# The dual space of locally integrable function space

I'm strongly interested in dual spaces. I learned the dual spaces of $L^p$, $L^{\infty}$, $C(X)$ and $M(X)$.

I wonder the dual space of locally integrable function space in $\mathbb{R}^n$. The topology on $L_{loc}^1$ is generated by seminorms

$$p_{K}(f-f_{0})=\int_{K}|f-f_{0}|dx$$ where $K$ is a compact set. I believe that this is standard topology of $L_{loc}^1$. I have some observations.

Observation 1.

Define $$L(f)=\int_{R}f(x)g(x)\chi_{K}(x) dx$$ where $K$ is a compact set, $f \in L_{loc}^1$, $g \in L^\infty$ and $\chi_{K}(x)$ is a characteristic function. Then it is a continuous linear functional.

Obeservation 2.

If $L(f)=\int f(x)g(x)dx$ is a linear functional, then $g(x)$ should be of the form of $h(x)\chi_{K}(x)$ for some $h \in L^\infty$ and some compact $K$.

The above observation can be proved by using reductio ad absurdum.

By the observations, I guess that $(L_{loc}^1)^*$ is the following space

$$\{ \chi_{K}g(x) : g \in L^\infty, compact \, K \}$$

Is it true? If not, what is the counterexample?, and is there any reference about dual space of $L_{loc}^1$?

p.s.

Observation 3.

$$C(\mathbb{R}^n) \subset L_{loc}^1$$ By the Hahn-Banach theorem $$L(f)=f(0)$$ is a continuous linear functional. Therefore, my candidate is wrong. Is there any good candidate for $(L_{loc}^1)^*$?

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The usual notation for your $\mathbb{LIF}$ space is $L^1_{\mathrm{loc}}$. – Giuseppe Negro Jul 16 '14 at 14:39
Oh, thanks. I forgot that. – preferman Jul 16 '14 at 14:41
I don't understand the logic in Observation 3. Yes, continuous functions are locally integrable. But $L(f)=f(0)$ is not a continuous linear functional on $L^1_{\rm loc}$. Inclusion of spaces does not imply inclusion of dual spaces. – Meta Jul 16 '14 at 16:42
@Thisismuchhealthier. You are right. I was stupid. – preferman Jul 16 '14 at 17:08

Every continuous functional $\phi$ on the space $L^1_{\rm loc}(\mathbb R)$ is of the form $\phi(f) = \int fg$ where $g\in L^\infty(\mathbb R)$ and the essential support of $g$ is bounded. Indeed,
1. If $\phi$ is bounded on $L^1_{\rm loc}(\mathbb R)$, then it is bounded on $L^1(\mathbb R)$, since the latter space continuously embeds into the former. It follows that the restriction of $\phi$ to $L^1(\mathbb R)$ is of the form $\phi(f) = \int fg$ with $g\in L^\infty(\mathbb R)$.
2. If the essential support of $g$ is unbounded, we can find a sequence of $L^1$ functions $f_n$ which converges in $L^1_{\rm loc}(\mathbb R)$, but $\phi(f_n)\to\infty$. Namely, $f_n = \sum_{k=1}^n M_k \chi_{I_k}$ where the numbers $M_k$ are large and the disjoint intervals $I_k$ are chosen so that they go off into infinity, and $\int_{I_k} g \ne 0$.
3. Having established that $g$ has bounded essential support, we notice that $f\mapsto \int fg$ is continuous on $L^1_{\rm loc}(\mathbb R)$. Since $L^1(\mathbb R)$ is a dense subspace in this space, continuity implies $\phi(f) = \int fg$ for all $f\in L^1_{\rm loc}(\mathbb R)$.
If instead of $\Bbb R$ I would work with a topological space $X$ without a concept of distance, would I have to modify "bounded" with "precompact"? (Probably choose $X$ paracompact, for argument (2) to work?) – Alex M. Sep 19 '15 at 6:41