Here is the question and the answer:
I would like to ask in the first and second line, how does $R(T)$ being closed in $l^2$ and the open mapping theorem imply the inverse mapping $S$ being bounded.
Asked
Active
Viewed 467 times
1
-
Since $T:\ell^2\rightarrow R(T)$ is open by the open mapping theorem, its inverse is continuous by definition of continuity (inverse images of open sets open) and thus bounded – Peter Melech Jun 06 '22 at 07:58
-
related: https://math.stackexchange.com/q/1108244/173147 – glS Sep 25 '24 at 12:32
2 Answers
2
A closed subspace of a Banach space is Banach. You can think of $T$ as a linear continuous bijection from $\ell^{2}$ onto $R(T)$ and open Mapping Theorem can be applied bacause $\ell^{2}$ and $R(T)$ are Banach spaces.
Kavi Rama Murthy
- 359,332
-
I see. But we already know that $l^2$ is Banach and $T$ is a surjection hence $R(T)$=$l^2$. It seems like $R(T)$ could be a Banach space without the condition of being closed. Doing the same approach again to show $R(T)$ is closed seems also work. – math noob Jun 06 '22 at 08:13
-
1@mathnoob $T$ is not a surjection. The sequence $(1,\frac 1 2 , \frac 1 3, ...)$ cannot be $Tx$ for any $x$: if it is $Tx$ then $\frac {x_n} n=\frac 1 n$ so $x_n=1$ for all $n$. But then $\sum |x_n|^{2}=\infty$. – Kavi Rama Murthy Jun 06 '22 at 08:17
-
$T:\ell^2\rightarrow R(T)$ with this image is a surjection and if $R(T)$ was closed this would be a bijection between Banach spaces and then the open mapping theorem would imply that it is open and thus the inverse bounded which it is not, so $R(T)$ cannot be closed – Peter Melech Jun 06 '22 at 08:19
-
@ geetha290krm Oh, I get it. So this is the point of introducing $R(T)$ to replace $l^2$ in order to get a surjection. Thank you very much. – math noob Jun 06 '22 at 08:24
-
-
I do not see a reason for applying such a strong tool as open mapping theorem. The range is a proper dense subspace of $\ell^2,$ hence it is not closed. – Ryszard Szwarc Jun 06 '22 at 11:17
-
@RyszardSzwarc I agree. But I thought OP had a specific question about the presented proof. – Kavi Rama Murthy Jun 06 '22 at 11:27
2
The range of the operator $T$ is dense as $e_n=T(ne_n).$ But the range is not equal $\ell^2$ as $\{1/n\}_{n=1}^\infty\notin T(\ell^2)$, (as observed by @gettha290krm) Indeed if $(Tx)_n={1/ n},$ then $x_n=1$ for any $n.$
Ryszard Szwarc
- 44,675
-
May I ask how to show the range of $T$ is dense as well? By $e_n$=$T(ne_n)$, I know $e_n$$\in$$R(T)$ and I wonder how to prove $\forall$ x$\in$$l^2$, $|x-e_n|$<$\epsilon$ $\forall$ $\epsilon$>0. – math noob Jun 06 '22 at 18:03
-
The range of $T$ contains all elements $e_n.$ Hence it contains all finite linear combinations of $e_n,$ i.e. the elements of the form $\sum_{k=1}^n a_ke_k.$ Given $x\in \ell^1.$ Then $x^{(n)}:=\sum_{k=1}^nx_ke_k$ belongs to the range of $T$ and $|x-x^{(n)}|=\sum_{k=n+1}^\infty |x_k| \to 0$ when $n\to \infty.$ – Ryszard Szwarc Jun 06 '22 at 18:14
-
I think I got some of the ideas. We are considering $l^1$ since it is greater than $l^2$. And $l^1$ is a subset of c_0, so $x_k$ → 0 as k→∞. Intuitively, it implies $\sum_{k=n+1} ^{\infty} |x_k|$→0 as n→∞ but I couldn't prove it rigorously. Would you give me more hints please? – math noob Jun 07 '22 at 11:16
-
@mathnoob Sorry. I have mixed up the norms in my previous comment. It should be $|x-x^{(n)}|^2=\sum_{k=n+1}^\infty |x_k|^2\to 0$ when $n\to \infty.$ – Ryszard Szwarc Jun 07 '22 at 11:21
-
-
@RyszardSzwarc, do we know the codimension of $T$? Is it just one? i.e., $\textrm{Ran}(T)\oplus {{1/n}_{n=1}^{\infty}} = l_2$ as vector space? – Mohith Nagaraju Mar 12 '24 at 09:55
-
The codimension is infinite as the sequences $v_a:={n^{-a}}$ do not belong to the range for $1<a\le {3/2}$ and the elements ${v_a}$ for different $a$ and $R(T)$ are linearly independent. – Ryszard Szwarc Mar 12 '24 at 10:30