= Solution
Let $K$ contain the support of $u\in\mathcal E'(\mathbb R^n)$. Choose a <cutoff function> that equals one near $K$. It makes
$$
\widehat u(\lambda)=\langle u(x),e^{-i\lambda\cdot x}\rangle
$$
well-defined, and differentiating the parameter under the pairing gives
$$
\partial_\lambda^\gamma\widehat u(\lambda)
=\langle u(x),(-ix)^\gamma e^{-i\lambda\cdot x}\rangle.
$$
Thus the <Fourier transform of a compactly supported distribution> is a <smooth function>. Since a compactly supported distribution has finite order, some $N$ and $C$ satisfy
$$
|\langle u,\psi\rangle|\leq C\sum_{|\alpha|\leq N}\sup_K|\partial^\alpha\psi|.
$$
Applying this estimate to the exponential yields $|\widehat u(\lambda)|\leq C'\langle\lambda\rangle^N$.
Now take $v\in\mathcal E'(X)$, multiply by a <cutoff function> supported in $X$ and equal to one near $\operatorname{supp}v$, and regard the result as an element of $\mathcal E'(\mathbb R^n)$. Choose $m$ so large that
$$
g(\lambda)=\langle\lambda\rangle^{-2m}\widehat v(\lambda)
$$
is <Lebesgue integrable function>[Lebesgue integrable]. The inverse <Fourier transform> $f=\mathcal F^{-1}g$ is a bounded <continuous function>, and the <Fourier transform of a derivative> gives
$$
v=(1-\Delta)^m f
$$
as a <distributional identity>. This is the <Bessel potential> proof of the <structure theorem for compactly supported distributions>.
The function $f$ itself need not have <compact support>. Choose another cutoff $\rho\in\mathcal D(X)$ equal to one near $\operatorname{supp}v$. Then $v=\rho(1-\Delta)^mf$. Repeatedly using
$$
\rho\,\partial^\alpha f
=\sum_{\beta\leq\alpha}(-1)^{|\alpha-\beta|}\binom{\alpha}{\beta}
\partial^\beta\bigl(f\,\partial^{\alpha-\beta}\rho\bigr)
$$
expresses $v$ as a finite sum $\sum_\beta\partial^\beta f_\beta$, where every coefficient $f_\beta$ is continuous and compactly supported in $X$.
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