Q
stringlengths
4
3.96k
A
stringlengths
1
3k
Result
stringclasses
4 values
For each \( {z}_{0} \in \mathbb{D} \), the function\n\n\[ \psi \left( z\right) = \frac{{z}_{0} - z}{1 - \overline{{z}_{0}}z} \]\n\nis an inner function and \( {\mathcal{M}}_{{z}_{0}} = \left\{ {f \in {\mathbf{H}}^{2} : f\left( {z}_{0}\right) = 0}\right\} = \psi {\mathbf{H}}^{2} \) .
The function \( \psi \) is clearly in \( {\mathbf{H}}^{\infty } \) . Moreover, it is continuous on the closure of \( \mathbb{D} \) . Therefore, to show that \( \psi \) is inner, it suffices to show that \( \left| {\psi \left( z\right) }\right| = 1 \) when \( \left| z\right| = 1 \) . For this, note that \( \left| z\righ...
Yes
Theorem 2.4.2. If \( {z}_{1},{z}_{2},\ldots ,{z}_{n} \in \mathbb{D} \) , \n\n\[ \n\mathcal{M} = \left\{ {f \in {\mathbf{H}}^{2} : f\left( {z}_{1}\right) = f\left( {z}_{2}\right) = \cdots = f\left( {z}_{n}\right) = 0}\right\} , \n\] \n\nand \n\n\[ \n\psi \left( z\right) = \mathop{\prod }\limits_{{k = 1}}^{n}\frac{{z}_{k...
Proof. It is obvious that a product of a finite number of inner functions is inner. Thus Theorem 2.4.1 above implies that \( \psi \) is inner. \n\nIt is clear that \( \psi {\mathbf{H}}^{2} \) is contained in \( \mathcal{M} \) . The proof of the opposite inclusion is very similar to the proof of the case of a single fac...
No
Corollary 2.4.3. Suppose that the inner function \( \phi \) has a zero of multiplicity \( s \) at 0 and also vanishes at the nonzero points \( {z}_{1},{z}_{2},\ldots ,{z}_{n} \in \mathbb{D} \) (allowing repetition according to multiplicity). Let\n\n\[ \psi \left( z\right) = {z}^{s}\mathop{\prod }\limits_{{k = 1}}^{n}\f...
Proof. Since \( \psi \) is a product of inner functions, \( \psi \) is inner. The function \( \phi \) is in the subspace \( \mathcal{M} \) of the preceding Theorem 2.4.2, so that theorem implies that \( \phi = {\psi S} \), where \( S \) is in \( {\mathbf{H}}^{2} \) . Moreover, \( \widetilde{\phi } = \widetilde{\psi }\w...
Yes
Theorem 2.4.4. If \( \phi \) is an inner function and \( \phi \left( 0\right) \neq 0 \), and if \( \left\{ {z}_{j}\right\} \) is a sequence in \( \mathbb{D} \) such that \( \phi \left( {z}_{j}\right) = 0 \) for all \( j \), then \( \left| {\phi \left( 0\right) }\right| < \mathop{\prod }\limits_{{j = 1}}^{n}\left| {z}_{...
Proof. For each natural number \( n \), let\n\n\[ \n{B}_{n}\left( z\right) = \mathop{\prod }\limits_{{j = 1}}^{n}\frac{{z}_{j} - z}{1 - \overline{{z}_{j}}z}.\n\]\n\nAs shown in Corollary 2.4.3, each \( {B}_{n} \) is an inner function and, for each \( n \), there is an inner function \( {S}_{n} \) such that \( \phi = {B...
Yes
If \( {z}_{k} = \frac{k}{k + 1} \) for natural numbers \( k \), there is no function \( f \) in \( {\mathbf{H}}^{2} \) whose set of zeros is exactly \( \left\{ {z}_{k}\right\} \) .
Proof. Suppose that \( f \) was such a function and let \( \phi \) be its inner part. In particular, \( \phi \left( 0\right) \neq 0 \) . By the previous theorem,\n\n\[ \left| {\phi \left( 0\right) }\right| < \mathop{\prod }\limits_{{j = k}}^{n}\left| {z}_{k}\right| \]\n\nfor every natural number \( n \) . But\n\n\[ \ma...
Yes
Corollary 2.4.7. If \( {\left\{ {z}_{k}\right\} }_{k = 1}^{\infty } \) are nonzero zeros of a function \( f \) in \( {\mathbf{H}}^{2} \) that is not identically zero, then\n\n\[ \mathop{\prod }\limits_{{k = 1}}^{\infty }\left| {z}_{k}\right| \;\text{ converges. } \]
Proof. If \( {p}_{n} = \mathop{\prod }\limits_{{k = 1}}^{n}\left| {z}_{k}\right| \), then \( \left\{ {p}_{n}\right\} \) is a decreasing sequence (since \( \left| {z}_{k}\right| < 1 \) for all \( k \) ) and hence converges to some \( P \geq 0 \) . It must be shown that \( P > 0 \) .\n\nIf \( f \) has a zero of multiplic...
Yes
Theorem 2.4.8. If \( 0 < {r}_{k} < 1 \) for all \( k \), then \( \mathop{\prod }\limits_{{k = 1}}^{\infty }{r}_{k} \) converges if and only if \( \mathop{\sum }\limits_{{k = 0}}^{\infty }\left( {1 - {r}_{k}}\right) \) converges.
Proof. Assume \( \mathop{\prod }\limits_{{k = 1}}^{\infty }{r}_{k} \) converges. Since \( \left\{ {\mathop{\prod }\limits_{{k = 1}}^{n}{r}_{k}}\right\} \) converges as \( n \rightarrow \infty \) to a number different from 0 , it follows that\n\n\[ \left\{ \frac{\mathop{\prod }\limits_{{k = 1}}^{n}{r}_{k}}{\mathop{\prod...
Yes
Theorem 2.4.9. Let \( 0 < {r}_{k} < 1 \) for all \( k \) . If \( \mathop{\prod }\limits_{{k = 1}}^{\infty }{r}_{k} \) converges, then\n\n\[ \left\{ {\mathop{\prod }\limits_{{k = m + 1}}^{n}{r}_{k}}\right\} \]\n\nconverges to 1 as \( n \) and \( m \) approach infinity.
Proof. Observe that the above sequence is just\n\n\[ \frac{\mathop{\prod }\limits_{{k = 1}}^{n}{r}_{k}}{\mathop{\prod }\limits_{{k = 1}}^{m}{r}_{k}} \]\n\nand hence, when \( m \) and \( n \) approach infinity, the sequence approaches 1 .
Yes
Corollary 2.4.10. If \( {\left\{ {z}_{k}\right\} }_{k = 1}^{\infty } \) are zeros of a function \( f \in {\mathbf{H}}^{2} \) and \( f \) is not identically zero, then\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }\left( {1 - \left| {z}_{k}\right| }\right) < \infty \]
Proof. It follows from Corollary 2.4.7 and Theorem 2.4.8 that the subseries obtained by including only the nonzero zeros converges. If the multiplicity of the zero at 0 is \( m \), adding in the terms corresponding to the zeros at 0 adds \( m \) to the sum of the subseries.\n\nThus if a function in \( {\mathbf{H}}^{2} ...
Yes
Theorem 2.4.12 (Hurwitz’s Theorem). Let \( \left\{ {g}_{n}\right\} \) be a sequence of functions that are analytic and have no zeros on a domain \( V \) . If \( \left\{ {g}_{n}\right\} \rightarrow g \) uniformly on compact subsets of \( V \), then either \( g \) has no zero in \( V \) or \( g \) is identically 0 on \( ...
Proof. Suppose that \( g \) is not identically 0 on \( V \) . We will show that assuming that \( g \) has a zero in \( V \) leads to a contradiction. Suppose, then, that \( g\left( {z}_{0}\right) = 0 \) for some \( {z}_{0} \in V \) . Choose \( r > 0 \) such that \( {z}_{0} \) is the only zero of \( g \) in the disk \( ...
Yes
Corollary 2.4.14. Suppose that the inner function \( \phi \) has a zero of multiplicity \( s \) at 0 and has nonzero zeros at the points \( {z}_{1},{z}_{2},{z}_{3},\ldots \) in \( \mathbb{D} \) (repeated according to multiplicity). Let\n\n\[ B\left( z\right) = {z}^{s}\mathop{\prod }\limits_{{k = 1}}^{\infty }\frac{\ove...
Proof. For each positive integer \( n \), let\n\n\[ {B}_{n}\left( z\right) = \mathop{\prod }\limits_{{k = 1}}^{n}\frac{\overline{{z}_{k}}}{\left| {z}_{k}\right| }\frac{{z}_{k} - z}{1 - \overline{{z}_{k}}z} \]\n\nand\n\n\[ {B}_{0}\left( z\right) = \mathop{\prod }\limits_{{k = 1}}^{\infty }\frac{\overline{{z}_{k}}}{\left...
Yes
There is a Blaschke product that is not analytic at any point of \( {S}^{1} \) .
Proof. Let \( \left\{ {c}_{n}\right\} \) be any sequence dense in \( {S}^{1} \) and, for each \( n \), let\n\n\[ \n{z}_{n} = \left( {1 - \frac{1}{{n}^{2}}}\right) {c}_{n} \n\]\n\nThen \( 1 - \left| {z}_{n}\right| = \frac{1}{{n}^{2}} \), so the \( \left\{ {z}_{n}\right\} \) are the zeros of a Blaschke product \( B \) .\...
Yes
Every continuous function on \( \left\lbrack {0,1}\right\rbrack \) is a uniform limit of polynomials whose exponents are prime numbers.
Proof. Euler's theorem (Theorem 2.5.2) and the Müntz-Szasz theorem (Theorem 2.5.1) immediately imply the corollary.
No
If \[ f\left( z\right) = \exp \left( \frac{z + 1}{z - 1}\right) \] then \( f \) is a singular inner function.
Proof. Recall that \( \left| {e}^{w}\right| = \left| {e}^{\operatorname{Re}w + i\operatorname{Im}w}\right| = \left| {e}^{\operatorname{Re}w}\right| = {e}^{\operatorname{Re}w} \) for every complex number \( w \) . Hence \[ \left| {f\left( z\right) }\right| = \exp \left( {\operatorname{Re}\left( \frac{z + 1}{z - 1}\right...
Yes
Corollary 2.6.6. If \( \phi \) is an inner function, then \( \phi \) can be written as \( \phi = {BS} \), where \( B \) is the Blaschke product formed from the zeros of \( \phi \) and \( S \) is a singular inner function given by an integral as in Theorem 2.6.5.
Proof. Given an inner function \( \phi \), let \( B \) be the Blaschke product formed from the zeros of \( \phi \). By Theorem 2.4.14, \( \phi /B \) is an inner function with no zeros in \( \mathbb{D} \). Thus \( \phi /B = S \) is a singular inner function and therefore has the form given in Theorem 2.6.5.
Yes
Theorem 2.6.10. Every collection of inner functions has a greatest common divisor. Every finite collection of inner functions has a least common multiple.
Proof. Let \( \left\{ {\phi }_{\alpha }\right\} \) be a collection of inner functions. Define\n\n\[ \mathcal{M} = \bigvee \left\{ {f : f \in {\phi }_{\alpha }{\mathbf{H}}^{2}\text{ for some }\alpha }\right\} \]\n\nIt is easily seen that \( \mathcal{M} \) is invariant under \( U \) . Thus \( \mathcal{M} = {\phi }_{g}{\m...
Yes
Corollary 2.6.11. If \( \mathcal{M} \) is an invariant subspace, other than \( \{ 0\} \), of the unilateral shift, then \( \mathcal{M} = \phi {\mathbf{H}}^{2} \), where \( \phi \) is the greatest common divisor of all the inner parts of all the functions in \( \mathcal{M} \) .
Proof. Since \( \mathcal{M} \) is an invariant subspace for the unilateral shift, Beurling’s theorem (Theorem 2.2.12) guarantees that there exists an inner function \( \phi \) such that \( \mathcal{M} = \phi {\mathbf{H}}^{2} \) . We will show that \( \phi \) is the greatest common divisor of all the inner parts of all ...
Yes
Theorem 2.6.13. Let \( A \) be a bounded operator on an infinite-dimensional separable Hilbert space. Suppose that \( \mathcal{M} \) and \( \mathcal{N} \) are in Lat \( A \) and \( \mathcal{M} \subset \mathcal{N} \) . If \( \mathcal{N} \ominus \mathcal{M} \) is infinite-dimensional, then the lattice\n\n\[ \n\{ \mathcal...
Proof. Let \( P \) be the projection onto the subspace \( \mathcal{N} \ominus \mathcal{M} \) . Define the bounded linear operator \( B \) on \( \mathcal{N} \ominus \mathcal{M} \) as \( B = {\left. PA\right| }_{\mathcal{N} \ominus \mathcal{M}} \) . We will show that Lat \( B \) is order-isomorphic to the lattice of the ...
Yes
Let\n\n\[ \phi \left( z\right) = \exp \left( \frac{z + 1}{z - 1}\right) \]\n\nand let \( \mathcal{M} = {\left( \phi {\mathbf{H}}^{2}\right) }^{ \bot } \) . Then Lat \( \left( {\left. {U}^{ * }\right| }_{\mathcal{M}}\right) \) is order-isomorphic to the closed unit interval \( \left\lbrack {0,1}\right\rbrack \) with its...
Proof. The function \( \phi \) is inner singular. The measure \( \mu \) defined by \( \mu \left( A\right) = {2\pi } \) for any Borel set \( A \) containing 0 and \( \mu \left( B\right) = 0 \) for Borel sets \( B \) that do not contain zero is the measure provided by Theorem 2.6.5.\n\nIt follows from Beurling’s theorem ...
Yes
Corollary 2.7.2 (The F. and M. Riesz Theorem). If \( f \in {\mathbf{H}}^{2} \) and the set \[ \left\{ {{e}^{i\theta } : \widetilde{f}\left( {e}^{i\theta }\right) = 0}\right\} \] has positive measure, then \( f \) is identically 0 on \( \mathbb{D} \) .
Proof. If the set \( \left\{ {{e}^{i\theta } : \widetilde{f}\left( {e}^{i\theta }\right) = 0}\right\} \) has positive measure, the function \( \log \left| {\widetilde{f}\left( {e}^{i\theta }\right) }\right| \) is not integrable on \( \left\lbrack {0,{2\pi }}\right\rbrack \) . Thus the previous lemma implies that \( f \...
Yes
Theorem 2.7.3. (i) For \( x > 0,\log x \leq x - 1 \) .
Proof of (i): If \( g\left( x\right) = x - 1 - \log x \), then \( g\left( 1\right) = 0 \) and \( {g}^{\prime }\left( x\right) = 1 - \frac{1}{x} \) for all \( x > 0 \) . Thus \( g \) is decreasing on \( \left( {0,1}\right) \) and increasing on \( \left( {1,\infty }\right) \), so \( g\left( x\right) \geq g\left( 1\right)...
Yes
Theorem 2.7.4. If \( f \) is in \( {\mathbf{H}}^{2} \) and \( f \) is not identically 0 on \( \mathbb{D} \), define\n\n\[ F\left( z\right) = \exp \left( {\frac{1}{2\pi }{\int }_{0}^{2\pi }\frac{{e}^{i\theta } + z}{{e}^{i\theta } - z}\log \left| {\widetilde{f}\left( {e}^{i\theta }\right) }\right| {d\theta }}\right) .\n\...
Proof. For each fixed \( z \in \mathbb{D} \) ,\n\n\[ \left| \frac{{e}^{i\theta } + z}{{e}^{i\theta } - z}\right| \]\n\nis a bounded function of \( {e}^{i\theta } \in {S}^{1} \) . Since \( \log \left| {\widetilde{f}\left( {e}^{i\theta }\right) }\right| \) is in \( {\mathbf{L}}^{1}\left( {{S}^{1},{d\theta }}\right) \) by...
Yes
Corollary 2.7.5. If \( f \) is in \( {\mathbf{H}}^{2}, f \) is not identically 0, and \( F \) is defined by\n\n\[ F\left( z\right) = \exp \left( {\frac{1}{2\pi }{\int }_{0}^{2\pi }\frac{{e}^{i\theta } + z}{{e}^{i\theta } - z}\log \left| {\widetilde{f}\left( {e}^{i\theta }\right) }\right| {d\theta }}\right) ,\]\n\nthen ...
Proof. Since \( F \) is in \( {\mathbf{H}}^{2} \) ,\n\n\[ \left| {\widetilde{F}\left( {e}^{i\theta }\right) }\right| = \mathop{\lim }\limits_{{r \rightarrow {1}^{ - }}}\left| {F\left( {r{e}^{i\theta }}\right) }\right| = \exp \left( {\mathop{\lim }\limits_{{r \rightarrow {1}^{ - }}}\frac{1}{2\pi }{\int }_{0}^{2\pi }{P}_...
Yes
Theorem 2.7.7. If \( f \) is in \( {\mathbf{H}}^{2} \) and \( F \) is defined by\n\n\[ F\left( z\right) = \exp \left( {\frac{1}{2\pi }{\int }_{0}^{2\pi }\frac{{e}^{i\theta } + z}{{e}^{i\theta } - z}\log \left| {f\left( {e}^{i\theta }\right) }\right| {d\theta }}\right) ,\]\nthen \( F \) is outer.
Proof. It was shown above (Theorem 2.7.4) that \( F \) is in \( {\mathbf{H}}^{2} \) . Thus \( F = {\phi G} \) for \( \phi \) an inner function and \( G \) an outer function (Theorem 2.3.4). It suffices to show that \( \phi \) is a constant function. Since \( \left| {\widetilde{\phi }\left( {e}^{i\theta }\right) }\right...
Yes
Corollary 2.7.8. The function \( G \) in \( {\mathbf{H}}^{2} \) is outer if and only if there exists a constant \( K \) of modulus 1 such that\n\n\[ G\left( z\right) = K\exp \left( {\frac{1}{2\pi }{\int }_{0}^{2\pi }\frac{{e}^{i\theta } + z}{{e}^{i\theta } - z}\log \left| {\widetilde{G}\left( {e}^{i\theta }\right) }\ri...
Proof. Since a nonzero constant times an outer function is outer, the previous theorem establishes that every function of the given form is outer.\n\nConversely, suppose \( G \) is any outer function and define\n\n\[ F\left( z\right) = \exp \left( {\frac{1}{2\pi }{\int }_{0}^{2\pi }\frac{{e}^{i\theta } + z}{{e}^{i\thet...
Yes
Corollary 2.7.9. Let \( f \) be a function in \( {\mathbf{H}}^{2} \) that is not identically 0, and let \( B \) denote the Blaschke product formed from its zeros. Then there exists a constant \( K \) of modulus 1 and a singular measure \( \mu \) on \( {S}^{1} \) such that \( f\left( z\right) \) has the form\n\n\[ \n{KB...
Proof. This follows immediately from Theorem 2.3.4, Corollary 2.4.14, Theorem 2.6.5, and Corollary 2.7.8.
No
Theorem 2.7.10. The function \( F \) in \( {\mathbf{H}}^{2} \) is outer if and only if\n\n\[ \log \left| {F\left( 0\right) }\right| = \frac{1}{2\pi }{\int }_{0}^{2\pi }\log \left| {\widetilde{F}\left( {e}^{i\theta }\right) }\right| {d\theta }. \]
Proof. If \( F \in {\mathbf{H}}^{2} \) is outer, then\n\n\[ F\left( z\right) = K\exp \left( {\frac{1}{2\pi }{\int }_{0}^{2\pi }\frac{{e}^{i\theta } + z}{{e}^{i\theta } - z}\log \left| {\widetilde{F}\left( {e}^{i\theta }\right) }\right| {d\theta }}\right) \]\n\nfor some \( K \) (Corollary 2.7.8).\n\nThen,\n\n\[ F\left( ...
Yes
Theorem 3.1.2. Let \( \phi \) be a function in \( {\mathbf{L}}^{\infty } \) with Fourier series\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\phi }_{n}{e}^{in\theta } \]\n\nThen the matrix of \( {M}_{\phi } \) with respect to the orthonormal basis \( {\left\{ {e}^{in\theta }\right\} }_{n = - \infty }^{\infty...
Proof. We compute, for each pair of integers \( \left( {m, n}\right) \),\n\n\[ \left( {{M}_{\phi }{e}_{n},{e}_{m}}\right) = \frac{1}{2\pi }{\int }_{0}^{2\pi }\phi \left( {e}^{i\theta }\right) {e}^{in\theta }\overline{{e}^{im\theta }}{d\theta } = \frac{1}{2\pi }{\int }_{0}^{2\pi }\phi \left( {e}^{i\theta }\right) {e}^{-...
Yes
Theorem 3.1.4. A bounded linear operator on \( {\mathbf{L}}^{2} \) is multiplication by an \( {\mathbf{L}}^{\infty } \) function if and only if its matrix with respect to the standard basis in \( {\mathbf{L}}^{2} \) is a Toeplitz matrix.
Proof. It was shown in Theorem 3.1.2 that every multiplication on \( {\mathbf{L}}^{2} \) has a Toeplitz matrix. To establish the converse, assume that \( A \) has a Toeplitz matrix.\n\nTo show that \( A = {M}_{\phi } \) for some \( \phi \in {\mathbf{L}}^{\infty } \), it suffices, by Theorem 2.2.5, to show that \( {AW} ...
Yes
Theorem 3.1.6. If \( \phi \in {\mathbf{L}}^{\infty } \), then \( \sigma \left( {M}_{\phi }\right) = \Pi \left( {M}_{\phi }\right) = \operatorname{ess}\operatorname{ran}\phi \) .
Proof. We prove this in two steps. We first show that ess \( \operatorname{ran}\phi \subset \Pi \left( {M}_{\phi }\right) \) , and then show that \( \sigma \left( {M}_{\phi }\right) \subset \) ess ran \( \phi \) . These two assertions together imply the theorem.\n\nLet \( \lambda \in \operatorname{essran}\phi \) . For ...
Yes
Theorem 3.2.2. The matrix of the Toeplitz operator with symbol \( \phi \) with respect to the basis \( {\left\{ {e}^{in\theta }\right\} }_{n = 0}^{\infty } \) of \( {\widetilde{\mathbf{H}}}^{2} \) is\n\n\[ \n{T}_{\phi } = \left( \begin{matrix} {\phi }_{0} & {\phi }_{-1} & {\phi }_{-2} & {\phi }_{-3} & \\ {\phi }_{1} & ...
Proof. This can easily be computed in the same way as the corresponding result for multiplication operators (Theorem 3.1.2). Alternatively, since \( P \) is the projection onto \( {\widetilde{\mathbf{H}}}^{2} \) and \( {T}_{\phi } \) is defined on \( {\widetilde{\mathbf{H}}}^{2} \), the matrix of \( {T}_{\phi } \) is t...
Yes
Theorem 3.2.4. If \( {T}_{\phi } \) is an analytic Toeplitz operator, then the matrix of \( {T}_{\phi } \) with respect to the basis \( {\left\{ {e}^{in\theta }\right\} }_{n = 0}^{\infty } \) is\n\n\[ \n{T}_{\phi } = \left( \begin{matrix} {\phi }_{0} & 0 & 0 & 0 & 0 & \\ {\phi }_{1} & {\phi }_{0} & 0 & 0 & 0 & \ddots \...
Proof. The Fourier coefficients of \( \phi \) with negative indices are 0 since \( \phi \) is in \( {\widetilde{\mathbf{H}}}^{2} \), so this follows immediately from Theorem 3.2.2.
Yes
Corollary 3.2.7. The operator \( T \) is a Toeplitz operator if and only if \( {U}^{ * }{TU} = \) \( T \), where \( U \) is the unilateral shift.
Proof. Note that, for nonnegative integers \( n \) and \( m \) ,\n\n\[ \left( {{U}^{ * }{TU}{e}_{n},{e}_{m}}\right) = \left( {{TU}{e}_{n}, U{e}_{m}}\right) = \left( {T{e}_{n + 1},{e}_{m + 1}}\right) . \]\n\nThus if \( T \) is a Toeplitz operator, \( \left( {{U}^{ * }{TU}{e}_{n},{e}_{m}}\right) = \left( {T{e}_{n},{e}_{m...
Yes
Theorem 3.2.8. The mapping \( \phi \mapsto {T}_{\phi } \) is an injective, bounded, linear, adjoint-preserving (i.e., \( {T}_{\phi }^{ * } = {T}_{\bar{\phi }} \) ) mapping from \( {\mathbf{L}}^{\infty } \) onto the space of Toeplitz operators regarded as a subspace of the algebra of bounded linear operators on \( {\wid...
Proof. The map is obviously linear, and\n\n\[ \begin{Vmatrix}{T}_{\phi }\end{Vmatrix} = \begin{Vmatrix}{P{M}_{\phi }}\end{Vmatrix} \leq \begin{Vmatrix}{M}_{\phi }\end{Vmatrix} = \parallel \phi {\parallel }_{\infty }, \]\n\nso the mapping is bounded. If \( {T}_{\phi } \) and \( {T}_{\psi } \) are equal, then comparing t...
Yes
Lemma 3.2.9. If \( {T}_{\psi } \) and \( {T}_{\phi } \) are Toeplitz operators and \( U \) is the unilateral shift, then\n\n\[ \n{U}^{ * }{T}_{\psi }{T}_{\phi }U - {T}_{\psi }{T}_{\phi } = P\left( {{e}^{-{i\theta }}\psi }\right) \otimes P\left( {{e}^{-{i\theta }}\bar{\phi }}\right) \;\text{ (see Notation 1.2.27),} \n\]...
Proof. Note that \( I = U{U}^{ * } + {e}_{0} \otimes {e}_{0} \), where \( {e}_{0} \otimes {e}_{0} \) is the orthogonal projection from \( {\widetilde{\mathbf{H}}}^{2} \) onto the constants. Therefore\n\n\[ \n{U}^{ * }{T}_{\psi }{T}_{\phi }U = {U}^{ * }{T}_{\psi }\left( {U{U}^{ * } + {e}_{0} \otimes {e}_{0}}\right) {T}_...
Yes
Theorem 3.2.11. For \( \psi \) and \( \phi \) in \( {\mathbf{L}}^{\infty },{T}_{\psi }{T}_{\phi } \) is a Toeplitz operator if and only if either \( {T}_{\psi } \) is coanalytic or \( {T}_{\phi } \) is analytic. In both of those cases, \( {T}_{\psi }{T}_{\phi } = {T}_{\psi \phi } \) .
Proof. That \( {T}_{\psi }{T}_{\phi } = {T}_{\psi \phi } \) when \( {T}_{\phi } \) is analytic is trivial, as remarked above. If \( {T}_{\psi } \) is coanalytic, then\n\n\[{\left( {T}_{\psi }{T}_{\phi }\right) }^{ * } = {T}_{\phi }^{ * }{T}_{\psi }^{ * } = {T}_{\bar{\phi }}{T}_{\bar{\psi }} = {T}_{\bar{\phi }\bar{\psi ...
Yes
Corollary 3.2.12. The product of two Toeplitz operators is 0 if and only if one of the factors is 0.
Proof. Assume that \( {T}_{\psi }{T}_{\phi } = 0 \) . Since 0 is a Toeplitz operator, the previous theorem implies that either \( {T}_{\psi } \) is coanalytic or \( {T}_{\phi } \) is analytic, and that \( {T}_{\psi }{T}_{\phi } = \) \( {T}_{\psi \phi } = 0 \) . Hence \( {\psi \phi } = 0 \) .\n\nIf \( {T}_{\phi } \) is ...
Yes
Corollary 3.2.14. If two Toeplitz operators commute with each other and neither is a linear combination of the identity and the other operator, then their product is a Toeplitz operator.
Proof. It follows from the previous theorem that either the first operator is analytic or the second operator is coanalytic. Theorem 3.2.11 finishes the proof.
No
Theorem 3.2.15. A Toeplitz operator is self-adjoint if and only if its symbol is real-valued almost everywhere.
Proof. This follows immediately from the fact that \( {T}_{\phi } = {T}_{\phi }^{ * } \) if and only if \( \phi = \bar{\phi } \)
Yes
Corollary 3.2.16. The Toeplitz operator \( {T}_{\phi } \) is normal if and only if there exist complex numbers \( c \) and \( d \) and a real-valued function \( \psi \) in \( {\mathbf{L}}^{\infty } \) such that \( \phi = {c\psi } + d \) a.e. That is, the only normal Toeplitz operators are affine functions of self-adjoi...
Proof. Since \( {T}_{\psi }^{ * } = {T}_{\psi } \), it is clear that every operator of the given form is normal.\n\nThe converse follows easily from Theorem 3.2.13. To prove it, suppose that \( {T}_{\phi } \) commutes with \( {T}_{\phi }^{ * } = {T}_{\bar{\phi }} \) . At least one of the three cases of Theorem 3.2.13 h...
Yes
Theorem 3.2.17. The only compact Toeplitz operator is 0.
Proof. Let \( {T}_{\phi } \) be a compact Toeplitz operator. Recall that compact operators map weakly convergent sequences to norm-convergent sequences [12, p. 173]. Therefore we have\n\n\[ \left\{ \left( {{T}_{\phi }{e}_{s + n},{e}_{t + n}}\right) \right\} \rightarrow 0\;\text{ as }n \rightarrow \infty ,\]\n\nsince \(...
Yes
Theorem 3.3.1 (The Spectral Inclusion Theorem). For all \( \phi \) in \( {\mathbf{L}}^{\infty } \) , the spectrum of \( {M}_{\phi } \) is contained in the spectrum of \( {T}_{\phi } \) . More precisely,\n\n\[ \n\text{ess}\operatorname{ran}\phi = \Pi \left( {M}_{\phi }\right) = \sigma \left( {M}_{\phi }\right) \subset \...
Proof. It has already been shown (Theorem 3.1.6) that ess ran \( \phi = \Pi \left( {M}_{\phi }\right) = \) \( \sigma \left( {M}_{\phi }\right) \) .\n\nAssume \( \lambda \in \Pi \left( {M}_{\phi }\right) \) . Then there exists a sequence \( \left\{ {f}_{n}\right\} \) of functions in \( {\mathbf{L}}^{2} \) with \( \begin...
No
Corollary 3.3.2. For \( \phi \) in \( {\mathbf{L}}^{\infty },\parallel \phi {\parallel }_{\infty } = \begin{Vmatrix}{M}_{\phi }\end{Vmatrix} = \begin{Vmatrix}{T}_{\phi }\end{Vmatrix} = r\left( {T}_{\phi }\right) \) (where \( r\left( {T}_{\phi }\right) \) is the spectral radius; see Definition 1.2.2).
Proof. We have already shown that \( \parallel \phi {\parallel }_{\infty } = \begin{Vmatrix}{M}_{\phi }\end{Vmatrix} \) (see Theorem 2.2.4).\n\nIt is an easy consequence of the spectral radius formula that the spectral radius of a normal operator is equal to its norm (see, for example, [41, p. 11] or \( \left\lbrack {{...
Yes
Corollary 3.3.3. The only quasinilpotent Toeplitz operator is the operator 0.
Proof. If \( r\left( {T}_{\phi }\right) = 0 \), the previous corollary gives \( \parallel \phi {\parallel }_{\infty } = 0 \), which implies that \( \phi = 0 \) a.e.; i.e., \( {T}_{\phi } = 0 \) .
Yes
Corollary 3.3.4. If \( \phi \) is in \( {\mathbf{L}}^{\infty } \) and \( K \) is compact, then \( \begin{Vmatrix}{{T}_{\phi } - K}\end{Vmatrix} \geq \begin{Vmatrix}{T}_{\phi }\end{Vmatrix} \) .
Proof. Since \( \begin{Vmatrix}{T}_{{e}^{-{in\theta }}}\end{Vmatrix} = 1 \) for each natural number \( n \) ,\n\n\[ \begin{Vmatrix}{{T}_{\phi } - K}\end{Vmatrix} \geq \begin{Vmatrix}{{T}_{{e}^{-{in\theta }}}\left( {{T}_{\phi } - K}\right) }\end{Vmatrix} \]\n\n\[ = \begin{Vmatrix}{{T}_{{e}^{-{in\theta }}\phi } - {T}_{{e...
Yes
There is a rank-one operator \( K \) such that \( \begin{Vmatrix}{{S}^{ * } - K}\end{Vmatrix} = \begin{Vmatrix}{S}^{ * }\end{Vmatrix} \) , where \( {S}^{ * } \) is the backward unilateral shift.
Proof. Let \( K = {e}_{0} \otimes {e}_{1} \) . Clearly, \( \begin{Vmatrix}{{S}^{ * } - K}\end{Vmatrix} = 1 = \begin{Vmatrix}{S}^{ * }\end{Vmatrix} \).
Yes
Theorem 3.3.6. For \( \phi \) in \( {\mathbf{L}}^{\infty } \), the following sets are identical:\n\n(i) the closed convex hull of \( \sigma \left( {T}_{\phi }\right) \) ;\n\n(ii) the closed convex hull of \( \sigma \left( {M}_{\phi }\right) \) ;\n\n(iii) the closure of the numerical range of \( {T}_{\phi } \) ;\n\n(iv)...
Proof. The closure of the numerical range of \( {M}_{\phi } \) is the convex hull of its spectrum, since \( {M}_{\phi } \) is normal (Theorem 1.2.12). By the spectral inclusion theorem (Theorem 3.3.1), the closed convex hull of \( \sigma \left( {M}_{\phi }\right) \) is contained in the closed convex hull of \( \sigma \...
Yes
Corollary 3.3.7. For every \( \phi \) in \( {\mathbf{L}}^{\infty } \), ess \( \operatorname{ran}\phi \) is contained in \( \sigma \left( {T}_{\phi }\right) \) and \( \sigma \left( {T}_{\phi }\right) \) is contained in the closed convex hull of ess \( \operatorname{ran}\phi \) . In particular, if ess ran \( \phi \) is c...
Proof. This is an immediate consequence of the previous theorem together with the spectral inclusion theorem (Theorem 3.3.1).
No
Theorem 3.3.8. If \( \phi \in {\mathbf{H}}^{\infty } \), then \( \sigma \left( {T}_{\phi }\right) \) is the closure of \( \phi \left( \mathbb{D}\right) \) .
Proof. For the proof of this theorem, it is convenient to regard \( {T}_{\phi } \) as acting on \( {\mathbf{H}}^{2} \) rather than on \( {\widetilde{\mathbf{H}}}^{2} \) . To establish one inclusion, suppose that \( \lambda = \phi \left( {z}_{0}\right) \) for some \( {z}_{0} \in \mathbb{D} \) . Then\n\n\[ \left( {\left(...
Yes
Corollary 3.3.9. If \( {T}_{\phi } \) is a coanalytic Toeplitz operator, and if \( \bar{\phi } \) is the function in \( {\mathbf{H}}^{\infty } \) whose boundary function is the complex conjugate of \( \phi \) a.e., then \( \sigma \left( {T}_{\phi }\right) \) is the closure of the set of complex conjugates of \( \bar{\p...
Proof. Recall that the adjoint of a Toeplitz operator is the Toeplitz operator whose symbol is the complex conjugate of the given one (Theorem 3.2.8). Since the complex conjugate of \( \phi \) is in \( {\widetilde{\mathbf{H}}}^{2} \), the result follows from the previous theorem and the fact that the spectrum of the ad...
Yes
Theorem 3.3.10 (The Coburn Alternative). If \( \phi \) is a function in \( {\mathbf{L}}^{\infty } \) other than 0, then at least one of \( {T}_{\phi } \) and \( {T}_{\phi }^{ * } \) is injective.
Proof. Suppose that \( {T}_{\phi }f = 0 \) for some \( f \neq 0 \) . Suppose also that\n\n\[ {T}_{\phi }^{ * }g = P\left( {\bar{\phi }g}\right) = 0. \]\n\nIt must be shown that \( g = 0 \) .\n\nWe are given \( P\left( {\phi f}\right) = 0 \), where \( P \) is the projection of \( {\mathbf{L}}^{2} \) onto \( {\widetilde{...
Yes
Corollary 3.3.11. A Toeplitz operator, other than 0, has dense range if it is not injective.
Proof. If \( {T}_{\phi } \) is not injective, it follows from the Coburn alternative (Theorem 3.3.10) that \( {T}_{\phi }^{ * } \) is injective. If the range of \( {T}_{\phi } \) were not dense, there would exist a \( g \) different from 0 such that \( \left( {{T}_{\phi }f, g}\right) = 0 \) for all \( f \in {\widetilde...
Yes
Corollary 3.3.12. For \( \phi \) a nonconstant function in \( {\mathbf{L}}^{\infty } \) , \[ {\Pi }_{0}\left( {T}_{\phi }\right) \bigcap \overline{{\Pi }_{0}\left( {T}_{\phi }^{ * }\right) } = \varnothing \] where \( \overline{{\Pi }_{0}\left( {T}_{\phi }^{ * }\right) } \) denotes the set of complex conjugates of the e...
Proof. Suppose that \( \lambda \in {\Pi }_{0}\left( {T}_{\phi }\right) \) . Then there is a function \( f \) other than zero such that \( \left( {{T}_{\phi } - \lambda }\right) f = \left( {T}_{\phi - \lambda }\right) f = 0 \) . Suppose that \( \left( {{T}_{\phi }^{ * } - \bar{\lambda }}\right) g = 0 \) ; it must be sho...
Yes
Corollary 3.3.13. If \( \phi \) is a real-valued nonconstant function in \( {\mathbf{L}}^{\infty } \), then \( {\Pi }_{0}\left( {T}_{\phi }\right) = \varnothing \) .
Proof. In this case, \( {T}_{\phi } \) is self-adjoint, so its spectrum is real. Thus if there was a \( \lambda \) in \( {\Pi }_{0}\left( {T}_{\phi }\right) ,\lambda = \bar{\lambda } \) would be in \( {\Pi }_{0}\left( {T}_{\phi }^{ * }\right) \), contradicting the previous corollary.
Yes
Lemma 3.3.17. Let \( \\beta \) and \( \\gamma \) be continuous functions on \( {S}^{1} \) neither of which assumes the value 0 . Then\n\n\[ \n{\\operatorname{Ind}}_{0}\\left( {\\beta \\gamma }\\right) = {\\operatorname{Ind}}_{0}\\beta + {\\operatorname{Ind}}_{0}\\gamma \n\]
Proof. See [9, p. 81].
No
Theorem 4.1.4. The Hankel operators \( {H}_{\phi } \) and \( {H}_{\psi } \) are equal if and only if \( \phi - \psi \) is in \( {e}^{i\theta }{\widetilde{\mathbf{H}}}^{2} \) .
Proof. Since the matrix of a Hankel operator depends only on the Fourier coefficients in nonpositive positions (Definition 4.1.3), two \( {\mathbf{L}}^{\infty } \) functions induce the same Hankel operator if and only if their Fourier coefficients agree for nonpositive indices. This is equivalent to the difference betw...
Yes
Theorem 4.1.7. The operator \( A \) has a Hankel matrix with respect to the standard basis of \( {\widetilde{\mathbf{H}}}^{2} \) if and only if it satisfies the equation \( {U}^{ * }A = {AU} \), where \( U \) is the unilateral shift.
Proof. First note that\n\n\[ \left( {{U}^{ * }A{e}_{n},{e}_{m}}\right) = \left( {A{e}_{n}, U{e}_{m}}\right) = \left( {A{e}_{n},{e}_{m + 1}}\right) .\n\]\n\nAlso,\n\n\[ \left( {{AU}{e}_{n},{e}_{m}}\right) = \left( {A{e}_{n + 1},{e}_{m}}\right) .\n\]\n\nTherefore,\n\n\[ \left( {{U}^{ * }A{e}_{n},{e}_{m}}\right) = \left( ...
Yes
Corollary 4.1.8. If A has a Hankel matrix with respect to the standard basis of \( {\widetilde{\mathbf{H}}}^{2} \), and \( U \) is the unilateral shift, then \( {U}^{ * }{AU} \) has a Hankel matrix.
Proof. This is easily seen by noticing the effect on the matrix of \( A \) of multiplying on the left by \( {U}^{ * } \) and on the right by \( U \) . Alternatively,\n\n\[ \n{U}^{ * }\left( {{U}^{ * }{AU}}\right) = {U}^{ * }\left( {{U}^{ * }A}\right) U \n\]\n\n\[ \n= {U}^{ * }\left( {AU}\right) U\;\text{ (by Theorem 4....
Yes
Theorem 4.1.9 (Douglas’s Theorem). Let \( \mathcal{H},\mathcal{K} \), and \( \mathcal{L} \) be Hilbert spaces and suppose that \( E : \mathcal{H} \rightarrow \mathcal{K} \) and \( F : \mathcal{H} \rightarrow \mathcal{L} \) are bounded operators. If \( {E}^{ * }E \leq {F}^{ * }F \), then there exists an operator \( R : ...
Proof. First of all, observe that the hypothesis \( {E}^{ * }E \leq {F}^{ * }F \) is equivalent to \( \parallel {Ex}\parallel \leq \parallel {Fx}\parallel \) for all \( x \in \mathcal{H} \), since \( \left( {{E}^{ * }{Ex}, x}\right) = \parallel {Ex}{\parallel }^{2} \) and \( \left( {{F}^{ * }{Fx}, x}\right) = \) \( \pa...
Yes
Lemma 4.1.10. Let \( \mathcal{H} \) and \( \mathcal{K} \) be Hilbert spaces and let \( B : \mathcal{H} \rightarrow \mathcal{K} \) be a bounded operator with \( \parallel B\parallel \leq 1 \) . Then \( {\left( I - {B}^{ * }B\right) }^{1/2}{B}^{ * } = {B}^{ * }{\left( I - B{B}^{ * }\right) }^{1/2} \) .
Proof. First of all, since \( \parallel B\parallel = \begin{Vmatrix}{B}^{ * }\end{Vmatrix} \leq 1 \), it follows that \( I - {B}^{ * }B \geq 0 \) and \( I - B{B}^{ * } \geq 0 \) . Thus \( {\left( I - {B}^{ * }B\right) }^{1/2} \) and \( {\left( I - B{B}^{ * }\right) }^{1/2} \) exist, since every positive operator has a ...
Yes
Theorem 4.1.11 (The Julia-Halmos Theorem). Let \( \mathcal{H} \) and \( \mathcal{K} \) be Hilbert spaces and let \( A : \mathcal{H} \rightarrow \mathcal{K} \) be a bounded operator with \( \parallel A\parallel \leq 1 \) . If \( U \) is the operator mapping \( \mathcal{K} \oplus \mathcal{H} \) into \( \mathcal{H} \oplus...
Proof. Applying Lemma 4.1.10 (with \( B = {A}^{ * } \) and with \( B = A \) ) gives\n\n\[ {\left( I - A{A}^{ * }\right) }^{1/2}A = A{\left( I - {A}^{ * }A\right) }^{1/2}\;\text{ and }\;{\left( I - {A}^{ * }A\right) }^{1/2}{A}^{ * } = {A}^{ * }{\left( I - A{A}^{ * }\right) }^{1/2}. \]\n\nMultiplying matrices shows that ...
Yes
Theorem 4.2.2. For a fixed element \( w \) of \( \mathbb{D} \), let \( {k}_{w} \) be defined by\n\n\[ \n{k}_{w}\left( z\right) = \frac{1}{1 - \bar{w}z} \n\]\n\n(see Definition 1.1.7). Then the rank-one operator \( {k}_{\bar{w}} \otimes {k}_{w} \) (see Notation 1.2.27) is the Hankel operator \( {H}_{{\breve{k}}_{\bar{w}...
Proof. To see that \( {k}_{\bar{w}} \otimes {k}_{w} \) has the stated matrix representation, fix any nonnegative integer \( n \) and compute\n\n\[ \n\left( {{k}_{\bar{w}} \otimes {k}_{w}}\right) {e}_{n} = \left( {{e}_{n},{k}_{w}}\right) {k}_{\bar{w}} = {w}^{n}\left( {{e}_{0} + w{e}_{1} + {w}^{2}{e}_{2} + {w}^{3}{e}_{3}...
Yes
Theorem 4.2.3 (Kronecker’s Theorem). Let\n\n\\[ \nH = \\left( \\begin{matrix} {a}_{0} & {a}_{1} & {a}_{2} & {a}_{3} & & \\cdots \\\\ {a}_{1} & {a}_{2} & {a}_{3} & & \\cdots & \\\\ {a}_{2} & {a}_{3} & & \\cdots & & \\\\ {a}_{3} & & \\cdots & & & \\\\ & \\cdots & & & & \\\\ \\vdots & & & & & \\\\ & & & & & \\\\ & & & & &...
Proof. Suppose that the columns of \\( H \\) are linearly dependent. Then there is a natural number \\( s \\) and complex numbers \\( {c}_{0},{c}_{1},{c}_{2},\\ldots ,{c}_{s} \\), not all zero, such that\n\n\\[ \n{c}_{0}\\left( \\begin{matrix} {a}_{0} \\\\ {a}_{1} \\\\ {a}_{2} \\\\ {a}_{3} \\\\ \\vdots \\end{matrix}\\r...
Yes
Corollary 4.2.4 (Kronecker’s Theorem). For \( \phi \in {\mathbf{L}}^{\infty } \), the Hankel operator \( {H}_{\phi } \) has finite rank if and only if the function\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{\phi }_{-k}}{{z}^{k}} \]\n\nis a rational function all of whose poles are in \( \ma...
Proof. Suppose \( {H}_{\phi } \) has finite rank. Then \( f \) is a rational function, by the version of Kronecker's theorem we have already established (Theorem 4.2.3). It must be shown that \( f \) has no poles outside \( \mathbb{D} \) .\n\nIf \( f \) had a pole outside \( \mathbb{D} \), say \( \lambda \), consider t...
Yes
Corollary 4.2.5 (Kronecker's Theorem). Let \( \mathcal{R} \) be the set of rational functions with poles inside \( \mathbb{D} \) . Then \( H \) is a bounded Hankel operator of finite rank if and only if \( H = {H}_{\psi } \) for some \( \psi \in {e}^{i\theta }{\widetilde{\mathbf{H}}}^{\infty } + \mathcal{R} \) .
Proof. The previous form of Kronecker's theorem (Corollary 4.2.4) immediately implies this form since Hankel operators are the same if and only if their symbols have identical coanalytic parts (Theorem 4.1.4).
No
Corollary 4.2.7. If the matrix of a Hankel operator with respect to the standard basis for \( {\widetilde{\mathbf{H}}}^{2} \) has only a finite number of entries different from 0, then there exists a function \( \psi \), continuous on \( {S}^{1} \), that is a symbol of the Hankel operator and satisfies \( \begin{Vmatri...
Proof. Since any finite Blaschke product is continuous on \( {S}^{1} \), the result follows from the previous theorem.
No
Corollary 4.3.3 (Hartman’s Theorem). The Hankel operator \( {H}_{\phi } \) is compact if and only if \( \phi \) is in \( {\widetilde{\mathbf{H}}}^{\infty } + \mathbf{C} \) .
Proof. If \( {H}_{\phi } \) is compact, the previous version of Hartman’s theorem (Theorem 4.3.1) implies that there is a continuous function \( \psi \) such that \( {H}_{\phi } = {H}_{\psi } \) . Then \( \phi \) and \( \psi \) have the same coanalytic parts, so \( \phi - \psi \) is a function in \( {\widetilde{\mathbf...
Yes
Theorem 4.4.2. If \( H \) is a Hankel operator, then \( {H}^{ * }{f}^{ * } = {\left( Hf\right) }^{ * } \) for every \( f \in {\widetilde{\mathbf{H}}}^{2} \) . In particular, \( \begin{Vmatrix}{{H}^{ * }{f}^{ * }}\end{Vmatrix} = \parallel {Hf}\parallel \) for every \( f \) in \( {\widetilde{\mathbf{H}}}^{2} \) .
Proof. Recall that the matrix of \( {A}^{ * } \) with respect to a given orthonormal basis can be obtained from that of \( A \) by taking the \
No
For every \( \phi \) in \( {\mathbf{L}}^{\infty },{H}_{\phi }^{ * } = {H}_{{\phi }^{ * }} \) . Moreover, if \( {H}_{\phi } \) is selfadjoint, then there exists a \( \psi \) in \( {\mathbf{L}}^{\infty } \) such that \( \psi = {\psi }^{ * } \) a.e. and \( {H}_{\phi } = {H}_{\psi } \) .
The fact that \( {H}_{\phi }^{ * } = {H}_{{\phi }^{ * }} \) follows from the matrix representation of \( {H}_{\phi } \) as indicated above. Alternatively, it is also a consequence of the following computation. For any \( f \) and \( g \) in \( {\mathbf{H}}^{2} \) ,\n\n\[ \left( {{H}_{\phi }^{ * }f, g}\right) = \left( {...
Yes
Lemma 4.4.4. If \( {H}_{\phi } \) and \( {H}_{\psi } \) are Hankel operators and \( U \) is the unilateral shift, then\n\n\[ \n{H}_{\phi }{H}_{\psi } - {U}^{ * }{H}_{\phi }{H}_{\psi }U = \left( {P\breve{\phi }}\right) \otimes \left( {P\bar{\psi }}\right) ,\n\]\n\nwhere \( P \) is the projection of \( {\mathbf{L}}^{2} \...
Proof. Note that\n\n\[ \n{H}_{\phi }{H}_{\psi } - {U}^{ * }{H}_{\phi }{H}_{\psi }U = {H}_{\phi }{H}_{\psi } - {H}_{\phi }U{U}^{ * }{H}_{\psi }\;\text{ (by Theorem 4.1.7) }\n\]\n\n\[ \n= {H}_{\phi }\left( {I - U{U}^{ * }}\right) {H}_{\psi }\n\]\n\nRecall that \( I - U{U}^{ * } \) is the projection of \( {\widetilde{\mat...
Yes
Theorem 4.4.5. The product of two nonzero Hankel operators is a Hankel operator if and only if both of the operators are constant multiples of the same Hankel operator of rank 1.
Proof. We have seen that every Hankel operator of rank 1 has the form \( {k}_{\bar{w}} \otimes {k}_{w} \) for some \( w \) in \( \mathbb{D} \), where \( {k}_{w} \) is the kernel function given by \( {k}_{w}\left( z\right) = \frac{1}{1 - \bar{w}z} \) (Theorem 4.2.2). Thus, to prove the first implication of the theorem, ...
Yes
Corollary 4.4.6. The product of two Hankel operators is a Toeplitz operator only if at least one of the Hankel operators is 0 .
Proof. By Lemma 4.4.4,\n\n\[ \n{H}_{\phi }{H}_{\psi } - {U}^{ * }{H}_{\phi }{H}_{\psi }U = \left( {P\breve{\phi }}\right) \otimes \left( {P\bar{\psi }}\right) .\n\]\n\nIf \( {H}_{\phi }{H}_{\psi } \) is a Toeplitz operator, then \( {U}^{ * }{H}_{\phi }{H}_{\psi }U = {H}_{\phi }{H}_{\psi } \) (by Corollary 3.2.7), so\n\...
Yes
Corollary 4.4.7. The product of two Hankel operators is 0 if and only if one of them is 0.
Proof. If the product of two Hankel operators is the Toeplitz operator 0, the previous corollary implies that at least one of the Hankel operators is zero.
No
Theorem 4.4.8. Let \( \phi \) and \( \psi \) be in \( {\mathbf{L}}^{\infty } \) and suppose that \( {H}_{\psi } \neq 0 \) . If \( {H}_{\phi } \) and \( {H}_{\psi } \) commute, then there exists a complex number \( c \) such that \( {H}_{\phi } = c{H}_{\psi } \) .
Proof. When \( {H}_{\phi } \neq 0 \), by Lemma 4.4.4,\n\n\[ \n{H}_{\phi }{H}_{\psi } - {U}^{ * }{H}_{\phi }{H}_{\psi }U = \left( {P\breve{\phi }}\right) \otimes \left( {P\bar{\psi }}\right) , \n\]\n\nand\n\n\[ \n{H}_{\psi }{H}_{\phi } - {U}^{ * }{H}_{\psi }{H}_{\phi }U = \left( {P\breve{\psi }}\right) \otimes \left( {P...
Yes
Corollary 4.4.9. Every normal Hankel operator is a multiple of a self-adjoint Hankel operator.
Proof. Let \( H \) be a normal Hankel operator; i.e., \( H{H}^{ * } = {H}^{ * }H \) . If \( H = 0 \), the result is trivial. In the other case, by the previous theorem, there is a constant \( c \) such that \( H = c{H}^{ * } \) . Since \( \parallel H\parallel = \begin{Vmatrix}{H}^{ * }\end{Vmatrix} \), we have \( \left...
Yes
Theorem 4.4.11. Every hyponormal Hankel operator is normal.
Proof. By Theorem 4.4.2, it follows that \( \begin{Vmatrix}{{H}^{ * }{f}^{ * }}\end{Vmatrix} = \parallel {Hf}\parallel \) for every \( f \in {\widetilde{\mathbf{H}}}^{2} \) , where \( {f}^{ * } \) is the vector whose coefficients are the conjugates of those of \( f \) (Notation 4.4.1). Applying this to \( {f}^{ * } \) ...
Yes
Theorem 4.5.1. Let \( \phi \) and \( \psi \) be in \( {\mathbf{L}}^{\infty } \) . Then\n\n\[ \n{H}_{{e}^{i\theta }\breve{\phi }}{H}_{{e}^{i\theta }\psi } = {T}_{\phi \psi } - {T}_{\phi }{T}_{\psi }\n\]
Proof. The flip operator, \( J \), and the projection onto \( {\widetilde{\mathbf{H}}}^{2}, P \), satisfy the following equation:\n\n\[ \n{JPJ} = {M}_{{e}^{i\theta }}\left( {I - P}\right) {M}_{{e}^{-{i\theta }}}.\n\]\n\n(This can easily be verified by applying each side to the basis vectors \( \left\{ {e}^{in\theta }\r...
Yes
Corollary 4.5.2. If \( \phi \) and \( \psi \) are in \( {\mathbf{L}}^{\infty } \), then\n\n\[ \n{H}_{\phi }{H}_{\psi } = {T}_{\breve{\phi }\psi } - {T}_{{e}^{i\theta }\breve{\phi }}{T}_{{e}^{-{i\theta }}\psi }.\n\]
Proof. By the previous theorem,\n\n\[ \n{H}_{{e}^{i\theta }\breve{\alpha }}{H}_{{e}^{i\theta }\beta } = {T}_{\alpha \beta } - {T}_{\alpha }{T}_{\beta }\n\]\n\nfor \( \alpha \) and \( \beta \) in \( {\mathbf{L}}^{\infty } \) . Let \( \alpha = {e}^{i\theta }\breve{\phi } \) and \( \beta = {e}^{-{i\theta }}\psi \) . Makin...
Yes
Corollary 4.5.3. If the product of two Hankel operators is Toeplitz, then at least one of the Hankel operators is 0.
Proof. If \( {H}_{\phi }{H}_{\psi } \) is a Toeplitz operator, then since the sum of two Toeplitz operators is Toeplitz, it follows from the previous corollary that \( {T}_{{e}^{i\theta }\breve{\phi }}{T}_{{e}^{-{i\theta }}\psi } \) is a Toeplitz operator. Thus either \( {e}^{-{i\theta }}\psi \) is analytic or \( {e}^{...
Yes
Theorem 4.5.4. Let \( \phi \) and \( \psi \) be in \( {\mathbf{L}}^{\infty } \) . Then\n\n\[ \n{T}_{\breve{\phi }}{H}_{{e}^{i\theta }\psi } + {H}_{{e}^{i\theta }\phi }{T}_{\psi } = {H}_{{e}^{i\theta }{\phi \psi }}.\n\]
Proof. This follows from a computation similar to that in the proof of Theorem 4.5.1. Using \( J{M}_{\phi }J = {M}_{\breve{\phi }} \) and \( {JPJ} = {M}_{{e}^{i\theta }}\left( {I - P}\right) {M}_{{e}^{-{i\theta }}} \), we get\n\n\[ \n{T}_{\breve{\phi }}{H}_{{e}^{i\theta }\psi } = \left( {P{M}_{\breve{\phi }}}\right) \l...
Yes
Corollary 4.5.5. (i) If \( \psi \) is in \( {\widetilde{\mathbf{H}}}^{\infty } \), then \( {H}_{\phi }{T}_{\psi } = {H}_{\phi \psi } \). (ii) If \( \psi \) is in \( {\widetilde{\mathbf{H}}}^{\infty } \), then \( {T}_{\breve{\psi }}{H}_{\phi } = {H}_{\psi \phi } \).
Proof. Recall from the previous theorem that, for \( \alpha \) and \( \beta \) in \( {\mathbf{L}}^{\infty } \), \[ {T}_{\breve{\alpha }}{H}_{{e}^{i\theta }\beta } + {H}_{{e}^{i\theta }\alpha }{T}_{\beta } = {H}_{{e}^{i\theta }{\alpha \beta }}. \] Taking \( \alpha = {e}^{-{i\theta }}\phi \) and \( \beta = \psi \) gives ...
Yes
Theorem 5.1.2. If \( {C}_{\phi } \) and \( {C}_{\psi } \) are composition operators then \( {C}_{\phi }{C}_{\psi } = {C}_{\psi \circ \phi } \) .
Proof. Note that\n\n\[ \left( {{C}_{\phi }{C}_{\psi }f}\right) \left( z\right) = \left( {{C}_{\phi }\left( {f \circ \psi }\right) }\right) \left( z\right) = \left( {f \circ \psi \circ \phi }\right) \left( z\right) = \left( {{C}_{\psi \circ \phi }f}\right) \left( z\right) ,\]\n\nand thus\n\n\[ {C}_{\phi }{C}_{\psi } = {...
Yes
Let \( \phi \left( z\right) = {z}^{2} \). If \( f \) is in \( {\mathbf{H}}^{2} \), then \( {C}_{\phi }f \) is in \( {\mathbf{H}}^{2} \) and, in fact, \( \begin{Vmatrix}{{C}_{\phi }f}\end{Vmatrix} = \parallel f\parallel \). Therefore \( {C}_{\phi } \) is an isometry mapping \( {\mathbf{H}}^{2} \) into itself.
Proof. If \( f \) has power series \( f\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}^{n} \), then \( {C}_{\phi }f \) has power series \( \left( {{C}_{\phi }f}\right) \left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}^{2n} \). Thus \( {C}_{\phi }f \in {\mathbf{H}}^{2} \) and al...
Yes
Lemma 5.1.4. If \( f \in {\mathbf{H}}^{2} \), then, for \( r{e}^{it} \in \mathbb{D} \), we have\n\n\[ \n{\left| f\left( r{e}^{it}\right) \right| }^{2} \leq \frac{1}{2\pi }{\int }_{0}^{2\pi }{P}_{r}\left( {\theta - t}\right) {\left| f\left( {e}^{i\theta }\right) \right| }^{2}{d\theta }. \n\]
Proof. Recall that, by the Poisson integral formula (Theorem 1.1.21), we have\n\n\[ \nf\left( {r{e}^{it}}\right) = \frac{1}{2\pi }{\int }_{0}^{2\pi }{P}_{r}\left( {\theta - t}\right) f\left( {e}^{i\theta }\right) {d\theta }. \n\]\n\nIf we define the measure \( {d\mu } \) by \( {d\mu }\left( \theta \right) = \frac{1}{2\...
Yes
Corollary 5.1.6. If \( {C}_{\phi } \) is a composition operator such that \( \phi \left( 0\right) = 0 \) then \( \begin{Vmatrix}{C}_{\phi }\end{Vmatrix} = 1 \) .
Proof. By the previous theorem and \( \phi \left( 0\right) = 0 \), we have \( \begin{Vmatrix}{C}_{\phi }\end{Vmatrix} \leq 1 \) . By the above observation, \( \begin{Vmatrix}{C}_{\phi }\end{Vmatrix} \geq 1 \) .
Yes
Corollary 5.1.8 (Littlewood’s Subordination Theorem). If \( f \) in \( {\mathbf{H}}^{2} \) is subordinate to \( g \) in \( {\mathbf{H}}^{2} \), then \( \parallel f\parallel \leq \parallel g\parallel \) .
Proof. Apply the previous corollary to \( f = {C}_{\phi }g \) .
No
Lemma 5.1.9. If \( {C}_{\phi } \) is a composition operator and \( {k}_{\lambda } \) is a reproducing kernel function, then \( {C}_{\phi }^{ * }{k}_{\lambda } = {k}_{\phi \left( \lambda \right) } \) .
Proof. For each \( f \) in \( {\mathbf{H}}^{2} \) , \n\n\[ \n\left( {f,{C}_{\phi }^{ * }{k}_{\lambda }}\right) = \left( {{C}_{\phi }f,{k}_{\lambda }}\right) = \left( {f \circ \phi ,{k}_{\lambda }}\right) = f\left( {\phi \left( \lambda \right) }\right) . \n\] \n\nBut also \n\n\[ \n\left( {f,{k}_{\phi \left( \lambda \rig...
Yes
Theorem 5.1.10. For every composition operator \( {C}_{\phi } \) , \[ \frac{1}{\sqrt{1 - {\left| \phi \left( 0\right) \right| }^{2}}} \leq \begin{Vmatrix}{C}_{\phi }\end{Vmatrix} \leq \frac{2}{\sqrt{1 - {\left| \phi \left( 0\right) \right| }^{2}}}. \]
Proof. Using the previous lemma with \( \lambda = 0 \) yields \[ {C}_{\phi }^{ * }{k}_{0} = {k}_{\phi \left( 0\right) } \] Recall (Theorem 1.1.8) that \[ {\begin{Vmatrix}{k}_{\lambda }\end{Vmatrix}}^{2} = \frac{1}{1 - {\left| \lambda \right| }^{2}} \] and therefore \( \begin{Vmatrix}{k}_{0}\end{Vmatrix} = 1 \) and \( \...
Yes
Corollary 5.1.11. The norm of the composition operator \( {C}_{\phi } \) is 1 if and only if \( \phi \left( 0\right) = 0 \) .
Proof. As indicated, we have already established (Corollary 5.1.6) that \( \begin{Vmatrix}{C}_{\phi }\end{Vmatrix} = \) 1 if \( \phi \left( 0\right) = 0 \) . Conversely, if \( \begin{Vmatrix}{C}_{\phi }\end{Vmatrix} = 1 \), then the inequality\n\n\[ \n\frac{1}{\sqrt{1 - {\left| \phi \left( 0\right) \right| }^{2}}} \leq...
Yes
Theorem 5.1.12. An operator \( A \) on \( {\mathbf{H}}^{2} \) is a composition operator if and only if \( {A}^{ * } \) maps the set of reproducing kernels into itself.
Proof. We showed above that \( {A}^{ * }{k}_{\lambda } = {k}_{\phi \left( \lambda \right) } \) when \( A = {C}_{\phi } \) . Conversely, suppose that for each \( \lambda \in \mathbb{D},{A}^{ * }{k}_{\lambda } = {k}_{{\lambda }^{\prime }} \) for some \( {\lambda }^{\prime } \in \mathbb{D} \) . Define \( \phi : \mathbb{D}...
Yes
Corollary 5.1.14. The operator \( A \) on \( {\mathbf{H}}^{2} \) is a composition operator if and only if it is multiplicative in the sense that \( \left( {Af}\right) \left( {Ag}\right) = A\left( {fg}\right) \) whenever \( f, g \) , and \( {fg} \) are all in \( {\mathbf{H}}^{2} \) .
Proof. It is clear that composition operators have the stated multiplicative property.\n\nConversely, if \( A \) has the multiplicative property then, in particular, \( A{e}_{n} = \) \( {\left( A{e}_{1}\right) }^{n} \) for all \( n \), so the fact that \( A \) is a composition operator follows from Theorem 5.1.13.
Yes
Theorem 5.1.15. The composition operator \( {C}_{\phi } \) is normal if and only if there exists \( \lambda \in \mathbb{C} \) such that \( \phi \left( z\right) = {\lambda z} \) and \( \left| \lambda \right| \leq 1 \) .
Proof. First note that \( \phi \left( z\right) = {\lambda z} \) implies that \( {C}_{\phi }{e}_{n} = {\lambda }^{n}{e}_{n} \) for all positive integers \( n \) . Hence \( {C}_{\phi } \) is a diagonal operator with respect to the canonical basis of \( {\mathbf{H}}^{2} \), and it is obvious that every diagonal operator i...
Yes
Theorem 5.1.16. If there exists a positive number \( s < 1 \) so that \( \left| {\phi \left( z\right) }\right| < s \) for every \( z \in \mathbb{D} \), then \( {C}_{\phi } \) is compact.
Proof. We show that \( {C}_{\phi } \) is compact by exhibiting a sequence of operators of finite rank that converge in norm to \( {C}_{\phi } \) . Observe that, since \( \left| {\phi \left( z\right) }\right| < s \) for all \( z \in \mathbb{D} \), we have, for each natural number \( k \), \[ \begin{Vmatrix}{\phi }^{k}\e...
Yes
Theorem 5.1.17. If \( {C}_{\phi } \) is compact, then \( \left| {\widetilde{\phi }\left( {e}^{i\theta }\right) }\right| < 1 \) a.e.
Proof. The sequence \( \left\{ {e}_{n}\right\} \) converges weakly to 0 . So, \( \left\{ {{C}_{\phi }{e}_{n}}\right\} \) converges to 0 in norm if \( {C}_{\phi } \) is compact (e.g.,[27, p. 95] or [12, p. 173]).\n\nIf \( \left| {\widetilde{\phi }\left( {e}^{i\theta }\right) }\right| \) was not less than 1 a.e., there w...
Yes
Theorem 5.2.1. The composition operator \( {C}_{\phi } \) is invertible if and only if \( \phi \) is a conformal mapping of \( \mathbb{D} \) onto itself. In this case, \( {C}_{\phi }^{-1} = {C}_{{\phi }^{-1}} \) .
Proof. If \( \phi \) is a conformal map, let \( {\phi }^{-1} \) be the inverse conformal map. Then \( {C}_{{\phi }^{-1}}{C}_{\phi } = {C}_{\phi }{C}_{{\phi }^{-1}} = I \) by Theorem 5.1.2. Hence \( {C}_{\phi }^{-1} = {C}_{{\phi }^{-1}} \) . To establish the converse, suppose that \( {C}_{\phi } \) is an invertible comp...
Yes
For a fixed \( \lambda \in \mathbb{D} \), define the function \( {\phi }_{\lambda } \) by\n\n\[ \n{\phi }_{\lambda }\left( z\right) = \frac{\lambda - z}{1 - \bar{\lambda }z}\n\]\n\nThen \( {\left( {C}_{{\phi }_{\lambda }}\right) }^{2} = I \) (i.e., \( {C}_{{\phi }_{\lambda }}^{-1} = {C}_{{\phi }_{\lambda }} \) ).
Proof. As indicated above, each such \( {\phi }_{\lambda } \) is a conformal mapping of \( \mathbb{D} \) into itself. An easy computation shows that \( {\phi }_{\lambda }\left( {{\phi }_{\lambda }\left( z\right) }\right) = z \) for all \( z \in \mathbb{D} \), from which the result follows.
No
Theorem 5.2.3. If the function \( \phi \) has a fixed point in \( \mathbb{D} \), then the operator \( {C}_{\phi } \) is similar to a composition operator \( {C}_{\psi } \) with the property that \( \psi \left( 0\right) = 0 \) .
Proof. Let \( \phi \left( \lambda \right) = \lambda \) for some \( \lambda \in \mathbb{D} \) . Let\n\n\[{\phi }_{\lambda }\left( z\right) = \frac{\lambda - z}{1 - \bar{\lambda }z}\]\n\nThen \( {C}_{{\phi }_{\lambda }}^{-1} = {C}_{{\phi }_{\lambda }} \) by Example 5.2.2, so\n\n\[{C}_{{\phi }_{\lambda }}^{-1}{C}_{\phi }{...
Yes
Theorem 5.2.5. If \( \phi \) has a fixed point in \( \mathbb{D} \), then the spectral radius of \( {C}_{\phi } \) is 1 .
Proof. By Theorem 5.2.3, \( {C}_{\phi } \) is similar to a composition operator whose defining function fixes the point 0 . Since similar operators have the same spectra, and therefore equal spectral radii, we may and do assume that \( \phi \left( 0\right) = 0 \) . By the spectral radius formula (Theorem 1.2.4), \[ r\l...
Yes
Corollary 5.3.2. If \( {C}_{\phi } \) is a composition operator and \( \phi \left( a\right) = a \) for some \( a \in \mathbb{D} \), then\n\n\[ \sigma \left( {C}_{\phi }\right) \supset \{ 1\} \cup \mathop{\bigcup }\limits_{{k = 1}}^{\infty }\left\{ {\left( {\phi }^{\prime }\left( a\right) \right) }^{k}\right\} . \]
Proof. This follows immediately from the previous theorem and Theorem 1.2.4.
No
Theorem 5.3.3. If \( \phi \) is a nonconstant analytic function mapping the disk into itself and satisfying \( \phi \left( a\right) = a \) for some \( a \in \mathbb{D} \), and if there exists a function \( f \) analytic on \( \mathbb{D} \) that is not identically zero and satisfies the Schröder equation\n\n\[ f\left( {...
Proof. The equations \( \phi \left( a\right) = a \) and \( f\left( {\phi \left( z\right) }\right) = {\lambda f}\left( z\right) \) yield \( f\left( a\right) = {\lambda f}\left( a\right) \) . If \( f\left( a\right) \neq 0 \), then clearly \( \lambda = 1 \) and the theorem is established in that case.\n\nSuppose \( f\left...
Yes