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My question is: For $f_n, g, h \in L^p(X)$, where $X$ is a finite measure space, if $f_n$ converges to $g$ a.e and $f_n$ converges to $h$ weakly in $L^p$, can we conclude any relationship between $g$ and $h$? Thanks!

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    Welcome to math.stackexchange! What does "$g$ converge to $h$" mean? Usually, we talk about convergence for a sequence (or a net), not for a single function.2012-01-25
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    Thanks, Davide! I have edited the question.2012-01-25
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    On which measured space do you work?2012-01-25
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    On any finite measure space.2012-01-25

2 Answers 2

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This may be a sledgehammer and is only a partial answer for the case when $1

Here are three facts:

1) Weakly convergent sequences are norm bounded (see for instance page 255 of this ).

2) For $1Real and Abstract Analysis).

3)Weak limits are unique.

Thus, if $f_n$ converges weakly to $h$ in $L_p$ for $1


More sledgehammers:

For $p=1$ with a finite measure space:

Weak convergence of $\{f_n\}$ insures that $\{f_n\}$ is uniformly integrable. This is known as the Dunford-Pettis theorem (see, e.g., pg. 59 here or IV.8.11 in Dunford and Schwartz' Linear Operators, Part 1, General Theory). By the Vitali Convergence Theorem (see also page 163 here), then, $\{f_n\}$ converges to $g$ in $L_1$. Then, since convergence in norm to $g$ implies weak convergence to $g$, we must have $g=h$ a.e. .

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    Thank you so very much, David!!2012-01-25
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    David. Just to make sure I understood your second(last) answer: Can we conclude $g=h$ a.e. if $ f_n$ converges to $g$ in $L^1$ and $f_n$ converges to $h$ a.e.? Thanks!2012-02-27
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    @SarahGrady I think that would be a different problem (but it's true, since norm convergence to $g$ implies the existence of an a.e. convergent subsequence to $g$). In the second argument above, it was shown that $f_n$ converges to $g$ in $L_1$. Then $f_n$ converges to $g$ weakly; and since weak limits are unique, we must have $g=h$ a.e..2012-02-27
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I will assume $10$, and $\delta>0$ such that if $\mu(A)\leq \delta$ then $\int_A |\phi|^q\leq \varepsilon^q$. Thanks to Egoroff's theorem, we can find $A_{\delta}$ such that $f_n\to g$ uniformly on $A_{\delta}$ and $\mu(A_{\delta}^c)\leq \delta$. So \begin{align*} \left|\int_X(g-h)\phi ~d\mu\right|&\leq \limsup_n \left(\left|\int_X(g-f_n)\phi ~d\mu \right| +\left|\int_X(f_n-h)\phi ~ d\mu \right| \right)\\\ &\leq \limsup_n\left|\int_X(g-f_n)\phi ~ d\mu \right|\\\ &\leq \limsup_n\left(\left|\int_{A_{\delta}}(g-f_n)\phi ~ d\mu \right|+\left|\int_{A_{\delta}^c}(g-f_n)\phi ~d\mu \right|\right)\\\ &\leq \limsup_n\:\sup_{A_{\delta}}|g-f_n|\left(\int_X|\phi|^qd\mu\right)^{\frac 1q}+\sup_k\lVert g-f_k\rVert_{L^p}\left(\int_{A_{\delta}}|\phi|^q\right)^{\frac 1q}\\\ &\leq \varepsilon\cdot \sup_k\lVert g-f_k\rVert_{L^p} \end{align*} so $\int_X(g-h)\phi d\mu=0$ for all $\phi \in L^q$: we conclude that $g=h$ almost everywhere.

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    Thank you very much Davide!!2012-01-25