Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a finite measure on a compact, take $f_n\in L^1$ with norms $\leq 1$ and suppose that $\int f_n g$ tends to a limit for all continuous $g$. Is it true that then $\int f_n g$ converge for any $g\in L^\infty$? How can one prove this?

share|improve this question
3  
mathoverflow.net/faq#whatnot Voting to close. –  Yemon Choi May 11 '11 at 6:33
    
The answer, by the way, is "no". Could you say a little more about where you came across this problem, which special cases you have already tried, and so forth? –  Yemon Choi May 11 '11 at 6:36
1  
Yemon, I agree this is a homework problem (or if not, it should be), but (possibly) at graduate level, so I think you're being a little harsh; it is conceivable that a non-(functional analyst) research mathematician, or even a (non-functional) analyst, might find this a bit tricky. There are many interesting special cases of $f_n$ where the answer is "yes"! –  Zen Harper May 11 '11 at 6:49
1  
In the other direction, asking what extra assumptions on $(f_n)$ and the measure will guarantee the answer "yes" seems to me like an interesting question, possibly connected with Tauberian theories. This is exactly the kind of question which can make research interesting, in my opinion, since I doubt there will be any precise necessary/sufficient characterisation from general theory which isn't just tautological. –  Zen Harper May 11 '11 at 7:29
1  
The needed "something extra" is a condition that guarantees that the sequence has weakly compact closure in $L^1$, such as no subsequence of $f_n$ is equivalent to the unit vector basis for $\ell^1$. –  Bill Johnson May 11 '11 at 21:16
show 10 more comments

1 Answer

If $(f_n)$ is a bounded sequence in $L^1(X,\mu)$, it is true that the set of $g\in L^\infty(X,\mu)$ such that $\int_X f_n g\, d\mu$ has a limit in $\mathbb{C}$ is a norm-closed linear subspace of $L^\infty(X,\mu)$. Thus, from a dense set of test functions $g$ you can infer the weak convergence of the sequence $(f_n)$ (to an element of ($L^\infty)^*$, of course, in general not in $L^1$). In conclusion, in general continuous functions are not enough (they are a closed subspace, usually proper, of $L^\infty$), but e.g. simple functions, which are uniformly dense, will do.

share|improve this answer
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.