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THEOREM : $U$ is an open subset of $\mathbb{R^n}$ and suppose $u \in C^2(U) $ is harmonic within $U$, then : $$u(x)= \def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{B(x,r)}u\,dy = \def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{\partial B(x,r)}u\,dS$$ for each ball $B(x,r) \subset U$

there's a nice proof in Evan's PDE book which only involves real analysis. since harmonic functions can be implicitly identified as the real or imagniary part of holomorphic functions, how does one prove or atleast reformulate this theorem using complex analysis notions ?

any comment, references will be greatly appreciated.

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  • $\begingroup$ I know that harmonic functions in $\Bbb R^2 \simeq \Bbb C$ are locally the real parts of holomorphic functions. Is that true for arbitrary dimensions? What would be a holomorphic function in $\Bbb R^3$? $\endgroup$
    – Martin R
    Feb 20, 2018 at 19:12
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    $\begingroup$ If $n=2$ and $u$ is given, can you proceed by finding $f$ with $f=u+iv$, write down the Cauchy integral formula for $f$, and then split into $f$ in integrand as $f=u+iv$?. When $n=2k$ is even, then $R^n$ = $C^k$, and maybe you can do the same thing. $\endgroup$
    – LucasSilva
    Feb 20, 2018 at 19:31
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    $\begingroup$ To follow up to my comment, check out Section 10.1 of Ullrich's book Complex Made Simple. $\endgroup$
    – LucasSilva
    Feb 20, 2018 at 19:37
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    $\begingroup$ @rapidracim Does it not work if you pick the boundary of the ball $B(x,r)$ as the contour? Also, I was thinking of the Cauchy integral formula ($f(a) = (2 \pi i)^{-1} \int_{\gamma} f(z)/(z-a)dz$), not the Cauchy theorem ($\int_{\gamma} f dz = 0$). But since the formula can be proved from the theorem, I guess it doesn't matter. $\endgroup$
    – LucasSilva
    Feb 20, 2018 at 19:44
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    $\begingroup$ As another reference, you can look at Chapter 11 of Rudin's Real and Complex Analysis. Rudin also gets the harmonic mean value theorem by the complex analysis route. Indeed, Ullrich's book is largely inspired by Rudin's treatment, as you can read in the preface. $\endgroup$
    – LucasSilva
    Feb 20, 2018 at 19:46

1 Answer 1

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For $n=2$ one can proceed as follows (this is essentially an elaboration of above comments):

If $u$ is harmonic in $ U \subset \Bbb C$ and $B(z_0, R) \subset U$ then $u = \operatorname{Re} f$ for some holomorphic function $f$ in $B(z_0, R)$. This is true in any simply-connected domain in $\Bbb C$, $f$ can for example be chosen as an anti-derivative of the (holomorphic) function $u_x - i u_y$.

The Cauchy integral formula then states that for $0 < r < R$ and the path $\gamma(t) = z_0 + re^{it}$, $0 \le t \le 2 \pi$, $$ f(z_0) = \frac{1}{2 \pi i} \int_\gamma \frac{f(\zeta)}{\zeta - z_0} \, d\zeta = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + re^{it}) \, dt \, . $$

Taking real parts gives the mean-value formula for $u$: $$ u(z_0) = \frac{1}{2 \pi} \int_0^{2 \pi} u(z_0 + re^{it}) \, dt \, . $$

Finally, the “area form” of the mean-value formula follows by integration over the radius: $$ \frac{1}{\pi r^2} \int_{B(z_0, r)} u(x, y)\, dx dy = \frac{1}{\pi r^2} \int_0^r \int_0^{2 \pi} u(z_0 + \rho e^{it}) \rho \, dt \, d\rho = \frac{1}{\pi r^2} \int_0^r 2 \pi u(z_0) \rho \, d\rho = u(z_0) \, . $$

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