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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 $z^{n-1}$.2011-06-23

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|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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    yes, it makes sense. That's a good way to look at it. Thanks!2011-06-23