# Show $L^2$ is not closed in $L^1$

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?

-
You 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 $x^{-1/2}$. – user29743 Apr 25 '12 at 7:52
As 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? – Nate Eldredge Apr 25 '12 at 13:01
Fixed, sorry haha. – Cardflow Apr 25 '12 at 14:12
@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 $1 \leq p \lt \infty$. – t.b. Apr 25 '12 at 14:46

## 1 Answer

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.

-