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?

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

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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.