Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
3  
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. –  Davide Giraudo Jan 25 '12 at 19:28
    
Thanks, Davide! I have edited the question. –  Sarah Grady Jan 25 '12 at 19:50
1  
On which measured space do you work? –  Davide Giraudo Jan 25 '12 at 19:58
1  
On any finite measure space. –  Sarah Grady Jan 25 '12 at 21:20
add comment

2 Answers

This may be a sledgehammer and is only a partial answer for the case when $1<p<\infty$.

Here are three facts:

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

2) For $1<p<\infty$, norm bounded sequences that converge pointwise a.e. are weakly convergent to their pointwise limit (see, e.g, Theorem 13.44 in Hewitt and Stromberg's, Real and Abstract Analysis).

3)Weak limits are unique.

Thus, if $f_n$ converges weakly to $h$ in $L_p$ for $1<p<\infty$ and converges pointwise to $g$ a.e, it follows that $f=g$ almost everywhere.


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. .

share|improve this answer
    
Thank you so very much, David!! –  Sarah Grady Jan 25 '12 at 21:26
    
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! –  Sarah Grady Feb 27 '12 at 18:21
    
@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.. –  David Mitra Feb 27 '12 at 18:50
    
Thanks, David!! –  Sarah Grady Feb 27 '12 at 20:08
add comment

I will assume $1<p<\infty$.Since we work on a finite measured space, we can use Egoroff theorem. Fix $\phi\in L^q$ where $q$ is the conjugate exponent (i.e. $p^{-1}+q^{-1}=1$). Fix $\varepsilon>0$, 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\right|d\mu&\leq \limsup_n\left|\int_X(g-f_n)\phi\right|d\mu+\left|\int_X(f_n-h)\phi\right|d\mu\\ &\leq \limsup_n\left|\int_X(g-f_n)\phi\right|d\mu\\ &\leq \limsup_n\left|\int_{A_{\delta}}(g-f_n)\phi\right|d\mu+\left|\int_{A_{\delta}^c}(g-f_n)\phi\right|d\mu\\ &\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.

share|improve this answer
    
Thank you very much Davide!! –  Sarah Grady Jan 25 '12 at 22:02
add comment

Your Answer

 
discard

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.