Is a continuous functional on continuous functions the restriction of a continuous functional on the space of all functions?

Question

As sets, we can consider the space $C(\mathbf{R}^n;\mathbf{R}^k)$ - of all continuous functions from $\mathbf{R}^n$ to $\mathbf{R}^k$ - to be a subset of the product space $(\mathbf{R}^k)^{\mathbf{R}^n}$ (because the latter is the set of all functions from $\mathbf{R}^n$ to $\mathbf{R}^k$)

Take $C(\mathbf{R}^n;\mathbf{R}^k)$ with the topology of uniform convergence of compact sets and suppose I have a continuous, bounded function $F$ on this space i.e. $F \in C_b\bigl( C(\mathbf{R}^n;\mathbf{R}^k)\bigr)$.

Is there some function $\widetilde{F} \in C_b\bigl( (\mathbf{R}^k)^{\mathbf{R}^n}\bigr)$ - i.e. a continuous, bounded function on the product space with the product topology - for which $$ \widetilde{F}(f) = F(f) $$ for every $f \in C(\mathbf{R}^n;\mathbf{R}^k)$?

Answer 1

$\newcommand\R{\mathbb R}$No. E.g., suppose that $n=k=1$ and $$F(f)=\min\Big(1,\int_0^2 |f(x)|\,dx\Big)$$ for $f\in C(\R;\R)$. Then $F$ is continuous and bounded on $C(\R;\R)$.

For natural $m$, let $$f_m:=f_{m^2;1/m},$$ where $f_{c;a}(x):=c\max(0,a-|x-a|)$ for real $x$. Then $f_m\in C(\R;\R)$ for each $m$, $f_m\to 0$ pointwise (as $m\to\infty$), but $F(f_m)=1\to1\ne0=F(0)$. So, $F$ cannot be extended to a continuous functional on $\R^\R$. $\quad\Box$