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Are harmonic functions continuous? I mean harmonic in the weak sense that second partial derivatives exist and $\Delta u=0$ on an open neighborhood. Many sources start with the assumption that harmonic functions are $C^2$ or are merely continuous, and I am wondering if it is possible to do away with these restrictions.

I've seen the proof that continous harmonic functions are the real part of holomorphic functions and so are $C^{\infty}.$ I am hoping that there is an easy way to get from $\Delta u=0$ to continuity. Maybe we can show a version of the mean value property and get continuity from there?

subrosar
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1 Answers1

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If partial derivative of $u(x,y)$ w.r.t both the variables $x,y$ exists then $u(x,y)$ is continuous in variable $x$ and variable $y$ separately then choose $\delta = \min(\delta_x,\delta_y)$.

For all $|(x',y')-(x,y)| < \delta \implies |x'-x| < \delta,|y'-y| < \delta$, $$|u(x',y')-u(x,y)|=$$ $$|u(x',y')-u(x',y)+u(x',y)-u(x,y)| \leq |u(x',y')-u(x',y)|+|u(x',y)-u(x,y)|$$

But this is as far as you can go. The above proves continuity only when same $\delta$ works for $|u(x',y')-u(x',y)| \leq \epsilon, |u(x',y)-u(x,y)| \leq \epsilon$. For a counter argument that $u$ need not be continuous but can be harmonic, Please see Do discontinuous harmonic functions exist?

Balaji sb
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