TITLE: Maximizing the ratio $\|p\|_\infty/\|p\|_2^2$ for a probability distribution $p$ on $n$ elements
QUESTION [7 upvotes]: Let $p$ be an arbitrary probability distribution on $[n]:= \{1,2,\dots,n\}$, for some $n\geq 2$. Equating $p$ with its probability mass function, it is not hard to show that its $\ell_r$ norms satisfy
$$
\|p\|_2^2 \leq \|p\|_\infty \leq \|p\|_2 \tag{1}
$$
and in particular
$$
1 \leq \frac{\|p\|_\infty}{\|p\|_2^2} \leq \sqrt{n} \tag{2}
$$
Further, the lower bound is tight in the worst case, as shown by taking $p$ to be the uniform distribution on $[n]$. I am trying to find for which extremal distribution $p$ the upper bound is tight. (Or asymptotically tight, considering a sequence $(p^{(n)})_{n\geq 2}$ of distributions, and letting $n\to \infty$.)
I am able to find a sequence of distributions $(p^{(n)})_{n\geq 2}$ such that
$$
\frac{\|p^{(n)}\|_\infty}{\|p^{(n)}\|_2^2} \operatorname*{\sim}_{n\to\infty} \frac{2\sqrt{n}}{\ln n} \tag{3}
$$
which is almost tight, up to a factor $\frac{2}{\ln n}$:
$$
p^{(n)}_k = \frac{1}{\sum_{\ell=1}^n \frac{1}{\sqrt{\ell}}}\cdot \frac{1}{\sqrt{k}}, \qquad 1\leq k\leq n
$$
as then
$
\frac{\|p^{(n)}\|_\infty}{\|p^{(n)}\|_2^2}  = \frac{\sum_{\ell=1}^n \frac{1}{\sqrt{\ell}}}{H_n}
$, giving (3).

For every $n\geq 2$, is there a distribution $p$ on $[n]$ achieving the upper bound of (2)? What about in the asymptotic sense above, with a sequence $(p^{(n)})_{n\geq 2}$? If not, can the upper bound in (2) be improved by a logarithmic factor (i.e., is (3) tight)?

REPLY [2 votes]: I think the answer by TheSimpliFire was only almost correct but not quite because it identified the wrong $p$ on the $(\|p\|_\infty,-\|p\|_2^2)$ Pareto frontier. Here's my attempt, which shows that $\lim_{n\to\infty}\max_{p\in\Delta^n}\frac{\|p\|_\infty}{\sqrt{n}\|p\|_2^2}=\frac12$, approaching it from above.
Let $\Delta^n$ denote the $(n-1)$-dimensional simplex. We are interested in $\max_{p\in\Delta^n}\frac{\|p\|_\infty}{\|p\|_2^2}$. The solution to this lives somewhere on the $(\|p\|_\infty,-\|p\|_2^2)$ Pareto frontier, which consists of solutions to $\max_{p\in\Delta^n}(\|p\|_\infty-\mu\|p\|_2^2)$ for some $\mu\in[0,\infty]$. The solutions to the latter are the same as permutations of solutions to $\max_{p\in\Delta^n}(p_1-\mu\|p\|_2^2)$ as TheSimpliFire noted. This later optimization problem is convex (i.e., max of concave objective over convex set) and permutation symmetric in entries $2$ through $n$. So an optimal $p$ must have the form $p_i=(1-p_1)/(n-1)~\forall i\geq2$ such that $p_1\geq 1/n$. Such $p$ has $\frac{\|p\|_\infty}{\|p\|_2^2}=\frac{p_1}{p_1^2+(n-1)(\frac{1-p_1}{n-1})^2}$. Now it's just some univariate calculus (aka Wolfram Mathematica) to get that
$$\max_{p\in\Delta^n}\frac{\|p\|_\infty}{\|p\|_2^2}=\max_{p_1\in[1/n,1]}\frac{p_1}{p_1^2+(n-1)(\frac{1-p_1}{n-1})^2}=\frac{\sqrt{n}+1}{2}.$$
Hence,
$\max_{p\in\Delta^n}\frac{\|p\|_\infty}{\sqrt{n}\|p\|_2^2}$ is both $\geq1/2$ and converges $\to1/2$ as $n\to\infty$.