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Suppose $1<p<\infty$ and $q$ is the conjugate exponent to $p$. Suppose $f_n\to f$ a.e. and $\sup_n||f_n||_p<\infty$. Prove that if $g \in L^q$, then $\lim_{n\to\infty}\int f_ng=\int fg$.

My attempt: Using Holder's inequality, I got $|\int f_ng|\le \int|f_ng| \le ||f_n||_p ||g||_q \le M||g||_q$ for some $M>0$.

However, I stuck here and I have no idea how to proceed. Does the boundedness of integral implies $\lim_{n\to\infty}\int f_ng=\int fg$? If not, what should I do?

bellcircle
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2 Answers2

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Let $M = \sup_{n}\|f_n\|_{p}$. From Fatou's lemma, note that $$\int |f|^p = \int \lim_{n} |f_n|^p\leq \liminf_{n}\int |f_n|^p \leq M^p$$ thus $\|f\|_{p}\leq M$. Now, let $\epsilon > 0$ and consider the set $X_N = \{x\in X: |g(x)| > 1/N\}$. Since $\int |g|^q < \infty$ ($g\in L^q$), there exists some $N$ such that $\int_{X\setminus X_n}|g|^{q} < \epsilon$ and by Chebyshev's inequality we have $\mu(X_N) < \infty$. By Egorov's theorem, there is a $Y\subset X_N$ such that $\int_{X_N\setminus Y}|g|^q < \epsilon$ and $f_n\to f$ uniformly on $Y$. We have

\begin{align*} \Bigg|\int fg - \int f_n g\Bigg| &\leq \int_{Y}|f - f_n||g| + \int_{X\setminus Y}|f - f_n||g|\\ &\leq \|f-f_n\|_{p}\|g\|_{q} + \|f - f_n\|_{p}\|g\|_{q}\\ &\leq \|f - f_n\|_{u}\mu(Y)^{1/p}\|g\|_{q} + (\|f\|_{p} + \|f_n\|_{p})(2\epsilon)^{1/q}\\ &\leq \|f - f_n\|_{u} \mu(Y)^{1/p}\|g\|_{q} + 2M(2\epsilon)^{1/q}\\ \end{align*} (Note that I applied Egorov to the measure $|g|^q d\mu$ and used Holder's inequality on second line). By uniform convergence on $Y$ we are done.

Wolfy
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