Show that for any $g \in L_{p'}(E)$, where $p'$ is the conjugate of $p$, $\lim_{k \rightarrow \infty}\int_Ef_k(x)g(x)dx = \int_Ef(x)g(x)dx$

Question

Let $1 < p < \infty, f_k \in L_p(E), k = 1, 2, ..., $ and $\lim_{k \rightarrow \infty}f_k(x) = f(x)$ a.e., $\sup_{1 \leq k<\infty}||f_k||_p \leq M$. Show that for any $g \in L_{p'}(E)$, where $p'$ is the conjugate of $p$, $\lim_{k \rightarrow \infty}\int_Ef_k(x)g(x)dx = \int_Ef(x)g(x)dx$

This is from a past qual. Not really sure what to do. Thought about using dominated convergence theorem. I did show that is $E$ is finite, $f_k \rightarrow f$ in $L_1(E)$ by vitali convergence theorem. I think I was able to extend this to $m(E) = \infty$. But, this doesn't show $f_k \rightarrow f$ in $L_p$ which is more useful. Any suggestions?

Answer 1

1. First reduction: considering $f_k-f$ instead of $f$, we can assume that $f=0$. Indeed, assume that we showed the result in the case $f=0$ and let see how to deal with the general one. We define $g_k:=f_k-f$. In this way, $g_k\to 0$ almost everywhere. Notice that by Fatou's lemma, $\lVert f\rVert\leqslant \liminf_n\lVert f_n\rVert$, hence for each $k$, $\lVert g_k\rVert\leqslant 2M$. If $g$ belongs to $L^{p'}$, we thus have (by the case $f=0$): $$\lim_{k\to \infty}\int g_kg\mathrm d\mu=0$$ and noticing that $\int g_kg\mathrm d\mu=\int f_kg\mathrm d\mu-\int fg\mathrm d\mu$ we get the wanted conclusion.

2. Second reduction: using an approximation of $g$ by simple functions and boundedness in $\mathbb L^p$ of $(f_k)$, we can reduce to the case where $g$ is the characteristic function of a set of finite measure, say $A$. Indeed, assume we can solve the problem when $g$ is a simple function. Fix $\varepsilon>0$ and take $g'$ a simple function such that $\lVert g-g'\rVert_{p'}\lt \varepsilon$. Then for each $k$, $$\left|\int f_kg\mathrm d\mu\right|\leqslant \left|\int f_kg'\mathrm d\mu\right|+\left|\int f_k(g-g')\mathrm d\mu\right|$$ and using Hölder's inequality and the assumption on the sequence $(f_k)$, the second term of the RHS is smaller than $M\varepsilon$. Since the problem is assumed to be solved when $g$ is simple, using this with $g'$ we get that $$\limsup_{k\to \infty}\left|\int f_kg\mathrm d\mu\right|\leqslant M\varepsilon.$$

3. We thus have to deal with the case $g=\chi_A$, with $A$ of finite measure and $f=0$. Notice that $\int |f_k|\chi\{|f_k|>R\}d\mu\leqslant \lVert f_k\rVert_p\mu\{|f_k|>R\}^{1/p'}\leqslant R^{-p/p'}M^{1+p'/p}$ and use the dominated convergence theorem for $\int_A |f_k|\chi\{|f_k|\leqslant R\}d\mu$.

Answer 2

This solution is based on reducing to the case of finite measure with $g$ is bounded, and then using Egorov's Theorem.

Observe that $$ \int |g|^{p'}1_{(1/n\leq |g|\leq n)}d\mu\rightarrow \|g\|_{p'}^{p'}. $$ Thus, you can pick $n\in\mathbb N$ such that $$ \| g 1_{(1/n\leq |g|\leq n)^c}\|_{p'}\leq \epsilon/6M. $$

Note now that $$ \mu(1/n\leq |g|\leq n)\leq\mu(1/n\leq |g|)\leq \int |g|^{p'}n^{p'}d\mu<\infty, $$ so the set $(1/n\leq |g|\leq n)$ has finite measure. By Egorov's Theorem, pick $F\subset (1/n\leq |g|\leq n)$ such that $f_k\rightarrow f$ uniformly on $F$ and $\mu ((1/n\leq |g|\leq n)\setminus F)<(\epsilon /6Mn)^{p'}$. Furthermore, pick $k$ such that, on $F$, $|f_k-f|\leq\epsilon/3\|g\|_{p'}\mu(F)^{1/p}$. Then \begin{align*} |\int (f-f_k)gd\mu|&\leq |\int (f-f_k)g1_{(1/n\leq |g|\leq n)^c}d\mu| + |\int (f-f_k)g1_{F}d\mu| \\ &\hspace{1cm}+ |\int (f-f_k)g1_{((1/n\leq |g|\leq n)\setminus F)}d\mu|\\ &\leq \|f-f_k\|_p\|g1_{(1/n\leq |g|\leq n)^c}\|_{p'} + \|(f-f_k)1_F\|_p\|g\|_{p'} \\ &\hspace{1cm} +n\|(f-f_k)\|_p\|1_{((1/n\leq |g|\leq n)\setminus F)}\|_{p'}\\ &\leq (2M)(\epsilon/6M) + (\epsilon/3\|g\|_{p'})(\|g\|_{p'})+(2MN)(\epsilon/6MN)\\ &=\epsilon. \end{align*}