Analytic functions (part 3)

Last time we showed that analytic functions are continuous. This time we’ll show that {f'} is also analytic, and with the same radius of convergence. Of course, it suffices to show power series; analytic functions being a bunch of “patched together” power series.

Theorem 1 Let {f(z)=a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots} for {a_n\in{\mathbb C}} be a power series centered at {z_0} with radius of convergence {R}. Then {f'} is also a power series, and has radius of convergence {R}.

Proof: We know that {f} is the pointwise limit of the polynomials

\displaystyle f_n(z)=a_0+a_1(z-z_0)+\cdots+a_n(z-z_0)^n.

By definition, {f'=\frac{d}{dz}f=\frac{d}{dz}\lim_{n\rightarrow\infty}f_n}. It would be nice to show that {f'} is the pointwise limit of {f'_n}. This comes down to showing that I can move the differential operator {\frac{d}{dz}} across the limit, so that

\displaystyle f'=\frac{d}{dz}\lim_{n\rightarrow\infty}f_n=\lim_{n\rightarrow\infty}\frac{d}{dz}f_n=\lim_{n\rightarrow\infty}f'_n.

In general I am not allowed to do this, but if {f_n\rightarrow f} uniformly, then I can. This is a standard theorem from analysis that I won’t prove. If you like, it’s theorem 7.17 in Baby Rudin. In general, {f'_n\rightarrow f'} need not be uniform, but on any ball with radius smaller than it’s radius of convergence, the limit is uniform. This is the same trick that we used yesterday to show that {f_n\rightarrow f} uniformly on smaller balls, giving continuity of {f}. This tells us that {f'} is exactly the derivative you would expect. You can differentiate a power series by differentiating each term in the series:

\displaystyle f'(z)=a_1+2a_2(z-z_0)+3a_3(z-z_0)^2+4a_4(z-z_0)^3+\cdots.

To compute the radius of convergence of {f'}, we’ll instead compute the radius of convergence of {(z-z_0)\cdot f'}. Clearly this converges if and only if {f'} does. The coefficient of {(z-z_0)^n} in {(z-z_0)\cdot f'} is {n\cdot a_n}, so the radius of convergence is given by

\displaystyle \left(\limsup_{n\rightarrow\infty}\sqrt[n]{n\cdot|a_n|}\right)^{-1}=\frac1{\displaystyle\lim_{n\rightarrow\infty}\sqrt[n]n}\cdot\left(\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}\right)^{-1}=1\cdot R.

This completes the proof. \Box

Corollary 2 Let {f:U\rightarrow{\mathbb C}} ({U} an open subset of {{\mathbb C}}) be analytic. Then {f\in{\mathcal H}(U)} (“yes”). Moreover, {f} is in fact infinitely differentiable on {U} (“HELL YES!”).

Proof: First, pick any {z\in U}. Locally, {f} is given by a power series centered at some point {z_0} with radius of convergence {R}, so that {|z-z_0|<R}. Then {f'} is locally given by a power series centered at {z_0} with the same radius of convergence, and so {f'} is defined at {z}. Our choice of {z\in U} was arbitrary, so {f'} is defined on all of {U}, and hence {f\in{\mathcal H}(U)}.

Moreover, {f'} is analytic, and so {f'\in{\mathcal H}(U)}. Also, {f''} analytic, so {f''\in{\mathcal H}(U)}. In general, {f^{(n)}} is analaytic, and so {f^{(n)}\in{\mathcal H}(U)}. Thus, {f} is infinitely differentiable.

\Box

We are now two-thirds done with that theorem about the equivalences:

Theorem 3 Let {f:U\rightarrow{\mathbb C}}. The following are equivalent:

  • {f} is differentiable (holomorphic)
  • {f} is infinitely differentiable
  • {f} is locally representable by power series (analytic)

The second obviously implies the first, and today we just showed that the third implies the second. To see that the first implies the third, we’ll need to build a bit more machinery, so integrals are next on the agenda.

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One Response to Analytic functions (part 3)

  1. Pingback: The theorem « Andy Soffer

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