Consider $ T \in\mathcal{D}'(\mathbb{R})$ defined as follows, $$\phi \to T(\phi) = \sum_{n=-\infty}^{+\infty} \phi(n)$$
Next, let $\psi \in \mathcal{D}(\mathbb{R})$ be a function with support in $[-1,1]$ such that $0 \le \psi(x) \le 1$ for all $x \in \mathbb{R}$ and such that $\psi(x) = 1$ for all $x \in [-0.5,0.5]$. Next, consider the sequence $$\phi_n(x) = \frac{1}{\sqrt{1+n}}\psi(\frac{x}{n})$$ Note that $\phi_n$ converges uniformly to zero as $n \to \infty$, however, $T(\phi_n) \to \infty$. Does this contradict the continuity of $T$?
My attempt:
Since the question states $T$ is a distribution, it must be continuous and so I'm led to believe that I must show why this isn't a contradiction.
My thoughts are that $\phi_n \notin \mathcal{D}(\mathbb{R})$, and so $T(\phi_n)$ not converging to zero isn't a contradiction. I believe that for each $n$, $\phi_n(x)$ is in $C^\infty$ and smooth, as $\phi^{(k)}_n(t) = \frac{1}{n^k}\frac{1}{\sqrt{1+n}}\psi^{(k)}(\frac{x}{n})$ and $\psi$ is in $\mathcal{D}(\mathbb{R})$ and the terms in front of it will simply be a constant.
However, we can also note that the support of $\phi_n$ is given by $[-n,n]$. Thus, as $n \to \infty$, the support of $\phi_n$ also goes to infinity and hence, in the limit $\phi_n$ does not have compact support and so is not in $\mathcal{D}(\mathbb{R})$.
I'm not too confident in my answer, but I've been reviewing this question for awhile and can't think of anything else. Does anyone have any ideas?
Uniform pointwise convergence is certainly necessary for convergence of $\varphi_n$ in test functions, but it is not sufficient. One further requirement of a Cauchy sequence (and also Cauchy net!) is that the supports be (uniformly) bounded. A sequence such as you constructed definitely does not have this property. So your $\varphi_n$ is not a Cauchy sequence in the space of test functions, so there's no paradox here.