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Suppose that $T$ denotes the unilateral right shift operator on $\ell^2(\mathbb{N})$. My argument here $$\| T \| = \sup_{\| f \|=1} \| Tf \| = \sup_{\| f \| =1} \| T \left( \sum_{k=1}^{\infty} \langle e_k, f \rangle e_k \right) \| = \sup_{\| f \| =1} \| \sum_{k=1}^{\infty} \langle e_k, f \rangle e_{k+1} \| = 1,$$ show that the norm of $T$ is 1. I want to show that there is a vector $v \in \ell^2(\mathbb{N})$ orthogonal to the range of $T - \lambda I$ if and only if $\left| \lambda \right| <1$. I've played around with inner products for a few days and haven't made any progress. Note that this is a self-study question.

Cheers.

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If $\langle (T - \lambda I)x, v \rangle = 0$ for all $x$ then $\langle x, (T - \lambda I)^\ast v \rangle = 0$ for all $x$. This is equivalent to the statement that $v$ is an eigenvector of the adjoint $T^\ast$ with eigenvalue $\bar{\lambda}$.

You have to do three things:

  1. compute the adjoint $T^\ast$ (you can write it down expliclitly).

  2. Show that every $\mu$ with $|\mu| < 1$ is an eigenvalue of $T^\ast$ by finding an eigenvector (you can write it down explicitly, too).

  3. Show that every $\mu$ with $|\mu| \geq 1$ is not an eigenvalue of $T^\ast$.

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    The adjoint of $T$ is clearly the left shift operator. Now, consider that $Te_k = \lambda e_k \implies e_{k+1} = \lambda e_k \implies (e_1, e_2, ...) = e_1(1, \lambda, \lambda^2, ...) \implies \lambda \in \sigma(T^{\ast}) \Leftrightarrow \left| \lambda \right| < 1$. Is that enough? – 2017-02-26
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    looks good. you seem to mix notation a bit: first $e_k$ is a vector (you can apply $T$ to it), then $e_k$ is a coordinate of a vector/scalar. also, $\sigma(T^\ast)$ usually means the entire spectrum, not just the eigenvalues. but that's just nitpicking. you got the main point, and I'm sure you can flesh out the details now. – 2017-02-26