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?

One Answer

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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