05:27
room topic changed to Functional analysis: For any discussions about functional analysis [functional-analysis]
Starting another room. Maybe it will be empty for most of the time. But it is still possible that some people interested in functional analysis might occasionally come here and discuss something and post here at least pointer to some interesting question.
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b) I might have misunderstood the problem but I don't think that $T$ necessarily even maps $\ell_2$ into $\ell_2$.
Just take $a=x=(1,1/2,1/3,\dots)$, i.e., $a_k=x_k=\frac1k$.
Clearly $a,x\in\ell_2$.
We have $(Tx)_n=\sum_{k=1}^n \frac1{k^2} = 1+\frac1{2^2}+\dots+\frac1{n^2}$. We have $\lim\lim...
Great counter-example, Martin! Many thanks for your help! I have a quick question though: the contradiction to "convergence criteria" is because of the fact that $l^2$ is a Hilbert space in this case. I wonder whether a counter-example exists if we replace $l^2$ by some other non-Hilbert and infinite-dimensional linear spaces?? — user177196 11 hours ago
I'd say that in works because of $x\in\ell_2$ $\Rightarrow$ $\sum x_n^2<+\infty$ $\Rightarrow$ $\lim\limits_{n\to\infty} x_n =0$. This is not directly related to the fact that $\ell_2$ is Hilbert space. What I'm using there is that $\ell_2\subseteq c_0$. (And also that it contains the sequence used in counterexample.) — Martin Sleziak 10 hours ago
I have to say that the question what happens when we replace $\ell_2$ by some other space is a bit unclear to me.
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Consider the two linear spaces:
$l^{2} = \left\{x = (x_1, x_2, . . . ) : \sum_{k=1}^{\infty} |x_k|^{2} < \infty\right\}$ with norm $||x||_{2} = (\sum_{k=1}^{\infty} |x_k|^{2})^{\frac{1}{2}}$, and $l^{\infty} = \left\{x = (x_1, x_2, . . . ) : \sup_{k} |x_{k}| < \infty\right\}$ with norm $||x||_{\i...
We fix $a\in\ell_2$ and define a function $T$ on $\ell_2$ as $(Tx)_{j} = \sum_{k=1}^{j} a_{k}x_{k}$.
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