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Let me expand Aaron's answer: there is a mean value theorem with any centrally symmetric surface. You integrate your harmonic function on the surface against the harmonic measure at the center, and y …

Equation $\Delta u=0$ is called the Laplace equation, btw. Edit. The answer to your question is no. Consider $f(z)=1/(z+i)$. On the real line it belongs to $L^p$ with any $p>1$. In the upper half-pl …

You have to say what is the meaning of $\Delta u$, and of $\int\Delta u$. For the integral to have a meaning, $u$ has to be a distribution and $\Delta u$ has to be a (signed, Radon) measure. Such dist …

To separate contribution of the "ends" and the lateral surface, write $w=u+v$, where $u$ has zero boundary conditions on the lateral surface, and $v$ is zero on the ends. The estimate $|u|\leq Ce^{-kt …

You must specify more exactly what do you mean by "singular analytic" (what singularities are allowed). Some version of this problem was investigated in arXiv:1309.5483, and in the literature mention …

If they are talking about the Laplace operator, this statement is true only in dimension 2. And this is only sufficient, not necessary. In general, for solvability of the classical Dirichlet problem, …

For every $n$ and for every regular region (in the sense of Dirichlet problem), it is easy to construct a harmonic function in $L^\infty$ that does not extend to any larger open set. Just solve the Di …

This is correct, and the same proof as in McOwen and Troyanov should work. In fact they were not the first who proved this result. The story begins with E. Picard, who wrote several papers on this (al …

First, some general background. For a bounded domain, the boundary value problem is solved by the Poisson formula, however an explicit form of the Poisson kernel for a cylinder of finite length is pro …

Just prove it yourself: Take $r=1$. Then $$w(x_0,2^{-n})\leq \lambda^n w(x_0,1)=:C\lambda^n.$$ To estimate $|u(x_0)-u(y)|$, where $y$ is close to $x_0$, choose $n$ so that $|x_0-y|\in[2^{-n-1},2^{-n} …

This question was addressed by Hansen and Nadirashvili in a series of papers, see, for example: MR1315353 Hansen, W., Nadirashvili, N., On Veech's conjecture for harmonic functions. Ann. Scuola Norm. …

This follows from the so-called (Eberhard) Hopf Minimum Principle. If you have a positive (super-) harmonic function $u$ in a ball, and $u(z_0)=0$ for some boundary point $z_0$, then the normal deriva …

No. Take any $u$ which is not zero, but compactly supported in $\Omega$. Then define $f=\Delta u$; it will be also compactly supported, and non-zero.

The answer is no. Let $M$ be a region in the upper half-space $x_1>0$ in $R^n$, (you can take $n=2$) and $\partial M$ contains an open piece $U$ of the plane $x_1=0$. Take $f=0$ in $U$. Then your harm …

Your guess is wrong. Take real $f$, for example, then harmonic extension is also real, and its derivative is singular. In general, I do not expect a simple answer. And certainly the answer cannot depe …