0
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ 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. $\endgroup$ Jul 5 at 18:20

1 Answer 1

1
$\begingroup$

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 $L^2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .