Suppose $f(z) = \sum_0^\infty a_nz^n$ converges for $|z| \le 1$.

a) Prove $\phi(z) = \sum_0^\infty \frac{a_n}{n!}z^n$ is entire and $|\phi(z)|\le Me^{|z|}$.

b) Prove $f(z) = \int_0^\infty e^{-s}\phi(sz)ds$. (Hint: Integrate by parts.)

Apply this result to $f(z) = \sum_0^\infty z^{2n}$, which converges in $|z| < 1$ and show that the integral provides an analytic continuation of $f(z)$.

I have proved part(a), but am stuck on part(b).

With integration by parts, I am currently at:

$$-\phi(sz)e^{-s}|_0^{\infty} + z\int_0^\infty e^{-s}\phi'(sz)ds$$


EDIT: if I ignore the convergence of $f(z)$ on the boundary, and just assume that the convergence is for $|z|<1$, then repeated integration by parts, and the fact that $\phi$ is infinitely differentiable, pushes out the terms

$a_0 + a_1z + a_2z^2 + ... = \sum_0^\infty a_nz^n = f(z)$, which is what we needed to prove for part(b). But, is it ok to ignore the convergence on the boundary? I.e., only consider $z$ such that $|z|<1$, but the I feel that we haven't proved that the integral is exactly $f(z)$. Or, I might just be interpreting the question incorrectly, too.

  • $\begingroup$ Hi @RandomVariable, I was able to show that the series f(z) = the integral for |z|<1, but I needed to use the upper bound assumption from part(a). Otherwise, I don't think the left-hand term of the integration by parts (see above) can be evaluated as a limit for s going to infinity? L'Hopital's rule would have to be applied infinitely many times, which doesn't seem correct. The answer given below is a very good one, but it seems to have just plugged in infinity to show the decay of the first term, resulting in $\phi(0)$ = $a_0$. $\endgroup$ – User001 Aug 9 '15 at 23:29
  • $\begingroup$ Also, you mention using the identity theorem, to show that, since f(z) agrees with the integral on an open set, namely |z|<1, then both functions must agree on the set |z|$\le$1, but I think that the identity theorem is applicable to domains, which by definition are open and connected -- can we talk about the closure of the set? I.e., I don't know whether we can actually conclude that f(z) must agree with the integral on |z|=1, too, by the identity theorem. What are your thoughts? Thanks, @RandomVariable. $\endgroup$ – User001 Aug 9 '15 at 23:29
  • $\begingroup$ Right, @RandomVariable. And, clearly, evaluating the product with the point at infinity doesn't seem valid either. I have posted a new question for this topic and hope to get some answers on it. Thanks for your time. $\endgroup$ – User001 Aug 10 '15 at 1:02

(a). Since $f(z) = \sum_0^\infty a_nz^n$ converges for $|z| \le 1$, $\varlimsup_{n\to\infty}{\sqrt[n]{|a_n|}}<1$. So $\exists \:0<r<1,\:M>0$ such that $|a_n|<Mr^n,\:\forall n$. Also $$ \lim_{n\to\infty}{\sqrt[n]{\frac{|a_n|}{n!}}}=\lim_{n\to\infty}{\frac{\sqrt[n]{|a_n|}}{n{\dfrac{\sqrt[n]{n!}}{n}}}}<\lim_{n\to\infty}\frac{e}{{n}}=0 $$ So $\phi(z) = \sum\limits_{n=0}^\infty \dfrac{a_n}{n!}z^n$ is entire, and $$ |\phi(z)|\leqslant \sum\limits_{n=0}^\infty \dfrac{|a_n|}{n!}|z|^n<M\sum\limits_{n=0}^\infty \dfrac{r^n|z|^n}{n!}<Me^{|z|} $$ (b). \begin{align} \int_0^\infty e^{-s}\phi(sz)ds&=-\phi(sz)e^{-s}\bigg|_0^{\infty} + z\int_0^\infty e^{-s}\phi'(sz)ds \\ &=-\sum\limits_{n=0}^\infty \dfrac{a_n}{n!}s^nz^ne^{-s}\bigg|_0^{\infty} + z\int_0^\infty e^{-s}\phi'(sz)ds \\ &=a_0+z\int_0^\infty e^{-s}\sum\limits_{n=1}^{\infty}\dfrac{a_n}{(n-1)!}(sz)^{n-1}ds \\ &=a_0+z\int_0^\infty e^{-s}\sum\limits_{n=0}^{\infty}\dfrac{a_{n+1}}{n!}(sz)^{n}ds \\ &=a_0+\sum\limits_{n=0}^{\infty}\dfrac{a_{n+1}}{n!}z^{n+1}\int_0^\infty e^{-s}s^{n}ds \\ &=a_0+\sum\limits_{n=0}^{\infty}\dfrac{a_{n+1}}{n!}z^{n+1}\Gamma(n+1) \\ &=f(z) \end{align} Now consider $f(z) = \sum_{n=0}^\infty z^{2n}$. For $|z|<1$ $$ f(z) = \sum_{n=0}^\infty z^{2n}=\frac1{1-z^2} $$ For $|z|\geqslant1$, since $\phi(z)=\sum_{n=0}^\infty \dfrac{z^{2n}}{2n!}$ is entire $$ \int_0^\infty e^{-s}\phi(sz)ds=\sum_{n=0}^\infty \frac{z^{2n}}{2n!}=\cosh{z} $$ Finally we prove that $\cosh{z}$ is the analytic continuation of $f(z)=\dfrac1{1-z^2}$ on $|z|<1$. \begin{align} \int_0^\infty e^{-s}\cosh(sz)ds&=\dfrac1{2}\int_0^\infty (e^{s(z-1)}+e^{-s(z+1)})ds \\ &=\dfrac1{2}\left(\dfrac{e^{s(z-1)}}{z-1}-\dfrac{e^{-s(z+1)}}{z+1}\right)\bigg |_0^{\infty} \\ &=\dfrac1{2}\left(\dfrac{e^{s(Re(z)-1)}e^{sIm(z)i)}}{z-1}-\dfrac{e^{-s(Re(z)+1)}e^{-sIm(z)i)}}{z+1}\right)\bigg |_0^{\infty} \\ &=\dfrac1{2}\left(\dfrac1{1-z}-\dfrac1{z+1}\right) \\ &=\dfrac1{1-z^2} \end{align} Note: $Re(z)-1<0$ on $|z|<1$ and so $e^{s(Re(z)-1)}\to0$ as $s\to\infty$. Also $Re(z)+1>0$ on $|z|<1$ and so $e^{-s(Re(z)+1)}\to0$ as $s\to\infty.$

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Jyrki Lahtonen Aug 10 '15 at 4:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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