Space of test functions defined by norms This is the problem assigned:

So I know that a locally convex Hausdorff space is defined by a vector space and a family of seminorms. So is part $a$ just wanting me to show that $\|\phi\|_m$ is in fact a seminorm? I'm a bit confused over what the question is asking me to actually show.
For part 2, I know that:
$$\|\phi_m - \phi_n\| = \| \sum_{j=m+1}^n \frac{\phi(x-j)}{j}\| \leq  \sum_{j=m+1}^n \frac{\|\phi(x-j)\|}{j}.$$
I would think this converges to $0$ as $m,n \rightarrow \infty$ which shows it's Cauchy. I don't know how to prove it doesn't converge to any function in $\mathcal D(\mathbb{R})$.
 A: For $(1)$, to prove that $\mathcal D(U)$ with $\{\| \cdot \|\}_{m \in \mathbb N}$ is a locally convex topological vector space, you need to show that $\{\|\cdot\|_m\}_{m \in \mathbb N}$ is a separating family of semi-norms on $\mathcal D(U)$. The question already says "norms" in it, so maybe you just have to show that it's separating, i.e., given $\phi \in \mathcal D(U)$, there exists $m \in \mathbb N$ such that $\| \phi \|_m \neq 0$. To prove that the resulting space is in fact a metric space, you have to provide the metric
$$
\text{(Hint: } \qquad d(\phi, \psi) = \sum_{m \in \mathbb N} \frac{1}{2^m} \frac{\|\phi - \psi\|_m}{1 + \|\phi - \psi\|_m} \text{ )}
$$
and LCTVS and metric topologies are equivalent (by restricting your attention to neighborhoods of $0$). The Hausdorff bit then follows from the metric space bit.
For $(2)$, convergence in $\mathcal D(\mathbb R)$ with the topology described by $(1)$ is defined by:
$$
\phi_n \to \phi \iff d(\phi_n, \phi) \to 0 \iff \|\phi_n - \phi\|_m \to 0 \quad\forall m \in \mathbb N \qquad \text{ as $n \to\infty$.}
$$
Cauchy-ness is actually defined by
$$
d(\phi_n, \phi_m) \to 0 \iff \|\phi_n - \phi_m \|_N \to 0 \quad \forall N \in \mathbb N \qquad \text{as $n, m \to \infty$}.
$$
because we have a family of semi-norms and/or a metric space, but not a normed space. Using these definitions, you need to show that the sequence
$$
\phi_m(x) = \sum_{j=1}^m \frac{\phi(x - j)}{j}
$$
defined in the problem is Cauchy and that there is no $\psi \in \mathcal D(\mathbb R)$ such that $\phi_m \to \psi$ as $m \to \infty$. Here's a picture showing $\phi(x)$ and $\phi_5(x)$ that should help you to see what's going on:

Since $\mathcal D (\mathbb R)$ is (I assume) $C_c^\infty(\mathbb R)$, $\phi_m$ cannot converge to an element of $\mathcal D(\mathbb R)$ since $\operatorname{supp} \phi_m = [1, m+1]$, which goes to $[1, \infty)$, a non-compact set, as $m \to \infty$.
UPDATE:
Cauchy-ness requires us to find, given $\varepsilon > 0$ and $N \in \mathbb N$, some $N_0 \in \mathbb N$ such that
$$
n, m \geq N_0 \implies \| \phi_n - \phi_m \|_N < \varepsilon.
$$
Assuming wlog $n > m$, we have
\begin{align*}
\| \phi_n - \phi_m \|_N = \left\| \sum_{j=m+1}^n \frac{\phi(x - j)}{j} \right\|_N &\leq \sum_{j=m+1}^n \frac{\| \phi(x - j) \|_N}{j} = \| \phi \|_N \sum_{j=m+1}^n \frac{1}{j}.
\end{align*}
After quite a bit of trying, I don't see how we can bound this (or the result from other approaches) by $\varepsilon$, because (in this case) the harmonic series $\sum_j 1/j$ diverges. However, if we instead defined
$$
\phi_m(x) = \sum_{j=1}^m \frac{\phi(x - j)}{j^2},
$$
as your instructor may have intended, then the above estimate would be $\| \phi_n - \phi_m \|_N \leq \|\phi\|_N \sum_{j=m+1}^n \frac{1}{j^2}$, and since the latter sum converges, its tail $\sum_{j=m+1}^n 1/j^2$ becomes arbitrarily small, even smaller than $\varepsilon/ (2 \|\phi\|_N)$, so we've proved $\| \phi_n - \phi_m\|_N < \varepsilon$ for arbitrary $N$, and the Cauchy condition follows.
