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I am reading about the Dirichlet Space right now. The definition of a Dirichlet space is the set of all holomorphic functions in the unit disc that are finite with respect to the semi-norm: $\mid \mid f \mid \mid$ = $\int _D \mid f^' (z)\mid ^2 dA(z)$

Here $dA$ is the normalized area measure, because we define $dA(x+iy)= \frac 1 \pi dx dy$. In the complex case, homorphicity and analyticity are equivalent, so we can say $f(z)=\sum^\infty_0a_nz^n$, then because $dA$ is an area normalizing measure, we can say $\mid \mid f \mid \mid$ = $\int _D \mid f^' (z)\mid ^2 dA(z)$ = $\int _D \mid f^' (z)\mid ^2 dA(z)$ = $\int_D \mid \sum_1^\infty na_n z^n-1 \mid^2 dA(z)$ = $\sum_1^\infty n\mid a_n \mid^2$. Question: Why is the last equality true?

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You want $\|f\|^2=\int_D|f'(z)|^2dA(z)$, not $\|f\|$. You can enclose exponents in { and }. E.g., $z^n-1$ gives you $z^n-1$, whereas $z^{n-1}$ gives you $z^{n-1}$. –  Jonas Meyer Jun 23 '11 at 7:05
    
After seeing the comment you posted in chat, I thought it would be worth mentioning that there is really no measure theory needed for this problem. Riemann integration suffices, since the functions are continuous. (Incidentally, the Town Hall Chat was for a discussion with candidates in the moderator election, and there is a separate chat room for general discussion.) –  Jonas Meyer Jun 23 '11 at 8:40
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up vote 3 down vote accepted

$$|f'(z)|^2=f'(z)\overline{f'(z)}=\left(\sum_1^\infty ma_mz^{m-1}\right)\left(\sum_1^\infty n\overline{a_n}\overline{z}^{n-1}\right) =\sum_{m,n=1}^\infty mna_m\overline{a_n}z^{m-1}\overline{z}^{n-1}.$$

Integration term-by-term is justified by uniform convergence on compact sets, and $\int_D z^{m-1}\overline{z}^{n-1} dA(z)$ is an easy computation using polar coordinates that will give you the desired sum.

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I should have clarified, by the unit disc I meant: the set $\mid z\mid <1$. This is not a compact set, and your uniform convergence argument doesn't hold –  r.g. Jun 23 '11 at 8:28
    
@Rohan: I know what you meant, but I'm glad you're looking for clarification since my answer wasn't clear. For $r<1$, let $D_r=\{z:|z|\leq r\}$. Then for each integrable $g:D\to\mathbb{C}$, $\int_D g(z)dA(z)=\lim_{r\to 1}\int_{D_r}g(z)dA(z)$. The series in question converge uniformly on each $D_r$. –  Jonas Meyer Jun 23 '11 at 8:35
    
yes, it makes sense. That's a good way to look at it. Thanks! –  r.g. Jun 23 '11 at 9:16
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