Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $I=[0, 1]$, $1\leq p<+\infty$ and let $A$ be a non empty subset of $\mathbb R^n$. Denote with $S_p(A)$ the set of functions $u\in L^p(I,\mathbb R^n)$ such that $u(x)\in A$ for almost every $x\in I$. Is it true that if $S_p(A)$ is weakly closed in $L^p(I,\mathbb R^n)$ then $A$ is closed and convex?

share|cite|improve this question
Doesn't $A=\{0,1\}$ satisfy the hypothesis? – tomasz Sep 5 '12 at 16:09
how do you prove that, with your choice of $A$, $S_p(A)$ is weakly closed in $L^p(I,\mathbb R^n)$? – guido giuliani Sep 5 '12 at 16:24
up vote 2 down vote accepted

$A$ is closed. If $x_n\in A$ converge to $x\in \mathbb R^n$, then the constant functions $f_n\equiv x_n$ converge in $L^p$ to $x$. Hence $x\in A$.

$A$ is midpoint-convex. Given $a,b\in A$, let $f_n(x)=a$ if $\lfloor nx\rfloor $ is even, and $f_n(x)=b$ otherwise. This sequence converges weakly in $L^p$ to the constant function $f\equiv (a+b)/2$. To prove this, first show that $\int_J f_n \to\int_J f$ for every subinterval $J\subset I$, and then use the density of step functions in $(L^p)^*=L^q$.

Since $A$ is closed and midpoint-convex, it is convex.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.