# How to prove that a power series in a unital Banach algebra converges normally in an open ball around 0

Let $$A$$ be an unital Banach algebra over a vector space $$X$$. Let us consider the power series:

$$\begin{equation*} \sum_{k=0}^{+\infty} c_k x^k \end{equation*}$$

with coefficients in $$\mathbb{K}$$ and the $$x$$ in $$X$$.

Suppose the series converges for some $$\bar{x}\neq 0$$. then the sequence $$c_k \bar{x}^k$$ converges to zero, which means that it is also bounded, so that there exists an $$L \in \mathbb{R}_+$$ such that $$|c_k \bar{x}^k |\leq L , \forall k \in \mathbb{N}$$. Now, let $$x$$ be such that $$|x|<|\bar{x}|$$. We have the following:

$$|c_k x^k| \leq |c_k| |x|^k = |c_k| |x|^k \left(\frac{|\bar{x}|}{|\bar{x}|}\right)^k = |c_k||\bar{x}|^k q^k$$

Having defined $$q = \frac{|x|}{|\bar{x}|}$$. I then want to say that the last term of the inequalities is less or equal to $$L q^k$$, with $$|q| < 1$$, so that the series has normal convergence by comparison with the geometric series. But that inequality doesn't follow from any property of the norm!! Am i missing some other way one could go about it?

• i know that $|c_k \bar{x}^k| \leq L$, but not that $|c_k||\bar{x}|^k \leq L$! And in fact $|c_k \bar{x}^k| \leq |c_k||\bar{x}|^k$ because the norm is submultiplicative. Apr 14, 2020 at 14:18

For example let $$X=M_2(\mathbb C)$$ with the operator norm, and consider the series $$\sum_{k=0}^\infty x^k.$$ Let $$\bar x=\begin{bmatrix} 0&3\\0&0\end{bmatrix}.$$ Since $$\bar x^2=0$$, the series converges at $$\bar x$$ (and is equal to $$1+\bar x$$). If now you take $$x=\begin{bmatrix} 1&1\\0&1\end{bmatrix},$$ then $$\|x\|=\sqrt{\frac{3+\sqrt5}2}\leq 3=\|\bar x\|.$$ But $$\sum_{k=0}^\infty x^k=\sum_{k=0}^\infty \begin{bmatrix} 1&k\\0&1\end{bmatrix}$$ does not converge.
• Thank you for your answer, i see the problem now. So this means that it’s not true, in general, that power series converge in balls, not even when the vector space is finite dimensional. I stumbled upon this proof because i wanted to find the most general setting possible to deal with power series, and to prove just once in that set the well-definedness of things like the exponential function, sine/cosine... and then prove things like $e^{x+y} = e^x e^y$ if $xy-yx=0$, or the trigonometry identities, without having to redo all the proofs for square matrices, complex numbers, etc. Apr 14, 2020 at 19:08
• Any tips on how to generalize, if not by proving this statement about the convergent sets? Also, shouldn’t $||x||$ be the square root of the one you wrote? Apr 14, 2020 at 19:12
• Series do converge in balls. You can still find the radius of convergence $R$ of the numerical series, and then the series in the Banach algebra will converge for any $x$ with $\|x\|<R$. What you cannot do, as the example shows, is to say that if the series converges for a certain element, it converges for all those with less radius. In the example, the radius of convergence of the series is $1$. Because the algebra has divisors of zero, there are elements with norm greater than 1 where the series still converges. Apr 14, 2020 at 20:02
• Got it. If we call $R$ the radius of convergence of the numerical series, and let $x \in X$ such that $|x|<R$ then, from the fact that the norm is submultiplicative follows that: \begin{equation*} \sum_{k=0}^{+\infty} |c_k x^k| \leq \sum_{k=0}^{+\infty} |c_k| |x|^k , \end{equation*} which means that convergence of the series on the right implies convergence of the series on the left (so, normal convergence). Now, because $X$ is a Banach space, normal convergence implies convergence. Apr 14, 2020 at 21:10