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An exercise I was doing asks (among other things) for the values of $z\in\mathbb{C}$ for which the following (operatorial) series converges absolutely: $$\sum_{n=0}^{\infty}z^nA^n$$ where $A$ is an operator in the Hilbert space $L^2(0,2\pi)$ such that $$(Af)(x)=\frac{1}{\pi}\int_0^{2\pi}[\cos(x)\cos(y)+\sin(x)\sin(y)]f(y)dy$$ I understand that $A$ is basically a projection operator in the form $Af = c\langle c,f\rangle+s\langle s,f\rangle$, where $s=\frac{1}{\sqrt{\pi}}\sin (x)$ and $c=\frac{1}{\sqrt{\pi}}\cos (x)$, so $||A||=1$ and $A^n = A$.

I also understand that, if I interpreted well, you should apply the Cauchy-Hadamard theorem to $\sum_{n=0}^{\infty}z^nA^n$ and search if it converges absolutely in the Banach space of all bounded operators between $L^2(0,2\pi)$ and itself. But with the Cauchy-Hadamard theorem you can conclude that the radius of convergence is $(\limsup||A^n||^{1/n})=1$. The answer to the exercise, however, is different and more cryptic:

"The series converges in norm when $||zA||\le 1$ ($|z|\le1$)"

How could you include the case $|z|=1$?

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As far as I see, your argument is correct and the answer to the exercise is wrong. – user31373 Jun 24 '12 at 21:08
The answer is wrong because of the simple case $z=1$. – Norbert Jun 24 '12 at 21:11
@Norbert Thank you! It is a bit frustrating when something in a book is wrong because you don't know if it your mistake or the author's. I just hope there aren't more exotic interpretations of the question for which the given answer would be correct... – Ralph Jun 24 '12 at 21:18
Well, I'll write an answer, to salvage this question from been unanswered :) – Norbert Jun 24 '12 at 21:25
up vote 1 down vote accepted

You solution is correct and solution in your book is wrong, because of the simple case $z=1$. For this case we have infinite sum of identical non-zero operators, more preciesly we have $$ \sum\limits_{n=1}^\infty 1^nA^n=\sum\limits_{n=1}^\infty A= ??? $$ Since $A^n=A$ our series is of the form $$ \sum\limits_{n=0}^\infty z^n A $$ which is absolutely convergent for $|z|<1$, because convergence depends only on the series $$ \sum\limits_{n=0}^\infty z^n $$

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