Pointwise convergence to $\exp(-x^2)$ implies uniform convergence in set of functions of compact support

Question

Let $C_c (\mathbb{R}) = \{f : \mathbb{R} \to \mathbb{R} \mid f$ is continuous and there exists a compact set $K$ such that $f = 0$ on $K^c\}$.

Let $g(x) = \exp(-x^2)$.

Is the following statement true?

If a sequence in $C_c (\mathbb{R})$ converges pointwise to $g$, then it must converge uniformly to $g$.

Answer 1

The statement is not true.

Consider $h_n$ a sequence who uniformly converges to $g$ and let $f_n=h_n+b_n$ where $b_n$ is a hat function with its maximum located at $n$ of value $1$ and a base of width $1$.

Then $\|f_n-g\|_{\infty} \to 1$ (hence we do not have uniform convergence) but $f_n$ converges pointwise to $g$.

Answer 2

No, it is not true. For $n \in \mathbb N_+$, let $f_n$ be the function which equals $0$ on $(-\infty, -n - 1]$ and $[n + 1, \infty)$, which equals $g$ on $[-n, n]$ and which is linear on the intervals $[-n-1, -n], [n, n+1]$.

Now let $g_n$ equal $f_n$ on all of $\mathbb R$ except $(0, 1/n)$. Let $g_n(1/(2n)) = 0$, and let $g_n$ be linear on both the intervals $[0, 1/(2n)]$ and $[1/(2n), 1/n]$. Then $(g_n)$ is a sequence in $C_c(\mathbb R)$ which converges pointwise to $g$, but not uniformly.