Let $V := C_c (\mathbb{R}^n)$ be the space of real-valued continuous functions on $\mathbb{R}^n$ with compact support. Define $\phi_n \rightarrow \phi$ in $V$ if the convergence is uniform and there exists a compact set $K \subseteq \mathbb{R}^n$ with spt$ \phi_n \subseteq K$ and spt$\phi \subseteq K$, where "spt" means support.
A linear functional $\mu$ on $V$ is called continuous if $\mu(\phi_n ) \rightarrow \mu(\phi)$ whenever $\phi_n \rightarrow \phi$ in $V$.
Suppose that $\mu_n$ is a sequence of continuous linear functionals on $V$ and suppose moreover that $\mu_n(\phi)$ is bounded by some constant (depending on $\phi$) for each $\phi \in V$.
Can we prove that the sequence $|\mu_n| :=\sup\{\mu_n (\phi) | $spt$ \phi \subseteq B_1 , |\phi(x)| \leq 1 \}$ is bounded? Here, $B_1$ is the unit open ball centered at the origin.
(I am self-studying some measure theory, and the above question appeared in a section concerning Radon measures. The book says, "use Banach-Steinhaus theorem", but I can't get it since I don't think that $V$ is metrizable into a Banach space.)
(You may use the stronger assumption that there exists a continuous linear $\mu$ such that $\mu_n (\phi) \rightarrow \mu(\phi)$ for each $\phi$.)
Any help will be appreciated.
Best Answer
Consider the restriction of each $\mu_n$ to $C_0(B_1) = \{\phi \in C(B_1): \phi(x) \to 0 \text{ as } |x| \to 1\}$ equipped with the $\sup$-norm. This is a Banach space and $\mu_n$ is a continuous linear functional on it for each $n$. Now you are in the setting where you can apply Banach-Steinhaus to conclude immediately.