TITLE: Derivatives distribution
QUESTION [9 upvotes]: Let $f$ be a distribution on $\mathbf{R}^n$ (in the Schwartz sense) such that
$$\frac{\partial f}{\partial x_i} = 0 \text{ for $i = 1, \ldots, n$.}$$
Then how to prove that $f$ is a constant? I had this exercise in a class last year but I couldn't find how to do the induction step (for $n = 1$ it is clear of course).

REPLY [6 votes]: If $f(x)$ is a smooth function, one gets that $f(x)$ is constant relatively easily as in Shai Covo's comment. For a general distribution, one can convolve with approximations to the identity: Let $\phi(x)$ be a fixed smooth function with compact support such that $\int \phi(x) = 1$, and let $\phi_k(x) =  k^n\phi(kx)$.
Recall $f \ast \phi_k$ is a smooth function (the integral of $<f, \phi_y>$ with respect to $y$, where $\phi_y$ is the translate of $\phi$ by $y$) and is defined by $<f \ast \phi_k, s> = <f(x) , s \ast \phi_k(x)>$ for $C_c(R^n)$ functions $s(x)$. Since $s \ast \phi_k(x)$ converges to $s(x)$ uniformly, with the same property holding for derivatives of $s(x)$, $f \ast \phi_k$ converges to $f$ as distributions.
But each $f \ast \phi_k(x)$ is a smooth function, and $<f \ast \phi_k, \partial_i s(x)>$ $=$ $<f(x), \partial_i s \ast \phi_k(x)>$ $=$ $<f(x), \partial_i (s \ast \phi_k(x))>$ $= 0$ by assumption. Hence by the smooth function case $f \ast \phi_k(x)$ is a constant. Thus $f(x)$ is the distributional limit of constants and must be constant itself.