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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 $p\neq 2$?

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Have you tried any examples? –  Anthony Quas Apr 25 '13 at 0:12
    
Try taking the derivative of $f(c):=\int_\Omega |u-c|^p\;d\mu$, and then think about justifying it later (dominated convergence, etc). Then you can see why $c$ should be the average of $u$ when $p=2$, and what you might expect otherwise. –  Daniel Spector Apr 25 '13 at 6:37
    
That's why I wrote "I'm interested in the case $p \neq 2$". I expect something different when $p \neq 2$ (when computing the Euler-Lagrange equations), but still I have some motivation to believe that the result is still true for $p \neq 2$. –  Josh Apr 25 '13 at 10:49
    
I have not tried any example. I do not like to try for counterexamples. But if you have a counterexample you can answer, instead of commenting :-) –  Josh Apr 25 '13 at 10:59
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up vote 1 down vote accepted

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.

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