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Products of convergent complex power series are represented by their Cauchy-product coefficients on the common disc

Statement

If $f (z) = \sum a_{n} (z - c)^{n}$ and $g (z) = \sum b_{n} (z - c)^{n}$ , then on their common open disc $f (z) g (z) = \sum_{n \geq 0} (\sum_{k = 0}^{n} a_{k} b_{n - k}) (z - c)^{n},$ and the product series converges locally uniformly.

Facts & Assumptions

Given: Two complex power series about $c$ and a point inside both radii.

[L1]

The Cauchy product of two absolutely convergent complex series converges absolutely and has the product of their sums as its sum (The Cauchy product of two absolutely convergent complex series converges absolutely to the product of their sums).

[L2]

Complex power series converge absolutely and uniformly on smaller closed subdiscs (A complex power series converges absolutely and uniformly on every closed subdisc strictly inside its disc of convergence).

[L3]

A complex function series dominated termwise by a convergent nonnegative real series converges absolutely pointwise and uniformly (Weierstrass M-test for complex-valued function series).

Proof

technique · direct

1.1

At a fixed point in the common disc, [L2] gives absolute convergence of both numerical series.

2.1

Apply [L1]; multiplying $(z - c)^{k} (z - c)^{n - k}$ gives $(z - c)^{n}$ , so the Cauchy coefficient is the displayed finite convolution. For $n = 0$ this is the one-term sum with $k = 0$ , not an empty sum.

step 1.1L1algebra

3.1

Fix a radius $r$ inside both original radii. The absolute Cauchy convolution has total sum $(\sum ∣ a_{n} ∣ r^{n}) (\sum ∣ b_{n} ∣ r^{n}) < \infty$ by [L1] and [L2], so [L3] gives uniform convergence of the product power series on $∣ z - c ∣ \leq r$ . This includes $r = 0$ and either input series being identically zero.

step 2.1L1L2L3algebra∎

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Sources

MIT 18.100C lecture notes on power series (standard reference, not scraped)