Poisson Integral Formula for complex in unit disc 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.
 A: 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.
A: 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.
A: 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?
