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5:27 AM
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.
For instructions how to render MathJax(TeX) in chat see this post on meta.
I wanted to continue here the discussion from the comments to my answer here:
A: Continuous Linear Operator

Martin Sleziakb) 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.
The original question is about $\ell_2$:
Q: Continuous Linear Operator

user177196Consider 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...

First I want to see whether I understand correctly the addition to the question.
So the question is, what happens in this situation:
We will work with $\ell_2$ and some sequence space $X$ (endowed with some norm).
We fix $a\in\ell_2$ and define a function $T$ on $\ell_2$ as $(Tx)_{j} = \sum_{k=1}^{j} a_{k}x_{k}$.
I.e. $Tx$ is the sequence of partial sums of $\sum a_kx_k$.
Then there are two questions:
Is $Tx\in X$ for each $x\in\ell_2$?
If yes, is the functional $T\colon \ell_2\to X$ continuous?
In the answers to the linked question we have seen this for $X=\ell_2$ and $X=\ell_\infty$.
I think that the counterexample in my post works whenever $X\subseteq c_0$ and the sequence $(1,1/2,1/3,\dots)$ belongs to $X$. (So for such spaces we have that $Tx\notin X$ for some $x\in\ell_2$.
@user177196 Can you at least confirm whether I interpreted your additional question correctly?
If not could you clarify.
I am not sure that I am able to add much more to what I have already said.
But by posting this here in chat, there is a chance that somebody will notice the question and provide some additional answer.
1 hour later…
6:37 AM
ok, I'm going offline (I have to grade some test)
But feel free to leave me a message here, I will see it some time later.

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