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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0I adjusted the title to be less confusing. – 2012-04-25
1 Answers
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There are closed subspaces of $L^1[0,1]$ consisting of square-integrable functions (e.g. every finite-dimensional subspace is closed). However, the closed graph theorem shows that on every $L^1$-closed subspace of $L^2$ the two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are eqivalent. This may help in concrete situations.