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Well, here is the thing. We know that the laplacian operator commutes with any partial derivative of a function, if the function is smooth. We also know that a harmonic function is infinitely differentiable, thus every partial derivative of a harmonic funtion is harmonic.

My problem is the following: $\frac 1 r$, where $r=(\sum_i x_i^2)^\frac 1 2$ is harmonic if $r\ne 0$. Is every partial derivative of this function going to be harmonic, even though the space where this function is harmonic is not compact nor simply connected?

Thank you for your help!

Lessa121
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2 Answers2

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The answer is yes. However, note that the domain of definition for $\frac{1}{r}$ is in fact connected.

EDIT:

To prove this let $u(x):D \subseteq \mathbb{R}^n \to \mathbb{R}$ be any harmonic function in some domain $D$, and let $\partial_i$ be any partial derivative operator. $u$ satisfies

$$\Delta u \equiv 0$$ for $x \in D$. Applying $\partial_i$ to both sides gives

$$\partial_i \Delta u \equiv \partial_i 0 \equiv 0 $$ for all $x \in D$. But since $\partial_i \Delta=\Delta \partial_i$, we have

$$\Delta(\partial_i u) \equiv 0 $$

for all $x \in D$, which means that $\frac{\partial u}{\partial x_i}$ is harmonic in the same domain.

user1337
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  • You are right! I'm sorry I meant simply connected, I'll change it now. One question though, why are the derivatives harmonic? Is there a theorem that assures that the property I mentioned still holds for any open space? – Lessa121 May 26 '15 at 13:43
  • $\mathbb{R}^3\setminus {0}$ is simply connected. – Willie Wong May 26 '15 at 14:10
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Clairaut's Theorem states that the order of taking mixed partial derivatives of a function $f(x_{1}, \ldots, x_{n})$ does not matter so long as $f$ is sufficiently smooth.

Also, note that $\frac{1}{r}$ is only harmonic in dimension $n = 3$. In general dimension, the function $f(x_{1}, \ldots, x_{n}) = r^{2-n}$ is harmonic, however.

Mark Hubenthal
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