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 $$ \underset{n\to +\infty}{\lim}\int_{\mathbb{R}}{f_ng_n}=\int_{\mathbb{R}}fg $$ ??


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 $$\int g_n f_n \mathrm{d} x \not\rightarrow 0 = \int f g \mathrm{d} x.$$

  • $\begingroup$ 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

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