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Let $f(z)=\sum_{n} a_n z^n$ has radius of convergence $R>0$ and $0<r<R$. Show:

$$\frac{1}{2\pi} \int_0^{2\pi} |f(re^{it})|^2 dt= \sum_{n}|a_n|^2r^{2n}$$

Use this equality to prove Liouvelle's Theorem.

Can anyone give me a hint to how to proceed with this?

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    $\begingroup$ Do you want a hint for how to prove the above equality, or for deducing Liouville's Theorem out of it? $\endgroup$ – sranthrop Mar 6 '16 at 16:50
  • $\begingroup$ I want a hint for proving the equality $\endgroup$ – mea43 Mar 6 '16 at 16:53
  • $\begingroup$ The formula is valid for any radius of convergence, but In the Liouvelle's Theorem the radius of convergence of the series is infinite. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 6 '16 at 18:28
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Note that by multiplying out and squaring the series for $f$, and using the relation $|w|^2=w\bar{w}$ we see that $$\frac{1}{2\pi} \int_0^{2\pi} |f(re^{it})|^2 dt = \frac{1}{2\pi} \int_0^{2\pi} \big(\sum_{j=0}^{\infty} c_jr^je^{ijt} \big)\big(\sum_{k=0}^{\infty} \bar{c_k} r^k e^{-ikt} \big) dt $$$$=\frac{1}{2\pi} \sum_{j,k=0}^{\infty}c_j\bar{c_k} r^{j+k} \int_0^{2\pi} e^{i(j-k)t} dt $$and in this last expression note that the integrand is zero unless $j=k$. I was purposefully informal with my infinite sums (there should be some limits in there).


For the second part of your question, suppose $f$ is a bounded entire function, say $|f| \leq M$.

Then the left hand side of the above equation is bounded above by $M^2$ for all $r>0$.

Now, for contradiction, assume that some $c_n \neq 0$. Then the right hand side is greater than or equal to $|c_n|^2r^{2n}$ which approaches to $\infty$ as $r \to \infty$.

Do you see a contradiction?

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  • $\begingroup$ How can that be a contradiction when $0<r<R$? $\endgroup$ – mea43 Mar 6 '16 at 17:32
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    $\begingroup$ Liouville's theorem : A bounded entire function is constant. So in part 2, we assune $R=\infty.$ $\endgroup$ – DanielWainfleet Mar 6 '16 at 19:12
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We have $f(z)=\sum_n c_nz^n$.

Substituting $z=re^{it}$ and evaluating,

$\left|f(re^{it})\right|^2=\sum_n\left|c_n\right|^2r^{2n}+\sum_n\sum_mg(n,m,r)\times e^{i(n-m)t}$

Integrating both sides from $t=0$ to $t=2\pi$ should do it.

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  • $\begingroup$ what is the function g here? $\endgroup$ – mea43 Mar 6 '16 at 17:11
  • $\begingroup$ Simply all possible combinations of $c_n\times \bar{c_m}\times r^{m+n}$ where $n\neq m$ $\endgroup$ – Mann Mar 6 '16 at 17:13

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