ${\cal C}$.
By counting in two ways triples $x,y,C$ with $C\in{\cal C}$ and $x,y$ linearly independent vectors in $C$, we infer that
\begin{equation}\label{eq:lambda}
(q^n-1)(q^n-q)\lambda = |{\cal C}| (q^k-1)(q^k-q).
\end{equation}
By counting in two ways the number of pairs $(x,y,C)$ with $C\in{\cal C}$ and $x,y$ linearly independent codewords of
 equal weight in $C$,
 we infer that
\begin{equation}
\sum_{C\in{\cal C}} \sum_{w\neq 0} A_w(C)(A_w(C)-(q-1)) =
\sum_{w\neq 0} {n\choose w}(q-1)^w \left( {n\choose w}(q-1)^w-(q-1) \right) \lambda.
\end{equation}
As a consequence,
\begin{equation} \frac{1}{|{\cal C}|} \sum_{C\in{\cal C}} \sum_{w\neq 0} A_w(C)(A_w(C)-(q-1)) <
\frac{\lambda}{|{\cal C}|} \sum_w {n\choose w}^2(q-1)^{2w}.
\end{equation}
Equation~ (\ref{eq:lambda}) implies that $\frac{\lambda}{|{\cal C}|} \leq q^{2k-2n}$, and hence,
 by using (\ref{eq:bound}),  we infer that
\begin{equation}
\frac{1}{|{\cal C}|} \sum_{C\in{\cal C}} \sum_{w\neq 0} A_w(C)(A_w(C)-(q-1)) < 2(q-1)^2.
\end{equation}
As a consequence, there is a $C\in{\cal C}$ such that $\sum_{w\neq 0} A_w(C)(A_w(C)-(q-1)) < 2(q-1)^2$.
According to Lemma~\ref{lem:suff}, $C$ is MWS. $\;\;\;\Box$

\begin{Lem}
For each integer $q$, there is constant $C_q$ such that for each positive integer $n$
\[ \sum_{w=0}^{n} {n\choose w}^2 (q-1)^{2w} < C_q \frac{q^{2n}}{\sqrt{n}} . \]
\end{Lem}
{\bf Proof} (sketch) For positive integers $n$ and $q$, let
\begin{equation} M(n,q) = \max\{ {n\choose w} (q-1)^w \mid 0\leq w\leq n\}.
\end{equation}
We obviously have that
\begin{equation}
 \sum_{w=0}^n {n\choose w}^2(q-1)^{2w} \leq \sum_{w=0}^n M(n,q) \cdot {n\choose w}(q-1)^w = M(n,q)q^n.
\end{equation}
As
\[ \frac{ {n\choose w}(q-1)^{w}}{{n\choose w-1}(q-1)^{w-1}} = \frac{(q-1)(n-w+1)}{w}, \]
it follows that ${n\choose w}(q-1)^{w}\geq {n\choose w-1}(q-1)^{w-1}$ if and only if  $w\leq\frac{q-1}{q}(n+1)$.
This implies that ${n\choose w}(q-1)^w$ is maximal for $w=w_{max}=\lfloor\frac{q-1}{q}(n+1)\rfloor$, so
$w_{max}\approx \frac{q-1}{q}n$.
Application of Stirling's formula yields that
\[  \log {n\choose \frac{q-1}{q}n} (q-1)^n \approx n\log q + \frac{1}{2}\log\left( \frac{q^2}{q-1}\frac{1}{n}\right) -\log(\sqrt{2\pi}), \]
so that $M(n,q)\approx q^n \cdot \sqrt{\frac{q^2}{q-1}\cdot \frac{1}{n}}. \;\;\;\Box$
\\ \\
By  combining the above lemmas, we obtain Theorem~\ref{Thm:4kexistence} (with $D_q= C_q/(2(q-1)^2) \approx
\frac{q}{2(q-1)^{5/2}})$.

\section{Conclusions and open problems}
By using a construction of MWS codes from QM codes, combined with the Gilbert-Varshamov bound, we obtained
MWS codes of length $n=2^{k\lambda_q}$,
  where $\lambda_q$ grows as $2q^3\log q$.
We provided a non-constructive existence proof  of  $[n,k]_q$ MWS codes of length $n=D_q q^{4k}$, where $D_q$ is
independent of $k$. \\
Combining the existence bound with (\ref{eq:lowbound}), we obtain that
\begin{equation}\label{eq:ineq} \frac{q}{2}\cdot \frac{q^k-1}{q-1} \leq n_{MWS,k,q} \leq D_q q^{4k} .
\end{equation}
It is an interesting problem to reduce the gap in the above inequality.
Another interesting problem is to determine, for a fixed $q$, the exisence or non-existence of the limit
\[ \lim_{k\rightarrow \infty} \frac{1}{k} \log_q (n_{MWS,k,q}).  \]
Note that (\ref{eq:ineq}) implies that for each prime power $q$
\[  \liminf_{k\rightarrow\infty} \frac{1}{k} \log _q(n_{MWS,k,q}) \geq 1
\mbox{ and } \limsup_{k\rightarrow\infty} \frac{1}{k} \log_q(n_{MWS,k,q})\leq 4 . \]

We finally remark that in \cite{Men}, Meneghetti studies the related problem of the existence
 of $[n,k]_q$ codes with an arbitrary number of weights.
He proves the existence for the binary case, and conjectures it for the $q$-ary case.


TITLE: Equivalence of Sobolev Norms via Fourier Transformation
QUESTION [1 upvotes]: I am working through a proof which shows that the usual Sobolev norm $\|\cdot\|_{k,2}$ on $W^{k,2}(\Omega)$ is equivalent to the norm
$$
\|u\|_{H^m} := \| (1 + |\cdot|^2)^{\frac{k}{2}}\mathcal{F}u \|_2
$$
At one point we use the inequality
$$
2^{-k} \leq \frac{1 + |\xi|^{2k}}{(1 + |\xi|^2)^k} \leq 2.
$$
Since the fraction always is smaller than one, the right inequality is not a problem.
But I can't find a way to prove the left inequality.
Is there an elementary proof to this inequality?

REPLY [2 votes]: Yes.  Distinguish the cases $|\xi|\leq 1$ and $|\xi|> 1$ and use crude bounds.


TITLE: Range of a function defined by a matrix
QUESTION [0 upvotes]: I am studying vector calculus, and I encountered a question, which I find the solution to this question somewhat colliding with my knowledge of linear algebra.
The question is this.
Consider the function $f:R^2\rightarrow R^3$ given by $f(x)=Ax$, where A=\begin{bmatrix}
2 & -1 \\
5 & 0 \\
-6 &3
\end{bmatrix} and the vector $x$ in $R^2$ is written as the 2x1 column matrix x=\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
Describe the range of f.

From what I learned in my linear algebra course, f is a linear transformation with a standard matrix as A, and therefore the range of A is a span of the column vectors of A. So I thought that since the column vectors are linearly independent, the range is a plane in $R^3$ spanned by two vectors (2, 5, -6) and (-1, 0, 3).
However, the solution manual had a different answer, it said that $$f(x)=(2x_1-x_2, 5x_1, -6x_1+3x_2)$$
and since $$f_3(x_1, x_2)=-6x_1+3x_2=-3(2x_1-x_2)=-3f_1(x_1, x_2)$$
The range of f would be all of $y=(y_1, y_2, y_3)$ in $R^3$ such that $y_3=-3y_1$.
 I am pretty much confused right now...can anybody explain to me about this result?

REPLY [1 votes]: Your answer and the solution manual's are different ways of describing the same plane.  Notice that your two spanning vectors (and any linear combination of them) have their third component equal to $-3$ times their first component; that is the solution manual's characterization of vectors in the solution plane.


TITLE: The laplace of the integral of Green's function
QUESTION [2 upvotes]: Let $G(\mathbf{x}, \mathbf r)$ be the Green’s function of the Dirichlet problem in a bounded normal domain $\Omega$ . Set $$u(\mathbf r) = \int_{\Omega}
G(\mathbf x, \mathbf r) d^3x.$$Prove that $\Delta u(\mathbf r) = −1 $ for $\mathbf r ∈ \Omega$.
I think that I should choose a mall ball, but I don't know how to do it. Please help me.

REPLY [0 votes]: Note that the Laplacian is with respect to $r$, while the integral is with respect to $x$, so the Laplacian moves straight in:
$$\Delta_r u =  \int_{\Omega} \Delta_rG(x,r)\,d^3x$$
Can you take it from there? What's the definition of the Green's function?
(As a side note, a word of warning: the definition of the Green's function that I assume you must be using is not universally standard; depending on conventions you might get +1 instead of -1.)


text:
\begin{document}
\allowdisplaybreaks
\parindent 0pt

\title{On the optimal constants in the two-sided Stechkin inequalities}

\author{Thomas Jahn}
\address[T. Jahn]{Faculty of Mathematics, Technische Universität Chemnitz, 09107 Chemnitz}
\email{thomas.jahn@mathematik.tu-chemnitz.de}
\thanks{}

\author{Tino Ullrich}
\address[T. Ullrich]{Faculty of Mathematics, Technische Universität Chemnitz, 09107 Chemnitz}
\email{tino.ullrich@mathematik.tu-chemnitz.de}
\thanks{}

\subjclass[2010]{41A25, 46E30}

\keywords{approximation space, Marcinkiewicz interpolation theorem, sparse approximation, Stechkin inequality}

\date{}

\begin{abstract}
We address the optimal constants in the strong and the weak Stechkin inequalities, both in their discrete and continuous variants.
These inequalities appear in the characterization of approximation spaces which arise from sparse approximation or have applications to interpolation theory.
An elementary proof of a constant in the strong discrete Stechkin inequality given by Bennett is provided, and we improve the constants given by Levin and Stechkin and by Copson.
Finally, the minimal constants in the weak discrete Stechkin inequalities and both continuous Stechkin inequalities are presented.
\end{abstract}

\maketitle

\section{Introduction}\label{chap:intro}
In the present paper, we address the minimal constants $c_1(q)$, $C_1(q)$, $c_{1,\infty}(q)$, $C_{1,\infty}(q)$, $\bar{c}_1(q)$, $\bar{C}_1(q)$, $\bar{c}_{1,\infty}(q)$, and $\bar{C}_{1,\infty}(q)>0$ in the inequalities
\begin{equation}
\frac{1}{c_1(q)}\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\leq\sum_{n=1}^\infty a_n\leq C_1(q)\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}},\label{eq:general-stechkin-q-intro}
\end{equation}
\begin{equation}
\frac{1}{c_{1,\infty}(q)}\sup_{n\in\NN}n\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\leq\sup_{n\in\NN} na_n\leq C_{1,\infty}(q)\sup_{n\in\NN}n\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}},\label{eq:general-weak-l1-q-intro}
\end{equation}
\begin{equation}
\frac{1}{\bar{c}_1(q)}\int_0^\infty\lr{\frac{1}{t}\int_t^\infty f(s)^q \dd s}^{\frac{1}{q}}\dd t\leq \int_0^\infty f(t)\dd t\leq \bar{C}_1(q)\int_0^\infty \lr{\frac{1}{t}\int_t^\infty f(s)^q \dd s}^{\frac{1}{q}}\dd t,\label{eq:general-stechkin-continuous-intro}
\end{equation}
and
\begin{equation}
\frac{1}{\bar{c}_{1,\infty}(q)}\sup_{t>0}t\lr{\frac{1}{t}\int_t^\infty f(s)^q \dd s}^{\frac{1}{q}}\leq\sup_{t>0} tf(t)\leq \bar{C}_{1,\infty}(q)\sup_{t>0}t\lr{\frac{1}{t}\int_t^\infty f(s)^q \dd s}^{\frac{1}{q}},\label{eq:general-weak-l1-continuous-intro}
\end{equation}
for sequences $(a_n)_{n\in\NN}$ with $a_1\geq a_2\geq\ldots\geq 0$, monotonically decreasing functions $f:(0,\infty)\to [0,\infty)$, and $1<q\leq\infty$.
These inequalities are henceforth referred to as the \emph{strong discrete}, the \emph{weak discrete}, the \emph{strong continuous}, and the \emph{weak continuous Stechkin inequality}, respectively.

The right-hand side inequalities of \eqref{eq:general-stechkin-q-intro}, \eqref{eq:general-weak-l1-q-intro}, and \eqref{eq:general-weak-l1-continuous-intro} also allow for $q=1$.
Note that in case $q=\infty$, the expressions $\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}$ and $\lr{\frac{1}{t}\int_x^\infty f(s)^q \dd s}^{\frac{1}{q}}$ are replaced by $\sup\setcond{a_k}{k\geq n}=a_n$ and $\sup\setcond{f(s)}{s\geq t}=f(t)$.

Copson \cite[Theorem~2.3]{Copson1928} proves that $C_1(q)\leq q^{\frac{1}{q}}$, cf. also Hardy, Littlewood, and Pólya \cite[Theorem~345]{HardyLiPo1934}.
Levin and Stechkin \cite[{\rus{Д}}.61]{HardyLiPo1948} improve Copson's result \cite[Theorem~2.3]{Copson1928} when $1<q<\infty$, showing that $C_1(q)=(q-1)^\frac{1}{q}$ when $3\leq q<\infty$ and giving upper bounds in the remaining cases.
Stechkin revisits \eqref{eq:general-stechkin-q-intro} for $q=2$ in \cite[\rus{Лемма}~3]{Stechkin1951} and \cite[\rus{Лемма}~1]{Stechkin1955} where it is proved that $C_1(2)\leq\frac{2}{\sqrt{3}}$.
Gao \cite[Theorem~1]{Gao2011} provides further improvement by showing that $C_1(q)=(q-1)^{\frac{1}{q}}$ even for $q_0\leq q<\infty$, where $q_0\approx 2.8855$ is a solution of the equation
$2^{\frac{1}{q-1}}\lr{(q-1)^{\frac{q}{q-1}}-(q-1)}-\lr{1+\frac{3-q}{2}}^{\frac{q}{q-1}}=0$.
De Bruijn \cite[p.~174]{DeBruijn1958} reports that $C_1(2)=1.1064957714$ \enquote{with an error of at most $9$ units at the last decimal place.}

Stechkin \cite[\rus{Лемма}~1]{Stechkin1955} is first to address the constant $c_1(2)$; he asserts $c_1(2)\leq 2$ and conjectures an improvement to $c_1(2)\leq\frac{\piup}{2}$ but proves neither claim; see also \cite[Section~7.4]{DungTeUl2018} for a historical discussion.
The existence of constants validating the inequalities \eqref{eq:general-stechkin-q-intro} and \eqref{eq:general-weak-l1-q-intro} is due to Pietsch \cite[Example~1 on p.~123]{Pietsch1981}, see also \cite[Theorem~4]{DeVore1998}.
Bennett \cite[Theorem~3]{Bennett1988} shows that
\begin{equation*}
c_1(q)=\frac{\piup}{q\sin(\frac{\piup}{q})}
\end{equation*}
in \eqref{eq:general-stechkin-q-intro}, thus confirming Stechkin's conjecture for $q=2$, see \cref{fig:left-stechkin-q} for an illustration.
Hardy, Littlewood, and Pólya \cite[Theorem~337]{HardyLiPo1934} prove
\begin{equation*}
\bar{C}_1(q)=(q-1)^{\frac{1}{q}}
\end{equation*}
in \eqref{eq:general-stechkin-continuous-intro}, depicted in \cref{fig:stechkin-continuous}.

The contribution of the present paper is as follows.
First, we give an alternative proof for the optimality of $c_1(2)=\frac{\piup}{2}$ in \eqref{eq:general-stechkin-q-intro} which uses an elementary insight from convex optimization.

Second, we extend the upper bound for $C_1(q)\leq 2\lr{2\frac{q}{q-1}-1}^{-\frac{q-1}{q}}$ proved by Levin and Stechkin for $\frac{5}{3}\leq q<3$ to $1<q<\infty$ via \cref{result:right-stechkin-q}.
A more detailed analysis of the same argument leads to $C_1(2)\leq 1.1086983$ in \cref{result:right-stechkin-upper-bound}.

Third, we improve the upper bounds for $C_1(q)$ from the literature when $1\leq q\leq\frac{2+\ln(2)}{2-\ln(2)}$ and $q\neq 2$.
Summarizing, the currently best known bounds for the constant $C_1(q)$ are
\begin{equation*}
C_1(q)\begin{cases}\leq\lr{\frac{\ee\ln(2)}{\sqrt{2}}}^{1-\frac{1}{q}},&1\leq q\leq\frac{2+\ln(2)}{2-\ln(2)},\\
\approx 1.1064957714,&q=2,\\
\leq 2\lr{2\frac{q}{q-1}-1}^{-\frac{q-1}{q}},&\frac{2+\ln(2)}{2-\ln(2)}<q< q_0,\\
=(q-1)^{\frac{1}{q}}&q_0\leq q\leq \infty
\end{cases}
\end{equation*}
with $q_0\approx 2.8855$, see \cref{result:right-stechkin-q,result:improved-right-stechkin-q} and \cref{fig:right-stechkin-q} for an illustration.

Fourth, we determine the optimal constants in \eqref{eq:general-weak-l1-q-intro} as
\begin{equation*}
c_{1,\infty}(q)=\zeta(q)^{\frac{1}{q}}\text{\quad and\quad}C_{1,\infty}(q)=\lr{\frac{1}{q}}^{-\frac{1}{q}}\lr{1-\frac{1}{q}}^{-\lr{1-\frac{1}{q}}}
\end{equation*}
where $\zeta(\cdot)$ denotes the Riemann zeta-function, see \cref{result:left-weak-q,result:right-weak-q,fig:left-weak-q,fig:right-weak-q}.
Next, we show that
\begin{equation*}
\bar{c}_1(q)=c_1(q)=\frac{\piup}{q\sin(\frac{\piup}{q})}
\end{equation*}
in \eqref{eq:general-stechkin-continuous-intro}, see \cref{result:left-stechkin-continuous,fig:stechkin-continuous}, as well as
\begin{equation*}
\bar{c}_{1,\infty}(q)=(q-1)^{-\frac{1}{q}}\text{\quad and\quad}\bar{C}_{1,\infty}(q)=C_{1,\infty}(q)=\lr{\frac{1}{q}}^{-\frac{1}{q}}\lr{1-\frac{1}{q}}^{-\lr{1-\frac{1}{q}}},
\end{equation*}
see \cref{result:left-weak-continuous,result:right-weak-continuous,fig:weak-l1-continuous}.

The inequalities discussed in this paper have applications in interpolation theory and nonlinear approximation.
On the one hand, \eqref{eq:general-stechkin-continuous-intro} can be used in the proof of the Marcinkiewicz interpolation theorem, see \cite[Theorem~1.3.1]{BerghLo1976}.
On the other hand \eqref{eq:general-stechkin-q-intro} and \eqref{eq:general-weak-l1-q-intro} play a role in the characterization of the approximation spaces $\mathcal{A}_r^\alpha(\mathcal{H})$, i.e., the set of elements $f$ of the infinite-dimensional separable Hilbert space $\mathcal{H}$ for which the quasi-norm
\begin{equation*}
\mnorm{f}_{\mathcal{A}_r^\alpha(\mathcal{H})}\defeq\begin{cases}
\lr{\sum_{n=1}^\infty(n^\alpha E_n(f)_{\mathcal{H}})^r \frac{1}{n}}^{\frac{1}{r}},&0<r<\infty,\\
\sup_{n\in\NN}n^\alpha E_n(f)_{\mathcal{H}},&r=\infty
\end{cases}
\end{equation*}
is finite.
Here, $\alpha>0$ and $E_n(f)$ denote the infimal distance of $f$ to elements of the form $\sum_{k\in\Lambda}\lambda_k e_k$, where $\Lambda\subset \NN$ is a set of cardinality $n-1$ and $(e_k)_{k=1}^\infty$ is an orthonormal basis of $\mathcal{H}$.
The consequences for the optimal constants in the inequalities stated by DeVore in \cite[Theorem~4]{DeVore1998} on sparse approximation in infinite-dimensional separable real Hilbert spaces are outlined in \cref{chap:devore}.

A recurring technique in this paper is that we prove the inequalities under consideration for finite sequences first, which then yields the general claim through a limiting process.
These finite-dimensional versions will be used to gain some geometric insight to \eqref{eq:general-stechkin-q-intro}.

\section{The strong discrete Stechkin inequality}\label{chap:stechkin}
In this section, we are concerned with the minimal constants $c_1(q)$ and $C_1(q)>0$ in the inequality
\begin{equation}
\frac{1}{c_1(q)}\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\leq \sum_{n=1}^\infty a_n\leq C_1(q)\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\label{eq:general-stechkin-q}
\end{equation}
for sequences $(a_n)_{n\in\NN}$ with $a_1\geq a_2\geq\ldots\geq 0$ and for $1<q\leq\infty$ with the appropriate modification for $q=\infty$, as indicated in \cref{chap:intro}.
The monotonicity assumption on $(a_n)_{n\in\NN}$ gives $\sup\setcond{a_k}{k\geq n}=a_n$, so $c_1(\infty)=C_1(\infty)=1$.
For $q=1$, we have $C_1(1)=1$ because
\begin{equation*}
\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty \sum_{n=1}^k \frac{a_k}{k}\leq \sum_{k=1}^\infty \sum_{n=1}^k \frac{a_k}{n}= \sum_{n=1}^\infty \frac{1}{n}\sum_{k=n}^\infty a_k
\end{equation*}
which holds as an equality when $a_n=0$ for all $n\geq 2$.

\subsection{On the optimal lower constant}
We give a rather elementary proof of the optimality of $c_1(2)=\frac{\piup}{2}$ in \eqref{eq:general-stechkin-q}, which is due to Bennett \cite[Theorem~3]{Bennett1988}.
Our proof uses an elementary insight from convex optimization.
\begin{Satz}\label{result:left-stechkin}
The minimal constant $c_1(2)>0$ for which
\begin{equation*}
\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^2}^{\frac{1}{2}}\leq c_1(2)\sum_{n=1}^\infty a_n
\end{equation*}
holds for all sequences $(a_n)_{n\in\NN}$ with $a_1\geq a_2\geq\ldots\geq 0$ is $c_1(2)=\frac{\piup}{2}$.
\end{Satz}
\begin{proof}
\emph{Step 1. We prove the claim for finite sequences.}
Consider
\begin{align*}
&\norel\sup\setcond{\frac{\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^2}^{\frac{1}{2}}}{\sum_{n=1}^\infty a_n}}{a_1\geq a_2\geq\ldots\geq 0,a_{N+1}=0,\sum_{n=1}^\infty a_n<\infty}\\
&=\sup\setcond{\frac{\sum_{n=1}^N\lr{\frac{1}{n}\sum_{k=n}^N a_k^2}^{\frac{1}{2}}}{\sum_{n=1}^N a_n}}{a_1\geq a_2\geq\ldots\geq 0,a_{N+1}=0}\\
&=\sup\setcond{\sum_{n=1}^N\lr{\frac{1}{n}\sum_{k=n}^N a_k^2}}{(a_n)_{n\in\NN}\in\Delta_N}.
\end{align*}
The set $\Delta_N\defeq\setcond{(a_n)_{n\in\NN}}{a_1\geq a_2\geq\ldots\geq 0,a_{N+1}=0,\sum_{n=1}^N a_n=1}$ is determined by a single linear equality and $n$ linear inequalities in $a_1,\ldots,a_n$ and thus is a $(n-1)$-dimensional simplex in its $N$-dimensional linear span $V_N$.
Therefore, the restriction of the convex function
\begin{align*}
V_N&\to\RR,\\(a_n)_{n\in\NN}&\mapsto\sum_{n=1}^N\lr{\frac{1}{n}\sum_{k=n}^N a_k^2}^{\frac{1}{2}}
\end{align*}
to $\Delta_N$ attains its supremum at one of the vertices of $\Delta_N$.
Similarly to the discussion in \cite[Section~2.1]{FoucartRa2013}, the vertices of $\Delta_N$ are precisely those points for which all but one of the defining inequalities are actually equalities.
This means that for each of the $N$ vertices of $\Delta_N$, there is a number $k_0\in\setn{1,\ldots,N}$ such that $a_1=\ldots =a_{k_0}> a_{k_0+1}=\ldots=a_N=0$.
Taking $\sum_{n=1}^Na_n=1$ into account, we obtain $a_1=\ldots =a_{k_0}=\frac{1}{k_0}$ and
\begin{equation*}
\sum_{n=1}^N\lr{\frac{1}{n}\sum_{k=n}^N a_k^2}^{\frac{1}{2}}=\sum_{n=1}^{k_0}\lr{\frac{1}{n}\sum_{k=n}^{k_0} \frac{1}{k_0^2}}^{\frac{1}{2}}=\sum_{n=1}^{k_0}n^{-\frac{1}{2}}\lr{\frac{k_0-n+1}{k_0^2}}^{\frac{1}{2}}.
\end{equation*}
Therefore
\begin{align*}
&\norel\sup\setcond{\frac{\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^2}^{\frac{1}{2}}}{\sum_{n=1}^\infty a_n}}{a_1\geq a_2\geq\ldots\geq 0,a_{N+1}=0,\sum_{n=1}^\infty a_n<\infty}\\
&=\sup\setcond{\sum_{n=1}^N\lr{\frac{1}{n}\sum_{k=n}^N a_k^2}^{\frac{1}{2}}}{(a_n)_{n\in\NN}\in\Delta_N}\\
&=\sup\setcond{\sum_{n=1}^{k_0}n^{-\frac{1}{2}}\lr{\frac{k_0-n+1}{k_0^2}}^{\frac{1}{2}}}{k_0\in\setn{1,\ldots,N}}.
\end{align*}
Clearly, this quantity is monotonically increasing in $N$ because not only the set $\setcond{(a_n)_{n\in\NN}}{a_1\geq a_2\geq\ldots\geq 0,a_{N+1}=0}$ is, but also $\Delta_N$ and its vertex set are.
We will show that the sequence $\lr{\sum_{n=1}^{k_0}n^{-\frac{1}{2}}\lr{\frac{k_0-n+1}{k_0^2}}^{\frac{1}{2}}}_{k_0\in\NN}$ is bounded above and the supremum is $\frac{\piup}{2}$.
Then we automatically know that
\begin{align*}
\frac{\piup}{2}&=\sup_{k_0\in\NN}\sum_{n=1}^{k_0}n^{-\frac{1}{2}}\lr{\frac{k_0-n+1}{k_0^2}}^{\frac{1}{2}}\\
&=\sup_{N\in\NN}\sup\setcond{\sum_{n=1}^{k_0}n^{-\frac{1}{2}}\lr{\frac{k_0-n+1}{k_0^2}}^{\frac{1}{2}}}{k_0\in\setn{1,\ldots,N}}\\
&=\sup_{N\in\NN}\sup\setcond{\frac{\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^2}^{\frac{1}{2}}}{\sum_{n=1}^\infty a_n}}{a_1\geq a_2\geq\ldots\geq 0,a_{N+1}=0}.
\end{align*}
The function $g:(0,k_0+1)\to [0,\infty)$, $g(t)\defeq t^{-\frac{1}{2}}\lr{\frac{k_0-t+1}{k_0^2}}^{\frac{1}{2}}$ is monotonically decreasing, so
\begin{equation}
f_{k_0}\defeq\int_1^{k_0+1}g(t)\dd t\leq\sum_{n=1}^{k_0} g(n) \leq g(1)+\int_1^{k_0} g(t)\dd t\eqdef h_{k_0}.\label{eq:sum-integral-estimate}
\end{equation}

For the computation of the antiderivative $\int t^{-\frac{1}{2}}(k_0-t+1)^{\frac{1}{2}}\dd t$, the change of variables $u=\sqrt{\frac{t}{k_0-t+1}}$ or $t=\frac{u^2(k_0+1)}{u^2+1}$ yields $\frac{\dd t}{\dd u}=\frac{2u(k_0+1)}{(u^2+1)^2}$ and
\begin{align*}
&\int t^{-\frac{1}{2}}(k_0-t+1)^{\frac{1}{2}}\dd t\\
&=2(k_0+1)\int \frac{1}{(u^2+1)^2}\dd u\\

&=2(k_0+1)\lr{\frac{u}{2(u^2+1)}+\frac{1}{2}\arctan(u)}\\

&=t^{\frac{1}{2}}(k_0-t+1)^{\frac{1}{2}}+(k_0+1)\arctan\lr{\lr{\frac{t}{k_0-t+1}}^{\frac{1}{2}}}.
\end{align*}
Plugging in the integration bounds, we arrive at
\begin{align*}
\int_1^{k_0+1} t^{-\frac{1}{2}}(k_0-t+1)^{\frac{1}{2}}\dd t&=(k_0+1)\frac{\piup}{2}-k_0^{\frac{1}{2}}-(k_0+1)\arctan\lr{k_0^{-\frac{1}{2}}},\\

\int_1^{k_0} t^{-\frac{1}{2}}(k_0-t+1)^{\frac{1}{2}}\dd t&=\lr{\arctan\lr{k_0^{\frac{1}{2}}}-\arctan\lr{k_0^{-\frac{1}{2}}}}(k_0+1).
\end{align*}
It follows that $\lim_{k_0\to\infty}f_{k_0}=\lim_{k_0\to\infty}h_{k_0}=\frac{\piup}{2}$.
Now, if we can show that the sequences $(f_{k_0})_{k_0\in\NN}$ and $(h_{k_0})_{k_0\in\NN}$ are monotonically increasing, then \eqref{eq:sum-integral-estimate} implies
\begin{align*}
\frac{\piup}{2}&=\lim_{k_0\to\infty}f_{k_0}=\sup_{k_0\in\NN}f_{k_0}\leq \sup_{k_0\in\NN}\sum_{n=1}^{k_0}n^{-\frac{1}{2}}\lr{\frac{k_0-n+1}{k_0^2}}^{\frac{1}{2}}\\
&\leq\sup_{k_0\in\NN}h_{k_0}=\lim_{k_0\to\infty}h_{k_0}=\frac{\piup}{2}
\end{align*}
and we are done.
Indeed, one has
\begin{align*}
\frac{\p}{\p x}\int_1^{x+1} \lr{\frac{x-t+1}{tx^2}}^{\frac{1}{2}}\dd t&=\frac{\p}{\p x}\frac{(x+1)\frac{\piup}{2}-x^{\frac{1}{2}}-(x+1)\arctan\lr{x^{-\frac{1}{2}}}}{x}\\
&=\frac{x^{\frac{1}{2}}-\arctan\lr{x^{\frac{1}{2}}}}{x^2}>0
\end{align*}
and
\begin{align*}
&\norel\frac{\p}{\p x}\lr{x^{-\frac{1}{2}}+\int_1^x \lr{\frac{x-t+1}{tx^2}}^{\frac{1}{2}}\dd t}\\
&=\frac{\p}{\p x}\lr{x^{-\frac{1}{2}}+\lr{\arctan\lr{x^{\frac{1}{2}}}-\arctan\lr{x^{-\frac{1}{2}}}}\frac{x+1}{x}}\\
&=\frac{x^{\frac{1}{2}}-4\arctan\lr{x^{\frac{1}{2}}}+\piup}{2x^2}>0
\end{align*}
for all $x\geq 1$.
For the latter claim, note that the function $(0,\infty)\ni x\mapsto \sqrt{x}-4\arctan\lr{\sqrt{x}}+\piup$ has a global minimizer at $x=3$ with minimum $\sqrt{3}-\frac{\piup}{3}>0$, which can be read off the signs of its derivative $x\mapsto\frac{x-3}{2\sqrt{x}(x+1)}$.

\emph{\qquad Step 2. We prove the claim for all sequences.}
It remains to show that
\begin{equation*}
\frac{\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^2}^{\frac{1}{2}}}{\sum_{n=1}^\infty a_n}\leq\frac{\piup}{2}
\end{equation*}
for all sequences $(a_n)_{n\in\NN}$ with $a_1\geq a_2\geq \ldots\geq 0$ and $\sum_{n=1}^\infty a_n\leq\infty$.
Let $\eps>0$.
Then there exists $N_1,N_2,N_3\in\NN$ such that
\begin{equation*}
\sum_{n=N+1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^2}^{\frac{1}{2}}<\frac{\eps}{2}
\end{equation*}
for all $N> N_1$, and
\begin{equation*}
\sum_{k=\floor{\frac{N}{2}}}^\infty a_k<\frac{\eps}{8}
\end{equation*}
for all $N> N_2$, and
\begin{equation*}
\sqrt{N}\floor{\frac{N}{2}}^{-\frac{1}{2}}\leq 2
\end{equation*}
for all $N>N_3$.
Also, for fixed $N\in\NN$ and $M\geq N+1$, apply \cite[Proposition~2.3]{FoucartRa2013} with $p=1$, $q=2$, and $s=\frac{N}{2}$ to the sequence $(a_n)_{n\geq\floor{\frac{N}{2}}}$ to obtain
\begin{equation*}
\lr{\sum_{k=N+1}^\infty a_k^2}^{\frac{1}{2}}\leq \floor{\frac{N}{2}}^{-\frac{1}{2}}\sum_{k=\floor{\frac{N}{2}}}^\infty a_k.
\end{equation*}
For $N>\max\setn{N_1,N_2,N_3}$, we conclude
\begin{align*}
\sum_{n=1}^N\lr{\frac{1}{n}\sum_{k=N+1}^\infty a_k^2}^{\frac{1}{2}}&\leq\sum_{n=1}^N n^{-\frac{1}{2}} \floor{\frac{N}{2}}^{-\frac{1}{2}}\sum_{k=\floor{\frac{N}{2}}}^\infty a_k\leq\int_0^N x^{-\frac{1}{2}}\dd x\floor{\frac{N}{2}}^{-\frac{1}{2}}\sum_{k=\floor{\frac{N}{2}}}^\infty a_k\\
&=2\sqrt{N}\floor{\frac{N}{2}}^{-\frac{1}{2}}\sum_{k=\floor{\frac{N}{2}}}^\infty a_k\leq 4\sum_{k=\floor{\frac{N}{2}}}^\infty a_k<\frac{\eps}{2}
\end{align*}
and
\begin{align*}
&\norel\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^2}^{\frac{1}{2}}\\
&=\sum_{n=1}^N\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^2}^{\frac{1}{2}}+\sum_{n=N+1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^2}^{\frac{1}{2}}\\
&\leq \sum_{n=1}^N\lr{\frac{1}{n}\sum_{k=n}^N a_k^2}^{\frac{1}{2}}+\sum_{n=1}^N\lr{\frac{1}{n}\sum_{k=N+1}^\infty a_k^2}^{\frac{1}{2}}+\sum_{n=N+1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^2}^{\frac{1}{2}}\\
&<\frac{\piup}{2}\sum_{n=1}^Na_n+\frac{\eps}{2}+\frac{\eps}{2}\\
&=\frac{\piup}{2}\sum_{n=1}^\infty a_n+\eps.
\end{align*}
Taking the limit $\eps\downarrow 0$ proves the assertion.
\end{proof}

The precise values for $c_1(q)$ from \cite[Theorem~3]{Bennett1988} are illustrated in \cref{fig:left-stechkin-q}.
\begin{figure}
\begin{center}
\begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=4.0cm,y=0.5cm]
\draw[black!50] (0,1)--(1,1);
\draw[black!50] (0,3)--(1,3);
\draw[black!50] (0,5)--(1,5);
\draw[black!50] (0.5,0)--(0.5,5);
\draw[black!50] (1,0)--(1,5);
\draw[line width=1pt,->] (0,0)--(1.2,0);
\draw[line width=1pt,->] (0,0)--(0,6.5);
\draw (0,6.5) node[left]{$c_1(q)$};
\draw (1.2,0) node[below]{$1/q$};
\draw (0,1) node[left]{$1$};
\draw (0,3) node[left]{$3$};
\draw (0,5) node[left]{$5$};
\draw (1,0) node[below]{$1$};
\draw (0.5,0) node[below]{$0.5$};

\draw[color=black,line width=1pt,smooth,samples=200,domain=0.001:0.8] plot(\x,{3.141592653589793*(\x)/sin((3.141592653589793*(\x))*180/pi)});
\end{tikzpicture}
\end{center}
\caption{The function $1/q\mapsto \frac{\piup}{q\sin(\frac{\piup}{q})}$.\label{fig:left-stechkin-q}}
\end{figure}

\begin{Bem}
Our proof of \cref{result:left-stechkin} already emphasizes the importance of finite sequences for our considerations which will be encountered again in the proof of \cref{result:right-weak-q}.
Let us draw a geometric picture.
Fix $N\in \NN$ and $q\in (1,\infty)$.
The quantities
\begin{align*}
\mnorm{a}_1&=\sum_{n=1}^N\abs{a_n}\\
\intertext{and}
\gamma_N(a)&\defeq\sum_{n=1}^N\lr{\frac{1}{n}\sum_{k=n}^N \abs{a_k}^q}^{\frac{1}{q}}
\end{align*}
define norms on $\RR^N$, whose closed unit balls shall be denoted by $B_1^N$ and $B_{1,q}^N$, respectively.
\cref{fig:q-balls} illustrates $B_{1,q}^N$ for several values of $q$.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=2cm,y=2cm]
\clip(-1.1,-0.75) rectangle (1.1,0.75);

\draw [dotted,domain=-1.1:1.1] plot(\x,\x);
\draw [dotted,domain=-1.1:1.1] plot(\x,{-\x});
\draw [dotted,domain=-1.1:1.1] plot(\x,0);
\draw [dotted,domain=-1.1:1.1] plot(0,\x);

\draw (0,0.70710678)--(1,0)--(0,-0.70710678)--(-1,0)--cycle;

\draw (1,0)--(0.5,0.5)--(-0.5,0.5)--(-1,0)--(-0.5,-0.5)--(0.5,-0.5)--cycle;

\draw plot[samples=250,domain=-1:1)] (\x,{-sqrt(4-2*(\x)^2)+sqrt(2)})--plot[samples=250,domain=1:-1)] (\x,{sqrt(4-2*(\x)^2)-sqrt(2)});

\begin{scope}[yscale=1,xscale=1]
\input{q3.tex}
\end{scope}
\begin{scope}[yscale=1,xscale=-1]
\input{q3.tex}
\end{scope}
\begin{scope}[yscale=-1,xscale=-1]
\input{q3.tex}
\end{scope}
\begin{scope}[yscale=-1,xscale=1]
\input{q3.tex}
\end{scope}

\begin{scope}[yscale=1,xscale=1]
\input{q6.tex}
\end{scope}
\begin{scope}[yscale=1,xscale=-1]
\input{q6.tex}
\end{scope}
\begin{scope}[yscale=-1,xscale=-1]
\input{q6.tex}
\end{scope}
\begin{scope}[yscale=-1,xscale=1]
\input{q6.tex}
\end{scope}

\begin{scope}[yscale=1,xscale=1]
\input{q1-4.tex}
\end{scope}
\begin{scope}[yscale=1,xscale=-1]
\input{q1-4.tex}
\end{scope}
\begin{scope}[yscale=-1,xscale=-1]
\input{q1-4.tex}
\end{scope}
\begin{scope}[yscale=-1,xscale=1]
\input{q1-4.tex}
\end{scope}
\end{tikzpicture}
\end{center}\caption{The unit balls of the norms $\gamma_N$ for $N=2$ and $q\in\setn{1,1.4,2,3,6,\infty}$.\label{fig:q-balls}}
\end{figure}

The inequality
\begin{equation}
\frac{1}{c_1(q)}\sum_{n=1}^N\lr{\frac{1}{n}\sum_{k=n}^N a_k^q}^{\frac{1}{q}}\leq\sum_{n=1}^N a_n\leq C_1(q)\sum_{n=1}^N\lr{\frac{1}{n}\sum_{k=n}^N a_k^q}^{\frac{1}{q}}\label{eq:finite-dimensional-general-stechkin}
\end{equation}
for all $a=(a_1,\ldots,a_N)\in\RR^N$ with $a_1\geq \ldots a_N\geq 0$ then translates to the following chain of set inclusions of convex bodies:
\begin{equation}
\frac{1}{c_1(q)}(B_1^N\cap K_N)\subset B_{1,q}^N\cap K_N\subset C_1(q)(B_1^N\cap K_N).\label{eq:optimal-containment-stechkin}
\end{equation}
Here $K_N\defeq \setcond{(a_1,\ldots,a_N)}{a_1\geq \ldots a_N\geq 0}$.
For understanding the shape of the convex bodies $B_{1,q}^N$, we note that norm $\gamma_N$ is the pointwise sum of the functions $\gamma_{n,N}:\RR^n\to\RR$ given by $\gamma_{n,N}(a)=\lr{\frac{1}{n}\sum_{k=n}^N a_k^q}^{\frac{1}{q}}$ for $n\in\setn{1,\ldots,N}$.
With ${}^\circ$ denoting the polar set with respect to the standard inner product, it follows that $B_{1,q}^N=\lr{B_{1,N}^\circ+\ldots+B_{N,N}^\circ}^\circ$, which is similar to the construction of the harmonic mean of convex bodies introduced by Firey in \cite{Firey1961}.
For $N=q=2$, the chain of set inclusions stated in \eqref{eq:optimal-containment-stechkin} with the optimal constants is illustrated in \cref{fig:optimal-containment-stechkin}.
This figure may also be used to convince oneself that for the left-hand side inequality in \eqref{eq:finite-dimensional-general-stechkin}, it is relevant to have $a\in K_N$, and thus monotonicity is also relevant for the left-hand side inequality in \eqref{eq:general-stechkin-q-intro}.
\end{Bem}

\begin{figure}[h!]
\begin{center}
\subfigure{
\begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=2cm,y=2cm]
\clip(-1.1,-1.1) rectangle (1.1,1.1);

\draw [dotted,domain=-1.1:1.1] plot(\x,\x);
\draw [dotted,domain=-1.1:1.1] plot(\x,{-\x});
\draw [dotted,domain=-1.1:1.1] plot(\x,0);
\draw [dotted,domain=-1.1:1.1] plot(0,\x);
\draw plot[samples=250,domain=-1:1)] (\x,{-sqrt(4-2*(\x)^2)+sqrt(2)})--plot[samples=250,domain=1:-1)] (\x,{sqrt(4-2*(\x)^2)-sqrt(2)});
\draw (0,1)--(1,0)--(0,-1)--(-1,0)--cycle;
\end{tikzpicture}
}\qquad
\subfigure{
\begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=3.666cm,y=3.666cm]
\clip(-0.1,-0.1) rectangle (1.1,1.1);

\draw [dotted,domain=-1.5:1.5] plot(\x,\x);
\draw [dotted,domain=-1.5:1.5] plot(\x,{-\x});
\draw [dotted,domain=-1.5:1.5] plot(\x,0);
\draw [dotted,domain=-1.5:1.5] plot(0,\x);
\draw (0,0)--plot[samples=250,domain={sqrt(2)/3}:1)] (\x,{sqrt(4-2*(\x)^2)-sqrt(2)})--cycle;
\draw (0,0)--(0.5,0.5)--(1,0)--cycle;
\draw[scale={2*sqrt(2)/3}] (0,0)--(0.5,0.5)--(1,0)--cycle;
\draw[scale={sqrt(2)*(sqrt(3)-1)}] (0,0)--(0.5,0.5)--(1,0)--cycle;
\end{tikzpicture}
}
\end{center}\caption{The unit balls of the norms $\gamma_N$ and $\mnorm{\cdot}_1$ for $N=q=2$ (left) and an optimally scaled version of their intersections with the cone $K_N$ (right).\label{fig:optimal-containment-stechkin}}
\end{figure}

\subsection{On the optimal upper constant}
In this section, we improve the known upper bounds on $C_1(q)$ for $1<q\leq\frac{2+\ln(2)}{2-\ln(2)}$.
The following result adapts Stechkin's proof technique in \cite[\rus{Лемма}~3]{Stechkin1951} and produces upper bounds for $C_1(q)$ from auxiliary sequences whose entrywise inverses in $\ell_{q^\prime}$ where $q^\prime$ is the Hölder conjugate of $q$.
\begin{Prop}\label{result:right-stechkin-q}
Let $1<q<\infty$, set $q^\prime=\frac{q}{q-1}$, and assume that $b=(b_n)_{n\in\NN_0}$ is a strictly monotonically increasing sequence with $b_0=0$ and $\sum_{n=1}^\infty\frac{1}{b_n^{q^\prime}}<\infty$.
Then for all $(a_n)_{n\in\NN}\in\ell_q$, we have
\begin{equation*}
\sum_{n=1}^\infty a_n\leq C_b(q)\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}
\end{equation*}
with $C_b(q)\defeq\sup\setcond{\lr{n^{\frac{q^\prime}{q}}(b_n-b_{n-1})^{q^\prime}\sum_{k=n}^\infty\frac{1}{b_k^{q^\prime}}}^{\frac{1}{q^\prime}}}{n\in\NN}$.
\end{Prop}
\begin{proof}
Consider
\begin{align*}
&\norel\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{a_n}{b_n}b_n=\sum_{n=1}^\infty\frac{a_n}{b_n}\sum_{k=1}^n(b_k-b_{k-1})\\
&=\sum_{n=1}^\infty(b_n-b_{n-1})\sum_{k=n}^\infty\frac{a_k}{b_k}\leq\sum_{n=1}^\infty(b_n-b_{n-1})\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\lr{n^{\frac{q^\prime}{q}}\sum_{k=n}^\infty\frac{1}{b_k^{q^\prime}}}^{\frac{1}{q^\prime}}\\
&=\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\lr{n^{\frac{q^\prime}{q}}(b_n-b_{n-1})^{q^\prime}\sum_{k=n}^\infty\frac{1}{b_k^{q^\prime}}}^{\frac{1}{q^\prime}}\leq C_b(q)\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}
\end{align*}
where $C_b(q)$ is chosen as stated in the assertion.
\end{proof}
We investigate the choice $b_k=(k(k+1))^p$ for $p\in (\frac{1}{2q^\prime},1]$.
If we set
\begin{equation*}
A_n\defeq n^{\frac{q^\prime}{q}}((n(n+1))^p-(n(n-1))^p)^{q^\prime}\sum_{k=n}^\infty\frac{1}{(k(k+1))^{q^\prime p}}
\end{equation*}
for $n\in\NN$, then $C(p,q)\defeq\sup\setconds{A_n^\frac{1}{q^\prime}}{n\in\NN}$ the constant defined in \cref{result:right-stechkin-q} for our particular choice of the sequence $(b_n)_{n\in\NN}$.
\begin{Lem}\label{result:upper-bound-q}
Let $p>\frac{1}{2q^\prime}$.
For all $n\in\NN$, we have
\begin{equation*}
A_n\leq\lr{\frac{(n+1)^p-(n-1)^p}{n^{p-1}}}^{q^\prime}\frac{1}{2q^\prime p-1}.\label{eq:upper-bound-q}
\end{equation*}
\end{Lem}
\begin{proof}
The assertion follows from
\begin{align*}
\sum_{k=n}^\infty\frac{1}{b_k^{q^\prime}}&=\sum_{k=n}^\infty\frac{1}{(k(k+1))^{q^\prime p}}=\sum_{k=n}^\infty\lr{\int_k^{k+1}x^{-2}\dd x}^{q^\prime p}\\
&\leq \sum_{k=n}^\infty\int_k^{k+1}x^{-2q^\prime p}\dd x=\frac{1}{2q^\prime p-1}n^{-2q^\prime p+1},
\end{align*}
for which $p>\frac{1}{2q^\prime}$ is crucial.
\end{proof}

We can say even more about the right-hand side of \eqref{eq:upper-bound-q}.
\begin{Lem}\label{result:monotonicity}
Let $0<p\leq 1$.
The sequence $(A_n^\prime)_{n\in\NN}$ defined by
\begin{equation*}
A_n^\prime\defeq\frac{(n+1)^p-(n-1)^p}{n^{p-1}}
\end{equation*}
is monotonically decreasing.
\end{Lem}
\begin{proof}
For $p=1$ the claim is trivial.
Otherwise, consider the functions $f_1,g_1:(0,\infty)\to\RR$ defined by $g_1(x)=x^p$ and
\begin{equation*}
f_1(x)=\frac{g_1(x+1)-g_1(x-1)}{x^{p-1}}.
\end{equation*}
Note that $f_1(n)=A_n^\prime$.
We will show that $f_1$ is monotonically decreasing on $[2,\infty)$ and that $A_1^\prime>A_2^\prime$.
The Taylor expansion of $g_1$ at $x\geq 2$ is given by
\begin{equation*}
g_1(x+h)=\sum_{k=0}^n \frac{h^k}{k!}x^{p-k} \prod_{m=0}^{k-1}(p-m)+R(x,h,n),
\end{equation*}
and the corresponding approximation error is
\begin{equation*}
R(x,h,n)\defeq \int_x^{x+h}\frac{(x+h-t)^n}{n!}t^{p-n-1}\prod_{m=0}^n(p-m)\dd t.
\end{equation*}
Next, note that
\begin{align*}
\abs{R(x,1,n)}&\leq\int_x^{x+1}\frac{\abs{x+1-t}^n}{n!}t^{p-n-1}\prod_{m=0}^n\abs{p-m}\dd t\\
&\leq\int_x^{x+1}t^{p-n-1}\dd t=\frac{1}{p-n}((x+1)^{p-n}-x^{p-n})\leq\frac{1}{n-p}\\
\intertext{and}
\abs{R(x,-1,n)}&\leq\int_{x-1}^x\frac{(x-1-t)^n}{n!}t^{p-n-1}\prod_{m=0}^n(p-m)\dd t\\
&\leq\int_{x-1}^x t^{p-n-1}\dd t=\frac{1}{p-n}(x^{p-n}-(x-1)^{p-n})\leq\frac{1}{n-p}.\\
\end{align*}
This follows from $\frac{1}{n!}\prod_{m=0}^n\abs{p-m}=p\prod_{m=1}^n\frac{m-p}{m}\leq 1$ and $\abs{x+h+t}^n \leq 1$ for $h\in\setn{-1,1}$ and $t$ between $x$ and $x+h$.
As $\lim_{n\to\infty}\frac{-1}{p-n}=0$, we know that $\lim_{n\to\infty}R(x,h,n)=0$ for $h\in\setn{-1,1}$, and we may write
\begin{align*}
g_1(x+1)&=\sum_{k=0}^\infty\frac{1}{k!}x^{p-k} \prod_{m=0}^{k-1}(p-m)\\
\intertext{and}
-g_1(x-1)&=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{k!}x^{p-k} \prod_{m=0}^{k-1}(p-m).
\end{align*}
Setting $h_n\defeq \frac{1}{(2n-1)!}\prod_{m=0}^{2n-2}(p-m)$ for $n\in\NN$ and noticing $h_n>0$, this gives
\begin{align*}
g_1(x+1)-g_1(x-1)&=2\sum_{n=1}^\infty\frac{1}{(2n-1)!}x^{p-(2n-1)} \prod_{m=0}^{2n-2}(p-m)\\
&=2px^{p-1}+2\sum_{n=2}^\infty\frac{1}{(2n-1)!}x^{p-(2n-1)} \prod_{m=0}^{2n-2}(p-m)\\
&=2px^{p-1}+2\sum_{n=2}^\infty h_n x^{p-(2n-1)}.
\end{align*}
As a function of $x$, the expression
\begin{equation*}
f_1(x)=\frac{g_1(x+1)-g_1(x-1)}{x^{p-1}}=2p+\sum_{n=2}^\infty\frac{h_n}{x^{2n-2}}
\end{equation*}
is thus monotonically decreasing on $[2,\infty)$.
In order to show that
\begin{equation}
A_1^\prime=2^p\geq \frac{3^p-1}{2^{p-1}}=A_2^\prime \label{eq:monotonicity-start}
\end{equation}
for all $p\in (0,1]$, consider the functions $f_2,g_2:\RR\to\RR$ defined by $f_2(x)=3^x-1$ and $g_2(x)=2^{2x-1}$.
Then $f_2^\prime(x)=3^x\ln(3)$, $g_2^\prime(x)=4^x\ln(2)$, $f(1)=g(1)$, and $f_2,g_2,f_2^\prime$, and $g_2^\prime$ are monotonically increasing.
Therefore $f_1^\prime(1)>g_2^\prime(1)$ shows that $f(x)\leq g(x)$ for all $x\leq 1$ (with equality only for $x=1$).
This implies \eqref{eq:monotonicity-start}.
\end{proof}
In \cref{result:improved-right-stechkin-q}, we will show how \cref{result:upper-bound-q,result:monotonicity} and an in some sense optimal choice of $p$ in $b_n=(n(n+1))^p$ yield $C_1(2)\leq \sqrt{\frac{\ee\ln(2)}{\sqrt{2}}}$ which is already an improvement to Stechkin's $C_1(2)\leq\frac{2}{\sqrt{3}}$.
More detailed analysis for a specific choice of $p$ in this construction yields further improvement when $q=2$, coinciding with de Bruijn's result \cite[p.~174]{DeBruijn1958} in the first two decimal places.
\begin{Kor}\label{result:right-stechkin-upper-bound}
The minimal constant $C_1(2)>0$ for which
\begin{equation*}
\sum_{n=1}^\infty a_n\leq C_1(2)\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^2}^{\frac{1}{2}}
\end{equation*}
holds for all $(a_n)_{n\in\NN}\in\ell_2$ is at most $1.1086983$.
\end{Kor}
\begin{proof}
For $N\in\NN$ with $N\geq 2$, \cref{result:upper-bound-q,result:monotonicity} give
\begin{align}
C(p,2)&\leq \max\setn{\max\limits_{n=1,\ldots,N-1}\sqrt{A_n},\sup\limits_{n\geq N}\frac{1}{\sqrt{4p-1}}\frac{(n+1)^p-(n-1)^p}{n^{p-1}}}\nonumber\\
&=\max\setn{\max\limits_{n=1,\ldots,N-1}\sqrt{A_n},\frac{1}{\sqrt{4p-1}}\frac{(N+1)^p-(N-1)^p}{N^{p-1}}}.\label{eq:improved-constant}
\end{align}
The last expression evaluated at $N=100$ and $p=0.88$ can be bounded above by $1.1086983$.
For the computation of $A_1,\ldots,A_{100}$, the series $\sum_{k=n}^\infty\frac{1}{(k(k+1))^{2p}}$ have been truncated to $\sum_{k=n}^M \frac{1}{(k(k+1))^{2p}}$ with $M=2\cdot 10^5$.
The proof of \cref{result:upper-bound-q} then shows $\sum_{k=M}^\infty\frac{1}{(k(k+1))^{2p}}\leq \frac{1}{4p-1}M^{-4p+1}$
For $n\in\NN$, we also have
\begin{equation*}
n((n(n+1))^p-(n(n-1))^p)^2=n^{4p-1}\lr{\frac{(n+1)^p-(n-1)^p}{n^{p-1}}}^2\leq n^{4p-1}
\end{equation*}
by \cref{result:monotonicity}.
The truncation error is therefore at most $\frac{4p-1}N^{4p-1}M^{-4p+1}\approx 4.8018\cdot 10^{-9}$, which can be neglected in the computation of the maximum in \eqref{eq:improved-constant}.
\end{proof}

For $p=1$, \cref{result:upper-bound-q} gives $A_n^{\frac{1}{q^\prime}}\leq 2(2q^\prime-1)^{-\frac{1}{q^\prime}}$, and \cref{result:right-stechkin-q} yields
\begin{equation}
\sum_{n=1}^\infty a_n\leq 2(2q^\prime-1)^{-\frac{1}{q^\prime}}\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\label{eq:stechkin-choice-bound-q}
\end{equation}
This extends the bound obtained by Levin and Stechkin \cite[{\rus{Д}}.61]{HardyLiPo1948} for $\frac{5}{3}\leq q< 3$ to arbitrary $1<q<\infty$.
The $q=2$ case $C_1(2)\leq \frac{2}{\sqrt{3}}$ has been addressed again in \cite[\rus{Лемма}~3]{Stechkin1951}.
For $1<q<\frac{2+\ln(2)}{2-\ln(2)}$, we can achieve better bounds by choosing the parameter $p$ optimally in $b_k=(k(k+1))^p$.
\begin{Satz}\label{result:improved-right-stechkin-q}
Let $1<q\leq\frac{2+\ln(2)}{2-\ln(2)}$.
The minimal constant $C_1(q)>0$ for which
\begin{equation*}
\sum_{n=1}^\infty a_n\leq C_1(q)\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}
\end{equation*}
holds for all $(a_n)_{n\in\NN}\in\ell_q$ is at most $\lr{\frac{\ee\ln(2)}{\sqrt{2}}}^{\frac{1}{q^\prime}}$.
\end{Satz}
\begin{proof}
Choose $p\in (\frac{1}{2q^\prime},1]$ and set $b_k=(k(k+1))^p$ in \cref{result:right-stechkin-q}.
Then \cref{result:upper-bound-q,result:monotonicity} show that
\begin{equation*}
\sum_{n=1}^\infty a_n\leq \frac{2^p}{(2q^\prime p-1)^{\frac{1}{q^\prime}}}\sum_{n=1}^\infty\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}
\end{equation*}
for all $(a_n)_{n\in\NN}\in\ell_q$.
For fixed $q\in (1,\frac{2+\ln(2)}{2-\ln(2)}]$, we find a minimizer of $p\mapsto\frac{2^p}{(2q^\prime p-1)^{\frac{1}{q^\prime}}}$.
Through the change of variables $\lambda\defeq q^\prime p$, this expression becomes $\lr{\frac{2^\lambda}{2\lambda-1}}^{\frac{1}{q^\prime}}$.
The latter is minimized at $\lambda=\frac{2+\ln(2)}{\ln(4)}$, so $\frac{1}{2q^\prime}<p=\frac{\lambda}{q^\prime}\leq 1$, and the minimum is $\frac{\ee\ln(2)}{\sqrt{2}}$.
\end{proof}

The conclude this section by comparing the bounds from \eqref{eq:stechkin-choice-bound-q} and \cref{result:improved-right-stechkin-q} to those from the literature.

Copson \cite[Theorem~2.3]{Copson1928} shows that $C_1(q)\leq q^{\frac{1}{q}}$.
This result is also reported by Hardy, Littlewood, and Pólya \cite[Theorem~345]{HardyLiPo1934}.
The bound from \eqref{eq:stechkin-choice-bound-q} improves the one from \cite[Theorem~2.3]{Copson1928} when $2(2q^\prime-1)^{-\frac{1}{q^\prime}}<q^{\frac{1}{q}}$.
As
\begin{align*}
\lim_{q\to 1}q^{\frac{1}{q}}=\lim_{q\to\infty}q^{\frac{1}{q}}&=1,\\
\lim_{q\to 1}2(2q^\prime-1)^{-\frac{1}{q^\prime}}=\lim_{q\to\infty}2(2q^\prime-1)^{-\frac{1}{q^\prime}}&=2,
\end{align*}
and
$2(2q^\prime-1)^{-\frac{1}{q^\prime}}<q^{\frac{1}{q}}$ at $q=2$, there are real numbers $q_1$ and $q_2$ such that $1<q_1<q_2$ and $2(2q^\prime-1)^{-\frac{1}{q^\prime}}<q^{\frac{1}{q}}$ for all $q\in(q_1,q_2)$.
Furthermore, the function $q\mapsto q^{\frac{1}{q}}$ is monotonically increasing on $(1,\ee)$ and monotonically decreasing on $(\ee,\infty)$.
Also, for $q_3\approx 1.7718$, the function $q\mapsto 2(2q^\prime-1)^{-\frac{1}{q^\prime}}$ is monotonically decreasing on $(1,q_3)$ and monotonically increasing on $(q_3,\infty)$.
Thus for $q$ sufficiently close to $1$ or $\infty$, the bound from \cite[Theorem~2.3]{Copson1928} is smaller than the one from \eqref{eq:stechkin-choice-bound-q}.
It turns out that $q_1$ and $q_2$ can be chosen such that $2(2q^\prime-1)^{-\frac{1}{q^\prime}}<q^{\frac{1}{q}}$ \emph{if and only if} $q\in(q_1,q_2)$.
Analytical expressions for $q_1$ and $q_2$ are not available through the inequality $2(2q^\prime-1)^{-\frac{1}{q^\prime}}<q^{\frac{1}{q}}$.
However, we have $q_1\approx 1.3229$ and $q_2\approx 4.4124$.

An improvement to \cite[Theorem~2.3]{Copson1928} is reported by Levin and Stechkin in their appendix to the Russian 1948 edition \cite{HardyLiPo1948} of Hardy, Littlewood, and Pólya's monograph.
Levin and Stechkin's bound \cite[{\rus{Д}}.61]{HardyLiPo1948} translates to our notation as
\begin{equation}
C_1(q)\begin{cases}\leq 2^{\frac{1}{q}-2}\lr{3-\frac{1}{q}}q(2-\frac{1}{q})^{\frac{1}{q}-1},&1<q<\frac{5}{3},\\
\leq 2\lr{2\frac{q}{q-1}-1}^{-\frac{q-1}{q}},&\frac{5}{3}\leq q<3,\\
=(q-1)^\frac{1}{q},&3\leq q<\infty.\end{cases}\label{eq:levin-bound}
\end{equation}
Note that at first glance, \eqref{eq:levin-bound} is not what is stated in \cite[{\rus{Д}}.61]{HardyLiPo1948} or its English translation \cite[D.61]{LevinSt1960}.
The mismatch is $2^{\frac{1}{q}-2}\lr{3-\frac{1}{q}}q(2-\frac{1}{q})^{\frac{1}{q}-1}$ versus $2^{\frac{1}{q}-2}\lr{3-\frac{1}{q}}q(1-\frac{1}{q})^{\frac{1}{q}-1}$ in the $1<q<\frac{5}{3}$ case but the latter would not be an improvement over \cite[Theorem~345]{HardyLiPo1934}, and it wouldn't fit in as a special case of the two-parameter inequality \cite[{\rus{Д}}.62]{HardyLiPo1948} or \cite[D.62(v)]{LevinSt1960}, regardless of the discrepancies between the formulas in the different versions and regardless of the fact that it is not the $r=-p$ special case as claimed by Levin and Stechkin but actually the $r=+p$ one.

The bound from \eqref{eq:stechkin-choice-bound-q} is larger than the one from \cite[{\rus{Д}}.61]{HardyLiPo1948} when $3<q<\infty$.
(In this case the latter result provides the optimal constant.)
The bounds coincide for $\frac{5}{3}\leq q\leq 3$.
In the remaining case $1<q<\frac{5}{3}$, our bound from \eqref{eq:stechkin-choice-bound-q} is monotonically decreasing in $q$ while the Levin and Stechkin's bound \cite[{\rus{Д}}.61]{HardyLiPo1948} is monotonically increasing, reversing orders at the boundaries.
Thus the bounds coincide at most once for $1<q<\frac{5}{3}$, and they do for $q_4\approx 1.3725$, which means that the bound from \eqref{eq:stechkin-choice-bound-q} is smaller for $q_4<q\leq\frac{5}{3}$ and larger for $1<q<q_4$ than the one from \cite[{\rus{Д}}.61]{HardyLiPo1948}.

By construction, the bound from \cref{result:improved-right-stechkin-q} outperforms the one from \eqref{eq:stechkin-choice-bound-q}.
Moreover, \cref{result:improved-right-stechkin-q} is an improvement over \cite[{\rus{Д}}.61]{HardyLiPo1948} for $1<q\leq \frac{2+\ln(2)}{2-\ln(2)}\approx 2.0608$.

Gao \cite[Theorem~1]{Gao2011} provides further improvement by showing that $C_1(q)=(q-1)^{\frac{1}{q}}$ not only for $3\leq q<\infty$ as stated in \cite[{\rus{Д}}.61]{HardyLiPo1948}, but even for $q_0\leq q<\infty$, where $q_0\approx 2.8855$ is a solution of the equation
\begin{equation*}
2^{\frac{1}{q-1}}\lr{(q-1)^{\frac{q}{q-1}}-(q-1)}-\lr{1+\frac{3-q}{2}}^{\frac{q}{q-1}}=0.
\end{equation*}

De Bruijn \cite[p.~174]{DeBruijn1958} reports that $C_1(2)=1.1064957714$ \enquote{with an error of at most $9$ units at the last decimal place.}
This is an improvement over Levin and Stechkin's bound $C_1(2)\leq\frac{2}{\sqrt{3}}\approx 1.1547$ established in \cite[{\rus{Д}}.61]{HardyLiPo1948}, and over our bound $C_1(2)\leq \sqrt{\frac{\ee\ln(2)}{\sqrt{2}}}\approx 1.1542$ from \cref{result:improved-right-stechkin-q}.
A visualization of the various upper bounds on $C_1(q)$ is given in \cref{fig:right-stechkin-q}.

\begin{figure}[h!]
\begin{center}
\begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=4.0cm,y=2.0cm]
\draw[black!50] (0,1)--(1,1);
\draw[black!50] (0,2)--(1,2);
\draw[black!50] (1,0)--(1,2);
\draw[black!50] (0.5,0)--(0.5,2);
\draw[line width=1pt,->] (0,0)--(1.2,0);
\draw[line width=1pt,->] (0,0)--(0,2.5);
\draw (1.2,0) node[below]{$1/q$};
\draw (0,2.6) node[left]{upper bounds};
\draw (0,2.4) node[left]{for $C_1(q)$};
\draw (0,1) node[left]{$1$};
\draw (0,2) node[left]{$2$};
\draw (1,0) node[below]{$1$};
\draw (0.5,0) node[below]{$0.5$};
\draw[densely dashed,line width=1pt,smooth,samples=200,domain=1.8713492058682709E-6:1] plot(\x,{(\x)^(-(\x))});
\draw[dotted,line width=1pt,smooth,samples=200,domain=0:0.9999982192917657] plot(\x,{2*((-(\x)-1)/((\x)-1))^((\x)-1)});
\draw[dotted,line width=1pt,smooth,samples=200,domain=0.4852511696625:1] plot(\x,{(2.718281828459045*ln(2)/sqrt(2))^(1-(\x))});

\draw[line width=1pt,smooth,samples=100,domain=0.0001:0.346552] plot(\x,{((1.0-(\x))/(\x))^((\x))});
\draw[line width=1pt,smooth,samples=200,domain=1/3:0.6] plot(\x,{2*((-(\x)-1)/((\x)-1))^((\x)-1)});
\draw[line width=1pt,smooth,samples=100,domain=0.6:1] plot(\x,{2.0^(1.0/(1.0/(\x))-2.0)*(3.0-1.0/(1.0/(\x)))*1.0/(\x)*(2.0-1.0/(1.0/(\x)))^(1.0/(1.0/(\x))-1.0)});
\end{tikzpicture}
\end{center}
\caption{The upper bounds on $C_1(q)$ given by Copson \cite[Theorem~2.3]{Copson1928} (dashed line), Levin and Stechkin \cite[{\rus{Д}}.61]{HardyLiPo1948} and Gao \cite[Theorem~1]{Gao2011} (solid line), and \eqref{eq:stechkin-choice-bound-q} and \cref{result:improved-right-stechkin-q} from the paper at hand (dotted line).\label{fig:right-stechkin-q}}
\end{figure}

\section{The weak discrete Stechkin inequality}\label{chap:weak}
Here we compute the optimal constants $c_{1,\infty}(q)$ and $C_{1,\infty}(q)>0$, for which the inequality
\begin{equation}
\frac{1}{c_{1,\infty}(q)}\sup_{n\in\NN}n\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\leq\sup_{n\in\NN} na_n\leq C_{1,\infty}(q)\sup_{n\in\NN}n\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\label{eq:general-weak-l1-q}
\end{equation}
holds true for all sequences $(a_n)_{n\in\NN}$ with $a_1\geq a_2\geq\ldots\geq 0$, when $1<q<\infty$.
With the modification indicated in \cref{chap:intro}, inequality \eqref{eq:general-weak-l1-q} holds true also for $q=\infty$.
The monotonicity assumption on $(a_n)_{n\in\NN}$ gives $\sup\setcond{a_k}{k\geq n}=a_n$, so $c_{1,\infty}(\infty)=C_{1,\infty}(\infty)=1$.
For $q=1$, we have $C_{1,\infty}(1)=1$ because
\begin{equation*}
\sup_{n\in\NN} na_n = \sup_{n\in\NN}\sum_{k=1}^n a_n\leq\sum_{k=1}^\infty a_k=\sup_{n\in\NN}\sum_{k=n}^\infty a_k,
\end{equation*}
which holds as an equality when $a_n=0$ for all $n\geq 2$.

Notable results in the direction of \eqref{eq:general-weak-l1-q} are the inequalities
\begin{equation*}
n^{1-1/q}\lr{\sum_{k=n}^\infty a_k^q}^{1/q}\leq\sum_{n=1}^\infty a_n
\end{equation*}
and
\begin{equation*}
n^{1-1/q}\lr{\sum_{k=n+1}^\infty a_k^q}^{1/q}\leq (q-1)^{-1/q}\sup_{n\in\NN}n a_n
\end{equation*}
proved in \cite[{\rus{Лемма}~IV.2.1}]{Temlyakov1986} and \cite[Proposition~2.11]{FoucartRa2013}, respectively.

The following estimate on the Riemann zeta-function is required in our calculation of $c_{1,\infty}(q)$.
\begin{Lem}\label{result:zeta-estimate}
Let $1<q<\infty$.
Then $\zeta(q)>\frac{1}{q-1}+\frac{1}{2}$.
\end{Lem}
\begin{proof}
Consider
\begin{align*}
\zeta(q)-\frac{1}{q-1}&=\zeta(q)-\int_1^\infty x^{-q}\dd x>\zeta(q)-\sum_{n=1}^\infty \frac{n^{-q}+(n+1)^{-q}}{2}\\
&=\zeta(q)-\frac{2\zeta(q)-1}{2}=\frac{1}{2}.
\end{align*}
\end{proof}
The estimate in \cref{result:zeta-estimate} is not best possible.
In fact, the constant $\frac{1}{2}$ can be replaced by the Euler--Mascheroni constant, see \cite[(2.1.16)]{Titchmarsh1986}.
Nonetheless this estimate enables the computation of the precise constant on the left-hand side of \eqref{eq:general-weak-l1-q}.
\begin{Satz}\label{result:left-weak-q}
Let $1<q<\infty$.
The minimal constant $c_{1,\infty}(q)>0$ for which
\begin{equation*}
\sup_{n\in\NN}n\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\leq c_{1,\infty}(q)\sup_{n\in\NN} na_n
\end{equation*}
holds for all $(a_n)_{n\in\NN}$ with $a_1\geq a_2\geq \ldots\geq 0$ is $c_{1,\infty}(q)=\zeta(q)^{\frac{1}{q}}$.
\end{Satz}
\begin{proof}
The supremum
\begin{equation}
\sup\setcond{\sup_{n\in\NN}n^{1-\frac{1}{q}}\lr{\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}}{a_1\geq a_2\geq\ldots\geq 0,\sup_{n\in\NN}na_n=1}\label{eq:left-weak-q}
\end{equation}
is attained at the sequence $(a_n)_{n\in\NN}$ defined by $a_n=\frac{1}{n}$ for all $n\in\NN$.
Indeed, for any sequence $(a_n)_{n\in\NN}$ with $a_1\geq a_2\geq\ldots\geq 0$ and $\sup_{n\in\NN}na_n=1$, we have $0\leq a_n\leq \frac{1}{n}$ for all $n\in\NN$.
Therefore, $\lr{\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\leq \lr{\sum_{k=n}^\infty\frac{1}{k^q}}^{\frac{1}{q}}$ for all $n\in\NN$ with equality if and only if $a_n=\frac{1}{n}$ for all $n\in\NN$.
Also, we have $\sup_{n\in\NN} n\frac{1}{n}=1$, and \eqref{eq:left-weak-q} evaluates to $\sup_{n\in\NN}n^{1-\frac{1}{q}}\lr{\sum_{k=n}^\infty\frac{1}{k^q}}^{\frac{1}{q}}=\zeta(q)^\frac{1}{q}$.
Note that the supremum $\sup_{n\in\NN}n^{1-\frac{1}{q}}\lr{\sum_{k=n}^\infty\frac{1}{k^q}}^{\frac{1}{q}}$ is attained at $n=1$ because
\begin{equation*}
\sum_{k=n}^\infty k^{-p}=n^{-p}+\sum_{k=n+1}^\infty k^{-p}\leq n^{-p}+\int_n^\infty x^{-p} \dd x=n^{-p}+\frac{n^{-p+1}}{p-1}
\end{equation*}
and \cref{result:zeta-estimate} give
\begin{equation*}
n^{1-\frac{1}{q}}\lr{\sum_{k=n}^\infty k^{-q}}^{\frac{1}{q}}\leq n^{1-\frac{1}{q}}\lr{n^{-q}+\frac{n^{-q+1}}{q-1}}^{\frac{1}{q}}=\lr{n^{-1}+\frac{1}{q-1}}^{\frac{1}{q}}\leq\zeta(q)^{\frac{1}{q}}.
\end{equation*}
\end{proof}

The result of \cref{result:left-weak-q} is depicted in \cref{fig:left-weak-q}.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=4cm,y=0.4cm]
\draw[black!50] (0,1)--(1,1);
\draw[black!50] (0,3)--(1,3);
\draw[black!50] (0,5)--(1,5);
\draw[black!50] (0,7)--(1,7);
\draw[black!50] (1,0)--(1,7);
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\draw[line width=1pt,->] (0,0)--(0,8.5);
\draw (0,8.5) node[left]{$c_{1,\infty}(q)$};
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\end{tikzpicture}
\end{center}
\caption{The function $1/q\mapsto \zeta(q)^{\frac{1}{q}}$.\label{fig:left-weak-q}}
\end{figure}

The optimal constant $C_{1,\infty}(q)$ in \eqref{eq:general-weak-l1-q} turns out to be invariant under taking Hölder conjugates.
\begin{Satz}\label{result:right-weak-q}
Let $1<q<\infty$.
The minimal constant $C_{1,\infty}(q)>0$ for which
\begin{equation*}
\sup_{n\in\NN} na_n\leq C_{1,\infty}(q)\sup_{n\in\NN}n\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}
\end{equation*}
holds for all $(a_n)_{n\in\NN}$ with $a_1\geq a_2\geq \ldots\geq 0$ is $C_{1,\infty}(q)=\lr{\frac{1}{q}}^{-\frac{1}{q}}\lr{1-\frac{1}{q}}^{-\lr{1-\frac{1}{q}}}$.
\end{Satz}
\begin{proof}
\emph{Step 1. We prove the claim for finite sequences.}
The infimum
\begin{equation*}
\inf\setcond{\sup_{n\in\NN}n\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}}{a_1\geq a_2\geq\ldots\geq 0,a_{N+1}=0,\sup_{n\in\NN}n a_n=1}
\end{equation*}
is attained at the sequence $(a_n)_{n\in\NN}$ given by $a_n=\frac{1}{N}$ for $n\in\setn{1,\ldots,N}$.
Its value is thus equal to
\begin{equation*}
\sup_{n=1,\ldots,N}n\lr{\frac{1}{n}\sum_{k=n}^N \frac{1}{N^q}}^{\frac{1}{q}}=\sup_{n=1,\ldots,N}\frac{1}{N}n^{1-\frac{1}{q}}(N-n+1)^{\frac{1}{q}}.
\end{equation*}
One can readily check that the derivative of the function $f:(0,N+1)\to \RR$, $f(x)=\frac{1}{N} x^{1-\frac{1}{q}} (N-x+1)^{\frac{1}{q}}$ is given by $f^\prime(x)=\frac{1}{Nq}x^{-\frac{1}{q}}(N-x+1)^{\frac{1}{q}-1}((N+1)(q-1)-qx)$.
Therefore, $f$ is maximized at $x=(N+1)\lr{1-\frac{1}{q}}$, the maximum is $\frac{N+1}{N}\lr{\frac{1}{q}}^{\frac{1}{q}}\lr{1-\frac{1}{q}}^{1-\frac{1}{q}}$, and $f$ is monotonically increasing for $x<(N+1)\lr{1-\frac{1}{q}}$ and monotonically decreasing for $x>(N+1)\lr{1-\frac{1}{q}}$.
Hence we have shown that
\begin{align*}
&\inf\setcond{\sup_{n\in\NN}n\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}}{a_1\geq a_2\geq\ldots\geq 0,a_{N+1}=0,\sup_{n\in\NN}n a_n=1}\\
&=\sup_{n=1,\ldots,N}n^{1-\frac{1}{q}}\lr{\frac{N-n+1}{N^q}}^{\frac{1}{q}}\\
&\leq\sup_{x\in (0,N+1)}x^{1-\frac{1}{q}}\lr{\frac{N-x+1}{N^q}}^{\frac{1}{q}}\\
&=\frac{N+1}{N}\lr{\frac{1}{q}}^{\frac{1}{q}}\lr{1-\frac{1}{q}}^{1-\frac{1}{q}}.
\end{align*}
Taking the infimum over $N\in\NN$, we see that
\begin{equation*}
\sup_{n\in\NN}n\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\leq \lr{\frac{1}{q}}^{-\frac{1}{q}}\lr{1-\frac{1}{q}}^{-\lr{1-\frac{1}{q}}}\sup_{n\in\NN}n a_n
\end{equation*}
for finite sequences $(a_n)_{n\in\NN}$ with $a_1\geq a_2\geq\ldots\geq 0$, which also shows that $C_{1,\infty}(q)\leq\lr{\frac{1}{q}}^{-\frac{1}{q}}\lr{1-\frac{1}{q}}^{-\lr{1-\frac{1}{q}}}$.
Next we show that also
\begin{equation*}
\sup_{n\in\NN}n\lr{\frac{1}{n}\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\geq \lr{\frac{1}{q}}^{-\frac{1}{q}}\lr{1-\frac{1}{q}}^{-\lr{1-\frac{1}{q}}}\sup_{n\in\NN}n a_n
\end{equation*}
for finite sequences $(a_n)_{n\in\NN}$ with $a_1\geq a_2\geq\ldots\geq 0$ and we do it for $q\geq 2$ first.
In this case, we have $x_N\defeq 1-\frac{1}{q}-\frac{1}{N+1}\in (0,1)$ for all $N\in\NN$.
The monotonicity properties of the function $g_N:(0,1)\to\RR$, $g_N(x)=f((N+1)x)=\frac{N+1}{N}x^{1-\frac{1}{q}}(1-x)^\frac{1}{q}$ yield
\begin{align*}
\sup\setcond{g_N(x)}{x=\frac{1}{N+1},\ldots,\frac{N}{N+1}}&\geq g_N\lr{\frac{1}{N+1}\floor{(N+1)\lr{1-\frac{1}{q}}}}\\
&\geq g_N(x_N).
\end{align*}
If we can show that $(g_N(x_N))_{N\geq 3}$ is monotonically decreasing, then we know that
\begin{align*}
\inf_{N\geq 3} g_N(x_N)=\lim_{N\to\infty}g_N(x_N)=\lr{\frac{1}{q}}^{\frac{1}{q}}\lr{1-\frac{1}{q}}^{1-\frac{1}{q}}
\end{align*}
and, in turn,
\begin{align*}
&\inf_{N\in\NN}\sup_{n=1,\ldots,N} n^{1-\frac{1}{q}}\lr{\frac{N-n+1}{N^q}}^{\frac{1}{q}}\\
&=\inf\setn{1,\sup\setn{\frac{1}{2}2^{\frac{1}{q}},\frac{1}{2}2^{1-\frac{1}{q}}},\inf_{N\geq 3}\sup_{n=1,\ldots,N} n^{1-\frac{1}{q}}\lr{\frac{N-n+1}{N^q}}^{\frac{1}{q}}}\\
&\geq\inf\setn{1,\sup\setn{\frac{1}{2}2^{\frac{1}{q}},\frac{1}{2}2^{1-\frac{1}{q}}},\inf_{N\geq 3}g_N\lr{x_N}}\\
&\geq\lr{\frac{1}{q}}^{\frac{1}{q}}\lr{1-\frac{1}{q}}^{1-\frac{1}{q}}.
\end{align*}
Indeed, let $a\defeq \frac{1}{q}\in (0,\frac{1}{2}]$, and consider the function $h:(0,1-a)\to\RR$, $h(t)=\lr{\frac{1-a-t}{1-t}}^{1-a}\lr{\frac{a+t}{1-t}}^a$.
Then $g_N(x_N)=h(\frac{1}{N+1})$, and it is sufficient to show that $h$ is monotonically increasing on $(0,\frac{1}{4})$ when $a\leq \frac{1}{2}$.
The derivative
\begin{align*}
h^\prime(t)&=(1-a)\lr{\frac{1-a-t}{1-t}}^{-a}\frac{-a}{(1-t)^2}\lr{\frac{a+t}{1-t}}^a\\
&\qquad+a\lr{\frac{a+t}{1-t}}^{a-1}\frac{1+a}{(1-t)^2}\lr{\frac{1-a-t}{1-t}}^{1-a}\\
&=\frac{1}{(1-t)^2}\lr{\frac{a+t}{1-a-t}}^{a-1}\lr{(1-a)\frac{a+t}{1-a-t}(-a)+a(1+a)}
\end{align*}
is $\geq 0$ if and only if $(1-a)\frac{a+t}{1-a-t}(-a)+a(1+a)\geq 0$ if and only if $a(a+t-1)(a+2t-1)\geq 0$.
For $t\in (0,1-a)$, we have $a(a+t-1)\leq 0$, so $h^\prime(t)\geq 0$ when $a+2t-1\leq 0$.
The latter is fulfilled when $t\leq\frac{1}{4}$.

For $q\leq 2$, we consider the function $(0,1)\ni x\mapsto g_N(1-x)$ instead of $g_N$.
This is the same as $g_N$ for the Hölder conjugate $q^\prime\geq 2$, and this is covered by the first case.

\emph{Step 2. We prove the claim for all sequences.}
It remains to show that
\begin{equation*}
\frac{\sup_{n\in\NN}n^{1-\frac{1}{q}}\lr{\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}}{\sup_{n\in\NN}n a_n}\geq\lr{\frac{1}{q}}^{\frac{1}{q}}\lr{1-\frac{1}{q}}^{1-\frac{1}{q}}
\end{equation*}
for all sequences $(a_n)_{n\in\NN}$ with $a_1\geq a_2\geq \ldots\geq 0$.
Let $\eps>0$.
Then, for all sequences $(a_n)_{n\in\NN}$ with $a_1\geq a_2\geq\ldots\geq 0$ and $\sup\setcond{na_n}{n\in\NN}=1$, there exists a number $N\in\NN$ such that $a_N\geq\frac{1-\eps}{N}$.
It follows that
\begin{align*}
&\sup_{n\in\NN}n^{1-\frac{1}{q}}\lr{\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}\geq\sup_{n=1,\ldots,N}n^{1-\frac{1}{q}}\lr{\sum_{k=n}^N \frac{1}{N^q}}^{\frac{1}{q}}(1-\eps)\\
&=\sup_{n=1,\ldots,N}n^{1-\frac{1}{q}}\lr{\frac{N-n+1}{N^q}}^{\frac{1}{q}}(1-\eps)\geq\lr{\frac{1}{q}}^{\frac{1}{q}}\lr{1-\frac{1}{q}}^{1-\frac{1}{q}}(1-\eps).
\end{align*}
Taking the limit $\eps\downarrow 0$ yields the desired inequality.
\end{proof}

The result of \cref{result:right-weak-q} is depicted in \cref{fig:right-weak-q}.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=4cm,y=2cm]
\draw[black!50] (1,0)--(1,2);
\draw[black!50] (0,1)--(1,1);
\draw[black!50] (0,2)--(1,2);
\draw[black!50] (0.5,0)--(0.5,2);
\draw[line width=1pt,->] (0,0)--(1.2,0);
\draw[line width=1pt,->] (0,0)--(0,2.3);
\draw (0,2.3) node[left]{$C_{1,\infty}(q)$};
\draw (1.2,0) node[below]{$1/q$};
\draw (0,1) node[left]{$1$};
\draw (1,0) node[below]{$1$};
\draw (0.5,0) node[below]{$0.5$};
\draw (0,2) node[left]{$2$};

\draw[line width=1pt,smooth,samples=200,domain=2.899999996260959E-8:0.9999990250810828] plot(\x,{1/((\x)^((\x))*(1-(\x))^(1-(\x)))});
\end{tikzpicture}
\end{center}
\caption{The function $1/q\mapsto \lr{\frac{1}{q}}^{-\frac{1}{q}}\lr{1-\frac{1}{q}}^{-\lr{1-\frac{1}{q}}}$.\label{fig:right-weak-q}}
\end{figure}

\section{The continuous Stechkin inequalities}
In this section, we are concerned with the optimal constants $\bar{c}_1(q)$, $\bar{C}_1(q)$, $\bar{c}_{1,\infty}(q)$, and $\bar{C}_{1,\infty}(q)>0$ in the inequalities
\begin{equation*}
\frac{1}{\bar{c}_1(q)}\int_0^\infty \lr{\frac{1}{t}\int_t^\infty f(s)^q \dd s}^{\frac{1}{q}}\dd t\leq \int_0^\infty f(t)\dd t\leq \bar{C}_1(q)\int_0^\infty \lr{\frac{1}{t}\int_t^\infty f(s)^q \dd s}^{\frac{1}{q}}\dd t
\end{equation*}
and
\begin{equation*}
\frac{1}{\bar{c}_{1,\infty}(q)}\sup_{t>0}t\lr{\frac{1}{t}\int_t^\infty f(s)^q\dd s}^{\frac{1}{q}}\leq\sup_{t>0} tf(t)\leq \bar{C}_{1,\infty}(q)\sup_{t>0}t\lr{\frac{1}{t}\int_t^\infty f(s)^q\dd s}^{\frac{1}{q}}
\end{equation*}
for monotonically decreasing functions $f:(0,\infty)\to [0,\infty)$.
In both cases, precise values for the optimal constants are available, either through the literature or shown here.
The proofs in this section are independent of their discrete counterparts, yet there is a strong resemblance in the case of the weak Stechkin inequalities.
(The arguments turn out to be less tedious in the continuous inequality, though.)

\subsection{The strong continuous Stechkin inequality}
Hardy, Littlewood, and Pól\-ya \cite[Theorem~337]{HardyLiPo1934} show that for $1<q<\infty$, the minimal constant $\bar{C}_1(q)>0$ for which
\begin{equation}
\frac{1}{\bar{c}_1(q)}\int_0^\infty \lr{\frac{1}{t}\int_t^\infty f(s)^q \dd s}^{\frac{1}{q}}\dd t\leq \int_0^\infty f(t)\dd t\leq \bar{C}_1(q)\int_0^\infty \lr{\frac{1}{t}\int_t^\infty f(s)^q \dd s}^{\frac{1}{q}}\dd t\label{eq:general-stechkin-continuous}
\end{equation}
holds true for all monotonically decreasing functions $f:(0,\infty)\to [0,\infty)$ is $\bar{C}_1(q)=(q-1)^{\frac{1}{q}}$ when $1<q<\infty$.
With the modification indicated in \cref{chap:intro}, inequality \eqref{eq:general-stechkin-continuous} holds true also for $q=\infty$.
The monotonicity assumption on $f:(0,\infty)\to [0,\infty)$ gives $\sup\setcond{f(s)}{s\geq t}=f(t)$, so $\bar{c}_1(\infty)=\bar{C}_1(\infty)=1$.
For $q=1$, the right-hand side of \eqref{eq:general-stechkin-continuous} holds in the sense that $\lr{\frac{1}{t}\int_t^\infty f(s)^q \dd s}^{\frac{1}{q}}\dd t$ diverges when $\int_0^\infty f(t)\dd t$ is finite.
Therefore $\lim_{q\to 1}\bar{C}_1(q)=0$.
For this, choose a function $f:(0,\infty)\to[0,\infty)$ with $\int_0^\infty f(t)\dd t\neq 0$.
Then $F(t)\defeq\int_t^\infty f(s)^q \dd s$ defines a monotonically decreasing function $F:(0,\infty)\to [0,\infty)$ for which there exist $\eps>0$ and $\delta>0$ such that $F(t)\geq \delta$ for all $t\in (0,\eps)$.
It follows that $\int_0^\infty \frac{1}{t}F(t)\dd t\geq \int_0^\eps\frac{1}{t}F(t)\dd t\geq\int_0^\eps\frac{1}{t}\delta\dd t=\infty$.

We complement these results by computing the minimal constant $\bar{c}_1(q)>0$ appearing in \eqref{eq:general-stechkin-continuous}.
\begin{Satz}\label{result:left-stechkin-continuous}
Let $1<q<\infty$.
The minimal constant $\bar{c}_1(q)>0$ for which
\begin{equation*}
\int_0^\infty \lr{\frac{1}{t}\int_t^\infty f(s)^q \dd s}^{\frac{1}{q}}\dd t\leq\bar{c}_1(q)\int_0^\infty f(t)\dd t
\end{equation*}
holds for all monotonically decreasing functions $f:(0,\infty)\to [0,\infty)$ is $\bar{c}_1(q)=\frac{\piup}{q\sin(\frac{\piup}{q})}$.
\end{Satz}

\begin{proof}
Lower bounds on the constant $\bar{c}_1(q)$ are given by the quotients
\begin{equation}
\frac{\int_0^t \lr{\frac{1}{t}\int_t^\infty f(s)^q \dd s}^{\frac{1}{q}}\dd t}{\int_0^\infty f(t)\dd t}\label{eq:left-stechkin-continuous-lower-bound}
\end{equation}
with $f:(0,\infty)\to [0,\infty)$ a monotonically decreasing function.
For $T\in (0,\infty)$, take $f=\frac{1}{T}\chi_{(0,T)}:(0,\infty)\to [0,\infty)$ in \eqref{eq:left-stechkin-continuous-lower-bound}, where $\chi_{(0,T)}$ denotes the function which is $1$ on $(0,T)$ and $0$ on $[T,\infty)$.
Then $\int_0^\infty f(t)\dd t=1$ and
\begin{align*}
&\int_0^\infty \lr{\frac{1}{t}\int_t^\infty f(s)^q \dd s}^{\frac{1}{q}}\dd t=\frac{1}{T}\int_0^\infty \lr{\frac{1}{t}\int_t^\infty \chi_{(0,T)}(s) \dd s}^{\frac{1}{q}}\dd t\\
&=\frac{1}{T}\int_0^T \lr{\frac{1}{t}\int_t^\infty \chi_{(0,T)}(s) \dd s}^{\frac{1}{q}}\dd t=\frac{1}{T}\int_0^T \lr{\frac{T-t}{t}}^{\frac{1}{q}}\dd t\\
&=B\lr{1-\frac{1}{q},1+\frac{1}{q}}=\frac{\piup}{q\sin(\frac{\piup}{q})}
\end{align*}
independently of $T$.
(Here $B(x,y)\defeq\int_0^1 t^{x-1}(1-t)^{y-1}\dd t$ denotes the beta function.)
This shows $\bar{c}_1\geq\frac{\piup}{q\sin(\frac{\piup}{q})}$.

Now fix a function $f:[0,\infty)\to [0,\infty)$ with $\int_0^\infty f(x)\dd x=1$.
For $\eps>0$, let $N\defeq\floor{\frac{1}{\eps}\sup_{t\in [0,\infty)}f(t)}\in \NN\cup\setn{\infty}$.
For $n\in\NN$ with $n\leq N$, let
\begin{equation*}
T_n\defeq \sup\setcond{T>0}{f(t)\geq \eps n \fall t\in (0,T)},
\end{equation*}
$\lambda_n\defeq \eps T_n$, and $g_n\defeq \frac{1}{\lambda_n}\eps\chi_{[0,T_n]}$.
Then $0\leq h_\eps(t)\defeq\sum_{n=1}^N \lambda_n g_n(t)\leq f(t)$ for all $t\in [0,\infty)$ and $\int_0^\infty g_n(t)\dd t =1$ for all $n$, so $\sum_{n=1}^N\lambda_n=\int_0^\infty \sum_{n=1}^N \lambda_n g_n(t)\dd t\leq \int_0^\infty f(t)\dd t=1$.
From the triangle inequality for integrals, it follows that
\begin{align*}
\int_0^\infty \lr{\frac{1}{t}\int_t^\infty h_\eps(s)^q\dd s}^{\frac{1}{q}}\dd t&\leq \sum_{n=1}^N \lambda_n \int_0^\infty \lr{\frac{1}{t}\int_t^\infty g_n(s)^q\dd s}^{\frac{1}{q}}\dd t\\
&=\sum_{n=1}^N \lambda_n \frac{\piup}{q\sin(\frac{\piup}{q})}\leq\frac{\piup}{q\sin(\frac{\piup}{q})}.
\end{align*}
With the abbreviations $H_\eps(t)\defeq \lr{\int_t^\infty h_\eps(s)^q\dd s}^{\frac{1}{q}}$ and $F(t)\defeq \lr{\int_t^\infty f(s)^q\dd s}^{\frac{1}{q}}$, we have $\lim_{k\to\infty} H_{2^{-k}}(t)=F(t)$ and $0\leq H_{2^{-k}}(t)\leq F(t)$ for all $t\in (0,\infty)$ and $k\in\NN$.
The dominated convergence theorem then yields
\begin{align*}
\int_0^\infty F(t)\dd t&=\int_0^\infty \lim_{k\to\infty}H_{2^{-k}}(t)\dd t=\lim_{k\to\infty}\int_0^\infty H_{2^{-k}}(t)\dd t\leq \frac{\piup}{q\sin(\frac{\piup}{q})}.
\end{align*}
This shows $\bar{c}_1\leq \frac{\piup}{q\sin(\frac{\piup}{q})}$ and completes the proof.
\end{proof}

The precise values for $\bar{c}_1(q)$ and $\bar{C}_1(q)$ given in \cref{result:left-stechkin-continuous} and \cite[Theorem~337]{HardyLiPo1934} are illustrated in \cref{fig:stechkin-continuous}.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=3cm,y=0.554cm]
\draw[black!50] (0,1)--(1,1);
\draw[black!50] (0,3)--(1,3);
\draw[black!50] (0,5)--(1,5);
\draw[black!50] (0.5,0)--(0.5,5);
\draw[black!50] (1,0)--(1,5);
\draw[line width=1pt,->] (0,0)--(1.2,0);
\draw[line width=1pt,->] (0,0)--(0,6.5);
\draw (0,6.5) node[left]{$\bar{c}_1(q)$};
\draw (1.2,0) node[below]{$1/q$};
\draw (0,1) node[left]{$1$};
\draw (0,3) node[left]{$3$};
\draw (0,5) node[left]{$5$};
\draw (1,0) node[below]{$1$};
\draw (0.5,0) node[below]{$0.5$};

\draw[color=black,line width=1pt,smooth,samples=200,domain=0.001:0.8] plot(\x,{3.141592653589793*(\x)/sin((3.141592653589793*(\x))*180/pi)});
\end{tikzpicture}
\qquad
\begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=3cm,y=2cm]
\draw[black!50] (1,0)--(1,1.5);
\draw[black!50] (0,1)--(1,1);
\draw[black!50] (0.5,0)--(0.5,1.5);
\draw[black!50] (0,0.5)--(1,0.5);
\draw[black!50] (0,1.5)--(1,1.5);
\draw[line width=1pt,->] (0,0)--(1.2,0);
\draw[line width=1pt,->] (0,0)--(0,1.8);
\draw (0,1.8) node[left]{$\bar{C}_1(q)$};
\draw (1.2,0) node[below]{$1/q$};
\draw (0,1) node[left]{$1$};
\draw (1,0) node[below]{$1$};
\draw (0.5,0) node[below]{$0.5$};

\draw[line width=1pt,smooth,samples=200,domain=0.001:0.9999996591791471] plot(\x,{(1/(\x)-1)^((\x))});
\end{tikzpicture}
\end{center}
\caption{The function $1/q\mapsto \frac{\piup}{q\sin(\frac{\piup}{q})}$ (left) and $1/q\mapsto (q-1)^{\frac{1}{q}}$ (right).\label{fig:stechkin-continuous}}
\end{figure}

\cref{result:left-stechkin-continuous} and \cite[Theorem~337]{HardyLiPo1934} can be transformed into a result on functions $f:\Omega\to\RR$ defined on a measure space $(\Omega,\mu)$.
For such a function $f$, the assignment $f^\ast(t)\defeq\inf\setcond{s>0}{\mu(\setcond{x\in\Omega}{\abs{f(x)}>s})\leq t}$ defines a function $f^\ast:(0,\infty)\to [0,\infty)$, called the \emph{non-increasing rearrangement} of $f$.
\begin{Kor}
Let $(\Omega,\mu)$ be a measure space and $1<q<\infty$.
The minimal constants $c(q),C(q)>0$ for which
\begin{align*}
\frac{1}{c(q)}\int_0^\infty \lr{\frac{1}{t}\int_t^\infty f^\ast(s)^q \dd s}^{\frac{1}{q}}\dd t&\leq \int_0^\infty \abs{f(x)}\dd \mu(x)\\
&\leq C(q)\int_0^\infty \lr{\frac{1}{t}\int_t^\infty f^\ast(s)^q \dd s}^{\frac{1}{q}}\dd t
\end{align*}
holds for all functions $f:\Omega\to \RR$ are $c(q)=\bar{c}_1(q)$ and $C(q)=\bar{C}_1(q)$.
\end{Kor}
\begin{proof}
Note that $\int_\Omega\abs{f(x)}\dd\mu(x)=\int_0^\infty f^\ast(t)\dd t$ and apply \cref{result:left-stechkin-continuous} and \cite[Theorem~337]{HardyLiPo1934} to $f^\ast$.
\end{proof}

\subsection{The weak continuous Stechkin inequality}
Here we compute the optimal constants $\bar{c}_{1,\infty}(q)$ and $\bar{C}_{1,\infty}(q)>0$, for which the inequality
\begin{equation}
\frac{1}{\bar{c}_{1,\infty}(q)}\sup_{t>0}t\lr{\frac{1}{t}\int_t^\infty f(s)^q\dd s}^{\frac{1}{q}}\leq\sup_{t>0} tf(t)\leq \bar{C}_{1,\infty}(q)\sup_{t>0}t\lr{\frac{1}{t}\int_t^\infty f(s)^q\dd s}^{\frac{1}{q}}\label{eq:general-weak-l1-continuous}
\end{equation}
holds true for all monotonically decreasing functions $f:(0,\infty)\to [0,\infty)$, when $1<q<\infty$.
With the modification indicated in \cref{chap:intro}, inequality \eqref{eq:general-weak-l1-continuous} holds true also for $q=\infty$.
The monotonicity assumption on $f$ gives $\sup\setcond{f(s)}{s\geq t}=f(t)$, so $\bar{c}_{1,\infty}(\infty)=\bar{C}_{1,\infty}(\infty)=1$.
For $q=1$, we have $\bar{C}_{1,\infty}(1)=1$ because
\begin{equation*}
\sup_{t>0} tf(t)=\sup_{t>0}\int_0^t f(t)\dd s\leq\int_0^\infty f(s)\dd s=\sup_{t>0}\int_t^\infty f(s)\dd s.
\end{equation*}

\begin{Satz}\label{result:left-weak-continuous}
Let $1<q<\infty$.
The minimal constant $\bar{c}_{1,\infty}(q)>0$ for which
\begin{equation*}
\sup_{t>0}t\lr{\frac{1}{t}\int_t^\infty f(s)^q\dd s}^{\frac{1}{q}}\leq \bar{c}_{1,\infty}(q)\sup_{t>0} tf(t)
\end{equation*}
holds for all monotonically decreasing functions $f:(0,\infty)\to [0,\infty)$ is $\bar{c}_{1,\infty}(q)=(q-1)^{-\frac{1}{q}}$.
\end{Satz}
\begin{proof}
The supremum of the expression
\begin{equation}
\sup_{t>0}t^{1-\frac{1}{q}}\lr{\int_t^\infty f(s)^q\dd s}^{\frac{1}{q}}\label{eq:left-weak-continuous}
\end{equation}
over the monotonically decreasing functions $f:(0,\infty)\to [0,\infty)$ with $\sup_{t>0}tf(t)=1$ is attained at the function $f:(0,\infty)\to [0,\infty)$ defined by $f(t)=\frac{1}{t}$.
Indeed, for any monotonically decreasing function $f:(0,\infty)\to [0,\infty)$ with $\sup_{t>0}tf(t)=1$, we have $0\leq f(t)\leq\frac{1}{t}$ for all $t>0$.
Therefore, $\lr{\int_t^\infty f(s)^q\dd s}^{\frac{1}{q}}\leq \lr{\int_t^\infty\frac{1}{s^q}\dd s}^{\frac{1}{q}}$ for all $t>0$ with equality if and only if $f(t)=\frac{1}{t}$ for all $t>0$.
Also, we have $\sup_{t>0} t\frac{1}{t}=1$, and \eqref{eq:left-weak-continuous} evaluates to $\sup_{t>0}t^{1-\frac{1}{q}}\lr{\int_t^\infty\frac{1}{s^q}\dd s}^{\frac{1}{q}}=(q-1)^{-\frac{1}{q}}$.
\end{proof}

The optimal constant $\bar{C}_{1,\infty}(q)$ in \eqref{eq:general-weak-l1-continuous} coincides with its discrete counterpart.
\begin{Satz}\label{result:right-weak-continuous}
Let $1<q<\infty$.
The minimal constant $\bar{C}_{1,\infty}(q)>0$ for which
\begin{equation*}
\sup_{t>0} tf(t)\leq \bar{C}_{1,\infty}(q)\sup_{t>0}t\lr{\frac{1}{t}\int_t^\infty f(s)^q\dd s}^{\frac{1}{q}}
\end{equation*}
holds for all monotonically decreasing functions $f:(0,\infty)\to [0,\infty)$ is $\bar{C}_{1,\infty}(q)=\lr{\frac{1}{q}}^{-\frac{1}{q}}\lr{1-\frac{1}{q}}^{-\lr{1-\frac{1}{q}}}$.
\end{Satz}
\begin{proof}
\emph{Step 1. We prove the claim for functions supported by an interval $(0,T)$ with $T>0$.}
The infimum of the expression
\begin{equation*}
\sup_{t>0}t\lr{\frac{1}{t}\int_t^\infty f(s)^q\dd s}^{\frac{1}{q}}
\end{equation*}
over the monotonically decreasing functions $f:(0,\infty)\to [0,\infty)$ with $f(T)=0$ and $\sup_{t>0}tf(t)=1$ is attained at the function $f=\frac{1}{T}\chi_{(0,T)}:(0,\infty)\to [0,\infty)$.
Its value is thus equal to
\begin{equation*}
\sup_{t\in (0,T)}t\lr{\frac{1}{t}\int_t^T\frac{1}{T^q}\dd s}^{\frac{1}{q}}=\sup_{t\in (0,T)}\frac{1}{T}t^{1-\frac{1}{q}}(T-t)^{\frac{1}{q}}.
\end{equation*}
One can readily check that the derivative of the function $F:(0,T)\to \RR$, $F(t)=\frac{1}{T} t^{1-\frac{1}{q}} (T-t)^{\frac{1}{q}}$ is given by $F^\prime(t)=\frac{q(T-t)+T}{q(T-t)T}\lr{\frac{T-t}{t}}^{\frac{1}{q}}$.
Therefore, $F$ is maximized at $t=T\lr{1-\frac{1}{q}}$, and the maximum is $\lr{\frac{1}{q}}^{\frac{1}{q}}\lr{1-\frac{1}{q}}^{1-\frac{1}{q}}$ independently of $T$.

\emph{Step 2. We prove the claim for all functions.}
It remains to show that
\begin{equation*}
\frac{\sup_{t>0}t^{1-\frac{1}{q}}\lr{\int_t^\infty f(s)^q\dd s}^{\frac{1}{q}}}{\sup_{t>0}tf(t)}\geq\lr{\frac{1}{q}}^{\frac{1}{q}}\lr{1-\frac{1}{q}}^{1-\frac{1}{q}}
\end{equation*}
for all monotonically decreasing functions $f:(0,\infty)\to [0,\infty)$.
Let $\eps>0$.
Then, for all monotonically decreasing functions $f:(0,\infty)\to [0,\infty)$ with $\sup\setcond{tf(t)}{t>0}=1$, there exists a number $T\in\NN$ such that $f(T)\geq\frac{1-\eps}{T}$.
It follows that
\begin{align*}
&\sup_{t>0}t^{1-\frac{1}{q}}\lr{\int_t^\infty f(s)^q\dd s}^{\frac{1}{q}}\geq \sup_{t\in (0,T)} t^{1-\frac{1}{q}}\lr{\int_t^T \frac{1}{T^q}\dd s}^{\frac{1}{q}}(1-\eps)\\
&=\sup_{t\in (0,T)}t^{1-\frac{1}{q}}\lr{\frac{T-t}{T^q}}^{\frac{1}{q}}(1-\eps)\geq\lr{\frac{1}{q}}^{\frac{1}{q}}\lr{1-\frac{1}{q}}^{1-\frac{1}{q}}(1-\eps).
\end{align*}
Taking the limit $\eps\downarrow 0$ yields the desired inequality.
\end{proof}

The precise values for $\bar{c}_{1,\infty}(q)$ and $\bar{C}_{1,\infty}(q)$ are illustrated in \cref{fig:weak-l1-continuous}.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=3cm,y=0.584cm]
\draw[black!50] (0,1)--(1,1);
\draw[black!50] (0,3)--(1,3);
\draw[black!50] (0,5)--(1,5);
\draw[black!50] (0.5,0)--(0.5,5);
\draw[black!50] (1,0)--(1,5);
\draw[line width=1pt,->] (0,0)--(1.2,0);
\draw[line width=1pt,->] (0,0)--(0,6.5);
\draw (0,6.5) node[left]{$\bar{c}_{1,\infty}(q)$};
\draw (1.2,0) node[below]{$1/q$};
\draw (0,1) node[left]{$1$};
\draw (0,3) node[left]{$3$};
\draw (0,5) node[left]{$5$};
\draw (1,0) node[below]{$1$};
\draw (0.5,0) node[below]{$0.5$};

\draw[line width=1pt,smooth,samples=200,domain=0.001:0.86] plot(\x,{((\x)/(1-(\x)))^((\x))});
\end{tikzpicture}
\qquad
\begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=3cm,y=1.6cm]
\draw[black!50] (1,0)--(1,2);
\draw[black!50] (0,1)--(1,1);
\draw[black!50] (0,2)--(1,2);
\draw[black!50] (0.5,0)--(0.5,2);
\draw[line width=1pt,->] (0,0)--(1.2,0);
\draw[line width=1pt,->] (0,0)--(0,2.3);
\draw (0,2.3) node[left]{$\bar{C}_{1,\infty}(q)$};
\draw (1.2,0) node[below]{$1/q$};
\draw (0,1) node[left]{$1$};
\draw (1,0) node[below]{$1$};
\draw (0.5,0) node[below]{$0.5$};
\draw (0,2) node[left]{$2$};

\draw[line width=1pt,smooth,samples=200,domain=2.899999996260959E-8:0.9999990250810828] plot(\x,{1/((\x)^((\x))*(1-(\x))^(1-(\x)))});
\end{tikzpicture}
\end{center}
\caption{The function $1/q\mapsto (q-1)^{-\frac{1}{q}}$ (left) and $1/q\mapsto \lr{\frac{1}{q}}^{-\frac{1}{q}}\lr{1-\frac{1}{q}}^{-\lr{1-\frac{1}{q}}}$ (right).\label{fig:weak-l1-continuous}}
\end{figure}

\section{Applications to sparse approximation}\label{chap:devore}
As mentioned in \cref{chap:intro}, the inequalities \eqref{eq:general-stechkin-q-intro} and \eqref{eq:general-weak-l1-q-intro} play an important role in nonlinear approximation.
More precisely, we will outline the connection of \cite[Theorem~4]{DeVore1998} and our results.
Let $\mathcal{H}$ be an infinite-dimensional separable real Hilbert space with inner product $\skpr{\cdot}{\cdot}_{\mathcal{H}}$ and norm $\mnorm{\cdot}_{\mathcal{H}}$.
The choice of an orthonormal basis $(e_k)_{k\in\NN}$ and Parseval's identity give an isometric isomorphism $\mathcal{H}\to\ell_2$, $f\mapsto (\skpr{f}{e_k}_{\mathcal{H}})_{k\in\NN}$.
Sparse approximation in $\mathcal{H}$ is implemented by defining the approximation error for $f\in\mathcal{H}$ as
\begin{equation*}
E_n(f)_{\mathcal{H}}=\setcond{\mnorm{f-g}_{\mathcal{H}}}{g\in\Sigma_{n-1}(\mathcal{H})}
\end{equation*}
where $\Sigma_{n-1}(\mathcal{H})\defeq\setcond{\sum_{k\in\Lambda} \lambda_k e_k}{\lambda_k\in\RR,\Lambda\subset\NN,\# \Lambda < n}$.
Then, for $\alpha>0$, the approximation space $\mathcal{A}_r^\alpha(\mathcal{H})$ is defined as the set of elements $f\in\mathcal{H}$ for which the quantity
\begin{equation*}
\mnorm{f}_{\mathcal{A}_r^\alpha(\mathcal{H})}\defeq\begin{cases}
\lr{\sum_{n=1}^\infty(n^\alpha E_n(f)_{\mathcal{H}})^r \frac{1}{n}}^{\frac{1}{r}},&0<r<\infty,\\
\sup_{n\in\NN}n^\alpha E_n(f)_{\mathcal{H}},&r=\infty
\end{cases}
\end{equation*}
is finite.
Approximation spaces are then subject to characterizations in terms of Lorentz sequence spaces $\ell_{p,r}$, i.e., the sets of bounded sequences for which the quantity
\begin{equation*}
\mnorm{(f_k)_{k\in\NN}}_{\ell_{p,r}}\defeq\begin{cases}
\lr{\sum_{n\in\NN}(n^{\frac{1}{p}}f_n^\ast)^r\frac{1}{n}}^{\frac{1}{r}},&0<r<\infty,\\
\sup_{n\in\NN} n^{\frac{1}{p}} f_n^\ast,&r=\infty
\end{cases}
\end{equation*}
is finite.
Here $0<p\leq\infty$, and $(f_k^\ast)_{k\in\NN}$ is the non-increasing rearrangement of the sequence $(\abs{f_k})_{k\in\NN}$.
Now, given $f\in\mathcal{H}$ and $f_k\defeq\skpr{f}{e_k}_{\mathcal{H}}$ for $k\in\NN$, DeVore \cite[Theorem~4]{DeVore1998} shows that
\begin{equation}
\mnorm{f}_{\mathcal{A}_r^\alpha(\mathcal{H})}\asymp\mnorm{(f_k)_{k\in\NN}}_{\ell_{\tau,r}},\label{eq:devore}
\end{equation}
meaning that there exist constants $c,C>0$ such that
\begin{equation*}
\frac{1}{c}\mnorm{f}_{\mathcal{A}_r^\alpha(\mathcal{H})}\leq\mnorm{(f_k)_{k\in\NN}}_{\ell_{\tau,r}}\leq C \mnorm{f}_{\mathcal{A}_r^\alpha(\mathcal{H})}.
\end{equation*}
Here $\tau=(\alpha+\frac{1}{2})^{-1}$.
Notable special cases of \eqref{eq:devore} are re-parameterized by the results from \cref{chap:stechkin,chap:weak}.
\begin{Satz}\label{result:devore}
With the definitions above, the following statements are true for the minimal constants $c,C>0$ in the inequality
\begin{equation}
\frac{1}{c}\mnorm{f}_{\mathcal{A}_r^\alpha(\mathcal{H})}\leq\mnorm{(f_k)_{k\in\NN}}_{\ell_{\tau,r}}\leq C\mnorm{f}_{\mathcal{A}_r^\alpha(\mathcal{H})},\label{eq:quasi-norm-equivalence}
\end{equation}
where $\tau=(\alpha+\frac{1}{2})^{-1}$.
\begin{enumerate}[label={(\roman*)},leftmargin=*,align=left,noitemsep]
\item{If $r=\tau$, then $c=c_1(2\alpha+1)^{\alpha+\frac{1}{2}}$ and $C=C_1(2\alpha+1)^{\alpha+\frac{1}{2}}$.\label{devore-strong-q}}
\item{If $r=\infty$, then $c=c_{1,\infty}(2\alpha+1)^{\alpha+\frac{1}{2}}$ and $C=C_{1,\infty}(2\alpha+1)^{\alpha+\frac{1}{2}}$.\label{devore-weak-q}}
\end{enumerate}
\end{Satz}

\begin{proof}
For $r=\tau$, inequality \eqref{eq:devore} becomes
\begin{align*}
\frac{1}{C_1}\lr{\sum_{n=1}^\infty\lr{n^\alpha\lr{\sum_{k=n}^\infty (f_k^\ast)^2}^{\frac{1}{2}}}^\tau \frac{1}{n}}^{\frac{1}{\tau}}&\leq\lr{\sum_{k=1}^\infty (f_k^\ast)^\tau}^{\frac{1}{\tau}}\\
&\leq C_2\lr{\sum_{n=1}^\infty\lr{n^\alpha\lr{\sum_{k=n}^\infty (f_k^\ast)^2}^{\frac{1}{2}}}^\tau \frac{1}{n}}^{\frac{1}{\tau}}.
\end{align*}
Setting $a_k=(f_k^\ast)^\tau$ and $q=\frac{2}{\tau}=2\alpha+1$ gives
\begin{equation*}
\frac{1}{C_1}\lr{\sum_{n=1}^\infty n^{-\frac{1}{q}}\lr{\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}}^{\frac{1}{\tau}}\leq\lr{\sum_{k=1}^\infty a_k}^{\frac{1}{\tau}}\leq C_2\lr{\sum_{n=1}^\infty n^{-\frac{1}{q}}\lr{\sum_{k=n}^\infty a_k^q}^{\frac{1}{q}}}^{\frac{1}{\tau}}.
\end{equation*}
Raising everything to the $\tau$th power shows \ref{devore-strong-q}.
Similarly, for $r=\infty$, inequality \eqref{eq:devore} becomes
\begin{equation*}
\frac{1}{C_1}\sup_{n\in\NN}n^\alpha\lr{\sum_{k=n}^\infty (f_k^\ast)^2}^{\frac{1}{2}}\leq\sup_{k\in\NN} k^{\frac{1}{\tau}} f_k^\ast\leq C_2\sup_{n\in\NN}n^\alpha\lr{\sum_{k=n}^\infty (f_k^\ast)^2}^{\frac{1}{2}}
\end{equation*}
Raising everything to the $\tau$th power gives
\begin{equation*}
\frac{1}{C_1}\sup_{n\in\NN}n^{\alpha\tau}\lr{\sum_{k=n}^\infty (f_k^\ast)^2}^{\frac{\tau}{2}}\leq\sup_{k\in\NN} k (f_k^\ast)^\tau\leq C_2\sup_{n\in\NN}n^{\alpha\tau}\lr{\sum_{k=n}^\infty (f_k^\ast)^2}^{\frac{\tau}{2}}.
\end{equation*}
Setting $a_k=(f_k^\ast)^\tau$ and $q=\frac{2}{\tau}=2\alpha+1$ shows \ref{devore-weak-q}.
\end{proof}

Stechkin \cite{Stechkin1951} considers the Hilbert space $\mathcal{H}=L_2(\mathbb{T}^d)$ of square-integrable functions on $\mathbb{T}^d = [0,2\pi]^d$.
An orthonormal basis in $L_2(\mathbb{T}^d)$ is given by $e_k(x)\defeq\frac{1}{\sqrt{2\piup}}\exp(\ii kx)$ for $k\in\ZZ$ and $x\in\mathbb{T}^d$.
Then \eqref{eq:devore} for $r=\tau=1$ shows that the approximation space $\mathcal{A}_1^{1/2}(L_2(\mathbb{T}^d))$ coincides with the Wiener algebra $\mathcal{A}(\mathbb{T}^d)\defeq\setcond{f\in C(\mathbb{T})}{\sum_{k\in\ZZ^d}\abss{\hat{f}(k)}<\infty}$ and, moreover, their quasi-norms are equivalent in the sense of \eqref{eq:quasi-norm-equivalence}.
\cref{result:devore} yields
\begin{equation*}
\frac{2}{\piup}\mnorm{f}_{\mathcal{A}_1^{1/2}(L_2(\mathbb{T}^d))}\leq\mnorm{f}_{\mathcal{A}(\mathbb{T}^d)}\leq 1.1064957714\mnorm{f}_{\mathcal{A}_1^{1/2}(L_2(\mathbb{T}^d))}
\end{equation*}
for $f\in L_2(\mathbb{T}^d)$, the constants being independent of $d$.

Following DeVore \cite[Remark~7.4]{DeVore1998}, if the Hilbert space $\mathcal{H}$ is chosen to be $L_2(\RR)$ with a wavelet orthonormal basis $(\psi_I)_{I\in\mathcal{D}}$, then the approximation space can be characterized in terms of Besov smoothness.
A multivariate version of this result using a tensorized wavelet basis is derived by Sickel and Ullrich in \cite[Theorem~2.7]{SickelUl2009}.
For further results in this direction, see the papers of Hansen and Sickel \cite{HansenSi2010,HansenSi2012}.

\textbf{Acknowledgements.} The authors would like to acknowledge support by the DFG Ul-403/2-1.
They would further like to thank Kai Lüttgen, Winfried Sickel, Vladimir Temlyakov, and Gerd Wachsmuth for insightful discussions.

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\href}[2]{#2}

config: arxiv


TITLE: Clarification wanted: Let $T$ be the set of all infinite sequences of $0$'s and $1$'s with finitely many $1$'s. Prove that $T$ is denumerable.
QUESTION [3 upvotes]: I think I'm misunderstanding the following proposition.

Let $T$ be the set of all infinite sequence of $0$'s and $1$'s with only finitely many $1$'s. Prove that $T$ is denumerable.

I'm also given the following lemma to use:

Let $\{A_{i}| i\in\mathbb{N}\}$ be a denumerable family of finite non-nonempty sets which are pairwise disjoint, then $\bigcup _{i\in\mathbb{N}}A_{i}$ is denumberable.

I'm not quite seeing how adding the condition that there are finitely many $1$'s makes this the set $T$ countable, since there is still a possibility of infintely many $0$'s. It still looks susceptible to the diagonalization argument.
Could anyone clear up why the the set is countable?

REPLY [0 votes]: For $b=(b_n)_{n\in N}\in T,$ if $b_n=0$ for all $n,$  then $f(b)=0.$
If $m_0$ is the least $n$ such that $b_n=1,$ and $m_1$ is the largest $n$ such that $b_n=1$  then $$f(b)=(2^{m_0-1})\sum_{j=0}^{m_1-m_0} b_{m_1-j} 2^j.$$ Then $f$ is a bijection from T to the non-negative integers.
The idea is to interpret $(b_{m_0},...,b_{m_1})$ as the base-two representation of an odd number $g(n)\in N,$ and let $ 2^{m_0-1} g(n)=f(n).$
An attempt to use a diagonal argument to try to prove that $T$ is uncountable will produce a binary sequence with infinitely $1$'s, which is not in $T$.