1
$\begingroup$

Let $f_n,g_n,f,g\in L^{2}(\mathbb{R})\cap C^{0}(\mathbb{R})$ with $$ f_n \overset{*}{\underset{n\to +\infty}\rightharpoonup} f $$ weakly in $L^{2}(\mathbb{R})$ $$ g_n \overset{*}{\underset{n\to +\infty}\rightharpoonup} g$$ weakly in $L^{2}(\mathbb{R}).$ If I had the hypothesis, $$f_n {\underset{n\to +\infty}\rightarrow} f $$ strongly in $L^{\infty}(K)$ on every $K$ compact subset of $\mathbb{R};$ if there a way to have $$ \underset{n\to +\infty}{\lim}\int_{\mathbb{R}}{f_ng_n}=\int_{\mathbb{R}}fg $$ ??

$\endgroup$
1
$\begingroup$

This statement is false. By a theorem of Radon and Riesz we know that for $1 < p < \infty$ the following conditions are equivalent:

  1. $f_n \rightarrow f$ in $L^p$-strongly.
  2. $f_n \rightarrow f$ in $L^p$ weakly and $\|f_n\|_p \rightarrow \|f\|_p$.

For the special choice $f_n = g_n$ we would already get strong-convergence in $L^p$.

Counterexample for the statement: Let $\phi \in C_c^\infty(\mathbb{R})$ be a non-trival bumpfunction with compact support in $[0,1]$ and let $f_n(x) = g_n(x) = \phi(x+n)$ and $f = g =0$. Then we have $f_n \rightarrow f$ uniformly on any compact set. Moreover, using Hölder-inequality, we have $$\Big| \int f_n h \, \mathrm{d} x \Big|^2 \leq \|\phi\|_\infty^2 \int I(n \leq x \leq n+1) h(x)^2 \mathrm{d}x.$$ The last term vanishes for $n \rightarrow \infty$ because of the dominated convergence theorem. Thus $f_n \rightarrow 0$ weakly. However, we see that $\int |f_n|^2 \mathrm{d} x$ is a constant independent of $n$ and also non-negative by the choice of $\phi$. Thus $$\int g_n f_n \mathrm{d} x \not\rightarrow 0 = \int f g \mathrm{d} x.$$

$\endgroup$
2
  • $\begingroup$ Ok, thanks for this example. Same result if I know that $f$ and $g$ are non-trivial? $\endgroup$
    – CintyL5
    Apr 21 '18 at 23:20
  • $\begingroup$ In the above example we could replace $f_n$, resp. $g_n$, by $\phi(x+n) + \phi(x)$ and $f =g= \phi$. Of course, we could construct similar examples with different $f_n$ and $g_n$. $\endgroup$
    – p4sch
    Apr 22 '18 at 9:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.