Let $\Omega$ be a measurable set having finite Lebesgue measure. Let $p\geq 1$ and $u\in L^p(\Omega)$. Is it true that the minimum value of the real function $$ c\in \mathbb{R}^n\mapsto\int_\Omega |u-c|^p \mathrm{d}\mu $$ is achieved when $c$ is the average of $u$? I'm interested in the case $p\neq 2$. Indeed for $p=2$ this is obvious.
By the way, if the average is no more a minimizer when $p\neq 2$, is it possible to find an explicit expression for a minimizer when $p\neq 2$?
Let us try an easy counter-example:
$\Omega=[0,1]^2, \ u(x)=x_1x_2\ $ so that $\ u_\Omega=\frac{1}{4}$ and $p=4$. Then the integral $\int_\Omega |x-c|^4dx$ is a non-negative polynomial in $c$ and has a (local) minimum whenever its derivative (in $c$) vanishes. Thus the local minima of the integral can be found when $c$ takes values among the roots of the polynomial
$$ p(c):=\int_0 ^1 \int_0 ^1 (x_1x_2-c)^3 dx_1dx_2 =-c^3+\frac{3}{4}c^2-\frac{1}{3}c+\frac{1}{16},$$ at least if my calculations are correct. Now it is easy to check that $p(\frac{1}{4} )=0.0104167\neq 0$ so $u_\Omega=1/4$ is not even a local minimum.
It doesn't look like it is possible to get a "general formula" for the minimizing $c$ since in the case that $p=2k$ is an even integer (say) you would have to solve a polynomial equation of degree $(2k-1)$ (with the integrals $\int_\Omega u^\ell $ appearing as coefficients), and thus no general solution can be written down in a closed form.
