# $f_k \rightarrow f$ in $L^p$ and that $g_k \rightarrow g$ weakly in $L^q$. Show that $f_k g_k \rightarrow fg$ weakly in $L^1$

I want to solve the following exercise, and I thankfully welcome some hints. Note that this is not homework.

Problem: Let $1 < p,q < \infty$ be conjugate exponents. Assume $f_k \rightarrow f$ in $(L^p, \| \cdot \|_p)$ and that $g_k \rightarrow g$ weakly in $(L^q, \| \cdot \|_q)$. Show that $f_k g_k \rightarrow fg$ weakly in $(L^1, \| \cdot \|_1)$.

I realize that normed convergence implies weak convergence and that $(L^q)^\ast$ can be identified with $L^p$. What I need to show is that

$$|\phi(f_k g_k) - \phi(f g)| \rightarrow 0, \ \ \forall \phi \in (L^1)^\ast$$

which follows if I can show $\| f_k g_k - f g \|_1 \rightarrow 0$, since

$$|\phi(f_k g_k) - \phi(f g)| \leq \|\phi \| \| f_k g_k - f g \|_1.$$

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Any assumption on the measure space? –  Davide Giraudo Dec 7 '12 at 19:18
No, unfortunately not. –  DoubleTrouble Dec 7 '12 at 20:05

In the $\sigma$-finite case, we can try this. The dual space of $L^1$ can be identified with $L^\infty$, so what you have to show is that for all $\phi\in L^\infty$, we have $$\int_X f_ng_n\phi\to \int_X fg\phi.$$ We have $$\left|\int_X f_ng_n\phi- \int_X fg\phi\right|\leqslant \lVert \phi\rVert_\infty\lVert f-f_n\rVert_p\sup_k\lVert g_k\rVert_q+\left|\int_X fg_n\phi- \int_X fg\phi\right|.$$ Now use the fact that $\sup_k\lVert g_k\rVert_q$ is finite and $f\phi\in L^p$.

As no assumption is done on the measure space we are working, we have to see how to solve the problem when the measure space is not assumed $\sigma$-finite.

The functions $\{f_n\}$ and $\{g_n\}$ are fixed. Define for positive integers $n,k$, $I_{n,k}:=\{|f_n|>k^{-1}\}$ and $J_{n,k}:=\{|g_n|>k^{—1}\}$. Each of these sets has finite measure by the assumptions of integrability of powers of these functions, and if we define $X':=\bigcup_{n,k\geqslant 1}I_{n,k}\cup J_{n,k}$ with the underlying measure, we still have a measure space, which is this time $\sigma$-finite. All the integrals in the first part taken on $X$ have the same values if these ones are taken on $X'$.

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Doesn't $(L^1)^* \cong L^\infty$ require some assumptions on the measure space? $\sigma$-finiteness suffices AFAIR. –  martini Dec 7 '12 at 19:16
Yes, you are right –  DoubleTrouble Dec 7 '12 at 20:03
@martini You are right. Maybe taking representents for $f_n,g_n$ and working on the sets where their doesn't vanish brings us to a $\sigma$-finite space. –  Davide Giraudo Dec 7 '12 at 20:07
@DavideGiraudo That should work. As $f_n \in L^p$, we have that $\{|f_n| > \frac 1k\}$ has finite measure for any $n,k$, likewise for the $g_n$. Now we can restrict ourselves to the $\sigma$-finite(!) union of these sets. –  martini Dec 7 '12 at 21:33
Good idea, I will try to work out the details, and then come back and accept your answer :) –  DoubleTrouble Dec 7 '12 at 22:04