This is my second attempt at calculating the spectrum of the left shift operator on a Hilbert space. I got stuck again and I would be grateful if someone could help. (You can find my previous (failed) attempt here).
Let $H$ be a Hilbert space with orthonormal base $e_i$ and let $L\in B(H)$ be the left shift operator, that is, $L(e_i) = e_{i-1}$.
In the following I will assume that if there exists a sequence $x_n\in H$ with $\|x_n\|=1$ and $Tx_n \to 0$ then $T$ is not invertible.
First note that $L$ is not invertible hence $0 \in \sigma (L)$.
Let $\lambda \in \mathbb C$ be non-zero.
I am stuck trying to find $x_n$ with $\|x_n\|=1$ and $(T-\lambda I)x_n \to 0$.
Edit
After doing some calculations I think that $L – \lambda I$ is never invertible for any $\lambda \in \mathbb C$ but I can't find $x \neq 0$ such that $(L-\lambda I)x = 0$. Could someone help me find such $x$ please? I'm sure that $\sigma(L) = \mathbb C$.
Best Answer
Since $\lVert L\rVert = 1$, it follows that $L-\lambda I$ is invertible for all $\lambda$ with $\lvert\lambda\rvert > 1$, we can see that using the Neumann series. For $\lvert \lambda\rvert > \lVert L\rVert$, the series
$$\sum_{n=0}^\infty \lambda^{-n}L^n$$
is absolutely convergent, and since $B(H)$ is a Banach space, it is convergent. One computes
$$(I-\lambda^{-1}L)\sum_{n=0}^\infty \lambda^{-n}L^n = \lim_{N\to\infty}(I-\lambda^{-1}L)\sum_{n=0}^N \lambda^{-n}L^n = \lim_{N\to \infty} I - \lambda^{-N-1}L^{N+1} = I,$$
and ditto for $\biggl(\sum\limits_{n=0}^\infty \lambda^{-n}L^n\biggr)(I-\lambda^{-1}L)$, and from that one sees that $L-\lambda I$ is invertible when $\lvert\lambda\rvert > \lVert L\rVert$, and the inverse is given by the series
$$(L-\lambda I)^{-1} = -\sum_{n=0}^\infty \lambda^{-n-1}L^n$$
then.
This argument uses no specific property of $L$, it works for all bounded linear operators on Banach spaces. The spectrum of a bounded linear operator $T$ on a Banach space is always conatined in the closed disk with radius $\lVert T\rVert$ and centre $0$.
For the left shift operator $L$, we have - using the separable Hilbert space $\ell^2(\mathbb{N})$ to get a convenient notation -
$$(L-\lambda I)x = (x_1 -\lambda x_0, x_2 - \lambda x_1, x_3 - \lambda x_2,\dotsc).$$
Thus $(L-\lambda I)x = 0$ if and only if we have $x_1 = \lambda x_0$, $x_2 = \lambda x_1$, and so on, which becomes $x_n = \lambda^n x_0$. For $\lvert \lambda\rvert < 1$, that defines a one-dimensional subspace of $H$, spanned by
$$\nu_\lambda = \sum_{n=0}^\infty \lambda^n\cdot e_n.$$
Thus we have $\ker (L-\lambda I) \neq \{0\}$ for $\lvert\lambda\rvert < 1$, and hence $\sigma(L)$ contains the open unit disk.
Since the spectrum of a bounded linear operator is compact, and we saw above that $\sigma(L)$ is contained in the closed unit disk, it follows that
$$\sigma(L) = \overline{\mathbb{D}} = \{ \lambda\in\mathbb{C} : \lvert \lambda\rvert \leqslant 1\}.$$