I am taking complex analysis. There's a question in the book when trying to prove the theorem, and the theorem goes like this:

If $f$ is analytic in the disk $|z-z_0|<R$,then the taylor series converges to $f(z)$ for all $z$ in the disk. Furthermore, the convergence of the series is uniform in any closed subdisk $|z-z_0| \leq R'< R$.

And the question is if we can prove the uniform convergence in every closed subdisk $|z-z_0| \leq R'< R$, then we will have point-wise convergence in the open disk $|z-z_0|<R$. Why is that ? I kind of know that it must be true but cannot give a rigorous proof. Or more precisely, I know that uniform convergence implies point-wise convergence, but how do you deal with the boundary $R$ ?

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    $\begingroup$ Any point in the open disk lies in some closed disk, and uniform convergence implies pointwise convergence. $\endgroup$ – Paul Siegel Apr 12 '14 at 19:13

Let $D_{R^\prime}$ denote the closed disk $|z-z_0|\leq R^\prime<R$, and let $D_R$ dentoe the open disk $|z-z_0|<R$. Then, since $R^\prime<R$, $D_{R^\prime}\subset D_R$. Uniform convergence implies pointwise convergence. Hence, uniform convergence in every closed subdisk $D_{R^\prime}\subset D_R$ implies pointwise convergence in $D_R$.


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