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?

  • $\begingroup$ 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. $\endgroup$
    – gent96
    Apr 14, 2020 at 14:18

1 Answer 1


The problem you found is essential. What you are trying to do works with numbers, but it doesn't necessarily work in a Banach algebra.

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.

  • $\begingroup$ 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. $\endgroup$
    – gent96
    Apr 14, 2020 at 19:08
  • $\begingroup$ 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? $\endgroup$
    – gent96
    Apr 14, 2020 at 19:12
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 14, 2020 at 20:02
  • 3
    $\begingroup$ So basically what fails in the Banach algebra is that the series will not necessarily diverge outside of the radius of convergence. $\endgroup$ Apr 14, 2020 at 20:02
  • $\begingroup$ 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. $\endgroup$
    – gent96
    Apr 14, 2020 at 21:10

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