# Duals of $C_c(\Omega)$ and $C_0(\Omega)$

Let $$C_c(\Omega)$$ denote the class of continuous functions with compact support and and $$C_0(\Omega)$$ denote the class of continuous functions that vanish at the boundary (for $$\Omega$$ bounded) or at $$\infty.$$

One can consider two topologies on these spaces namely the topology induced by the sup-norm (also called uniform topology, $$\tau_{sup}$$) and inductive limit topology ($$\tau_{ind}$$). The inductive limit topology is finer than the sup-norm topology. I have the following doubts:

1. Clearly $$C_0(\Omega)$$ is the closure of $$C_c(\Omega)$$ under sup norm topology. Does this result hold under the inductive limit topology?

2. What are the duals of these two spaces under two different topologies? More precisely what are the topological duals of the topological spaces $$(C_0(\Omega),\tau_{ind}),$$ $$(C_0(\Omega),\tau_{sup}),$$ $$(C_c(\Omega),\tau_{ind})$$ and $$(C_c(\Omega),\tau_{sup})?$$

P.S: I have seen some books mentioning the space of radon measure as the dual of $$C_c(\Omega)$$ but the underlying topology is not clearly mentioned. I guess it is under inductive limit topology as the radon measures do not define continuous linear functional on $$C_c(\Omega)$$ under uniform topology. So maybe it is an algebraic dual but not topological dual I guess. I see some results on math stack exchange but are not precisely answering my doubts. Any help is appreciated.

• When you say inductive limit topology, what is the corresponding inductive system of topological spaces? Commented Oct 1, 2021 at 11:17

$$C_0(\Omega)$$ is the closure of $$C_c(\Omega)$$ only with respect to the sup topology. The inductive limit topology on $$C_c(\Omega)$$ gives you a complete topological vector space.

The dual of $$(C_0(\Omega),\tau_{sup})$$ is the space of bounded (also called finite) Radon measures. The dual of $$(C_c(\Omega),\tau_{ind})$$ is the space of locally bounded Radon measures. At least, if $$\Omega$$ is $$\sigma$$-locally compact. The two spaces coincide if $$\Omega$$ is compact.

The space $$(C_0(\Omega),\tau_{ind})$$ does not make sense, because the inductive limit topology depends on the sequence of spaces you use to cover the entire functional space and, in general, you cannot cover $$C_0(\Omega)$$ through $$C_c(U)$$ spaces with $$U$$ included in $$\Omega$$.

Finally, the space $$(C_c(\Omega),\tau_{sup})$$ makes sense but bounded linear functionals on this space are naturally identified to bounded linear functionals on $$(C_0(\Omega),\tau_{sup})$$.

If you want to go deep into these questions you have to take a look to these references:

• Irene Fonseca, Giovanni Leoni. Modern Methods in the Calculus of Variations: $$L^p$$ Spaces. Springer Science & Business Media, 2007.
• Charalambos D. Aliprantis, Kim C. Border. Infinite Dimensional Analysis: A Hitchhiker's Guide. Springer-Verlag Berlin and Heidelberg GmbH & Company KG, 2013.
• Nelson Dunford, Jacob T. Schwartz. Linear operators, part 1: general theory. Vol. 10. John Wiley & Sons, 1988.
• The dual is w.r.t inductive limit topology? What is the dual if we consider norm topology on $C_c(\Omega)$ and $C_0(\Omega)$? Commented Oct 1, 2021 at 11:36
• I edited the topologies I am referring to.
– Kosh
Commented Oct 1, 2021 at 11:49
• Due to density of $C_c(\Omega)$ in $C_0(\Omega)$ under uniform topologly, I guess the dual of $(C_c(\Omega),\tau_{sup})$ is same as $(C_0(\Omega),\tau_{sup}).$ Isn't it? Commented Oct 1, 2021 at 11:55
• yes, up to identifications. I gave you possible references.
– Kosh
Commented Oct 1, 2021 at 11:57