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I have a Banach space X and infinite sequence of bounded linear operators $P_n$ such that for every of them there exists $x_n$ such that $$ ||P_n(x_n)||=2^{2n}||x_n|| $$

These elements $x_n$ converge to zero. What is the norm of linear operator $$ Q=\sum^\infty 2^{-n} P_n $$ ?

Because there is a very good reason to think it is continous. There is a bit strange norm for these operators in which

$$ Q_N=\sum^N 2^{-n} P_n $$

is a Cauchy sequence and I believe this norm is complete. But then I can prove that Q is continous, meaning that if there is some $y_n\in X$ converging to y, then

$$ ||Q y_n - Qy||\leq ||Q y_n -Q_N y_n|| + ||Q_N y_n - Q_N y|| + ||Q_N y - Q y||$$ and all of these three converge to zero.

But the problem is that operator is bounded iff it is continous.

Any clue??

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    $\begingroup$ I don't really understand what you are asking here? $\endgroup$ – copper.hat Oct 5 '13 at 7:03
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    $\begingroup$ The series does not converge in the operator norm. "There is a bit strange norm..." What norm? $\endgroup$ – Jonas Meyer Oct 5 '13 at 7:07
  • $\begingroup$ I will provide additional details in the moment. $\endgroup$ – Michał Migacz Oct 5 '13 at 7:18
  • $\begingroup$ Oh, sorry, it is NOT a norm. A metric. It is $sup_{\mu \in \text{probability measures on space D}} ||P_n \mu - P\mu||_W$, where $|| ||_W$ means any metric equivalent to weak convergence of norms. It is complete for the space R of operators $P$ that are linear and they they preserve $P\mu(D)=\mu(D)=1$ However, in afterthought I see that these $Q_N$ do not belong to this space,'cause they do not belong to this space R. If you somehow push them into R, for example dividing by $\sum^N 2^-n$, then it is no longer a Cauchy sequece.Sorry for redundant question then.But I really cracked my mind on it. $\endgroup$ – Michał Migacz Oct 5 '13 at 8:04
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In general $Q$ is not even well defined. Indeed, consider $P_n$ defined by equality $P_n(x)=2^{2n}x$. Then $Q_N=(2^{N+1}-1)x$, and obviously $Q(x)=\lim\limits_{n\to\infty}Q_N(x)$ does not exist for any $x\in X$.

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