A problem I've been stumped on for a while. Suppose $(f_{n})$ is some sequence in $L^{p}([0,1],\mu)$, where $\mu$ is the Lebesgue measure, such that $f_{n}\to f\in L^{p}$ pointwise a.e., and there exists some $M$ such that $\|f_{n}\|_{p}\leq M$. I'm trying to show that for all $g\in L^{q}([0,1],\mu)$, $\frac{1}{p}+\frac{1}{q}=1$, we have $$\lim_{n\to\infty}\int_{[0,1]}f_{n}g\,d\mu=\int_{[0,1]}fg\,d\mu.$$ It is deceptively challenging; my first instinct was just to use Holder, but I was only able to get our conclusion if we assume $f_{n}\to f$ in $L^{p}$ convergence, not pointwise a.e. convergence.
$\begingroup$ $\endgroup$
1 
- 1$\begingroup$ math.stackexchange.com/q/3966526 $\endgroup$Sine of the Time– Sine of the Time2024-11-10 20:49:18 +00:00Commented Nov 10, 2024 at 20:49
Add a comment |
2 Answers
$\begingroup$ 
$\endgroup$
The claim is false for $p=1,\infty$. For $p \in (1,\infty)$, $L^p$ is reflexive so a bounded sequence admits a weakly convergent subsequence. Let $(f_n)_{n \in M}$ be a subsequence of $(f_n)$. By reflexivity there exists a further subsequence $(f_n)_{n \in L \subset M}$ that converges weakly to some $\tilde f$ in $L^p$. Since $f_n \to f$ a.e., this implies that $f = \tilde f$; see this or this . Hence, every subsequence of $(f_n)$ has a further subsequence that converges weakly to $f$. This implies that $f_n \to f$ weakly.
answered Nov 10, 2024 at 20:23
Evangelopoulos Foivos
4,5441 gold badge7 silver badges28 bronze badges
$\begingroup$ $\endgroup$
4 I don't think this is true. I propose the following counterexample for $p=1$ (which should be a standard thing to consider): $$f_n(x) = \begin{cases} 0 & \text{for } x \leq \frac {n-1} n \\ nx - (n-1) & \text{for }x \geq \frac{n-1}{n}\end{cases}$$ Then $||f_n||_1 = \int_0^1 f_n(x)dx = 1$. Also $(f_n)_n$ converges pointwise almost everywhere (on the interval $[0,1)$)to the zero function. However (for $g(x) = 1$) you have $$1 = \lim_{n\to \infty} \int_0^1 f_n(x) dx \neq \int_0^1 0dx = 0.$$
- $\begingroup$ Hmm doesn't $\int_0^1 f_n$ tend to 0 as well? $\int_0^1 f_n = \int_0^{1/n} nx=\frac12 nx^2|_{x=0}^{x=1/n}=1/2n$ (?) $\endgroup$Al.G.– Al.G.2024-11-10 20:23:02 +00:00Commented Nov 10, 2024 at 20:23
- $\begingroup$ $\| f_n\|_1$ does not seem constant $\endgroup$Sine of the Time– Sine of the Time2024-11-10 20:34:34 +00:00Commented Nov 10, 2024 at 20:34
- $\begingroup$ That's my point, too. The argument would work for the sup norm, as $\|f_n\|_\infty = 1$ for all $n$, while still pointwise $f_n\to 0$. $\endgroup$Al.G.– Al.G.2024-11-10 20:37:03 +00:00Commented Nov 10, 2024 at 20:37
- $\begingroup$ Most likely, you meant $2n(nx-(n-1))$ instead of just $nx-(n-1)$, to make the height of the triangle grow linearly with $n$. Or maybe you meant $p=\infty$. You should correct it and then leave a comment to each of the commenters so we can delete them. $\endgroup$Brian Moehring– Brian Moehring2024-11-10 20:49:38 +00:00Commented Nov 10, 2024 at 20:49