# Prove that the functional in $C_c^0(Omega)$ is a Radon measure

Mathematics Asked on January 7, 2022

Let $$Omega subset mathbb{R}^n$$ be an arbitrary open set and $$(x_n)_{n inmathbb{N}} subset Omega$$ a sequence. Let $$(a_n)_{n inmathbb{N}} subset mathbb{C}$$ be a sequence such that
$$sum_{j=1}^{infty} |a_j| < infty.$$

I want to prove that the functional $$T:C_c^0(Omega) longrightarrow mathbb{C}$$
given by
$$T(varphi)=sum_{j=1}^{infty} a_jvarphi(x_j),; forall ; varphi in C_c^0(Omega)$$
is a Radon meausure in $$Omega$$, that is, I want to prove that $$T$$ is continuous when $$C_c^0(Omega)$$ is equipped with the topology inductive limit of the spaces $$C_c^0(K)$$, here $$K subset Omega$$ is an arbitrary compact subset of $$Omega$$.

I thought of the following: it is enough I prove that $$T$$ is continuous in $$C_c^0(K)$$, for all $$K subset Omega$$ compact, where the space $$C_c^0(K)$$ is (in particular) a Banach space with the norm
$$|f|_K=sup_{x in K} |f(x)|, ; forall ; f in C_c^0(K).$$

That’s what I thought true? This is the way?

Your thought is in the right direction.

You can also consider $$b_n=max(a_n,0)$$ and $$c_n=-min(a_n,0)$$.

Then $$T_+phi=sum_n b_nphi(x_n)$$ and $$T_- phi=sum_n c_nphi(x_n)$$. Both $$T_+$$ and $$T_-$$ are nonenagtive bounded operators on $$C_{00}(Omega)$$. The Reisz-Markov representation implies that $$T_+$$ and $$T_-$$ can be represented as positive finite measures on $$(Omega,mathscr{B}(Omega))$$. $$T=T_+-T_-$$ and $$|T|=T_+ +T_-$$ (this last part requires a small justification, but it is not complicated), thus $$T$$ is represented by a Radon measure ($$|T|$$ is the variation measure of $$T$$).

Answered by Oliver Diaz on January 7, 2022

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