Does $L^1(R)\cap L^2(R)$ have finite or infinite corank in $L^2(R)$? I guess the latter is the case but I have never seen this discussed, and would like to see a simple proof.
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3$\begingroup$ The family $f_s(x)=(1+|x|)^{-s}$ for $\frac12 <s<1$ spans an infinite-dimensional subspace with trivial intersection. $\endgroup$– user1688Commented May 29, 2018 at 9:19
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It is a truth universally acknowledged, that a Banach space in possession of a continuous and dense inclusion in a strictly larger Banach space must have infinite co-dimension in it. This is also the case of $$L^1(\mathbb{R})\cap L^2(\mathbb{R}),\ \|\cdot\|_1+\|\cdot\|_2 \longrightarrow L^1(\mathbb{R}) ,\ \|\cdot\|_1 . $$
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$\begingroup$ (a consequence of the Open Mapping Theorem) $\endgroup$ Commented May 29, 2018 at 11:14