Maximization of $\ell^2$-norm

Question

Consider for $r,c>0$ the set $$X_{r,c}=\{x \in \ell^1(\mathbb{N}) \mid \|x\|_1=r,\, \forall i \in \mathbb{N}: |x_i|<c\}.$$ Then I can show that $\inf_{x \in X_{r,c}} \|x\|_2 = 0.$ But is it possible to compute $$s_{r,c}:=\sup_{x \in X_{r,c}} \|x\|_2 \text{ ?}$$ Does at least $\lim_{c \to 0} s_{r,c}=r$ hold?

Answer 1

$\newcommand{\n}{\lfloor{r/c}\rfloor}$We have \begin{equation*} s_{r,c}=\sqrt{\n c^2+(r-\n c)^2}. \end{equation*}

Indeed, by continuity and interchangeability of the coordinates, \begin{equation*} s_{r,c}^2=\sup_{a\in A_{r,c}} \sum_{i\ge1}a_i^2, \end{equation*} where \begin{equation*} A_{r,c}:=\Big\{a\in[0,\infty)^\infty\colon \sum_{i\ge1}a_i=r,\ c\ge a_1\ge a_2\ge\cdots\Big\}. \end{equation*}

For natural $i$, let
\begin{equation*} b_i:=b_{r,c;i}:= c\,1(i\le n)+(r-nc)\,1(i=n+1), \end{equation*} where $$n:=n_{r,c}:=\n.$$ Then $b\in A_{r,c}$.

Take any $a\in A_{r,c}$. Take any real $t>0$. Then for some nonnegative integer $n_t=n_{a;t}$ and all natural $i$ we have \begin{equation*} a_i\ge t \text{ if }i\le n_t\quad\text{and}\quad a_i<t \text{ if }i>n_t. \end{equation*} So, if $n_t\le n$ (where $n=\n$, as above), then, letting $u_+:=\max(0,u)$, we have
\begin{equation*} \sum_{i\ge1}(a_i-t)_+=\sum_{1\le i\le n_t}(a_i-t) \le\sum_{1\le i\le n_t}(c-t) \\ =\sum_{1\le i\le n_t}(b_i-t) \le\sum_{i\ge1}(b_i-t)_+ \end{equation*} and, if $n_t>n$, then \begin{equation*} \sum_{i\ge1}(a_i-t)_+=\sum_{1\le i\le n_t}(a_i-t) \\ \le r-n_t t = \sum_{1\le i\le n_t}(b_i-t) \le\sum_{i\ge1}(b_i-t)_+. \end{equation*} So, in any case, \begin{equation*} \sum_{i\ge1}(a_i-t)_+\le\sum_{i\ge1}(b_i-t)_+ \end{equation*} for all real $t>0$.

Noting now that $u^2=2\int_0^\infty dt\,(u-t)_+$ for real $u\ge0$, we have \begin{equation*} \sum_{i\ge1}a_i^2=2\int_0^\infty dt\,\sum_{i\ge1}(a_i-t)_+ \\ \le2\int_0^\infty dt\,\sum_{i\ge1}(b_i-t)_+=\sum_{i\ge1}b_i^2. \end{equation*} So, $b$ is a maximizer of $\sum_{i\ge1}a_i^2$ over $a\in A_{r,c}$.

Thus, \begin{equation*} s_{r,c}=\sqrt{\sum_{i\ge1}b_i^2}=\sqrt{nc^2+(r-nc)^2} \\ =\sqrt{\n c^2+(r-\n c)^2}. \quad\Box \end{equation*}

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In particular, $s_{r,c}\sim\sqrt{rc}\to0$ as $c\downarrow0$. So, it is not true that $s_{r,c}\to r$ as $c\downarrow0$.