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Let $f \colon U \to \mathbb{R}$ be a twice differentiable function, where $U$ is an open subset of $\mathbb{R}^n$. Here twice differentiable means that all the second partial derivatives $\frac{\partial}{\partial x_i} (\frac{\partial}{\partial x_j} f)$ exist (however they are not necessarily continuous). Suppose $\Delta f = 0$, i.e. $\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2} f = 0$. Does this imply that $f$ is harmonic, i.e. that $f$ is twice continuously differentiable? (For $n = 1$ it does, so let's assume that $n \ge 2$.)

Remark: I have read somewhere that if $f$ is weakly harmonic, then it is harmonic. However I think the second derivatives of $f$ here are not necessarily the same as the weak second derivatives. So my guess is that there is a counterexample, but I was not able to find one. Such a counterexample would be an obstruction to the extension of the Lebesgue integral to more general functions, such that the Fubini theorem and the Newton-Leibniz theorem both remain true.

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    $\begingroup$ This isn't really a satisfying answer, so i've made it a comment, but the Laplacian is an elliptic operator, so if $\Delta f\in C^\infty$ then so is $f$, at least for square integrable $f$. $\endgroup$ – D. Kelleher Apr 16 '13 at 16:35
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Here's a counterexample in $n\ge3$ dimensions. With $\lVert\cdot\rVert$ denoting the Euclidean norm, set $$ f(x)=\begin{cases} x_1x_2x_3\lVert x\rVert^{-n-4},&{\rm if\ }x\not=0,\cr 0,&{\rm if\ }x=0. \end{cases} $$ This is harmonic on $x\not=0$, has first and second order derivatives at the origin, but $f$ along with all its derivatives are discontinuous at the origin.

Note that, $f=c\frac{\partial^3(\lVert x\rVert^{-n+2})}{\partial x_1\partial x_2\partial x_3}$ is a third order derivative of a harmonic function, so is harmonic (away from the origin). As $f$ vanishes whenever all but two coordinates are zero, it along with its first order derivatives all vanish on the coordinate axes. So, $f$ along with all its first and second order derivatives vanish at the origin.

Next, here's a counterexample in $n=2$ dimensions. Using $i=\sqrt{-1}$, define $f\colon\mathbb{R}^2\to\mathbb{R}$ by, $$ f(x,y)=\begin{cases} \Re\left[\exp\left(-(x+iy)^{-4}\right)\right],&{\rm if\ }(x,y)\not=0,\cr 0,&{\rm if\ }x=y=0. \end{cases} $$ As the real part of an analytic function, $f$ is harmonic away from the origin. Also, $f$ along with its derivatives to any order are bounded by a multiple of a power of $\lVert x\rVert$ mutiplied by $\exp(-\lVert x\rVert^{-4})$ on any bounded subset of the coordinate axes (outside of the origin). So, $f$ together with all partial derivatives to all orders exist and are zero at the origin. On the other hand, $f$ is not continuous at the origin. Note, this also gives an alternative counterexample to the first one above in $n\ge3$ dimensions by making $f$ independent of $x_3,\ldots,x_n$.

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  • $\begingroup$ But is this a counterexample to what was asked? $\endgroup$ – Deane Yang Apr 16 '13 at 22:39
  • $\begingroup$ @Deane: Yes, I think so. It satisfies all of the properties asked for in the question but is not even continuous (so not harmonic). What do you think is missing? $\endgroup$ – George Lowther Apr 16 '13 at 22:59
  • $\begingroup$ It is not a solution at the origin. $\endgroup$ – Deane Yang Apr 17 '13 at 0:50
  • $\begingroup$ It is. As $f$ vanishes on the coordinate axes we have $\frac{\partial^2f}{\partial x_i^2}=0$ at the origin, so $\Delta f=0$ there. $\endgroup$ – George Lowther Apr 17 '13 at 0:59
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    $\begingroup$ You produced a function which is harmonic in $\mathbb{R}^n\setminus 0$ hence is smooth there. Your function is homogeneous of degree$2-n$ and its $k$-th order derivatives are homogeneous of degree $2-n-k$. A nonzero function on $\mathbb{R}^n\setminus 0$ which is homogeneous of degree $\alpha$ extends to a continuous function on $\mathbb{R}^n$ if and only if $\alpha\geq 0$. $\endgroup$ – Liviu Nicolaescu Apr 17 '13 at 12:21
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Here is a general result. If $f$ is a distribution on $U$ and $\Delta f\in C^\infty(U)$, then $u\in C^\infty(U)$. More generally, you can replace $\Delta$ with any properly supported elliptic pseudodifferential operator on $U$. In particular, any elliptic operator with smooth coefficients is a properly supported elliptic pseudodifferential operator.

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If $U$ is an open subset of $\mathbb R^n$, $f$ is a distribution on $U$ such that $\Delta f$ is analytic on $U$, then $f$ is analytic on $U$. This hypoellipticity result is true for any elliptic differential operator with analytic coefficients.

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