Let $f_n\in L^4(\mathbb{R})\cap L^3(\mathbb{R})$. Suppose that $$ f_n\rightarrow f \mbox{ weak in } L^3$$ $$ f_n\rightarrow \tilde{f} \mbox{ weak in } L^4$$ Can i conclude that $f=\tilde{f}$ almost everywhere on $\mathbb{R}$?

This is not clear, because weak convergence doesn't imply convergence a.e along a subsequence. I'm interested also in the case where both convergence are weak*, and in the case where one convergence is weak and the other is weak*.

Thank you for any suggestions.


Hint: It follows that $$\int\phi f=\int\phi\tilde f$$for every $\phi\in L^{4/3}\cap L^{3/2}$.

  • $\begingroup$ Thank you, now it's clear. And what about weak* convergence? the same trick doesn't work right? $\endgroup$ – Capublanca Sep 4 '16 at 15:51
  • $\begingroup$ @Capublanca I don't see why weak* convergence is any different. $\endgroup$ – David C. Ullrich Sep 4 '16 at 16:14
  • $\begingroup$ OK, for sure it's the same in the case $L^4$ and $L^3$. But what if we have $L^1$ (or $L^{\infty})$? $\endgroup$ – Capublanca Sep 4 '16 at 16:25
  • $\begingroup$ @Capublanca There's no such thing as weak* convergence in $L^1$, since $L^1$ is not a dual space. What if we have weak convergence in $L^1$ and weak* convergence in $L^\infty$? That's no different - we still have $\int \phi f=\int\phi \tilde f$ for every $\phi\in L^1\cap L^\infty$... $\endgroup$ – David C. Ullrich Sep 4 '16 at 16:37
  • $\begingroup$ @Capublanca One might be concerned about the case where we have weak convergence in $L^\infty$, since we don't have a good handle on the dual of $L^\infty$. But that doesn't matter either; it's still true that $L^1\subset (L^\infty)^*$. $\endgroup$ – David C. Ullrich Sep 4 '16 at 16:40

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