2015年丘赛分析组个人赛试题

Proof.(Weingarten) $L^2(\mathbf R)$ is a Hilbert space, and $\|f_n\|_{L^2}\to\|f\|_{L^2}$ as $n\to\infty$ , so we only need to prove $f_n\in f$ weakly in $L^2(\mathbf R)$, that is, for each $g\in L^2(\mathbf R)$, there holds

\[\lim_{n\to\infty}\int_{\mathbf R}f_ng=\int_{\mathbf R} fg.\]

For each $\epsilon>0$, there exists $R>0$, such that \[\int_{|x|>R}|g|^2<\epsilon^2,\]

and by the absolute continuity of integration of $g$, there exists a positive $\delta$, such that: for any Lebesgue measurable subset $E$ of $\mathbf R$ with $m(E)<\delta$, there holds \[\int_E|g|^2<\epsilon^2.\]

By the Egoroff's thoerem, there exists a subset $E_\delta$ of $(-R,R)$ with $m((-R,R)\setminus E_\delta)<\delta$, such that the convergence $\lim\limits_{n\to\infty}f_n=f$ is uniform on the $E_\delta$, so there exists $N\in\mathbf Z_+$, such that \[(\forall n>N,x\in E_\delta),(|f_n-f|<\epsilon/\sqrt{2R}).\]

Assume $M=\|g\|_{L^2}+\|f\|_{L^2}+\sup\limits_n\|f_n\|_{L^2}$, hence $\forall n>N$, we have the following estimations

\begin{align*}\int_{\mathbf R}|f_n-f|\cdot|g|&=\left(\int_{E_\delta}+\int_{(-R,R)\setminus E_\delta}+\int_{|x|>R}\right)|f_n-f|\cdot|g|\\&\leq\sqrt{\int_{E_\delta}|f_n-f|^2\cdot\int_{E_\delta}|g|^2}+\sqrt{\int_{(-R,R)\setminus E_\delta}|f_n-f|^2\cdot\int_{(-R,R)\setminus E_\delta}|g|^2}\\&+\sqrt{\int_{|x|>R}|f_n-f|^2\cdot\int_{|x|>R}|g|^2}\\&\leq M\epsilon+2\sqrt{2}M\epsilon.\end{align*}

the proof is finished.