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I know that if $(f_n)_{n\in\mathbb{N}}$ is a sequence of $L^p(\mathbb{R}^d)$ functions than if $f_n\xrightarrow{wk}f$ (weak convergence), we also have $\int_{\mathbb{R}^d}f_n g \to \int_{\mathbb{R}^d}f g$. But is the same true for weak-$\star$ convergence? If $f_n\xrightarrow{wk\star}f$ then $\int_{\mathbb{R}^d}f_n g \to \int_{\mathbb{R}^d}f g$? (here $1/p+1/q=1$).

roi_saumon
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  • Where is $g$? Is it in $L^q$ where $q$ is the conjugate of $p$? Is $p>1$ or $p \geq 1$? – N. S. Nov 29 '20 at 18:04
  • thank you, I edited. I guess I am interested in what would happen if $p>1$ as well as $p=1$ and maybe even $p=\infty$ – roi_saumon Nov 29 '20 at 18:06
  • For $1 <p <\infty$, this follows from the reflexivity of $L^p$ spaces, which implies weak and weak star convergence are equivalent. This is not true in $L^\infty $. – rubikscube09 Nov 29 '20 at 18:10

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Note When $p >1$ then $L^p$ is a reflexive Banach space, and therefore the weak and weak-* topologies coincide.

For $p=1$, $L^1$ is not the dual of anything, see for example the link @rubikscube09 suggested here or this MO post. This means we cannot talk about the weak* topology on $L^1$.

N. S.
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  • To see why $L^1$ has no predual, the following answer is helpful: https://math.stackexchange.com/questions/137677/what-is-the-predual-of-l1 – rubikscube09 Nov 29 '20 at 18:12
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    @rubikscube09 Thank you, I realised that my argument was not true. – N. S. Nov 29 '20 at 18:14
  • Tank you. Is the result still true for $(f_n)_{n\in\mathbb{N}}\subset L^{\infty}(\mathbb{R^d})$? Why? (the space is not reflexive this time) – roi_saumon Nov 29 '20 at 18:47
  • @roi_saumon I think no for the following reason. The unit ball is weak* compact, but it is not weak compact (as weak compactness of the unit ball is equivalent to reflexivity). Moreover, the weak topology on the unit ball $B$ is metrisable by Eberlein-Smulain theorem. Finally, the weak* topology is metrisable on $B$ since $L^1$ is separable. – N. S. Nov 29 '20 at 19:37
  • ... This implies that there exists a sequence $f_n$ in the unit ball which is weak* convergent but not weak convergent. This sequence converges in the weak* topology but I think doesn't satisfy the given relation (not sure) – N. S. Nov 29 '20 at 19:41