Suppose $X \subset L^1_{([0,1])}$ is the subspace consisting of all square-integrable functions. I have to show that $X$ is not a closed subset of $L^1_{([0,1])}$. How do I go about doing this? What exactly do I need to prove?
Show $L^2$ is not closed in $L^1$
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real-analysis
functional-analysis
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3You need to find a sequence of square-integrable functions whose limit in the $L^1$ norm is not square integrable. A useful function to think about here is $x^{-1/2}$. – 2012-04-25
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0http://meta.math.stackexchange.com/questions/3286/how-do-i-accept-an-answer – 2012-04-25
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0As Jochen's answer shows, this is not true as stated. Did you mean $X$ to be *the* subspace consisting of *all* square-integrable functions, or at least that $X$ be infinite-dimensional? – 2012-04-25
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0Fixed, sorry haha. – 2012-04-25
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2@Nate: $X$ infinite-dimensional would not be good enough: If $Y_n$ is a sequence of independent Gaussians, all $L^p$-norms on their closed linear span $X$ in $L^2$ are constant multiples of the $L^2$-norm, so $X$ is a closed subspace of all $L^p$-spaces for $1 \leq p \lt \infty$. – 2012-04-25
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0I adjusted the title to be less confusing. – 2012-04-25