Does there exists a sequence $T_n$ such that $\|T_n-T\|\rightarrow 0$

Asked Jan 28 '23 at 15:53

Active Jan 28 '23 at 15:59

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Let $(a_n)_n\in \ell^\infty$ and define $$T:\ell^2\rightarrow \ell^2: (x_n)\mapsto (a_nx_n)$$ Assume that $a_n\rightarrow 0$. I want to check if there exists a sequence $T_n$ such that $T_n\rightarrow T$

My guess would be $T_n((x_n)_n)=(a_1x_1,...,a_nx_n,0,0,.....)$ then $$\begin{align}\|T_n-T\|^2&=\sup_{\|x\|_2\leq 1}\sum_{k>n} |a_kx_k|^2\\&\leq \sup_{\|x\|_2\leq 1}\|x\|_2^2\sup_{k>n}|a_k| \\&\leq \sup_{k>n}|a_k|\rightarrow 0 \end{align}$$

Is this correct or am I doing something wrong?

edited Jan 28 '23 at 15:59

asked Jan 28 '23 at 15:53

user1294729

2,008

Your answer is correct but your question is "wrong" ;-) Of course you do not just want a sequence $T_n\to T,$ otherwise you'd just have to take $T_n=T.$ – Anne Bauval Jan 28 '23 at 15:57
@AnneBauval no I think also my answer is wrong since $T_n-T$ is not the sum I wrote above – user1294729 Jan 28 '23 at 15:58
Oh sorry I did not read carefully. You are right (to say it is wrong): change all $k\le n$'s to $k>n.$ – Anne Bauval Jan 28 '23 at 15:59
@AnneBauval yes you are right I got it also in this moment – user1294729 Jan 28 '23 at 16:00
@AnneBauval now I think it works – user1294729 Jan 28 '23 at 16:01
Yes, and note you've thus proved that $T$ is a compact operator. – Anne Bauval Jan 28 '23 at 16:02
@AnneBauval really? I thought for compactness we also need $\dim(T_n(\ell^2))<\infty$ – user1294729 Jan 28 '23 at 16:03
Yes, here $ \dim(T_n(\ell^2))=n.$ Btw, I found a kind of duplicate: Is the multiplication operator on $\ell^p$ compact? – Anne Bauval Jan 28 '23 at 16:04
@AnneBauval sorry it does not, is the dimension here finite? – user1294729 Jan 28 '23 at 16:05
@AnneBauval how do I get this? I don't see how that one can compute the dimension of an operator? – user1294729 Jan 28 '23 at 16:06
It is not the "dimension of an operator" but of its range. The range of $T_n$ is a subspace of sequences $(x_i)_{i\in\Bbb N}$ s.t. $x_k=0$ for all $k>n.$ – Anne Bauval Jan 28 '23 at 16:08
@AnneBauval but in this subspace couldn't I have infinitely many sequences with this property? – user1294729 Jan 28 '23 at 16:10
Of course, the finite-dimensional space $\Bbb R^m$ is an infinite set (for $m>0$) – Anne Bauval Jan 28 '23 at 16:11
@AnneBauval sorry I am still a bit confused so how can we immediately see that the dimension of its range is $n$? – user1294729 Jan 28 '23 at 16:13
In fact it might be smaller if some of the numbers $a_1,\dots,a_n$ are $0.$ But assume for simplicity it is not the case. Then the range is exactly $\Bbb R^n\times{(0,0,\dots)},$ isomorphic to $\Bbb R^n.$ If you only want to prove it is finite dimensional, you only need to see it is a subspace of that space. – Anne Bauval Jan 28 '23 at 16:15
@AnneBauval ah okey I see perfect thanks! – user1294729 Jan 28 '23 at 16:17

Does there exists a sequence $T_n$ such that $\|T_n-T\|\rightarrow 0$

0 Answers0