$\mathcal D$ is the space of compactly supported smooth functions on $\mathbb R^n$. Show $\mathcal D'$ is complete with respect to weak$^*$-topology. The topology on $\mathcal D$ is the locally convex topology induced by seminorms like $\|\partial_\alpha f\|_\infty$, where $\alpha=(d_1,\ldots,d_n) $ and $\partial_\alpha=\partial_{x_1}^{d_1}\ldots\partial_{x_n}^{d_n}$.
$\mathcal D'$ is its continuous dual space.
The weak^$^*$ topology on $\mathcal D'$ is induced by seminorms like $\|\phi\|_f=|\phi(f)|$, where $f\in \mathcal D$.
It is easy to define the limit of a Cauchy net on $\mathcal D'$: $\Phi(f)=\lim_{\lambda}\phi(f)$.
However, I don't know how to show $\Phi$ is continuous, that is, $\Phi(f_\eta)\to 0$ whenever $\|\partial_\alpha f_\eta\|\to 0$ for all $\alpha\in \mathbb N^n$.
Could someone please give me some hints?
 A: Perhaps to clarify some often-muddled things:
The (continuous) dual $V^*$ of a topological vector space $V$ is a well-defined vector space, and/but has no canonical topology.
It is an old theorem that the weak-star dual of a Frechet space is quasi complete. This means that a bounded (in the TVS sense) Cauchy sequence converges. This implies sequential competeness. ("Bounded" of a set $E$ here means that, given an open nbd $U$ of $0$, there is $t_o>0$ such that, for every $t\ge t_o$, $tU\supset U$.)
The fullest-possible notion of "complete" would be that every Cauchy net converges. In metric spaces, this is provably equivalent to sequential completeness. Generally, although weak-star duals of Frechet spaces are quasi-complete, they are not "complete" in this fuller sense. This phenomenon already occurs for weak-star duals of infinite-dimensional Hilbert spaces. (The proof is not trivial...)
EDIT: to be clear, "continuous dual" of $V$ means continuous linear maps to scalars. This is completely well-defined as a set, and as a vector space (over whatever scalars, real or complex...)
There is no obligatory topology on this vector space. Choice of a topology depends on context. One extreme choice is "weak-star", which has the feature that it is the weakest-possible topology for which $\lambda\to \lambda(v)$ (for $v\in V$, and $\lambda$ in the dual) is continuous. And the continuous-dual of $V^*$ with this topology is $V$, again. So, in this qualified sense, we can make $V$ "reflexive". :)
But there are stronger topologies that are useful, and/or of interest. Already we can see this with the "strong" dual topology for Hilbert spaces, so that the strong dual is a Hilbert space. Similarly for Banach spaces. This becomes more complicated for Frechet and fancier TVSs.
And, to be clear, "continuous dual" does not refer to a topology on the dual. "Weak-star dual" is one choice of a topology on the continuous dual. :)
A: The problem lies within the claim implicit in your question - $\mathcal{D}'$ is not complete in the weak-* topology.
It is beneficial to set this question within the the general framework of dual pairings of vector spaces, as in e.g. Section 8.1 of the book of H. Jarchow, Locally Convex Vector Spaces (B. G. Teubner, 1981). A dual pairing is a triple $(E,F,\mathsf{B})$, where $E,F$ are vector spaces over the field $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$ and $\mathsf{B}:F\times E\rightarrow\mathbb{K}$ is a non-degenerate bilinear form, that is,

*

*If $\mathsf{B}(y,x)=0$ for all $y\in F$, then $x=0$;

*If $\mathsf{B}(y,x)=0$ for all $x\in F$, then $y=0$.

Of course, one may slightly abuse notation and simply call $\mathsf{B}$ the dual pairing between $E$ and $F$, since all the information about the triple is encoded in $\mathsf{B}$ - we shall do so from now on. One can then define the weak topology $\sigma(E,F)$ on $E$ associated with $\mathsf{B}$  as the locally convex vector topology defined by the seminorms $x\mapsto|\mathsf{B}(y,x)|$, $y\in F$ (the dual pairing is implicit from the context and omitted from the notation for the weak topology). The non-degeneracy of $\mathsf{B}$ guarantees that one may identify $F$ with a vector subspace of the algebraic dual $E^*$ of $E$ through the (by 2.) injective linear map $F\ni y\mapsto\mathsf{B}(y,\cdot)\in E^*$ (that, by the way, explains the "weird" order of the arguments of $\mathsf{B}$) and that $\sigma(E,F)$ is Hausdorff since (by 1.) the above seminorms separate the points of $E$. The importance of the weak topology is that $\sigma(E,F)$ is the weakest Hausdorff locally convex vector topology on $E$ such that the topological dual of $E$ with respect to it is $F$ (via the identification $y\leftrightarrow\mathsf{B}(y,\cdot)$). More generally, we say that a locally convex vector topology $\tau$ on $E$ is compatible with $\mathsf{B}$ if $F$ is the topological dual of the locally convex vector space $(E,\tau)$ (under the same identification). Any such topology is finer than the weak topology, of course.
The immediate case that comes to mind is when $F=E^*$ and $\mathsf{B}(y,x)=y(x)$ is the canonical dual pairing between $E$ and $E^*$, but the above setup also encompasses other important cases:

*

*$F$ is a vector subspace of $E^*$ - particularly, if $E$ is a locally convex vector space and $F=E'$ is the topological dual of $E$, and $\mathsf{B}$ is the restriction of the canonical dual pairing to $F\times E$. If $F=E'$, one may also call $\mathsf{B}$ the canonical dual pairing between $E$ and $E'$;

*$E=F^*$ or a vector subspace thereof (e.g. $F$ is a locally convex vector space and $E=F'$), and $\mathsf{B}(y,x)=x(y)$.

The point is that the weak topology $\sigma(E,E')$ of a locally convex vector space $E$ and the weak-* topology $\sigma(E',E)$ of $E'$ (watch out for the exchange of the roles of $E$ and $F$ in this case!) do not depend on the original topology of $E$, only on the dual pairing between $E'$ and $E$ (and vice-versa, respectively), since the weak and weak-* seminorms themselves clearly only depend on that. In these cases, the non-degeneracy with respect to the argument in $E$ is obvious, whereas the non-degeneracy with respect to the argument in $E'$ is a consequence of the Hahn-Banach theorem. That, by the way, is the reason why the duality theory of locally convex vector spaces is much more powerful than that of general topological vector spaces. Just think of how poor is the theory of $L^p$ spaces for $0<p<1$ as compared with the case $1\leq p\leq\infty$.
Going back to the question, it is a standard result (see e.g. Proposition 8.1.5, pp. 147-148 of Jarchow's book) that if $\mathsf{B}$ is a dual pairing between two vector spaces $E,F$, then $\sigma(E,F)$ is complete if and only if $E=F^*$. Phrased differently, $F^*$ endowed with the topology $\sigma(F^*,F)$ is a completion of $E$ if the latter is endowed with the topology $\sigma(E,F)$. This shows that the weak-* topology on the topological dual $E'$ of a locally convex vector space $E$ is usually not complete if $E$ is infinite dimensional, for it is rather rare that $E'=E^*$ in this case. A typical counterexample if $E$ is normed and infinite dimensional is the following: let $B_0=\{e_1,e_2,\ldots,\}\subset E$ be a countably infinite linearly independent subset such that $\|e_m\|=1$ for all $m\in\mathbb{N}$. Embed $B_0$ into an algebraic (Hamel) basis $B\supset B_0$ of $E$, and define the linear functional $u:E\rightarrow\mathbb{K}$ as $u(e_m)=m$ for each $n\in\mathbb{N}$ and $u(e)=0$ if $e\in B\smallsetminus B_0$. By construction, $u$ is not continuous since it is unbounded. Of course, for this counterexample to work one is relying on the existence of an algebraic basis of $E$ containing $B_0$ and therefore ultimately on the axiom of choice.
This is also the case if $F=\mathcal{D}$ - we know that $F'(=\mathcal{D}')\neq F^*$, therefore $\mathcal{D}'$ cannot be complete in the weak-${}^*$ topology. More precisely, there is an example of a discontinuous linear functional on $\mathcal{D}$ similar to the one shown above in the normed case, using a suitable countably infinite linearly independent subset $B_0\subset\mathcal{D}$ - take e.g. (for $n=1$) a sequence of closed (mutually disjoint) intervals $I_m=[a_m,b_m]$ contained in $K=[0,1]$ with $0<a_{m+1}<b_{m+1}<a_m<b_m<1$ for all $m\in\mathbb{N}$. Set $c_m=\frac{a_m+b_m}{2}$, $d_m=\frac{b_m-a_m}{2}$. Consider $g\in\mathcal{D}([-1,1])\subset\mathcal{D}$ taking values in $[0,1]$ - for instance, $$g(t)=\begin{cases} 0 & |t|\geq \frac{1}{2} \\ \exp\left(\frac{4t^2}{4t^2-1}\right) & |t|<\frac{1}{2} \end{cases}\ ,$$ and set $$g_m(t)=g\left(\frac{t-c_m}{d_m}\right)\ ,\quad f_m(t)=\frac{1}{m\|g_m\|_{m,K}}g_m(t)\ ,$$ where $$\|f\|_{m,K}=\sum^m_{k=0}\sup_{t\in K}|f^{(k)}(t)|\ ,\quad m\in\mathbb{N}$$ is a separating and increasing sequence of seminorms defining the topology of $\mathcal{D}(K)$. By construction, the $f_m$'s have mutually disjoint supports and therefore are linearly independent. More precisely, we have that $\mathrm{supp}\ \!f_m\subset\mathrm{int}\ \!I_m\subset K$ and $\|f_{m'}\|_{m,K}\leq\frac{1}{m'}$ for all $m'\geq m$ - therefore, $\lim_{m\rightarrow\infty}f_m=0$ in $\mathcal{D}(K)$ and (thus) in $\mathcal{D}$. Set now $B_0=\{f_m\ |\ m\in\mathbb{N}\}\subset\mathcal{D}(K)\subset\mathcal{D}$, and now proceed just like in the normed case - embed $B_0$ into an algebraic basis $B\supset B_0$ of $\mathcal{D}$ and define the linear functional $u:\mathcal{D}\rightarrow\mathbb{R}$ by setting $u|_{B_0}\equiv 1$, $u|_{B\smallsetminus B_0}\equiv 0$. Since $\lim_{m\rightarrow\infty}u(f_m)=1\neq 0$, $u$ is discontinuous. It is straightforward to extend the above construction to $n>1$.
That being said, it turns out that the strong topology $\beta(\mathcal{D}',\mathcal{D})$ on $\mathcal{D}'$ is indeed complete. Recall that this is the topology of uniform convergence on bounded subsets of $\mathcal{D}$, defined by the seminorms $p_B(u)=\sup_{f\in B}|u(f)|$, where $B$ is any nonvoid bounded subset of $\mathcal{D}$ (recall that any such $B$ is a bounded subset of $\mathcal{D}(K)$ for some $K\subset\mathbb{R}^n$ compact). The proofs of this fact usually rely on abstract functional analysis and not on a direct argument. Here is one such proof: since $\mathcal{D}$ is the strict inductive limit of the sequence of Fréchet spaces $\mathcal{D}(K_m)$, where $\varnothing\neq K_m$ is compact, $K_m\subset\mathrm{int}\ \!K_{m+1}$ and $\cup^\infty_{m=1}K_m=\mathbb{R}^n$ and every metrizable locally convex vector space is bornological (i.e. any bounded linear map from it into another locally convex vector space is continuous), we conclude that $\mathcal{D}$ is bornological. Since the strong topology on the topological dual of any bornological locally convex vector space is complete, we conclude that $\mathcal{D}'$ is complete with respect to its strong topology $\beta(\mathcal{D}',\mathcal{D})$. Check e.g. Sections 13.1 and 13.2 of Jarchow's book for more details. The lack of weak-* completeness is no big deal, though, since it can be shown that $\mathcal{D}$ is weak-* sequentially dense in $\mathcal{D}'$. By the Banach-Steinhaus theorem, this can be upgraded to the strong topology since it implies that any weak-* convergent sequence in $\mathcal{D}'$ is also strongly convergent (more precisely, this is a consequence of the fact that $\mathcal{D}$ is reflexive and therefore $\mathcal{D}'$ is barrelled, see e.g. Theorem 11.1.3, pp. 220 of Jarchow's book).
