How to prove the following inequality involving complex power series

Question

Let $f(z) = \sum_{n=0}^\infty a_n z^n$, $g(z) = \sum_{n=0}^\infty b_nz^n$ be convergent complex power series.

Let $f_N(z) = a_0 + a_1z + \dots + a_N z^N$, $g_N(z) = b_0 + b_1z + \dots + b_N z^N$.

Prove that the following inequality holds:

\begin{equation} \lvert (fg)(z) - f_N(z)g_N(z) \rvert \leq \sum_{n=N+1}^\infty \sum_{k=0}^n \lvert a_k \rvert \lvert b_{n-k} \rvert \lvert z \rvert^n \end{equation}

If you want some context, check Serge Lang's Complex Analysis, pages 60-61, proof for Theorem 3.1 (in the book you'll find $(fg)_N(z)$ instead of $(fg)(z)$, but judging from the context it's probably just a typo).

Answer 1

For each $n\in\mathbb Z_+$, let$$c_n=\sum_{k=0}^na_kb_{n-k}\text{ and let }C_n=\sum_{k=0}^n\lvert a_k\rvert\lvert b_{n-k}\rvert.$$So, what you want to prove is that$$\bigl\lvert f(z)g(z)-f_N(z)g_N(z)\bigr\rvert\leqslant\sum_{n=N+1}^\infty C_n\lvert z\rvert^n.\tag1$$But$$f(z)g(z)=\sum_{n=0}^\infty c_nz^n$$and, if$$f_N(z)g_N(z)=\sum_{n=0}^\infty d_nz^n,$$then:

• $d_n=c_n$ if $n\leqslant N$;
• $d_n$ is the sum of some of the numbers $a_0b_n,a_1b_{n-1},\ldots,a_nb_0$ (whereas the sum of all these numbers is $c_n$) if $N<n\leqslant2N$;
• $d_n=0$ if $n>2N$.

So,$$f(z)g(z)-f_N(z)g_N(z)=\sum_{n=N+1}^\infty(c_n-d_n)z^n,$$and, by the properties mentioned above,$$\lvert c_n-d_n\rvert\leqslant\sum_{n=0}^N\lvert a_k\rvert\lvert b_{n-k}\rvert=C_n,$$and therefore we have $(1)$.