# The topology of $L_\mathrm{loc}^2 (\mathbb{R})$

In fact I am reading the book of Ohsawa, Analysis of Several Complex Variables, and I came across this line on page 13,

... $L^{2}_\mathrm{loc}(\Omega)$ with respect to the topology induced by the $L^2$ convergence on compact sets.

Yet I am not clear how to induce a topology from. I have a guess. In the following I am considering the case $\Omega=\mathbb{R}$. Define $||f||_n=\int_{-n}^n |f|$ for $f\in L^2_\mathrm{loc}(\mathbb{R}), n=1,2,3,\ldots$. Put $$||f||=\sum\limits _{n=1}^\infty\frac{||f||_n}{1+||f||_n} \frac{1}{2^n}.$$ Then $||\cdot||$ is a metric for $L_\mathrm{loc}^2 (\mathbb{R})$. Is it this topology? Or something else? Would someone be kind enough to give me some hints on this? Thank you very much.

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If your intention is $\|f\|_n = \left(\int_{-n}^{n} |f|^{\color{red}{2}}\right)^{\color{red}{1/2}}$ then you're correct. The way you write it you'd get the topology of $L_{\operatorname{loc}}^{\color{red}{1}}(\mathbb{R}) \supset L_{\operatorname{loc}}^2(\mathbb{R})$. For general $\Omega$ just exhaust it by compact sets such that $K_{n+1}$ contains $K_{n}$ in its interior and check that the topology does not depend on the exhaustion. – t.b. Sep 18 '11 at 14:46
@t.b. Thank you very much for the hints. – user14242 Sep 19 '11 at 13:08
@t.b.:something not on mathemmatics:how did you edit to get the equation concerning $||f||$ in a single line?And how did you get my question into several paragraphs?(this looks indeed nice). – user14242 Sep 19 '11 at 13:15
To get a formula displayed like this $$f(x) = \int_{0}^x e^{-t}\,dt$$ simply enclose it in double dollar signs $$f(x) = \int_{0}^{x} e^{-t}\,dt$$. To get paragraphs you need to hit enter twice. You can see what I did in detail by clicking here (you get there by clicking on edited xx time ago above my user name next to yours). – t.b. Sep 19 '11 at 16:09
@t.b.:thanks again!:) – user14242 Sep 20 '11 at 15:49

While it is nice to have a metric, it's often not very useful. For example if $\Omega \subseteq \mathbb R^d$ is open, $C^\infty(\Omega)$ can be given a metric that turns it into a Fréchet space, but it's much simpler (and often more useful) to state what the mode of convergence on the space is: uniform convergence of $f_n$ and all its derivatives on compact subsets of $\Omega$.
In the language of topological vector spaces, we can say that we're specifying the topology on $C^\infty(\Omega)$ as the initial topology with respect to the seminorms $\lVert f\rVert_{K,\alpha} := \sup_{x \in K}|D^\alpha f(x)|$ , where $K$ ranges over the compact subsets of $\Omega$ and $\alpha$ ranges over multiindices in $\mathbb N_0^d$.
Contrast this with how hard it is to work with a metric on the space: first we have to find a countable family of nested compact sets $(K_n)$ that exhaust $\Omega$, which there is no canonical way of doing. Then we have to define our metric as (something like) $$d(f,g) := \sum_{n=1}^\infty 2^{-n}\min\{\sup\limits_{x\in K_n} \max\limits_{|\alpha| \leq n} \left|D^\alpha (f-g)\right|,1\},$$
and I think we can all see that proving that the topology produced by this metric is independent of the choice of the $K_n$ is rather irksome.
In other words, while your construction does indeed work, it's kind of missing the point: Just saying that $L^2_{\mathrm{loc}}$ has the topology of $L^2$-convergence on compact subsets of $\mathbb R^n$ is enough to simultaneously specify the topology and describe its most useful feature.