A proof of conjectures of Esterle and Ransford on negative powers of contractions
January 01, 1970
Abstract
Building on work of Ransford, we prove that whenever \(E\) is a closed subset of the unit circle of Lebesgue measure zero, there exists a positive sequence \(u_n\to\infty\) such that if \(T\) is a contraction on a Hilbert space with \(\sigma(T)\subset E\) and \(\|T^{-n}\|=O(u_n)\), then \(T\) is unitary. This confirms conjectures of Esterle and Ransford. Our main new idea is a spikes-in-collars principle for positive subharmonic functions.
1 Introduction↩︎
Let \(T\) be a bounded linear operator on a complex Hilbert space, and write \(\sigma(T)\) for its spectrum. We say that \(T\) is a contraction if \(\|T\|\le1\). We establish the following conjecture of Esterle [1].
Theorem 1 (Esterle’s conjecture). Let \(E\subset\mathbb{T}\) be closed and have Lebesgue measure zero. Then there exists a positive sequence \(u_n\to\infty\) such that if \(T\) is a contraction on a Hilbert space, \(\sigma(T)\subset E\), and \(\|T^{-n}\|=O(u_n)\), then \(T\) is unitary.
The problem belongs to a line of work connecting negative powers of contractions, spectral synthesis phenomena, and uniqueness properties of closed subsets of the circle. Ransford proved this under the additional finite-rank assumption \(\operatorname{rank}(I-T^*T)<\infty\) [1], which entered through a determinant reduction. Moreover, 1 is known to be sharp, in the sense that one cannot replace the measure zero and Hilbert space assumptions [1]. Weaker versions of this theorem and related results were known for various sets \(E\) [2]–[5]. See [1] for more information on the history of this problem.
We remove the finite-rank hypothesis of Ransford with new potential-theoretic input which may be of independent interest. In finite dimensions, one can reduce the problem to a scalar singular-inner function whose singular measure is supported on \(E\). Instead of trying to replace the determinant in infinite dimensions, we use the subharmonic function \[w_\Theta(z)=\log\|\Theta(z)^{-1}\|.\]
Let \(E\subset\mathbb{T}\) be closed and have Lebesgue measure zero. We say that a subharmonic function \(w\) on \(\mathbb{D}\) is carried by \(E\) if \(w\) is locally bounded in a neighbourhood of \(\overline{\mathbb{D}}\setminus E\) and has continuous boundary value zero on \(\mathbb{T}\setminus E\). We show that such a function must have sufficiently large spikes somewhere in shrinking collars over that set.
Theorem 2. Let \(E\subset\mathbb{T}\) be closed and have Lebesgue measure zero. Then there exists a continuous nonincreasing function \(L:(0,1]\to[1,\infty)\) such that \(L(t)\to\infty\) and \(tL(t)\to 0\) as \(t\to 0\), with the following property.
Suppose \(w\) is a nonzero, nonnegative subharmonic function carried by \(E\). Then for any \(A>0\) and \(\eta>0\), there exists a point \(z=re^{i\theta}\in\mathbb{D}\) such that \[0<1-r<\eta, \qquad \operatorname{dist}(e^{i\theta},E)<\eta,\] and \[w(z)\ge A L(1-r).\]
Remark↩︎
The author had the idea for this proof after a presentation of Thomas Ransford on [1] at the Conference on Classical Analysis in Memory of Paul Koosis. Ransford independently came up with another proof [6] following this event, based on a method suggested by Fedja Nazarov.
2 Proof of 2↩︎
For \(\eta>0\), put \[E_\eta=\{\zeta\in\mathbb{T}:\operatorname{dist}(\zeta,E)\le \eta\},\] and define the collar over \(E_\eta\) by \[K_\eta=\{re^{i\theta}\in\mathbb{D}:1-\eta\leq r<1,\;e^{i\theta}\in E_\eta\}.\] Let \(\Omega_\eta\) be the connected component of \(\mathbb{D}\setminus K_\eta\) containing the origin, and put \[\Sigma_\eta=\partial\Omega_\eta\cap\mathbb{D}.\] Thus \(\Sigma_\eta\) is the inner boundary of the removed collar. We write \(\omega_a^{\Omega_\eta}\) for harmonic measure in \(\Omega_\eta\) from \(a\in\Omega_\eta\).
2.1 Harmonic measure estimates↩︎
For \(0<R<1\), define \[p_R(\eta)= \sup_{|a|\le R}\omega_a^{\Omega_\eta}(\Sigma_\eta), \qquad \eta<1-R,\] and for fixed \(\eta>0\) and \(0<R<1-\eta\), define \[q_{R,\eta}(s)= \sup_{|a|\le R} \omega_a^{\Omega_\eta}\bigl(\Sigma_\eta\cap\{1-|z|\le s\}\bigr).\]
Proof. Write \(u_\eta(a)=\omega_a^{\Omega_\eta}(\Sigma_{\eta})\). If \(0<\eta'<\eta\), then \(K_{\eta'}\subset K_\eta\), and hence \(\Omega_\eta\subset\Omega_{\eta'}\). Thus \(u_{\eta'}\) is harmonic on \(\Omega_{\eta}\). On \(\Sigma_\eta\), we have \(0\le u_{\eta'}\le 1=u_\eta,\) while on the outer boundary \(\mathbb{T}\setminus E_\eta\), both functions have boundary value \(0\). By the maximum principle, \(u_{\eta'}\le u_\eta\) in \(\Omega_\eta.\) In particular, \(u_\eta(0)\) decreases as \(\eta\to 0\).
By Harnack’s principle, \(u_\eta\) converges locally uniformly as \(\eta\to 0\) to a bounded harmonic function \(u\) on \(\mathbb{D}\). Let \(J\) be a closed arc contained in \(\mathbb{T}\setminus E\). For all sufficiently small \(\eta\), the collar \(K_\eta\) is disjoint from a neighbourhood of \(J\), and \(u_\eta\) has boundary value \(0\) on \(J\). Since \(0\le u\le u_\eta\), it follows that \(u\) has boundary value \(0\) on \(J\). As \(J\subset\mathbb{T}\setminus E\) was arbitrary and \(|E|=0\), we deduce that \(u\equiv0\) and therefore that \(u_\eta(0)\to 0\). Harnack’s inequality gives the desired uniform convergence.
For the second assertion, for fixed \(\eta\), the set \(E_\eta\) is a finite union of arcs. Hence \(\Omega_\eta\) is a finitely connected domain with piecewise smooth boundary, and the sets \(\Sigma_\eta\cap\{1-|z|\le s\}\) decrease, as \(s\to 0\), to a finite set of boundary points. Harmonic measure has no atoms at boundary points of such a domain, so \[\omega_0^{\Omega_\eta}(\Sigma_\eta\cap\{1-|z|\le s\})\xrightarrow[]{s\to 0} 0.\] Another application of Harnack’s inequality on compact subsets of \(\Omega_\eta\) gives the desired uniformity. ◻
Let \(0<R_1<R_2<\cdots\uparrow1\). We now choose a slowly increasing weight.
Proof. We first construct a step function with the desired property, and then smooth it slightly. Choose \(\eta_j\to 0\) recursively. At stage \(j\), impose the finitely many conditions \[\begin{align} j\,p_{R_m}(\eta_j)&\le 2^{-j-m}, && m\le j, \tag{1}\\ q_{R_m,\eta_i}(\eta_j)&\le 2^{-j-i-m}, &&i<j,\;m\le i, \tag{2} \end{align}\] and also arrange \(\eta_j <1-R_j\) and \(j\eta_j\to0\). This is possible by [lem:pR].
Define a step function \(L_0:(0,1]\to[1,\infty)\) by \(L_0(t)=j\) for \(\eta_{j+1}<t\le\eta_j\),and extend it as a bounded positive function on \([\eta_1,1]\). Then \(L_0(t)\to\infty\) and \(tL_0(t)\to0\).
Fix \(m\), \(j\ge m\), and \(|a|\le R_m\). Let \(\nu_{j,a}\) be the image of \(\omega_a^{\Omega_{\eta_j}}|_{\Sigma_{\eta_j}}\) under the map \(z\mapsto 1-|z|\). Thus \[\nu_{j,a}((0,s]) = \omega_a^{\Omega_{\eta_j}} \bigl(\Sigma_{\eta_j}\cap\{1-|z|\le s\}\bigr).\] Since \(L_0(t)=j\) on \((\eta_{j+1},\eta_j]\), \(L_0(t)=k\) on \((\eta_{k+1},\eta_k]\) for \(k>j\), and the total mass is supported in \((0,\eta_j]\), summation by parts gives \[\begin{align} \int_{\Sigma_{\eta_j}} L_0(1-|z|)\, d\omega_a^{\Omega_{\eta_j}}(z) &\le j\,\nu_{j,a}((0,\eta_j]) + \sum_{k>j}\nu_{j,a}((0,\eta_k]). \end{align}\] The first term is bounded by \[j\,p_{R_m}(\eta_j)\le 2^{-j-m}.\] For \(k>j\), condition 2 , applied at stage \(k\) with \(i=j\), gives \[\nu_{j,a}((0,\eta_k]) \le q_{R_m,\eta_j}(\eta_k) \le 2^{-k-j-m}.\] Therefore \[\int_{\Sigma_{\eta_j}} L_0(1-|z|)\, d\omega_a^{\Omega_{\eta_j}}(z) \le 2^{-j-m} + \sum_{k>j}2^{-k-j-m},\] which tends to \(0\) as \(j\to\infty\), uniformly for \(|a|\le R_m\).
Finally choose a continuous nonincreasing function \(L:(0,1]\to[1,\infty)\) such that \[L_0(t)\le L(t)\le L_0(t)+1, \qquad 0<t\leq 1.\] This can be done by smoothing the jumps of \(L_0\) on the intervals immediately above the points \(\eta_j\). Then we still have \(L(t)\to\infty,\) and \(tL(t)\to 0\) as \(t\to 0\). Moreover, \[\begin{align} \int_{\Sigma_{\eta_j}} L(1-|z|)\, d\omega_a^{\Omega_{\eta_j}}(z) &\le \int_{\Sigma_{\eta_j}} L_0(1-|z|)\, d\omega_a^{\Omega_{\eta_j}}(z) + \omega_a^{\Omega_{\eta_j}}(\Sigma_{\eta_j}). \end{align}\] The first term tends to \(0\) uniformly for \(|a|\le R_m\), as shown above, and the second tends to \(0\) uniformly by [lem:pR]. Hence \(L\) has the required properties. ◻
2.2 Proof of 2↩︎
Choose \(a\in\mathbb{D}\) with \(w(a)>0\), and choose \(m\) such that \(|a|\le R_m\). Suppose the conclusion fails. Then there exist \(A>0\) and \(\eta_0>0\) such that \[w(re^{i\theta})<A L(1-r)\] whenever \[0<1-r<\eta_0, \qquad \operatorname{dist}(e^{i\theta},E)<\eta_0.\]
Choose \(j\ge m\) so large that \(\eta_j<\min\{\eta_0,1-|a|\}.\) Then \(a\in\Omega_{\eta_j}\). Moreover, by the definition of \(\Sigma_{\eta_j}\), every \(z=re^{i\theta}\in\Sigma_{\eta_j}\) satisfies \(0<1-r\le \eta_j\) and \(\operatorname{dist}(e^{i\theta},E)\le \eta_j\). Hence \(\Sigma_{\eta_j}\) is contained in the above collar, and therefore \(w(z)<A L(1-|z|)\) for \(z\in\Sigma_{\eta_j}\).
Fix \(0<s<\eta_j\) and truncate \(\Omega_{\eta_j}\) near the unit circle by considering the component \(\Omega_{\eta_j,s}\) of \[\Omega_{\eta_j}\cap\{|z|<1-s\}\] which contains \(a\). On the part of the inner collar boundary belonging to \(\partial\Omega_{\eta_j,s}\), we have \(w(z) < A L(1-|z|)\). On the remaining boundary near \(\mathbb{T}\setminus E_{\eta_j}\), the function \(w\) is uniformly small as \(s\to 0\), because \(\mathbb{T}\setminus E_{\eta_j}\) is compactly contained in \(\mathbb{T}\setminus E\) and \(w\) has continuous boundary value \(0\) there. Thus, there exists \(\varepsilon_s\to 0\) as \(s\to 0\) such that \[\label{eq:harm1} w(a) \le A\int_{\Sigma_{\eta_j,s}} L(1-|z|) \,d\omega_a^{\Omega_{\eta_j,s}}(z) + \varepsilon_s,\tag{3}\] by the subharmonic maximum principle in \(\Omega_{\eta_j,s}\). Domain monotonicity of harmonic measure further implies \[\label{eq:harm2} \int_{\Sigma_{\eta_j,s}} L(1-|z|) \,d\omega_a^{\Omega_{\eta_j,s}}(z) \le \int_{\Sigma_{\eta_j}} L(1-|z|) \,d\omega_a^{\Omega_{\eta_j}}(z).\tag{4}\] Combining 3 and 4 and letting \(s\to 0\) gives \[w(a) \le A\int_{\Sigma_{\eta_j}} L(1-|z|) \,d\omega_a^{\Omega_{\eta_j}}(z).\] Now let \(j\to\infty\). Since \(|a|\le R_m\), [lem:weighted-collar] implies \(w(a)\le0\), contradicting \(w(a)>0\). This proves the Theorem.
3 Proof of Esterle’s conjecture↩︎
Let \(F,F'\) be Hilbert spaces and let \(\Theta:\mathbb{C}_\infty\setminus E\to\mathcal{L}(F,F')\) be holomorphic. We say that \(\Theta\) is unitary-valued on \(\mathbb{T}\setminus E\) if \(\Theta(\zeta)\) is a unitary operator from \(F\) onto \(F'\) for each \(\zeta\in\mathbb{T}\setminus E\). For \(n\ge1\), set \[\delta_n(\Theta)= \inf_{z\in\mathbb{D}} \max\bigl\{|z|^n,\|\Theta(z)^{-1}\|^{-1}\bigr\},\] with the convention that \(\|\Theta(z)^{-1}\|^{-1}=0\) if \(\Theta(z)\) is not invertible. The quantity \(\delta_n\) appeared in [1] as the central quantity to control. In fact, Ransford formulated the following Theorem as a conjecture, except with the requirement that \(\Theta(z)\) is purely contractive in \(\mathbb{D}\), meaning that \(\|\Theta(z) x\|<\|x\|\) for all \(z\in\mathbb{D}\), \(x\in F \setminus\{0\}\). We only require the weaker assumptions that \(\Theta\) is contractive in \(\mathbb{D}\) and not constant unitary.
Theorem 3. Let \(E\subset\mathbb{T}\) be closed and have Lebesgue measure zero. There exists a positive sequence \(\varepsilon_n\to 0\) such that the following holds. Suppose that \(\Theta:\mathbb{C}_\infty\setminus E\to\mathcal{L}(F,F')\) is holomorphic, contractive in \(\mathbb{D}\), unitary-valued on \(\mathbb{T}\setminus E\), and not constant unitary. Then \[\liminf_{n\to\infty}\frac{\delta_n(\Theta)}{\varepsilon_n}=0.\]
To prove this theorem, we need the following consequence of the work in Section 2
Proof of Lemma [thm:collar-uniformity]. Let \(L\) be the weight from [lem:weighted-collar]. Put \[M(t)=\frac{L(t)}{4t}.\] Then \(M(t)\to\infty\) as \(t\to 0\). For \(n\ge1\), define \[\alpha_n=\inf\{L(t):0<t<1,\;M(t)\ge n\}.\] The set in the infimum is nonempty. For example, if \(t=1/(4n)\), then \(M(t)=nL(t)\ge n\). Also \(\alpha_n\to\infty\). Indeed, if \(M(t_j)\ge n_j\to\infty\) while \(L(t_j)\) stayed bounded, then \(t_j\to0\), contradicting \(L(t)\to\infty\). Set \[\beta_n=\frac{1}{10}\sqrt{\inf_{m\geq n}\alpha_m}, \qquad \varepsilon_n=e^{-\beta_n}.\] Clearly \(\varepsilon_n\) is nonincreasing and tends to zero. Moreover \(\varepsilon_n=e^{-o(n)}\). Indeed, with \(t_n=1/(4n)\), the admissibility above gives \(\alpha_n\le L(1/(4n)).\) Since \(tL(t)\to0\), this implies \(\alpha_n=o(n)\), and hence \[\beta_n\le \frac{1}{10}\sqrt{\alpha_n}=o(n).\]
Let \(w\ge0\) be nonzero and carried by \(E\). By 2 with \(A=1\), there are points \(z_k\in\mathbb{D}\) such that \[t_k=1-|z_k|\to0, \qquad w(z_k)\ge L(t_k).\] Set \[n_k=\left\lfloor\frac{L(t_k)}{4t_k}\right\rfloor.\] Since \(M(t_k)\to\infty\), we have \(n_k\to\infty\). Passing to a subsequence if necessary, assume that \(n_k\) is strictly increasing. For all large \(k\), \(n_k t_k\geq L(t_k)/8.\) Therefore \[\begin{align} d_{n_k}(w) &\le \max\{|z_k|^{n_k},e^{-w(z_k)}\} \\ &\le \max\{(1-t_k)^{n_k},e^{-L(t_k)}\} \\ &\le e^{-cL(t_k)} \end{align}\] for an absolute constant \(c>0\). Since \(n_k\le M(t_k)\), the point \(t_k\) is admissible in the definition of \(\alpha_{n_k}\). Hence \[\alpha_{n_k}\le L(t_k), \qquad \beta_{n_k}\le \frac{1}{10}\sqrt{L(t_k)}.\] Thus \[\frac{d_{n_k}(w)}{\varepsilon_{n_k}} =d_{n_k}(w)e^{\beta_{n_k}} \le \exp\big(-cL(t_k)+\frac{1}{10}\sqrt{L(t_k)}\big) \longrightarrow0.\] The result follows. ◻
Proof of 3. If \(\Theta(z_0)\) is not invertible for some \(z_0\in\mathbb{D}\), then \(\delta_n(\Theta)\le |z_0|^n\). Since \(\varepsilon_n=e^{-o(n)}\), it follows that \(\delta_n(\Theta)/\varepsilon_n\to0\). Assume therefore that \(\Theta(z)\) is invertible for every \(z\in\mathbb{D}\), and put \[w_\Theta(z)=\log\|\Theta(z)^{-1}\|.\] Then \(w_\Theta\) is subharmonic. Since \(\Theta(z)\) is contractive, \(\|\Theta(z)^{-1}\|\ge1\), and so \(w_\Theta\ge0\). Also, since \(\Theta\) extends holomorphically through \(\mathbb{T}\setminus E\) and is unitary-valued there, \(w_\Theta\) has continuous boundary value zero on \(\mathbb{T}\setminus E\). Thus \(w_\Theta\) is carried by \(E\), unless it is identically zero.
If \(w_\Theta\not\equiv0\), then the conclusion follows from [thm:collar-uniformity] since \(d_n(w_\Theta)=\delta_n(\Theta)\). It remains to exclude the case \(w_\Theta\equiv0\). Then \(\|\Theta(z)^{-1}\|=1\) for all \(z\in\mathbb{D}\). Since \(\Theta(z)\) is contractive, for every \(x\in F\), \[\|\Theta(z)x\|\le\|x\|, \qquad \|x\|=\|\Theta(z)^{-1}\Theta(z)x\| \le \|\Theta(z)x\|.\] Thus \(\Theta(z)\) is a surjective isometry for every \(z\in\mathbb{D}\), hence unitary. For each fixed \(x\in F\), the Hilbert-space-valued holomorphic function \(z\mapsto\Theta(z)x\) has constant norm \(\|x\|\), and is therefore constant. Hence \(\Theta\) is constant unitary, contrary to the hypothesis. ◻
Proof of 1. We now explain how 3 gives 1, following [1]. Let \((\varepsilon_n)\) be the sequence from Lemma [thm:collar-uniformity], and set \(u_n=\varepsilon_n^{-1}.\) Then \(u_n\to\infty\). Let \(T\) be a non-unitary Hilbert-space contraction with \(\sigma(T)\subset E\). We show that \[\limsup_{n\to\infty}\frac{\|T^{-n}\|}{u_n}=\infty.\] This is enough to prove the theorem. By a standard reduction as in Ransford’s work, using the Sz.-Nagy–Foias functional-model theory for contractions, one may pass to a completely non-unitary contraction \(T_1\) on a separable Hilbert space such that \(\sigma(T_1)\subset\sigma(T)\) and \(\|T_1^{-n}\|\le\|T^{-n}\|\) for \(n\geq 1\) [1]. Since \(\sigma(T_1)\subset\mathbb{T}\) has Lebesgue measure zero, both \(T_1^n\) and \(T_1^{*n}\) converge strongly to zero [1]. The Sz.-Nagy–Foias model theorem then realizes \(T_1\) as a compressed shift \(S_\Theta\) associated with an operator-valued inner function \(\Theta\). Moreover, \(\Theta\) may be chosen so that it is purely contractive in \(\mathbb{D}\), extends holomorphically to \(\mathbb{C}_\infty\setminus E\), and is unitary-valued on \(\mathbb{T}\setminus E\). (See [1] for more details.) The hypotheses of 3 are therefore satisfied. It follows that along a subsequence, \(\varepsilon_n/\delta_n(\Theta)\to\infty\). Finally, [1] gives \[\|S_\Theta^{-n}\| \ge \frac{1}{2}\Big(\frac{1}{\delta_n(\Theta)}-1\Big), \qquad n\geq 1.\] Hence \[\begin{align} \limsup_{n\to\infty}\frac{\|T^{-n}\|}{u_n} &\ge \limsup_{n\to\infty}\frac{\|T_1^{-n}\|}{u_n} \\ &= \limsup_{n\to\infty}\varepsilon_n\|S_\Theta^{-n}\| \\ &\ge \frac{1}{2}\limsup_{n\to\infty} \Big(\frac{\varepsilon_n}{\delta_n(\Theta)}-\varepsilon_n\Big) =\infty, \end{align}\] as desired. ◻
Acknowledgments↩︎
The author would like to thank Marcu-Antone Orsoni for his comments, and Thomas Ransford for helpful discussions and for sharing his manuscript [6].