Is the $L^2$ limit of a sequence of symmetric functions also symmetric?

Question

I want to show that the subspace of $L^2(\mathbb{R}^2)$ containing all functions which are symmetric in their arguments (i.e. $f(x,y) = f(y,x)$ for almost every $x$ and $y$) is a Hilbert space. As $L^2(\mathbb{R}^2)$ is itself Hilbert, the only thing that remains is to show the said subspace is closed in $L^2$-norm. How do I prove this?

Thanks.

Answer 1

The linear operator $S:L^2(\Bbb R^2)\to L^2(\Bbb R^2)$ given by $[Sf](x,y):=f(y,x)$ is an isometry, hence continuous. The subset of symmetric functions is $\ker(I-S)$, which is closed.