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If $u$ is a distribution in open set $\Omega\subset \mathbb R^n$ such that ${\partial ^i}u = 0$ for all $i=1,2,\ldots,n$. Then is it necessarily that $u$ is a constant function?

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up vote 3 down vote accepted

It's true if we assume that $\Omega$ is connected. We will show that $u$ is locally constant, and the connectedness will allow us to conclude that $u$ is indeed constant. Let $a\in\Omega$ and $\delta>0$ such that $\overline{B(a,2\delta)}\subset \Omega$. Consider a test function $\varphi\in\mathcal D(\Omega)$ such that $\varphi=1$ on $B(a,2\delta)$, and put $S=\varphi u$, which is a distribution with compact support. Let $\{\rho_k\}$ a mollifier, i.e. non-negative functions, of integral $1$, with support contained in the ball $\overline{B\left(0,\frac 1k\right)}$, and consider $S_k=\rho_k* S$. It's a distribution which can be associated to a test function. We have $\partial^i S=\partial^i\varphi u$, hence $\partial^iS=0$ on $B(a,2\delta)$. $\partial^iS_k=\rho_k*(\partial^iS)$ is $0$ if $\frac 1k<\delta$, hence $S_k$ is constant. Since $S_k$ converges to $S$ in $\mathcal D'$, $S$ is constant on $B(a,\delta)$ and so is $u$.

This topic answers the cases $\Omega=\mathbb R^n$.

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