The shift operator on $\ell^2(\mathbb{N})$

Keywords: shift · ell2 · isometry · unitary operator · infinite dimension

Exercise 1 : The shift operator on $\ell^2(\mathbb{N})$

Question 1

Recall that

\ell^2(\mathbb{N})

denotes the space of sequences

(x_n)_{n \in \mathbb{N}}

of complex numbers satisfying

\sum_{n=0}^{\infty} |x_n|^2 < \infty

, equipped with the inner product

\langle x, y \rangle = \sum_{n=0}^{\infty} x^*_n y_n

We define the right-shift operator $T \colon \ell^2(\mathbb{N}) \to \ell^2(\mathbb{N})$ by:

T(x_0, x_1, x_2, ...) = (0, x_0, x_1, x_2, ...).

Question 2

Show that

T

is a linear isometry.

Solution

Linearity. For

\alpha, \beta \in \mathbb{C}

and

x, y \in \ell^2(\mathbb{N})

T(\alpha x + \beta y) = (0,\, \alpha x_0 + \beta y_0,\, \alpha x_1 + \beta y_1,\, ...) = \alpha\, T(x) + \beta\, T(y).

Isometry. We verify that the norm is preserved:

\|T(x)\|^2 = |0|^2 + \sum_{n=0}^{\infty} |x_n|^2 = \|x\|^2.

Thus $T$ is an isometry, and in particular $T$ is injective (since $\|T(x)\| = 0 \Rightarrow \|x\| = 0 \Rightarrow x = 0$ ).

Question 3

T

a Hilbert-space isomorphism (a unitary operator)?

Solution

T

is not a Hilbert-space isomorphism. A Hilbert-space isomorphism must be bijective. However,

T

is not surjective: the sequence

y = (1, 0, 0, ...) \in \ell^2(\mathbb{N})

has no preimage under

T

, because the first component of every

T(x)

0

Thus, $T$ is a non-unitary isometry. This example illustrates a phenomenon specific to infinite dimension: in finite dimension, every linear isometry is automatically surjective (by a dimension argument), which is no longer true in infinite dimension.

The shift operator on ℓ2(N)\ell^2(\mathbb{N})ℓ2(N)

The shift operator on $\ell^2(\mathbb{N})$