If $f_n\to f$ a.e. and are bounded in $L^p$ norm, then $\int f_n g\to \int fg$ for any $g\in L^q$

Question

Suppose $p>1$ and $q$ is its conjugate exponent. Suppose $f_n \rightarrow f~a.e.$ and $\sup_n\|f_n\|_p < \infty $. prove that if $g \in L^q,$ then $\lim_{n \rightarrow \infty} \int f_ng=\int fg.$ Does this extend to the case where $p=1$ and $q=\infty$? If not, give a counter example.

Progress

@Prahlad Vaidyanathan how? $sup_n||f_n||_p< \infty$ tells us $f_n \in L^p$, then get the dominated function $f_N$? even I find the dominated function g, then use the dominated convergence theorem in $L^p$ to get $f_n \rightarrow f $in $L^p$? — aliyun, Mar 08 '15 at 02:24
Is the measure of the underlying space finite? Any conditions on the underlying space? — copper.hat, Mar 08 '15 at 04:15
I found answer in this website：http://math.stackexchange.com/questions/910117/show-that-for-any-g-in-l-pe-where-p-is-the-conjugate-of-p-lim/910209#910209 — mathsara, Mar 09 '15 at 02:25
These answers are incomplete, but this question has been answered fully at http://math.stackexchange.com/questions/442550/pointwise-convergence-and-boundedness-in-norm-imply-weak-convergence. — Tomo, Oct 30 '15 at 02:06

copper.hat · Accepted Answer · 2015-03-08T04:24:46.683

3

You have $\int |f|^p = \int \liminf_n |f_n|^p \le \liminf_n \int |f_n|^p < \infty$, so $f \in L^p$.

Counterexample for $p=1$: Let $X=[0,1]$, $f_n=n1_{[0,{1 \over n}]}$, $\|f_n\|_1 = 1$, $f_n(x) \to 0$ ae., and let $g=1$. Then $\int f_n g = 1$ for all $n$, but $\int 0 g = 0$.

edited Mar 08 '15 at 04:24

answered Mar 08 '15 at 02:52

copper.hat

172,524

how to show $\int |f|^p = \int \liminf_n |f_n|^p$ – aliyun Mar 08 '15 at 02:56
You are given $f_n(x) \to f(x)$ ae., so $|f_n(x)| \to |f(x)|$ ae., so $\liminf_n |f_n(x)|^p = \lim_n |f_n(x)|^p = |f(x)|^p$ ae. – copper.hat Mar 08 '15 at 03:00
We are using Fatou's lemma here. – copper.hat Mar 08 '15 at 03:05
Yes,the second equality is by Fatou lemma. After we know $f \in L^p$, then how to continue it? – aliyun Mar 08 '15 at 03:08
@aliyun: You don't need to accept the answer, I haven't answered all of it. I think one could use Egorov's theorem to answer this, but this requires the underlying space to have finite measure. – copper.hat Mar 08 '15 at 19:00

score 0 · Answer 2 · answered Jul 30 '15 at 14:18

For $p> 1$ it is true that $\int{f_ng}\rightarrow\int{fg}$. This follows from Theorem 3. Because the cited theorem has a different notation, I will write the body of it with your notation (in terms of $p$, $L_p$ ) and then apply it.

Theorem 3.

Let $M$ be a positive constant and $(X,\mu)$ be a complete measure space. Suppose that $0<p\leq \infty$ and that $f,\,f_1,\,f_2,....$ is a sequence of $\mu-$measurable functions, such that $\|f_n\|_p\leq M,\,\,n=1,2,...$ and $\lim\limits_{n\rightarrow\infty}{f_n(x)}=f(x)\quad$ $\mu-$a.e $x\in X$. Then $\lim\limits_{n\rightarrow\infty}{\|f_ng-fg\|_s}=0$ for every $s\in(0,p)$ and every $g\in L_q(X)$, where $\frac{1}{q}+\frac{s}{p}=1$

Now you apply this theorem with $p>1$, $s=1\in(0,p)$ and $\frac{1}{q}+\frac{1}{p}=1$. You get that $\lim\limits_{n\rightarrow\infty}{\|f_ng-fg\|_1}=0\Leftrightarrow \int\limits_{X}{|f_ng-fg|d\mu}=0$. Consequently $|\int\limits_{X}{(f_ng-fg)d\mu}|\leq \int\limits_{X}{|f_ng-fg|d\mu}\rightarrow 0$

If $f_n\to f$ a.e. and are bounded in $L^p$ norm, then $\int f_n g\to \int fg$ for any $g\in L^q$

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