Math Problems Vault

Problem 79

If $A$ is normal, then $\Gamma (A) = \Pi_0(A)$ (and therefore $\operatorname {spec} A = \Pi (A)$ ). Alternative formulation: the residual spectrum of a normal operator is always empty.

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Remark.

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Recall that the residual spectrum of $A$ is $\Gamma (A) - \Pi_0(A)$ .

Problem 80

For each diagonal operator, find its point spectrum, compression spectrum, and approximate point spectrum.

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Problem 81

For each multiplication, find its point spectrum, compression spectrum, and approximate point spectrum.

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Problem 82

What is the spectrum of the unilateral shift, and what are its parts (point spectrum, compression spectrum, and approximate point spectrum)? What are the answers to the same questions for the adjoint of the unilateral shift?

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Remark.

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The unilateral shift is the operator $U$ on $l^2$ defined by

U\left\langle \xi _0, \xi _1, \xi _2, \ldots \right\rangle = \left\langle 0, \xi _0, \xi _1, \xi _2, \ldots \right\rangle .

It is an isometry whose range is the proper subspace of $l^2$ of vectors with vanishing first coordinate. If $e_n$ is the vector $\left\langle \xi_0, \xi_1, \xi_2, \ldots \right\rangle$ for which $\xi_n = 1$ and $\xi_i = 0$ whenever $i \neq n$ ( $n = 0, 1, 2, \ldots$ ), then the $e_n$ 's form an orthonormal basis for $l^2$ , and $Ue_n = e_{n+1}$ ( $n = 0, 1, 2, \ldots$ ). Since, in $\mathbf{H}^2$ (see Problem 33), $e_n(z) = z^n$ , the effect of shifting forward by one index is the same as the effect of multiplication by $e_1$ ; in other words, the unilateral shift is the same as the multiplication operator on $\mathbf{H}^2$ defined by $(Uf)(z) = zf(z)$ .

Problem 83

Is there a non-scalar operator whose set of eigenvectors has a non-empty interior?

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Problem 84

What is the spectrum of the bilateral shift, and what are its parts (point spectrum, compression spectrum, and approximate point spectrum)? What are the answers to the same questions for the adjoint of the bilateral shift?

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Remark.

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Let $\mathbf{H}$ be the Hilbert space of all two-way (bilateral) square-summable sequences; the elements of $\mathbf{H}$ are most conveniently written in the form

\left\langle \ldots , \xi _{-2}, \xi _{-1}, (\xi _0), \xi _1, \xi _2, \ldots \right\rangle ,

the term in parentheses indicating the one corresponding to the index $0$ . The bilateral shift is the operator $W$ on $\mathbf{H}$ defined by

W\left\langle \ldots , \xi _{-2}, \xi _{-1}, (\xi _0), \xi _1, \xi _2, \ldots \right\rangle = \left\langle \ldots , \xi _{-3}, \xi _{-2}, (\xi _{-1}), \xi _0, \xi _1, \ldots \right\rangle .

It is an isometry, and since its range is all of $\mathbf{H}$ , it is even unitary. If $e_n$ is the vector $\left\langle \ldots , \xi_{-1}, (\xi_0), \xi_1, \ldots \right\rangle$ for which $\xi_n = 1$ and $\xi_i = 0$ whenever $i \neq n$ ( $n = 0, \pm 1, \pm 2, \ldots$ ), then the $e_n$ 's form an orthonormal basis for $\mathbf{H}$ , and $We_n = e_{n+1}$ ( $n = 0, \pm 1, \pm 2, \ldots$ ).

Problem 85

A necessary and sufficient condition that an operator $A$ on a Hilbert space $\mathbf{H}$ be representable as a multiplication on a functional Hilbert space is that the eigenvectors of $A^{*}$ span $\mathbf{H}$ .

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Remark.

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A functional Hilbert space is a Hilbert space $\mathbf{H}$ whose elements are complex-valued functions on some set $X$ , such that for each $x \in X$ evaluation at $x$ is a bounded linear functional on $\mathbf{H}$ ; by the Riesz representation theorem there is then, for each $x$ , a vector $K_x \in \mathbf{H}$ (the reproducing kernel at $x$ ) such that $f(x) = \left\langle f, K_x \right\rangle$ for every $f \in \mathbf{H}$ , and the $K_x$ 's always span $\mathbf{H}$ . A multiplication on a functional Hilbert space $\mathbf{H}$ (with domain $X$ ) is an operator $A$ for which there is a function $\varphi$ (the multiplier) on $X$ such that $(Af)(x) = \varphi (x)f(x)$ for every $f \in \mathbf{H}$ and every $x \in X$ .

Caution: as the facts for multiplications on $L^{2}$ spaces show (cf. Solution 81) this characterization is applicable to functional Hilbert spaces only. The result seems to be due to P. R. Halmos and A. L. Shields.