Is $L^2_{\text{loc}}$ a Hilbert space?

Question

My question is in the title. I know that $L^2$ is a Hilbert space, but I'm not sure about $L^2_{\text{loc}}$. Is it even an inner product space? Thanks.

Answer 1

Unfortunately $\mathbb L_{\mathrm{loc}}^2(\mathbb R)$ (and in general $\mathbb L_{\mathrm{loc}}^2(\Omega)$, where $\Omega\subset\mathbb R^n$, open) is not, and it can not become a Hilbert space, or an inner product space or even a normed space.

Its topology is determined by a countable family of semi-norms. Let $K_i\subset K_{i+1}\subset \Omega$, compact sets, such that $\cup_{i\in\mathbb N} K_i=\Omega$, and for every $f\in \mathbb L_{\mathrm{loc}}^2(\Omega)$ define $$ p_i(f)=\left(\int_{K_i}|f|^2\,dx\right)^{1/2}, $$ then the $p_i$'s are semi-norms, and $$ d(f,g)=\sum_{i=1}^\infty \frac{2^{-i}p_i(f-g)}{1+p_i(f-g)}, $$ defines a metric in $\mathbb L_{\mathrm{loc}}^2(\Omega)$, which makes $\mathbb L_{\mathrm{loc}}^2(\Omega)$ a Fréchet space.