Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Show that a function analytic in the disc $\left | z \right |<1+\epsilon$ for some $\epsilon>0$ satisfies $\displaystyle f(z)=i Im(f(0))+\frac{1}{2\pi}\int_0^{2\pi}\frac{e^{it}+z}{e^{it}-z}Re(f(e^{it}))dt$

Suppose that the function above also satisfies $Re(f(e^{it}))\geq0, \forall t\in[0,2\pi]$, prove that: $\displaystyle \frac{1-\left | z \right |}{1+\left | z \right |}Re(f(0))\leq Re(f(0))\leq\frac{1+\left | z \right |}{1-\left | z \right |}Re(f(0)),\forall z: \left | z \right |<1$

I don't even know where to begin ._.

share|improve this question
add comment

1 Answer

The first line is the Schwarz integral formula, and it follows directly from the Poisson integral formula: The real part is the Poisson integral of the real part of $f$ on the circle, and the integral is analytic as a function of $z$, so the difference of $f$ and the integral is an imaginary constant. Evaluating both sides for $z=0$ you find that it has to be the imaginary part of $f(0)$.

In your line of inequalities the term in the middle should have argument $z$, not $0$. After that, the inequalities follow easily from $$ \frac{1-|z|}{1+|z|} \le \left|\frac{e^{it}+z}{e^{it}-z}\right| \le \frac{1+|z|}{1-|z|}$$ and $$u(0) = \frac1{2\pi} \int_0^{2\pi} u(e^{it}) \, dt,$$ where $u$ is the real part of $f$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.