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Let $f(z)$ be continuous in the closed unit disc and analytic in the unit disk. Prove the poisson integral formula: $$ f(a)=\frac{1}{2\pi} \int_0^{2\pi} f(e^{i\theta}) \frac{1-r^2}{1-2r \cos (t-\theta) +r^2} \, d\theta $$ $(a= re^{it}$, $0\leq r <1)$

I have done this problem by using the Cauchy Integral formula but my professor says that we can not use Cauchy because $f(z)$ is not analytic on the boundary of the unit disk.

Help will be appreciated.

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    $\begingroup$ So you've proved it is true in cases where $f$ is analytic on the boundary and in the interior. The remaining case could involve some functions that behave very badly on the boundary. $\endgroup$ – Michael Hardy Oct 3 '14 at 17:52
  • $\begingroup$ On the boundary also function is continuous. You mean the spikes on the boundary ? $\endgroup$ – user178061 Oct 7 '14 at 5:15
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    $\begingroup$ That it is continuous on the boundary was given. I had in mind that, for example, it might be nowhere differentiable on the boundary. $\endgroup$ – Michael Hardy Oct 7 '14 at 16:08
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Consider $f_\alpha (z):=f(\alpha z)$ where $0<\alpha<1$. Now $f_\alpha $ is analytic on the unit circle.Use the Cauchy integral formula to obtain the following: $$ I_\alpha:=\frac1{2\pi} \int_0^{2\pi} f_\alpha(e^{i\theta}) P(r,\theta-t) d\theta =f(\alpha a)$$ As $\alpha \to 1$ , $f(\alpha a) \to f(a)$ and $I_\alpha \to I$ ( by the boundedness of $P$ and uniform convergence of $f_\alpha $ to $f$ on the unit circle)

So we get the Poisson formula.

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I might try something like this: Define a function $g(a)$ to be that integral. Prove that $g$ is analytic in the interior (maybe Morera's theorem can be used for that, and quite possibly other methods). Conclude that $g-f$ is analytic in the open disk. If you can show that whenever $a$ approaches a point on the boundary, then $g(a)\to f(a)$, then define $g$ on the boundary by continuity. Then you have a function $g-f$ that is zero on the boundary, analytic in the interior, and continuous on the whole closed disk. Then see if you can use that to show that $g-f$ is everywhere zero.

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The Cauchy formula is still valid if $f$ is continuous on the closed unit disk and analytic inside. Why not prove that too, using the standard Cauchy for disks slightly smaller?

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  • $\begingroup$ If you use the disk slightly smaller how do you get circumference of of the disk 2 pi? $\endgroup$ – user178061 Oct 7 '14 at 5:17
  • $\begingroup$ Oh, I see, take a look here en.wikipedia.org/wiki/Cauchy%27s_integral_formula $\endgroup$ – Orest Bucicovschi Oct 7 '14 at 6:38
  • $\begingroup$ It says we can do what you said. It makes sense. I had the same feeling but professor made me confuse.Thanks anyway $\endgroup$ – user178061 Oct 10 '14 at 3:09

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