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I am studying complex analysis, I knew that in $\mathbb{C}$, every convergent power series can do termwise integration or differentiation in their radii of convergence. And I vaguely remember that for a sequence of function, if $f_n\to f$ in $L^1$ on $I$, then

$$\lim_{n\to\infty}\int_If_ndx=\int_I\lim_{n\to\infty}f_ndx$$

What's the relationship between these two conditions?

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1 Answer 1

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Taylor's theorem for holomorphic functions tells you that $P_nf(z)$, the $n$th degree Taylor polynomial for $f$ about $z_0$ with radius of convergence $R$, is such that $|P_nf(z) - f(z)| \rightarrow 0$ uniformly for every $z$ in a disc $D(z_0,r)$ with $r<R$. This satisfies the criterion for convergence that you gave above, and hence for any contour $\Gamma$ in this region of convergence,

$$ \lim_{n \rightarrow \infty} \int_{\Gamma} P_nf(z) \, dz = \int_{\Gamma} \lim_{n \rightarrow \infty} P_nf(z) \, dz = \int_{\Gamma} f(z) \, dz$$

As each integral $\int P_nf(z) \, dz$ is just the integral of a polynomial, then you can integrate it termwise as usual. This justifies the fact that you can integrate holomorphic functions termwise.

Termwise differentiation can be obtained from the result that $f'(z)$ is itself holomorphic, and then writing it as another Taylor series.

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oh I see, $P_n f(z)\to f(z)$ in $L^\infty$ within its radius of convergence, thanks!~ –  Ziqian Xie Sep 25 '12 at 2:19

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