If the sequence $\{f_n\}$ in $L^2(\mathbb{R})$ converges to $f$ in $L^2(\mathbb{R})$ and $\int_{\mathbb{R}}f_n(t)dt\leq h$ for some $h>0$ and all $n\in \mathbb{N}$; does $f$ satisfy $\int_{\mathbb{R}}f(t)dt\leq h$?
Is the set $\{f: \int_{\mathbb{R}} f \leq h\}$ closed in $L^2(\mathbb{R})$?
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real-analysis
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0@Siminore is right, or you either need a bound $f_n\geq - g$ for some non-negative integrable function. – 2012-08-26
1 Answers
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Not necessarily. Try $f_n=2\,\mathbf 1_{(0,1)}-\frac1n\mathbf 1_{(1,n+1)}$, and $f=2\,\mathbf 1_{(0,1)}$. Then the integral of $(f_n-f)^2$ is $\frac1n$ hence $f_n\to f$ in $L^2$, but the integral of $f_n$ is $1$ and the integral of $f$ is $2$.
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0Great, I think this works! – 2012-08-26