1
$\begingroup$

It's clear, that in the finite dimension there is the identity matrix which has this feature:

$$ AA^{-1} = A^{-1}A = I $$

I know, that it's not supposed to be generally true in the normed linear spaces of the infinite dimension.

So, could you, please, provide me any example of normed linear space and continuous linear mappings $f$ and $g$ which have the following feature?

$$ f \circ g = I\\ g \circ f \neq I $$

I'm not very experienced with this, so I really can't think of any example right now.

$\endgroup$
  • $\begingroup$ You might want to look at left and right inverses. $\endgroup$ – Nigel Overmars Sep 24 '16 at 20:47
1
$\begingroup$

The canonical example is given by the unilateral shift. You take $X=\ell^2(\mathbb N)$, and consider $$ S(a_1,a_2,\ldots)=(0,a_1,a_2,\ldots) $$ and $$ T(a_1,a_2,a_3,\ldots)=(a_2,a_3,\ldots). $$ Then $TS=I$ but $$ ST(a_1,a_2,\ldots)=(0,a_2,a_3,\ldots). $$ Both $S,T$ are bounded with $\|S\|=\|T\|=1$. In fact, $S$ is an isometry: $\|Sa\|=\|a\|$ for all $a$. It is an example of what is called a proper isometry.

$\endgroup$
  • $\begingroup$ Thank you! To be honest, I've only heard about unilateral shift so far. Could you, please, provide me any source to read a little about it? $\endgroup$ – Eenoku Sep 24 '16 at 20:55
  • $\begingroup$ Not really, sorry. It is usually mentioned in an exercise where you are asked to prove the things I 've mentioned above (which are all straightforward, together with the fact that $T=S^*$). $\endgroup$ – Martin Argerami Sep 24 '16 at 20:57
  • $\begingroup$ Nevermind :-) And BTW, is T operator really supposed to "return" always $a_2$? Or is it supposed to return $(a_1, a_2, a_3,...)$? $\endgroup$ – Eenoku Sep 24 '16 at 22:49
  • $\begingroup$ That would be the identity. Here the operator $T$ is the "reverse shift". The problem is that because there is no space, it "kills" $a_1$. $\endgroup$ – Martin Argerami Sep 25 '16 at 1:42
  • $\begingroup$ Great, I understand it now... Then I'll propose a little correction from $(a_2, a_2, ...)$ to $(a_2, a_3, ...)$. And I have one last question - why is it needed to have $X = l^2(\mathbb{N})$ ? $\endgroup$ – Eenoku Sep 25 '16 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.