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If $f$ and $f_k$ are integrable functions to $\mathbb{R}$ on an closed interval and $\{f_k\}$ converges pointwise to $f$ with $\sup_{k\in\mathbb{N}}\Vert f_k\Vert_\infty<\infty$, I think $$\lim_{k\rightarrow\infty}\Vert f_k-f\Vert_{L^2}=0$$ holds.

But I can't get the proof. If there is no boundedness condition, I know it would not converges in $L^2$.

Moreover, there is a solution of this problem at Pointwise convergence implies $L^{2}$ convergence, but I need a proof with a basic analysis, not including several concepts in the measure theory or other advanced courses.

Is there anyone who can give me a proof or hints?

tinlyx
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Analysis
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  • I would try using the norm bound to approximate the functions by continuous functions on a sufficiently small set, and then using the fact that continuous functions coverging pointwise on compact sets converge uniformly. – Zach L. Dec 13 '13 at 04:19

1 Answers1

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This one requires the Lebesgue bounded convergence theorem: If $\{ g_{n}\}$ is a sequence of measurable functions on $I$ with $|g_{n}| \le M$ on $I$ for all $n$, and if $\{ g_{n}\}$ converges pointwise to $g$ on your closed interval $I$, then $g$ is measurable, bounded by $M$, and $\lim_{n} \int_{I}g_{n}\,dx = \int_{I}g\,dx.$

In your case, $M=\sup_{n}\|f_{n}\|_{\infty} < \infty$, and $|f| \le M$ because $\{ f_{n}\}$ converges pointwise to $f$. Therefore, $g_{n}=|f_{n}-f|^{2} \le 4M^{2}$ and converges pointwise to 0. So the Lebesgue bounded convergence theorem applies to give $\lim_{n}\|f_{n}-f\|_{L^{2}}=0$.

Disintegrating By Parts
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