Suppose $T$ is right shift operator on $l^{2}\mathbb(N)$, $A$ is bounded operator on $l^{2}\mathbb(N)$ and $||A-T||<1$, prove that $A$ is not invertible.
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5Welcome to MSE. You need to show what you have tried or where you are stuck in order to get immediate feedbacks. – Leyla Alkan Apr 02 '18 at 10:59
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Some information I got from the conditions are (i) since set of invertible elements in Banach algebra form an open set so if A is invertible then there exists neighbourhood around A where all elements are invertible, but problem is pushing T in the neighbourhood to get the contradiction, (ii) A is bounded below and range of A is closed from the conditions – Apr 02 '18 at 11:20
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If $A$ is invertible, then $Ay=(1,0,\ldots)$ for some $y \in l^2(\mathbb{N})$. Thus $(A-T)y = (1,y_1,y_2,\ldots)$ and $\|(A-T)y\|^2=1+\|y\|^2$. Why we must have $\|A-T\| \geq 1$?
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