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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).

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up vote 5 down vote accepted

For $\partial f/\partial x_i$ to vanish, then $f$ has to vanish on the space of test functions of the form $\partial \phi_i/\partial x_i$. You effectively need to show the subspace spanned by all of these (as $i$ varies) is of codimension one in $C_c(\mathbb{R}^n)$. This subspace has to be the space of all test functions integrating to zero. To go from $n-1$ to $n$, take a test function $\phi$ with integral $0$ and let $\psi(x_1,\ldots,x_{n-1})=\int\phi(x_1,\ldots,x_n)dx_n$. Then $\psi$ is a test function on $\mathbb{R}^{n-1}$ with integral $0$ so is a sum of functions $\partial \psi_i/\partial x_i$ where the $\psi_i$ are test functions on $\mathbb{R}^{n-1}$. Extend to test functions $\phi_i$ on $\mathbb{R}^n$ by letting $\phi_i(x_1,\ldots,x_n) =\psi_i(x_1,\ldots,x_{n-1})h(x_n)$ where $h$ is a fixed bump function with integral $1$. Then $\psi-\sum\partial\phi_i/\partial x_i$ has zero integral along all lines in the $x_n$-direction, so is the $x_n$-partial derivative of a test function.

Added I should add that the above is effectively the proof that $H_c^n(\mathbb{R}^n)=\mathbb{R}$ where $H_c^*$ denotes de Rham cohomology with compact supports.

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This is actually the type of argument I was looking for. Thanks. – Jonas Teuwen Nov 16 '10 at 16:59

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, 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.

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Ah, I didn't think of this, I was trying some approach similar to Robin Chapman's. However his approach uses less machinery (like that the approximated sequence converges in $L^p$). I'll see which one gets the most upvotes to see which one I will accept since both are good. – Jonas Teuwen Nov 16 '10 at 16:19
It is usually what I try when I need a fact about distributions that is easy to show for the case of smooth functions. – Zarrax Nov 16 '10 at 16:43

Suppose that $(x_1,\ldots,x_n) \neq (y_1,\ldots,y_n)$. If we fix $x_2,\ldots,x_n$, then $f(x_1,x_2,\ldots,x_n)$ is constant with respect to $x_1$. Hence, $f(x_1,x_2,\ldots,x_n) = f(y_1,x_2,\ldots,x_n)$. Next, fix $y_1,x_3,\ldots,x_n$. Then, $f(y_1,x_2,\ldots,x_n)$ is constant with respect to $x_2$. Hence, $f(y_1,x_2,\ldots,x_n) = f(y_1,y_2,\ldots,x_n)$. Continue this way to conclude that $f(x_1,\ldots,x_n) = f(y_1,\ldots,y_n)$, that is, $f$ is constant.

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It is quite non obvious, but I am pretty sure that the question is talking about distributions in the sense of Schwartz, see – Mariano Suárez-Alvarez Nov 16 '10 at 14:50
Thank you, I agree. – Shai Covo Nov 16 '10 at 15:07
I'm sorry, I thought it would be clear, I'll modify the question. – Jonas Teuwen Nov 16 '10 at 15:24
I should have realized from the tags that it isn't a trivial question... – Shai Covo Nov 16 '10 at 15:53

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