The mean of $f$ minimizes $||f-\xi||_p$ over all $\xi\in\Bbb R$. [Reference request]

Question

I would like to know where I can find a reference for the following fact:

Let $\Omega$ be a set of finite measure, $1<p<\infty$ and $f\in L^p(\Omega,\mu)$. Denote $\bar f:=\frac 1{|\Omega|} \int_\Omega f d\mu$. Then $$ \bar f = \arg\min \{ ||f-\xi||_p : \xi\in\Bbb R \}. $$

For $p=2$ the result is extremely well-known whereas for $p=1$ the above doesn't hold.

I am surprise that I couldn't find a reference for the above result for a general $p>1$ since I assumed it to be somewhat well-known as well. Does anyone know a book where this theorem is stated?

Answer 1

There is no such reference, because the result is not true for $p\ne 2$.

Let $\Omega=(0,3)$, $f = \chi_{(0,1)}$. Then for $s\in \mathbb R$, $$ g(s):=\int_\Omega |f-s|^p = |s-1|^p + 2|s|^p. $$ Now clearly $s<0$ is not a minimum, as well as $s>1$. For $s\in (0,1)$, the derivative of $g$ is given by $$ g'(s) = p( 2s^{p-1} - (1-s)^{p-1} ). $$ The integral mean of $f$ is $1/3$, so $$ g'(1/3)= p\ 3^{1-p}\ (2 - \cdot 2^{p-1}) = p\ 3^{1-p}\ 2(1-2^{p-2}), $$ which is zero if and only if $p=2$.

For $p=1$ the minimum is at $s=0$. Does this qualify as median?

Answer 2

This result only holds for $p = 2$. Indeed, if we denote $$J(\xi) = \int_\Omega |f - \xi|^p \, \mathrm d\mu,$$ we get (at least formally) $$J'(\xi) = \int_\Omega p \, |f - \xi|^{p-2} \, (f - \xi) \, \mathrm d\mu.$$ For $p = 2$, this is $0$ iff $\xi$ is the mean of $f$. But this is not true for $p \ne 2$.