Infimum of p-norms given unit 1-norm

Question

I am trying to prove that over the vectors $x \in \mathbb{R}^n_+ + \{0\}$ such that $\lVert x \rVert_1 = 1$ and for (positive?) real $p$ that $\inf \lVert x \rVert_p = \sqrt[p]{1/n^{p-1}}$.

I feel really daft because I feel like this should be very straightforward, but I only have a hand-wavy proof for integer $p$ and I need something rigorous.

Edit: I do not think $1/p$ is correct, I found a mistake in my proof and am trying to correct it... sorry. I still don't know how to go about this.

Answer 1

Answer for $1<p<\infty$: Let $q=\frac p {p-1}$ so that $\frac 1 p+\frac 1 q=1$. By holder's inequality $\|x\|_1=1$ implies $1 \leq \|x\|_p\|x\|_q$. Now $\|x\|_q \leq n^{1/q}$ because $|x_i| \leq 1$ for each $i$. Hence $ n^{-1/q}\leq \|x\|_p$. The lower bound $ n^{-1/q}$ is attained when $x_i=\frac 1n$ for each $i$. Hence the required infimum is $n^{-1/q}=n^{-p/{p-1}}=(\frac 1{n^{p-1}})^{1/p}$.

Answer 2

Given non-negative real numbers $x_i$ with

$$\sum x_i = 1,$$

you want to minimize

$$\sum x^p_i$$

with $p > 0$.

You didn't give your "heuristic" argument, but here is a way to make some of those arguments rigorous. The region you are interested in is compact, and the function you want to minimize is continuous, so you know maximum is obtained. But now you can assume you have a minimum, take small perturbations, and make deductions about the parameters. Let $a = x_1$ and $b = x_2$. It follows that $a^p + b^p$ is minimized subject to the constraint that $a,b \ge 0$ and $s:=a+b$ is fixed.

So you want to minimize $a^p + (s - a)^p$ for $a \in [0,s]$. Take the derivative and set to zero, you get $a = s/2$, so the minimum must be taken from $a \in \{0,s/2,s\}$, for which you get the values $s^p, 2^{1-p} s^p, s^p$ respectively. Hence:

1. If $p \ge 1$, then the minimum is $a = b$. It follows that any global minimum must have $x_i = x_j$ for all pairs (because the minimum exists).

2. If $0 < p < 1$, then the minimum is $a=0$ or $b=0$. It follows that any global minimum must have at least one of $x_i$ and $x_j$ equal to zero for any pair, and thus all but one of the $x_i$ non-zero. Hence the minimum is obtained for

$$(1/n,1/n,\ldots,1/n)$$

for $p > 1$ and

$$(1,0,\ldots,0)$$

for $0 < p < 1$. Only the case $p \ge 1$ corresponds to an actual norm.