# weak-strong convergence in $L^2$ together with bound in $L^\infty$

I want to prove the following result:
Let $$\Omega\subset\mathbb{R}^n$$ be an open bounded set and let $$\{f_n\}\subset L^\infty(\Omega)$$ be a sequence such that $$\lVert f\rVert_{\infty}\leq C$$ and $$f_n\to f$$ strongly in $$L^2(\Omega)$$. Let $$\{g_n\}\subset L^2(\Omega)$$ be another seuqence such that $$g_n\to g$$ weakly in $$L^2(\Omega)$$. Then $$f_n g_n\to fg$$ weakly in $$L^2(\Omega)$$.
My try:
First since $$\{f_n\}$$ is bounded in $$L^\infty(\Omega)$$, then the $$L^2$$ limit $$f$$ should be in $$L^\infty(\Omega)$$ and also bounded by $$C$$. (By paasing the a.e. convergent subsequence, we can easily see this.)
Then I take arbitrary $$\phi\in L^2(\Omega)$$.
\begin{align*} &|\int_{\Omega}f_ng_n\phi dx-\int_{\Omega}fg\phi dx|\\ \leq&|\int_{\Omega}(f_n-f)g_n\phi dx|+|\int_{\Omega}f(g_n-g)\phi dx| \end{align*} the second term goes to zero because of the weak convergence but what about the first term? If I can use Holder inequality, it seems like to be easy, but I do not think $$g_n\phi$$ is in $$L^2(\Omega)$$.
Can someone maybe help?

• For the first term use Cauchy-Schwarz to bound by $\Vert g_n\Vert_2 \cdot \Vert (f-f_n)\phi\Vert_2.$ As $(g_n)_n$ concerges weakly, we have $(\Vert g_n\Vert_2)_n$ bounded. Furthermore, $\Vert f-f_n\Vert_\infty\leq C$ uniformly in $n$ and thus by dominated convergence $\Vert (f-f_n)\phi\Vert_2$ tends to zero. Jul 5 at 18:20

Here is another, more abstract approach: From $$f_n \to f$$ in $$L^2$$ and $$g_n \rightharpoonup g$$ in $$L^2$$, we get $$f_n g_n \rightharpoonup f g$$ in $$L^1$$. Thus, the functionals $$f_n g_n$$ are bounded in $$(L^2)^*$$ and on the dense subspace $$L^\infty$$ of $$L^2$$ they converge to $$f g$$. This implies convergence of all of $$L^2$$, which can be seen as follows:
Let $$\phi \in L^2$$ and $$\varepsilon > 0$$ be given. By density, there exists $$\phi_\varepsilon \in L^\infty$$ with $$\|\phi_\varepsilon - \phi\|_{L^2} \le \varepsilon$$. By weak convergence in $$L^1$$, there exists $$N \in \mathbb N$$ with $$\left| \int (f_n g_n - f g) \phi_\varepsilon \, \mathrm{d} x \right| \le \varepsilon\qquad\forall n \ge N.$$ Thus, $$\left| \int (f_n g_n - f g) \phi \, \mathrm{d} x \right| \le \left| \int (f_n g_n - f g) \phi_\varepsilon \, \mathrm{d} x \right| + \left| \int (f_n g_n - f g) (\phi_\varepsilon - \phi) \, \mathrm{d} x \right| \le \varepsilon + \|f_n g_n - f g\|_{L^2} \varepsilon \qquad \forall n \ge N.$$ This shows $$f_n g_n \rightharpoonup f g$$ in $$L^2$$.