# Example of non-commutative linear mappings compound which gives identity in one direction

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.

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

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.
• 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 $T=S^*$). – Martin Argerami Sep 24 '16 at 20:57
• Nevermind :-) And BTW, is T operator really supposed to "return" always $a_2$? Or is it supposed to return $(a_1, a_2, a_3,...)$? – Eenoku Sep 24 '16 at 22:49
• That would be the identity. Here the operator $T$ is the "reverse shift". The problem is that because there is no space, it "kills" $a_1$. – Martin Argerami Sep 25 '16 at 1:42
• 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 $X = l^2(\mathbb{N})$ ? – Eenoku Sep 25 '16 at 16:45