Real Analysis – Dual Space of Locally Integrable Function Space

Question

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

Obeservation 2.

If $f \mapsto L(f)=\int f(x) \ g(x) \ dx$ is a linear functional, then $g$ should be of the form of $h \ \chi_{K}$ 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

$$\{ \chi_{K}\ g : g \in L^\infty, \text{ compact } K \}$$

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

Answer 1

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 $f\in L^1_{\rm loc}(\mathbb R)$.