Vtyjkyuk

For $x=(x_n) in l_2$, define

$T(x)=(0,x_1,x_2, cdots)$ and $S(x)=(x_2,x_3, cdots)$.

Which of the following statements are true?

$(a)$ $|T| = |S|=1$.

$(b)$ If $A: l_2 longrightarrow l_2$ is a continuous linear operator such that $|A-T|<1$, then $SA$ is invertible.

$(c)$ If $A$ is as above, then $A$ is not invertible.

$(a)$ is true which is easy to show. But I have no clue regarding $(b)$ and $(c)$. Would anybody please help me in this regard?

Thank you very much.

1. 
   

   $lVert Tx rVert = lVert x rVert$, thus $lVert T rVert le 1$. Using $x = (1, 0, dots)$ we have $lVert x rVert = 1$ and $lVert Tx rVert = 1$ as well. Hence $lVert T rVert = 1$.

   
   
   

   $ST = Id Rightarrow 1 = lVert ST rVert le lVert S rVert 1 Rightarrow lVert S rVert ge 1$. We also have $lVert Sx rVert le lVert x rVert$, hence $lVert S rVert = 1$.

   
   


2. 
   

   4-ier gave the solution in the comments: If $C$ is invertible and $lVert C - D rVert < lVert C^-1 rVert^-1$, then $D$ is invertible as well.

   
   
   

   Apply this on the invertible operator $ST$: $|SA-ST|=|S(A-T)| leq |S||A-T| < |S| = |I| = |(ST)^-1|^-1$

   
   
   

   You can find a proof outline at https://math.stackexchange.com/a/1508092/571891.

   
   



3. If $A$ were invertible, then per (2) $SAA^-1 = S$ would be invertible as well. However, $S$ is clearly not injective.