# Thread: Convergence

1. ## Convergence

Suppose {f_n} is a uniformly bounded sequence of holomorphic functions in D ( n=1... $). I wrote$ for infinity.
Suppose that exist such constant 0<C<$and |f_n(z)|<=C for every n from N and every z from D, and for every z from D exist limit lim f_n(z)=f(z) for n->$.

Prove that the convergence ({f_n}->f)is uniform on every compact subset of D.
Hint: Apply the dominated convergence theorem to the Cauchy formula.

I don't know how to apply the dominated convergence theorem to the Cauchy formula.

Thank you for helping me...

2. Originally Posted by naty
Hint: Apply the dominated convergence theorem to the Cauchy formula.

I really had to trouble myself to see how one can actually use the ...hint!
This result is easier to obtain with standard complex analysis arguments. And even if that is not satisfactory, there's always the Arzela-Ascoli theorem. But anyhow, can't show disrespect by ignoring advice.

Now for the proof. For a fixed $z\in D$, consider a circle $\gamma_z$ around $z$, the radius being so small as $\gamma_z\subset D$. Using the Cauchy integral formula,

$f_n(z)-f(z)=\frac{1}{2\pi{\rm i}}\int_{\gamma_z}\frac{f_n(w)-f(w)}{w-z}dw$

and dominated convergence grants us

$\bigg|\int_{\gamma_z}\frac{f_n(w)-f(w)}{w-z}dw\bigg|\rightarrow 0$,

so that, for any ${\epsilon}>0, \exists n_0$ such that

$\bigg|\int_{\gamma_z}\frac{f_n(w)-f(w)}{w-z}dw\bigg|\leq \epsilon, \ \forall n>n_0$. (1)

(Note: This $n_0$ is independent of $z$! Dominated Convergence rules )

Now we show ${\rm lim}_n[{\rm sup}_{z\in D}|f_n(z)-f(z)|]=:{\rm lim}\omega_n=0$, proving that the convergence is uniform. Using (1), forall $n>n_0$ we have

$\omega_n={\rm sup}_{z\in D}|f_n(z)-f(z)|={\rm sup}_{z\in D}\left(\bigg|\int_{\gamma_z}\frac{f_n(w)-f(w)}{w-z}dw\bigg|\right)\leq \epsilon$

from which follows ${\rm lim}\omega_n=0$.

ps. I still think the non-Lebesgue argument is sweeter.

3. Sidenote: Arzela-Ascoli for analytic functions is called Montel's Theorem.