Sequence of continuous functions $(f_k)$ which converge to $f$ in $C(\Bbb R)$. Does it imply that $f_k$ will converge to $f$ in $\mathcal D'(\Bbb R)$ as well? Or we need something more to make this assertion?
[Here $\mathcal D'(\Bbb R)$ is a linear continuous functional on $\mathcal D(\Bbb R)$ ; collection of all linear maps: from $\mathcal D(\Bbb R)$ to $\Bbb R$)
and we say $f\in$ $\mathcal D'(\Bbb R)$, is continuous in the sense that for all $\phi_n\to \phi$ in $\mathcal D(\Bbb R)$ => $(f,\phi_n)\to(f,\phi)$ in $\mathcal D'(\Bbb R)$]
It seems that the type of convergence considered is: $f_n\to f$ in $C(\mathbb{R})$ if for every fixed compact $K\subset \mathbb{R}$ we have that $f_n\to f$ uniformly in $K$. By using this definition we have that $f_n\to f$ as a distribution, i.e. if we define $T_n(\phi)=\int_\mathbb{R} f_n\phi$ and $T(\phi)=\int_\mathbb{R}f\phi$, $\phi\in \mathcal{D}(\mathbb{R})$ then $T_n\to T$ in $\mathcal{D}'(\mathbb{R})$
To prove the claim in the last paragraph, let $K=\overline{\operatorname{support}(\phi)}$, where by definition $K$ is a compact set. Hence, $f_n\to f$ uniformly in $K$, which implies that
$T_n(\phi)=\int_\mathbb{R}f_n\phi=\int_K f_n\phi\to \int_Kf\phi=\int_\mathbb{R} f\phi=T(\phi)$ for every $\phi\in \mathcal{D}(\mathbb{R})$