# A converse to Jensen’s inequality for sums of powers

2023-09-28

Abstract

We find a suitable converse to a certain instance of Jensen’s Inequality involving sums of powers.

# Introduction

Fix any $$x \in \mathbb R^d$$. By Jensen’s inequality for the square, $\left(\sum_{i=1}^d x_i^6\right) \left(\sum_{i=1}^d x_i^2\right) ~\ge~ \left(\sum_{i=1}^d x_i^4\right)^2 .$ In this post we are looking for a converse. That is, we want to find a function $$f(d)$$ such that for all $$x \in \mathbb R^d$$ we have $\left(\sum_{i=1}^d x_i^6\right) \left(\sum_{i=1}^d x_i^2\right) ~\le~ f(d) \left(\sum_{i=1}^d x_i^4\right)^2 .$ What type of inequality is this? We are asked to prove $\|x^3\| \cdot \|x\| ~\le~ \sqrt{f(d)} \|x^2\|^2 ,$ where all powers of $$x$$ are understood component-wise, and the norm is Euclidean. Equivalently, we may read this as a statement about the relation between $$p$$-norms for $$p \in \{2,4,6\}$$, namely $\|x\|_6^6 \cdot \|x\|_2^2 ~\le~ f(d) \|x\|_4^8 .$ We can further observe that $$f(d) \ge 1$$ is necessary just by plugging in any constant $$x_i=c$$ for all $$i$$. It must be that $$f$$ is increasing as we can pad $$x$$ with zeroes to increase its dimension without changing the requirement. Finally, the question is invariant to scaling: scaling $$x$$ by a constant $$\alpha$$ scales both sides by $$\alpha^8$$.

## First step with max $$\to$$ sum $$\to$$ max

To see that the problem is not hopeless (i.e. forcing $$f(d)=\infty$$), we can use \begin{equation}\label{eq:max.and.sum} \max_i x_i^2 ~\le~ \sum_i x_i^2 ~\le~ d \max_i x_i^2. \end{equation} We find $\left(\sum_i x_i^6\right) \left(\sum_i x_i^2\right) ~\le~ \left(d \max_i x_i^6\right) \left(d \max_i x_i^2\right) ~=~ d^2 \left(\max_i x_i^4\right)^2 ~\le~ d^2 \left(\sum_i x_i^4\right)^2$ showing that $$f(d) = d^2$$ works. But is this tight? We should have some worry, as the two inequalities in $$\eqref{eq:max.and.sum}$$ cannot both hold with equality for the same $$x$$.

## Hölder

We may apply Hölder with $$\|\cdot\|_1$$ and $$\|\cdot\|_\infty$$ to find $\left(\sum_i x_i^6\right) \left(\sum_i x_i^2\right) ~\le~ \left(\sum_i x_i^4\right) \left( \max_i x_i^2 \right) d \max_i x_i^2 ~=~ d \left(\sum_i x_i^4\right) \left( \max_i x_i^4 \right) ~\le~ d \left(\sum_i x_i^4\right)^2$ Already a factor $$d$$ won! But are we there now? Are there $$x$$ for which this factor $$d$$ actually manifests? What do hard $$x$$ for our inequality look like?

# Optimum

Let’s abbreviate $$y_i = x_i^2$$ and consider the function defined on $$\mathbb R_+^d$$ by \begin{equation}\label{eq:g} g(y) ~=~ \frac{\left(\sum_i y_i^3\right) \left(\sum_i y_i\right)}{\left(\sum_i y_i^2\right)^2} . \end{equation} With this notation, the optimal answer is $$f(d) = \max_{y \in \mathbb R_+^d} g(y)$$.

An optimal point $$y$$ for $$g$$ needs to cancel the derivative, i.e. we must have for each $$i$$ $0 ~=~ 3 y_i^2 M_1 M_2 -4 y_i M_3 M_1 + M_3 M_2$ where $$M_r = \sum_i y_i^r$$. Hence the solution must satisfy $y_i ~=~ \frac{2 M_3}{3 M_2} \pm \sqrt{\left(\frac{2 M_3 }{3 M_2}\right)^2 - \frac{M_3}{3 M_1}}$ Cool, so there are only two levels in an optimal $$y$$. Let’s call $$d_+$$ and $$d_-$$ the numbers of each type. To further find an optimal $$y$$, we must then have \begin{align*} M_r ~=~ d_+ \left( \frac{2 M_3}{3 M_2} + \sqrt{\left(\frac{2 M_3 }{3 M_2}\right)^2 - \frac{M_3}{3 M_1}} \right)^r + d_- \left( \frac{2 M_3}{3 M_2} - \sqrt{\left(\frac{2 M_3 }{3 M_2}\right)^2 - \frac{M_3}{3 M_1}} \right)^r \end{align*} Let’s call these two levels $$x_i^2 \in \{1,y\}$$. How many of each do we need? Let’s optimise over the fraction $$p = \frac{d_+}{d} \in [0,1]$$. We have $\max_p \frac{\left(p + (1-p) y^3\right) \left(p + (1-p) y\right)}{\left(p + (1-p) y^2\right)^2} ~=~ \frac{(1+y)^2}{4 y} \qquad \text{at} \qquad p ~=~ \frac{y^2}{1+y^2}$ Now an optimal $$p,y$$ pair will hit the boundaries of $$p \in [1/d,1-1/d]$$. For $$y \in [1, \sqrt{d-1}]$$ the optimal $$p \le 1-1/d$$, and the best value obtainable for such $$y$$ is $\frac{1}{4} \left( \sqrt{d-1} + \frac{1}{\sqrt{d-1}} + 2 \right).$ If we go higher with $$y$$, we need to clamp $$p$$ to $$p=1-1/d$$. And indeed, we need to solve for $\max_{y \ge 0}~ \frac{\left(d-1 + y^3\right) \left(d-1 + y\right)}{\left(d-1 + y^2\right)^2}$ We can bound this using $$\max\{a,b\} \le a+b \le 2 \max \{a,b\}$$ by $\max_{y \ge 0} \frac{ 4 \max \{d-1,y^3\} \max \{d-1,y\} }{ \max\{d-1,y^2\}^2 } ~=~ 4 \sqrt{d-1}$ which is maximised precisely at $$y = \sqrt{d-1}$$ where it takes the value on the right. If we instead use the max vs sum relation to obtain a lower bound, we find that it is at least $$\frac{1}{4} \sqrt{d-1}$$. So we sandwiched $$f(d)$$ up to a small constant factor. Nice.