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Let $S \in L(\ell^2(\mathbb{N}))$ be the unilateral shift defined by $S\delta_k = \delta_{k+1},k \in \mathbb{N}$. How do we show that if $T$ is an invertible bounded linear operator on $\ell^2 (\mathbb{N})$, then $\|S − T\| \geq 1$

nonpara
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1 Answers1

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Recall that $S^*$ is the "right shift" operator, taking a sequence $(\delta_0, \delta_1, \delta_2, \ldots)$ to $(0, \delta_0, \delta_1, \ldots)$. We have that $SS^* = I$, hence $S^*$ is an isometry, and so $\|S^*\| = 1$.

Now, suppose $T$ is invertible such that $\|S - T\| < 1$. Then, $$\|I - TS^*\| = \|SS^* - TS^*\| \le \|S - T\| \|S^*\| < 1.$$ Any operator with distance less than $1$ to $I$ is invertible, hence $TS^*$ is invertible. Since $T$ is invertible, this would imply $S^*$ is invertible. But this is a contradiction, as $S^*$ is not surjective.

user780985
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  • Pardon, but I'm pretty sure that the last word in your answer should be "injective". The unilateral shift $S$ as defined in the OP is injective, but not surjective . Its adjoint $S^$ is surjective, but not injective since it has a $1$-dimensional kernel spanned by the first element in the ONB. It appears you are swapping $S^$ and $S$. – Hanno Jan 10 '22 at 14:06