Let $ \Omega \subset \mathbb{R}^d $ be a bounded and open set. Suppose $ \{f_n\} \subset L^{\infty} (\Omega) , f \in L^{\infty} (\Omega) $. Prove that

$ f_n \rightharpoonup^* f \ \ \text{in} \ \ L_{\infty} (\Omega) \Longleftrightarrow \left( \sup\limits_{n \in \mathbb{N} } \| f_n \|_{L^{\infty}} < + \infty \ \ \text{and} \ \ { \displaystyle \forall_{ V \subset \Omega} \ \lim\limits_{n \to \infty} \ \int\limits_V \ (f_n(x) - f(x)) =0} \right) \\ \text{where } V= \Pi^d_{i=1}(a_i,b_i). $

Can anyone give me any clue? I would appreciate any help.

  • 1
    $\begingroup$ First prove this for hypercubes $\Omega = \prod_{i=1}^d (a_i, b_i)$, then try to generalise using that every bounded open set in $\mathbb R^d$ can be exhausted by open hypercubes. $\endgroup$
    – AlexR
    Dec 18 '14 at 21:18

$\Rightarrow$: a weak* convergent sequence is bounded. For the remaining part, use the definition of weak* convergence with the characteristic function of $V$.

$\Leftarrow$: if $g$ is integrable on $\Omega$, we can approximate in $\mathbb L^1(\Omega)$ this function by linear combinations of sums of charactistic functions of cubes.

  • 2
    $\begingroup$ Worth mentioning in $\impliedby$ that boundedness is used to carry through the approximation. $\endgroup$
    – fourierwho
    Jul 22 '17 at 0:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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