A distribution $u$ such that all of its derivatives are of order zero is smooth

Question

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.

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 $f \in \mathcal{D}(K)$

equivalently we can say

$u$ extends to a continuous linear map on $C^0_c(\Omega)$

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?

Answer 1

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.