Suppose that $f\in L^2(\mathbb{R}/2\pi\mathbb{Z})$ takes the form $$f(\theta)=\sum_{n=1}^\infty a_ne^{in\theta}.$$ The function $$F(z)=\sum_{n=1}^\infty a_nz^n$$ converges in $|z|<1$. How can I evaluate the integral $$\int\int_{|z|<1}\left|\frac{\partial}{\partial r}F(z)\right|^2(1-|z|)dxdy$$ in terms of the coefficients $a_n$?

  • $\begingroup$ Have you tried polar coordinates? $\endgroup$
    – Jose27
    Dec 13, 2013 at 7:48

1 Answer 1


Related technique: (I), (II). Here is how you advance. We find the partial derivative w.r.t $r$ of $F(r,\theta)$

$$ F(z)=\sum_{n=1}^\infty a_n r^n e^{in\theta} \implies F_r = \sum_{n=1}^\infty n a_n r^{n-1} e^{in\theta}.$$

$$ \Bigg| \sum_{n=1}^\infty n a_n r^{n-1} e^{in\theta} \Bigg|^2 = \sum_{n=1}^\infty n a_n r^{n-1} e^{ni\theta}\sum_{m=1}^\infty m \bar{a_m } r^{m-1} e^{-im\theta}. $$

$$ 1-|z| = 1-|re^{i\theta}|=1-r $$

Gathering the above, the integral becomes

$$ \sum_{n=1}^\infty \sum_{m=1}^\infty n m a_n \bar{a_m}\int_{0}^{2\pi}\int_{0}^{1} r^{n+m-2}(1-r)e^{i(n-m)\theta} r dr d\theta\,. $$

Now, you just need to integrate and simplify. Note that, the

$$ \int_{0}^{2\pi} e^{i(n-m)\theta}d\theta $$

equals $0$ if $n\neq m$ and $2\pi$ if $n=m$.

Note: We used the fact

$$ |w|^2= w\, \bar {w}. $$

  • $\begingroup$ Didn't you rather mean $|w|^2=w\cdot\overline{w}$ ? $\endgroup$
    – Lucian
    Dec 14, 2013 at 4:46
  • $\begingroup$ @Lucian: It is multiplication of two complex numbers. $\endgroup$ Dec 14, 2013 at 5:03
  • $\begingroup$ Yes. $(a-bi)(a+bi)=a^2-(ib)^2=a^2+b^2=|a\pm bi|^2$. $\endgroup$
    – Lucian
    Dec 14, 2013 at 5:06

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