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I wonder if $L^2_{loc}(\mathbb{R}^n)$ is a Hilbert space. It is easy to see that it is an inner product vector spaces, but I am having trouble showing that it is a complete metric space.

MathLearner
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    What is the inner product on $L^2_{\text{loc}}(\mathbb{R}^n)$ you are referring to? In general, $f \in L^2_{\text{loc}}(\mathbb{R}^n)$ only belongs to $L^2(K)$ for compact sets $K \subset \mathbb{R}^n$. Consequently, $\langle f,f \rangle_{L^2(\mathbb{R}^n)}$ may be infinite. – rolandcyp Mar 27 '19 at 15:50
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    Consider the function $f(x) := (1+|x|)^{-1/2}$. Then, $f \in L^2_{\text{loc}}(\mathbb{R})$ but $\langle f,f \rangle_{L^2(\mathbb{R})} = \infty$. – rolandcyp Mar 27 '19 at 15:56
  • You may also find this answer helpful: https://math.stackexchange.com/questions/654597/is-l2-textloc-a-hilbert-space – rolandcyp Mar 27 '19 at 15:57
  • Thank you rolandcyp. – MathLearner Mar 27 '19 at 16:12

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$L^2_{loc}(\mathbb R^n)$ is not a normed space, because it has no bounded neighbourhoods of $0$. Any neighbourhood $V$ of $0$ must contain, for some compact $K$ and $\epsilon > 0$, $U(K,\epsilon) = \{f \in L^2_{loc}(\mathbb R^n): \int_K |f(x)|^2 \; d^n x < \epsilon\}$. But no positive scalar multiple of that will be contained in $U(K', \epsilon)$ where $K \subset \text{Interior}(K')$, so $V$ is not bounded.

Robert Israel
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