I'm reading Demailly's *Complex Analytic and Differential Geometry* In Section I.2.D.4 he uses the following fact: Suppose $u \in \mathcal{D}'(\Omega)$, where $\Omega \subset \mathbb{R}^n$ is a distribution such that all of its derivatives are of order zero, i.e.



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> for every compact $K \subset \Omega$ there is a $C\_K > 0$ s.t. $ |\langle u, \partial^\alpha f \rangle| \leq C\_K \| f \|\_\infty $ for all $f \in \mathcal{D}(K)$
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equivalently we can say



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> $u$ extends to a continuous linear map on $C^0\_c(\Omega)$
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Then the claim is that $u$ is actually given by a smooth function. I am not sure how to prove this and would appreciate any help or references.


These are my ideas so far:


* distributions of order zero are always given by integration agains a measure - I feel like combining this with the Lebesgue fundamental theorem of calculus could give a proof for $n=1$.
* I know that if $u \in \mathcal{D}'$ has a *continous* derivative then $u$ is $C^1$, so it's sufficient to prove continuity of $u$ instead of smoothness. It seems like maybe every distribution with a zero-order second derivative must be continuous?



A distribution $u$ such that all of its derivatives are of order zero is smooth
*Step 1.* With no loss of generality we may assume that $u$ is compactly supported (just multiply $u$ by a test function if it is not). Let $B = [-M,M]^n$ be a box that contains the support of $u$.


*Step 2.* Let $v = \partial^{(2,2,\ldots,2)} u$. Since $|\langle v, f\rangle| = |\langle u, \partial^{(2,2,\ldots,2)} f\rangle| \leqslant C \|f\|\_\infty$, the distribution $v$ extends to a continuous linear functional on the space of continuous functions vanishing at infinity, and hence $v$ corresponds to a finite complex-valued measure $\mu$.


*Step 3.* Now define $$\tilde u(x) = \int\_{(-\infty, x\_1) \times \ldots \times (-\infty, x\_n)} (x\_1 - y\_1) \cdots (x\_n - y\_n) \mu(dy).$$ Clearly, $\tilde u$ is a continuous function, and by Fubini's theorem and the fact that $\int\_t^\infty (s - t) f''(s) ds = f(t)$, we easily find that
$$ \int \tilde u(x) \partial^{(2,2,\ldots,2)}g(x) dx = \int g(y) \mu(dy) = \langle v, g\rangle = \langle u, \partial^{(2,2,\ldots,2)}g \rangle. $$
for every test function $g$.


*Step 4.* In particular, the distributional derivative $\partial^{(2,2,\ldots,2)} \tilde{u}$ is equal to zero in the complement of $B$. Since $\tilde{u}(x) = 0$ when $x\_1,x\_2,\ldots,x\_n < -M$, we find that in fact $\tilde{u}(x) = 0$ in the complement of $B$.


*Step 5.* Since $\partial^{(2,2,\ldots,2)}g$ can be equal to an arbitrary test function $f$ on $B$, and both $u$ and $\tilde{u}$ are supported in $B$, we have
$$ \int \tilde u(x) f(x) dx = \langle u, f \rangle $$
for every test function $f$. That is, $u$ corresponds to a continuous function.


*Step 6.* Apply the above to the derivatives of $u$ to find out that $u$ is in fact smooth.


