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There are some theorems that say in a unital C* algebra $A$ when one can deduce that the functional calculus of a continuous function f is continuous as map from some subset of $A$ to $A$. In the proof of one of these theorems it assumes that $A \mapsto A^n$ is continuous, and that $A \mapsto p(A, A^*)$ is continuous for any polynomial in $z, \bar z$. I actually suspect that $A \mapsto A^n$ is continuous in a more general setting where the latter statement wouldn't make sense, namely Banach algebras or even normed algebras. Please let me know why this is true, or at any greater or lesser level of generality, as long as it contains the C* algebra case. Thanks.

Incidentally, I am aware that raising to powers can be not continuous for powers more than $1$. If the integer powers give rise to continuous maps as I've conjectured above, I'd appreciate an example that illuminates why real positive powers can be so different.

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It is true in any topological ring. Not sure if that's what you're asking about. – tomasz Aug 19 '12 at 0:52
Actually it's strange, the same thing just occurred to me regarding my first question because you view it as $A \mapsto (A, A, A,...., A) \mapsto \times(A, A, A, ..., A)$. What's so strange to me is that a calculation involving norms and estimates does not seem to give way to this. If everybody would know a proof of that form, I'd be interested. Also, I still don't know the answer to my second question concerning real powers. – Jeff Aug 19 '12 at 0:57
Jeff, what exactly is your question about real powers? – Jonas Meyer Aug 19 '12 at 6:04
Actually nothing. After understanding this and some results that were obtainable afterwards, I now understand that raising to a given real power is a continuous operation on any bounded set within the positive cone, hence the entire positive cone. (My original question was about the continuity of this operation, and it's only definable on the positive cone, so this is as much as I could hope for.) – Jeff Aug 19 '12 at 8:40
up vote 3 down vote accepted

For $A\mapsto A^n$ to be continuous, it is enough for multiplication to be continuous. Since multiplication is continuous in any normed algebra (it's an obvious consequence of submultiplicativity of the norm), the map is continuous in any normed algebra.

To show that use induction with respect to $n$.

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Yeah this is basically what I wrote in the comments, but I'm just really thrown off by the fact that there's no sequence of estimates that seems to yield this result. If only A is commutative, then I'd be able to expand $x^n-y^n$ in the usual high school way, bound one factor, and the other factor is $x-y$. It strikes me as extremely odd that nothing like this is possible. – Jeff Aug 19 '12 at 1:36
@Jeff: It might be just that I'm really not into real analysis, but I would be really put off by a proof of this fact using a sequence of estimates. That said, if you really wanted to, this should be, in principle, possible to translate this proof into a sequence of estimates (inductively). But I don't expect it would be anything short of incredibly ugly. – tomasz Aug 19 '12 at 1:40
Actually, by translating between the abstract proof we've given so far and what it means when it's norms and not topologies. Just write $x^n-y^n$ as the telescoping sum $x^n-x^{n-1}y+x^{n-1}y...$ We just have different intuitions I guess, but I wanted to do something like the high school commutative case with that usual expansion, and when I couldn't I thought that only some minor modification would be needed, when it turns out that you actually just have to change strategies. – Jeff Aug 19 '12 at 1:42

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