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I'm trying to prove the following exercise:

Suppose that $f,f_n,g\in\mathcal{L}^2(\mathbb{R})$, $n=1,2,...$, $f_n$ converges to $f$ $\mu-$almost everywhere, and $$ \int_{\mathbb{R}}|f_n(x)|^2 dx\leq 1 , n=1,2,... $$ then we have: $$ \lim_{n\rightarrow \infty}\int_{\mathbb{R}}|f_n(x)-f(x)||g(x)|dx=0 $$ Here are my attempts:

use the Cauchy inequality, we have: $$ \left(\int_{\mathbb{R}}|f_n(x)-f(x)||g(x)|dx\right)^2\leq \int_{\mathbb{R}}|f_n(x)-f(x)|^2dx\int_{\mathbb{R}}|g(x)|^2dx $$ and since $g\in \mathcal{L}^2(\mathbb{R})$, $\int_{\mathbb{R}}|g(x)|^2dx<\infty$, it suffices to prove that: $$ \lim_{n\rightarrow \infty}\int_{\mathbb{R}}|f_n(x)-f(x)|^2dx=0 $$ but I was stucked there since I can't find a dominating function and apply the Lebesgue Dominated Convergence Theorem. I also tried Vitali Convergence Theorem, but in vain as well. Can anybody give me some hints?

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  • $\begingroup$ Your last statement isn't true. $f_n(x) = \sqrt{n} \chi_{[0,1/n]}(x), f(x) = 0, g(x) = \chi_{[0,1]}(x)$ where $\chi_S$ is the characteristic function $1$ if $x \in S$ and $0$ if $x \notin S$. $\endgroup$
    – aschepler
    Commented May 6, 2022 at 14:44

1 Answer 1

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What is asked to show is that boundedness in $\mathbb L^2$ combined with almost everywhere convergence implies that $\lvert f_n-f\rvert$ converges weakly in $\mathbb L^2$ to $0$. In general, as the example given by aschepler $f_n(x) = \sqrt{n} \chi_{[0,1/n]}(x), f(x) = 0$, strong convergence in $\mathbb L^2$ does not hold.

Here is a way to solve the question.

  1. Replacing $f_n$ by $\lvert f_n-f\rvert$, it suffices to treat the case $f=0$ and $f_n$ non-negative.

  2. Using the fact that linear combinations of indicator functions of sets of measure $0$ are dense in $\mathbb L^2$, boundedness of $\left(\lVert f_n\rVert_{\mathbb L^2}\right)_{n\geqslant 1}$ and Cauchy-Schwarz inequality, it suffices to show that $\int f_n(x)\mathbf{1}_A(x)dx\to 0$ for each set $A$ of finite measure.

  3. To do so, for a fixed $\varepsilon$, let $B_n=\{x\in\mathbb R\mid f_n(x)>\varepsilon\}$. Then $$ \int f_n(x)\mathbf{1}_A(x)dx=\int f_n(x)\mathbf{1}_A(x)\mathbf{1}_{B_n}(x)dx+\int f_n(x)\mathbf{1}_A(x)\mathbf{1}_{\mathbb R\setminus B_n}(x)dx\leqslant \lVert f_n\rVert_{\mathbb L^2}\lambda\left(A\cap B_n\right)^{1/2}+\varepsilon\lambda(A). $$ To conclude, notice that the first terms is a bounded sequence multiplied by something which converges to $0$ hence for each positive $\varepsilon$, $$ \limsup_{n\to\infty}\int f_n(x)\mathbf{1}_A(x)dx\leqslant \varepsilon\lambda(A). $$

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