# minimal $L^2$ norm with $L^1$ norm fixed to one [closed]

maybe this is a stupid question, but I dare it anyway: Let $\Omega$ be some bounded domain in ${\mathbb R}^n$. Then under all $L^1(\Omega)$ functions $f$ of fixed $L^1$-norm one, the constant function $\frac 1 {|\Omega|} 1_{\Omega}$ minimises the $L^2(\Omega)$ norm, as a quick Cauchy-Schwarz argument shows.

Question: Is this the only minimizing function or are there others?

-

## closed as too localized by Yemon Choi, quid, Willie Wong, Bill Johnson, Ryan BudneyAug 17 '11 at 16:46

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

The proof of the Cauchy-Schwarz theorem tells you precisely when equality can occur... which should be enough to answer your question. I wouldn't say it's stupid, but unless I've missed something it is not really appropriate for MO, and would have belonged better on math.stackexchange.com – Yemon Choi Aug 17 '11 at 10:14

There are lots of others -- think of the function equal to $1/|\Omega|$ on one half of $\Omega$, and to $-1/|\Omega|$ on the other half. However if you restrict to non-negative functions then it is the only minimizer, as the equality case in Cauchy-Schwarz shows.