Product of two $2$-variables Taylor series

Question

Using the standard multi-index notation, suppose we have the two Taylor series $$ f(\theta) := \sum_{|\alpha|=0}^{\infty} a_{\alpha} \theta^{\alpha} $$ and $$ g(\theta) := \sum_{|\alpha|=0}^{\infty} b_{\alpha} \theta^{\alpha} $$ where $a_{\alpha} \in \mathbb{R}$, $\theta := (\theta_1, \theta_2)^T \in \mathbb{R}^2$ and $\alpha \in \left( \mathbb{Z}_{\geq 0} \right)^2$.

Question: On the convergence domain of $fg$, does $$ (fg)(\theta) = \sum_{|\alpha|=0}^{\infty} c_{\alpha} \theta^{\alpha} $$ where $$ c_{\alpha} := \sum_{\substack{\beta_1 + \gamma_1 = \alpha_1 \\ \beta_2 + \gamma_2 = \alpha_2\\}} a_{\beta} b_{\gamma}? $$

Answer 1

The proof is the same as in the one-variable case.

Suppose $f,g$ both converge at a point $(S_1,S_2)$ with $S_1,S_2\neq 0$ (if convergence requires $S_2=0$ then you are in a one-variable case). Let $C$ be such that $|a_\alpha.S^\alpha|,|b_\alpha.S^\alpha| \leq C$ for all $\alpha$ (this exists by convergence).

Now let $0<r<1$ and let $T_i=r|S_i|$. We show that the series for $fg$ converges uniformly absolutely in $[-T_1,T_1]\times[-T_2,T_2]$.

For this, consider the series

$$(*)\quad\sum_{\alpha,\beta} a_\alpha b_\beta \theta^\alpha \theta^\beta.$$

Putting absolute values we get the series

$$\sum_{\alpha,\beta} |a_\alpha b_\beta \theta^\alpha \theta^\beta| \leq \sum_{\alpha,\beta} |a_\alpha S^\alpha||b_\beta S^\beta|r^{|\alpha+\beta|}$$ $$\leq C^2 \sum_{\alpha,\beta} r^{|\alpha+\beta|} = C^2 \left(\sum_{n=0}^\infty r^n\right)^4 = \frac{C^2}{(1-r)^4}.$$

It follows that the series $(*)$ converges absolutely and uniformly in the stated domain. To see that it converges to $fg$ note that $$\left(\sum_{|\alpha|\leq N}a_\alpha \theta^\alpha\right) \left(\sum_{|\beta|\leq N}b_\beta \theta^\beta\right) $$ is a sequence of partial sums of $(*)$ which exhausts all summands. Now take $N\to\infty$. On the one hand you get $fg$ and on the other hand the sum of $(*)$.

Finally, given that the series $(*)$ converges absolutely to $fg$, we can arbitrarily reorder it without changing either the fact of convergence or the value of the sum. In particular, we can order the terms by the value of $\alpha+\beta$. Then the partial sums of your $\sum_\alpha c_\gamma \theta^\gamma$ are a subsequence of the partial sums of the rearrangement, so again they converge and to the same value.