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This post is not short. However I'm sure that a guy who good handle these concepts, could read and answer in five minutes. I only want to write my attempt, in order to understand where I'm wrong.

Let $\Omega\in\Bbb C$ be a domain; $\varphi:\Omega\to[-\infty,+\infty[$ upper semicontinous (i.e. $\varphi(z_0)\ge\limsup_{z\to z_0}\varphi(z)\;\;\forall z_0\in\Omega$).

I need to show, given $\bar\Delta_{z_0,r}\Subset\Omega$ (the unitary disk of radius $r$) that $$ \varphi(z_0)\le\frac1{2\pi r}\int_{\partial\Delta_{z_0,r}}\varphi(s)\;ds $$ implies $$ \varphi(z_0)\le\frac i{2\pi r^2}\int_{\Delta_{z_0,r}}\varphi(z,\bar z)\;dz\wedge d\bar z\;\;. $$

Now what everybody would do is to rewrite the first inequality as

$$ (2\pi t)\varphi(z_0)\le\int_{\partial\Delta_{z_0,t}}\varphi(s)\;ds $$ and then integrate over $]0,r]$ wrt the variable $t$.

In this way LHS become easily $\pi r^2\varphi(z_0)$.

My problem is with RHS. Roughly I'd write $$ \int_0^r\int_{\partial\Delta_{z_0,t}}\varphi(s)\,ds\,dt =\int_{\Delta_{z_0,r}}\varphi(z)\,dz $$

but I'm not sure it has some sense. So I started form the other side, and this is what I got: \begin{align*} \int_{\Delta_{z_0,r}}\varphi(z,\bar z)\;dz\wedge d\bar z &=-2i\int_{\Delta_{z_0,r}}\varphi(x,y)\,dx\wedge dy\\ &=-2i\int_{\Delta_{z_0,r}}\varphi(x,y)\,dx\wedge dy \end{align*}

this could be seen in two ways: writing $dz\wedge d\bar z=(dx+idy)\wedge(dx-idy)=-2i(dx\wedge dy)$ or changing variable via the isomorphism $\beta:\Bbb R^2\stackrel{\simeq}{\to}\Bbb C$ defined by $(x,y)\mapsto(x+iy,x-iy)$.

What we have, then, is the integral of a $2$-form on a $2$-parametric manifold. So I consider now $$ \alpha:]0,r[\times[0,2\pi[\longrightarrow\Delta_{z_0,r}\setminus\{z_0\} $$ defined by $$ (t,\theta)\longmapsto (\Re z_0+t\cos\theta,\Im z_0+t\sin\theta) $$

from which we have $$ -2i\int_{\Delta_{z_0,r}}\varphi(x,y)\,dx\wedge dy=\\ =-2i\int_{]0,r[\times[0,2\pi[}\varphi (\Re z_0+t\cos\theta,\Im z_0+t\sin\theta) \underbrace{[\partial_t\alpha_1\partial_{\theta}\alpha_2-\partial_t\alpha_2\partial_{\theta}\alpha_1]}_{=\partial\alpha_1\wedge\partial\alpha_2(\partial_t\alpha,\partial_{\theta}\alpha)=t}\,dt\,d\theta\\ =-2i\int_0^rt\underbrace{\int_0^{2\pi}\varphi (\Re z_0+t\cos\theta,\Im z_0+t\sin\theta)\,d\theta}_{=:A(t)}\,dt $$

And till here it seems (to me!) to have made no mistakes.

Now I thought I could work on $A(t)$. Using $\beta^{-1}$ I got $$ A(t)=\frac i2\int_0^{2\pi}\varphi(z_0+te^{i\theta})\,d\theta $$ The problem comes now: how to proceed? I would change variable to this last integral $s=z_0+te^{i\theta}$: in this way it's ALMOST (and almost in Mathematics means wrong) equal to the wanted integral, except for a $ie^{i\operatorname{arg}(s)}$ or something similar (I erased it from my notebook).

Where am I wrong? How can I conclude?

Many thanks!

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  • $\begingroup$ "a complex function can be integrated over a path, not over a surface" Why? What prohibits us from integrating a complex function, like $f(z)=e^z$ over some region, like the unit disk or a square? It's the usual integral with respect to a measure, the measure being two-dimensional Lebesgue. $\endgroup$ – user147263 Feb 1 '15 at 22:19
  • $\begingroup$ You're right, thanks. However the identity which preceeds this sentence doesn't convince me. $\endgroup$ – Joe Feb 1 '15 at 22:21
  • $\begingroup$ You could write $\int_0^r\int_{\partial\Delta_{z_0,t}}\varphi(s)\,ds\,dt=\int_{\Delta_{z_0,t}} \varphi(z)\,d \text{ "some symbol for 2d Lebesgue measure"}$, and it would be correct. Integration in polar coordinates is an instance of Fubini's theorem; its legitimacy is justified by measure theory, not by complex analysis. $\endgroup$ – user147263 Feb 1 '15 at 22:23
  • $\begingroup$ Ok, but how can this agree with the integral of a $2$-form I wrote at the beginning? $\endgroup$ – Joe Feb 1 '15 at 22:25
  • $\begingroup$ I haven't read all this, but note that the inequality you wish to prove needs to be fixed. On the RHS, you probably need a factor of $i/2$, to give the usual area integral. $\endgroup$ – Ted Shifrin Feb 1 '15 at 23:06
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You didn't quite edit correctly. The $i/2$ belongs in the inequality $$\varphi(z_0)\le \frac1{2\pi} \frac i2\int_{D(z_0,r)} \varphi(z)dz\wedge d\bar z.$$ Although I'm a huge fan of using $dz$ and $d\bar z$, I think it's easier just to use a polar coordinates double integral on this problem.

Given that $$\varphi(z_0)\le \frac1{2\pi r}\int_{\partial D(z_0,r)} \varphi (s)ds = \frac1{2\pi}\int_0^{2\pi}\varphi(z_0+re^{i\theta})d\theta$$ for all $0\le r\le R$, we proceed as follows: \begin{align*} \frac{R^2}2\varphi(z_0) &= \int_0^R \varphi(z_0) r\,dr \le \frac1{2\pi}\int_0^R\left(\int_0^{2\pi}\varphi(z_0+re^{i\theta})d\theta\right)r\,dr \\ &= \frac1{2\pi}\int_{D(z_0,R)} \varphi(z) r\,drd\theta = \frac1{2\pi}\frac i2\int_{D(z_0,R)} \varphi(z)\,dz\wedge d\bar z. \end{align*} This gives $\varphi(z_0) \le \displaystyle{\frac 1{\pi R^2}\frac i2\int_{D(z_0,R)} \varphi(z)\,dz\wedge d\bar z}$, as requested.

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  • $\begingroup$ First of all: THANK YOU! Second: can you please explain the last integral equality? Why should I have $\int_{D(z_0,R)}\varphi(z)rdrd\theta=\frac i2\int_{D(z_0,R)}\varphi(z)dz\wedge d\bar z$? $\endgroup$ – Joe Feb 2 '15 at 0:45
  • $\begingroup$ Because $r\,dr\wedge d\theta=dx\wedge dy =\dfrac i2 dz\wedge d\bar z$. $\endgroup$ – Ted Shifrin Feb 2 '15 at 0:46
  • $\begingroup$ (I'm at first sights to differential forms, sorry) but in the penultimate integral you wrote $drd\theta$, not $dr\wedge d\theta$. Why? How does this work? $\endgroup$ – Joe Feb 2 '15 at 0:48
  • $\begingroup$ You need to learn the definitions. Once orientations are correct, $\int_R f dx\wedge dy = \int_R f\,dx\,dy$. $\endgroup$ – Ted Shifrin Feb 2 '15 at 0:51
  • $\begingroup$ I studied that the first integral you just wrote is the integral of a 2-differential form (on a parametric manifold of dimension 2) but the second should be the usual Lebesgue integral. I'm studying hard these first tools about differential forms but I can't see the link between these two objects. $\endgroup$ – Joe Feb 2 '15 at 1:00
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I can rewrite $A(t)$ as $$ \int_0^{2\pi}\varphi(z_0+te^{i\theta})\,d\theta $$ WITHOUT adding $-2i$ or $i/2$ factors (which are the $\Bbb R^2\longleftrightarrow\Bbb C$ passage) because I didn't change variable, I simply rewrote $(\Re z_0+t\cos\theta,\Im z_0+t\sin\theta)$ in another way.

In this way I came to $$ -2i\int_0^rt\int_0^{2\pi}\varphi(z_0+te^{i\theta})\,d\theta\,dt\;\;. $$ Now we change variable $s=z_0+te^{i\theta}$ from which $ds=|ite^{i\theta}|d\theta=td\theta$ (because $s$ is the element of arc). Thus the last integral is equal to $$ -2i\int_0^r\int_{\partial\Delta_{z_0,t}}\varphi(s)\,ds\,dt\;. $$ Hence \begin{align*} \frac{1}{-2i}\int_{\Delta_{z_0,r}}\varphi(z,\bar z)dz\wedge d\bar z &=\int_0^r\int_{\partial\Delta_{z_0,t}}\varphi(s)\,ds\,dt\\ &\ge\int_0^r\varphi(z_0)2\pi t\,dt\\ &=\pi\varphi(z_0)r^2 \end{align*}

which concludes this little "odissey".

Thanks to all, once again I learned something new. I'm going to bed, here it's 5am.

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