Let $\Omega \subset \mathbb{R}^n$ be an open set and let $1 < p < \infty$.
Consider a sequence $(f_n)_n \subset L^p(\Omega)$ and $f \in L^p(\Omega)$ such that $f_n$ converges to $f$ weakly in $L^p(\Omega)$.
Let $g_n: \Omega \rightarrow \mathbb{R}$ and $g: \Omega \rightarrow \mathbb{R}$ be measurable functions such that $g_n \rightarrow g$ in measure and $||g_n||_{L^\infty} \le C$, $\forall n$ and for some constant $C > 0$.
Prove that $g \in L^\infty (\Omega)$ and $f_n g_n \rightharpoonup fg$ in $L^p(\Omega)$.
$\textbf{Note}$: $g_n \rightarrow g$ in measure if $\forall \epsilon > 0$, $\mu\{x \in \Omega : |g_n(x) – g(x)| > \epsilon\} \rightarrow 0$, where $\mu$ in this case denotes Lebesgue measure.
Here is my attempt.
Since $g_n \rightarrow g$ in measure, there exists a subsequence $(g_{n_k})_k$ such that $g_{n_k} \rightarrow g$ a.e. in $\Omega$. It follows that $|g_{n_k}(x)| \rightarrow |g(x)|$ for a.e. $x \in \Omega$. By hypothesis, $|g_{n_k}(x)|\le ||g_{n_k}||_{L^{\infty}} \le C$ for all $k \in \mathbb{N}$ $\Rightarrow$ $|g(x)| \le C$ for a.e. $x \in \Omega$. This implies that $g \in L^{\infty}(\Omega)$.
We aim to prove that $f_n g_n \rightharpoonup fg$ in $L^p(\Omega)$ $\Longleftrightarrow$ $\forall \phi \in L^q(\Omega)$ (where $\frac{1}{p} + \frac{1}{q} = 1$) , $\int_{\Omega} f_n g_n \phi dx \rightarrow \int_{\Omega} fg \phi dx$.
First of all, let us check $f_n g_n \in L^p(\Omega)$.
$||f_n g_n||_{L^p}^p = \int_{\Omega} |f_n g_n|^p dx \le ||g_n||_{L^{\infty}}^p ||f_n||_{L^p}^p < \infty$ since, by hypothesis, $f_n \in L^p(\Omega), g_n \in L^\infty(\Omega), \, \forall n$. (The same argument holds true also for $fg$).
Note that $f_n g_n – fg = (f_n -f)g_n + f (g_n – g)$. Then
$|\int_{\Omega} f_n g_n \phi dx – \int_{\Omega} fg \phi dx| = |\int_{\Omega} (f_n g_n -fg) \phi dx| \le |\int_{\Omega} (f_n-f) g_n \phi dx| + |\int_{\Omega} f(g_n -g) \phi dx|$.
Let us focus on the first integral on the right hand-side of the expression above.
$|\int_{\Omega} (f_n-f) g_n \phi dx| \le ||g_n||_{L^{\infty}} |\int_{\Omega} (f_n-f) \phi dx| \le C |\int_{\Omega} (f_n-f) \phi dx| \rightarrow 0$, since $f_n \rightharpoonup f $.
Let $\epsilon > 0$ and define $A_{\epsilon} := \{x \in \Omega: |g_n(x) – g(x)| < \epsilon\}$. Convergence in measure implies that $|\int_{\Omega} f(g_n -g) \phi dx| \le |\int_{A_{\epsilon}} f(g_n – g) \phi dx| + |\int_{\Omega \setminus A_{\epsilon}} f(g_n -g) \phi dx| < \epsilon \int_{A_{\epsilon}} |f| \phi dx + \int_{\Omega \setminus A_{\epsilon}} |f(g_n -g) \phi| dx$ $< \epsilon \int_{\Omega} |f| \phi dx + \int_{\Omega \setminus A_{\epsilon}} |f(g_n -g) \phi| dx$. The second integral tends to zero for $n \rightarrow \infty$ because $g_n \rightarrow g$ in measure (which means that $\mu(\Omega \setminus A_{\epsilon}) \rightarrow 0$, the thesis is a consequence of the absolute continuity of the integral), while $\int_{\Omega} |f| \phi dx$ is merely a constant.
If anyone could check the above reasoning, it would be greatly appreciated.
Best Answer
As pointed out by r9m, the step $$|\int_{\Omega} (f_n-f) g_n \phi dx| \le ||g_n||_{L^{\infty}} |\int_{\Omega} (f_n-f) \phi dx|$$ may not be valid, for example, if $f=0$, $f_n=\mathbf{1}_{[n,n+1)}-\mathbf{1}_{(-n-1,-n]}$ and $g_n=\frac{1}{n}f_n$, the right hand side is zero if $\phi$ is even.
It is actually better to slip in the following way: $$ \int f_ng_n\phi-\int fg\phi=\int f_n\left(g_n-g\right)\phi+\int f_ng\phi-\int fg\phi $$ because one can directly exploit the weak convergence assumption to get the convergence to $0$ of the difference of the last two integrals. It remains to check that $a_n:=\int f_n\left(g_n-g\right)\phi\to 0$. To do so, define for a fixed $\varepsilon$ and a fixed $n$ the set $A_{n,\varepsilon}=\{\lvert g_n-g\rvert>\varepsilon\}$. Then $$ \lvert a_n\rvert\leqslant \left\lvert \int_{A_{n,\varepsilon}}f_n\left(g_n-g\right)\phi \right\rvert+\left\lvert \int_{A_{n,\varepsilon}^c}f_n\left(g_n-g\right)\phi \right\rvert\leqslant 2C\int_{A_{n,\varepsilon}}\lvert f_n\rvert\lvert \phi\rvert +\varepsilon\int_{A_{n,\varepsilon}^c}\lvert f_n\rvert\lvert \phi\rvert $$ and by Hölder's inequality, we get $$\lvert a_n\rvert\leqslant 2C\sup_{N\geqslant 1}\lVert f_N\rVert_p\lVert \phi\mathbf{1}_{A_{n,\varepsilon}}\rVert_q+\varepsilon \sup_{N\geqslant 1}\lVert f_N\rVert_p\lVert \phi \rVert_q.$$ Since $\mu(A_{n,\varepsilon})\to 0$ for each positive $\varepsilon$, it follows that $\lVert \phi\mathbf{1}_{A_{n,\varepsilon}}\rVert_q\to 0$ hence $\limsup_n \lvert a_n\rvert\leqslant \varepsilon \sup_{N\geqslant 1}\lVert f_N\rVert_p\lVert \phi \rVert_q$, giving the result.