Question regarding $l_2$-space.

Question

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.

Answer 1

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.