= Solution
Use complex-linear <distributions>, so the pairing contains no complex conjugation. The <space of smooth functions> is $\mathcal E(X)=C^\infty(X)$, with <seminorms>
$$
p_{K,r}(f)=\max_{|\alpha|\leq r}\sup_{x\in K}|\partial^\alpha f(x)|,\qquad K\Subset X.
$$
Thus $f_j\to f$ means uniform convergence of every <derivative> on every <compact subset> of $X$. A cofinal <compact exhaustion> with $K_j\subset\operatorname{int}K_{j+1}$ makes $p_j=p_{K_j,j}$ an increasing defining family. Such an exhaustion exists for every <open set>; for example, use bounded sets staying a positive distance from its complement and enlarge them slightly. This is a <Fréchet space>.
The <compactly supported distribution space> is the <continuous dual space> $\mathcal E'(X)$. We use its <weak-star topology>: $u_j\to u$ precisely when $\langle u_j,f\rangle\to\langle u,f\rangle$ for every $f\in\mathcal E(X)$. A stronger standard choice is the <strong dual topology>, which requires uniform convergence on bounded subsets of $\mathcal E(X)$; specifying the weak convention avoids conflating these definitions.
A continuous linear functional necessarily takes null sequences to zero. Conversely, suppose a linear functional $u$ fails to be continuous. For each $j$, it is unbounded on $\{f:p_j(f)\leq1\}$; otherwise scaling would give a continuity estimate. We can therefore choose $f_j$ with
$$
p_j(f_j)\leq\frac1j,\qquad |u(f_j)|\geq1.
$$
For each fixed $l$, $p_l(f_j)\leq p_j(f_j)\to0$ once $j\geq l$. This contradicts the assumed null-sequence property. Hence \b[sequential continuity characterizes the continuous dual]:
$$
\boxed{u\in\mathcal E'(X)\iff f_j\to0\text{ in }\mathcal E(X)\Longrightarrow u(f_j)\to0.}
$$
This is the <sequential continuity criterion in a metrizable vector space> applied to a linear functional.
Continuity also yields one compact $K\Subset X$, an integer $m$ and a constant $C$ such that $|u(f)|\leq Cp_{K,m}(f)$. Thus $u$ has <compact support> in $K$ and finite <order of a distribution>. Conversely, a <compactly supported distribution> extends to $\mathcal E(X)$ by $u(f)=u(\eta f)$, where a <cutoff function> $\eta$ equals one near its <support of a distribution>. The local finite-order estimate makes this extension continuous and independent of $\eta$. This identifies the dual definition with compactly supported <distributions>.
Here is an explicit <compact continuous-derivative representation of a distribution>. Extend $u$ by zero to $\mathbb R^n$ using a <cutoff function> inside $X$, retaining finite order $m$. Put $r=m+2$ and
$$
E(x)=\prod_{j=1}^n\frac{(x_j)_+^{r-1}}{(r-1)!},\qquad\gamma=(r,\ldots,r).
$$
This locally $C^m$ function satisfies $\partial^\gamma E=\delta_0$. Indeed, each one-dimensional factor has $r$th <distributional derivative> equal to the <Dirac delta distribution>. The <convolution> $F=u*E$ is continuous: the finite-order estimate extends $u$ to $C^m$ functions near its <compact support>, and translated $E$ varies continuously in their $C^m$ norms. Equivalently, mollify $E$ and use the estimate to obtain local uniform convergence of the convolved functions. Distributional differentiation gives $\partial^\gamma F=u$.
Choose $\chi\in C_c^\infty(X)$ equal to one near $\operatorname{supp}u$. Since $\chi u=u$, repeated <Leibniz rule> gives
$$
u=\chi\partial^\gamma F=\sum_{\beta\leq\gamma}(-1)^{|\beta|}\binom\gamma\beta\partial^{\gamma-\beta}\big((\partial^\beta\chi)F\big).
$$
Every coefficient function on the right is continuous and compactly supported in $X$. Thus \b[the required representation is finite]:
$$
\boxed{u=\sum_\alpha\partial^\alpha f_\alpha,\qquad f_\alpha\in C_c(X).}
$$
The equality first holds on <test functions> and then on all <smooth functions> after inserting a common cutoff, so it holds in $\mathcal E'(X)$.
\b[A general <distribution> need not admit one finite such sum], even if <compact support> is not required of the <continuous functions>. On $X=\mathbb R$, consider
$$
v=\sum_{j=1}^\infty\delta_j^{(j)}.
$$
The sum is locally finite, so it defines a <distribution>. If it were a finite sum $\sum_{l=0}^M\partial^l g_l$ with all $g_l$ continuous, then on every fixed compact set its action would be bounded by test <derivatives> through order $M$. Near an integer $j>M$, however, it is exactly $\delta_j^{(j)}$, whose order is $j$. To see the contradiction directly, choose a <test function> $\psi$ with $\psi^{(j)}(0)\ne0$ and put $\psi_\varepsilon(x)=\varepsilon^j\psi((x-j)/\varepsilon)$. All <derivative> norms through order $M$ tend to zero, while $\langle\delta_j^{(j)},\psi_\varepsilon\rangle=(-1)^j\psi^{(j)}(0)$. This is a <distribution of unbounded order>. In dimensions $n>1$, the same example uses point masses at $(j,0,\ldots,0)$ and <derivatives> in the first coordinate.
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