Infinitesimal generators of semigroups with prescribed boundary fixed points

Abstract

We study infinitesimal generators of one-parameter semigroups in the unit disk $\mathbb {D}$ having prescribed boundary regular fixed points. Using an explicit representation of such infinitesimal generators in combination with Krein–Milman Theory we obtain new sharp inequalities relating spectral values at the fixed points with other important quantities having dynamical meaning. We also give a new proof of the classical Cowen–Pommerenke inequalities for univalent self-maps of $\mathbb {D}$.

Appendix

For completeness, we give proofs of some elementary facts used in the paper.

Proof of Lemma 7.2

Let $\varphi \in {{\mathfrak {U}}}_{\tau }[F]$. Fix some $\sigma _j,\sigma _k\in F$, $j\ne k$. Choose any $C^1$-smooth Jordan arc $\Gamma \subset {\overline{\mathbb {D}}}\setminus \{\tau \}$ joining $\sigma _j$ with $\sigma _k$ and orthogonal to $\partial \mathbb {D}$ at these points.

Note that ${\widetilde{\Gamma }}:=\varphi (\Gamma )$ satisfies the same requirements imposed on $\Gamma$. Let us show that $\Gamma$ and ${\widetilde{\Gamma }}$ are homotopic relative to end-points in ${\mathbb {C}}\setminus \{\tau \}$. Denote by $D_1$ and $D_2$ the two connected components of $\mathbb {D}\setminus \Gamma$, with $\tau \in D_2$, and let $C_1$ and $C_2$ be the two complementary arcs of $\partial \mathbb {D}$ such that $C_j\subset \partial D_j$, $j=1,2$. In particular, $\Gamma$ is homotopic in ${\mathbb {C}}\setminus \{\tau \}$ relative to the end-points to $C_1$.

Furthermore, let ${\widetilde{D}}_1$ and ${\widetilde{D}}_2$ stand for the two connected components of ${\mathbb {D}\setminus {\widetilde{\Gamma }}}$, numbered in such a way that $C_j\subset \partial {\widetilde{D}}_j$, $j=1,2$. Using the conformality at $\sigma _j$ of $\varphi$ restricted to a Stolz angle, we see that the $\varphi (D_1)$ intersects ${\widetilde{D}}_1$, and $\varphi (D_2)$ intersects ${\widetilde{D}}_2$. Since $\varphi :\mathbb {D}\rightarrow \mathbb {D}$ is a homemorphism onto its image, it follows that $\varphi (D_j)\subset {\widetilde{D}}_j$, $j=1,2$. In particular, $\tau =\varphi (\tau )\in \varphi (D_2)\subset {\widetilde{D}}_2$ and hence $\tau \not \in {\widetilde{D}}_1$. It follows that relative to end-points ${\widetilde{\Gamma }}$ is homotopic in ${\mathbb {C}}\setminus \{\tau \}$ to $C_1$ and hence to $\Gamma$. Therefore, for $\Phi (z):=(\varphi (z)-\tau )/(z-\tau )$ we have

$$\begin{aligned} \int _\Gamma \frac{\Phi '(z)}{\Phi (z)}\,\mathrm {d}z=\int _{{\widetilde{\Gamma }}}\frac{\,\mathrm {d}w}{w-\tau }-\int _{\Gamma }\frac{\,\mathrm {d}z}{z-\tau }=0. \end{aligned}$$

All the above integrals exist because $\tau \not \in \Gamma \cup {\widetilde{\Gamma }}$ and because $\varphi$ is of class $C^1$ on $\Gamma$ including the end-points.

The above argument is valid for any two distinct points $\sigma _j,\sigma _k\in F$. Taking into account that $\Phi$ is holomorphic and non-vanishing in $\mathbb {D}$, it follows that $\Phi '/\Phi$ admits an antiderivative in $\mathbb {D}$ that has vanishing angular limits at every point of F, and this is the desired single-valued branch of $\log \Phi$. This proves part (A).

To prove part (B), note that continuity of $t\mapsto \varphi _t(z)$ for each $z\in \mathbb {D}$ is equivalent to continuity of $t\mapsto \varphi _t\in \mathsf {Hol}(\mathbb {D},\mathbb {C})$ in the open-compact topology because all holomorphic self-maps of $\mathbb {D}$ form a normal family. Therefore, for any $t_0\in I$ the limit

$$\begin{aligned} \lim _{I\ni t\rightarrow t_0}\exp F_t,\quad \text {where }F_t:=\Psi [\varphi _t]-\Psi [\varphi _{t_0}], \end{aligned}$$

exists in the open-compact topology and equals 1 identically in $\mathbb {D}$. Moreover, note that

$$\begin{aligned} \lim _{r\rightarrow 1^-} F_t(r\sigma _{k_0})=0\quad \text {for all }t\in I. \end{aligned}$$

(8.1)

Since by the hypothesis, for some $\delta >0$, the function ${t\mapsto \varphi '_t(\sigma _{k_0})}$ is bounded on $I_\delta :=I\cap (t_0-\delta ,t_0+\delta )$, using Julia’s Lemma 2.1 we see that for any $\varepsilon >0$ there exists $r_\varepsilon \in (0,1)$ such that $\gamma _\varepsilon :=[r_\varepsilon \sigma _{k_0},\sigma _{k_0})$ and $\varphi _t(\gamma _\varepsilon )$ lie in the disk $D_\varepsilon :=\{z:|z-(1-\varepsilon )\sigma _{k_0}|\leqslant \varepsilon \}\subset \mathbb {D}$ for all $t\in I_\delta$.

Clearly we can choose $\varepsilon >0$ small enough, so that $\tau \not \in D_\varepsilon$ and

$$\begin{aligned} \max _{z,w\in D_\varepsilon }\big |\log (z-\tau )-\log (w-\tau )\big |=:C<\pi \end{aligned}$$

(8.2)

for some (and hence any) choice of the single-valued branch of $\log (z-\tau )$ in $D_\varepsilon$.

Combining (8.1) and (8.2), we see that $|F_t(r_\varepsilon \sigma _{k_0})|\leqslant 2C<2\pi$ for all $t\in I_\delta$. Recalling that $\exp F_t(z)\rightarrow 1$ locally uniformly in $\mathbb {D}$ as $I\ni t\rightarrow t_0$, we conclude that $F_t\rightarrow 0$ locally uniformly in $\mathbb {D}$ as $I\ni t\rightarrow t_0$, which completes the proof of (B). $\square$

Next lemma shows that the harmonic mean is concave. We include its proof for the sake of completeness.

Lemma 8.1

For any $n\in \mathbb {N}$ the function $Q(x_1,\ldots x_n):=\Big (\sum _{j=1}^n x_j^{-1}\Big )^{-1}$ is concave on $(0,+\infty )^n$.

Proof

The assertion of the lemma holds trivially for $n=1$. So we suppose that $n\geqslant 2$.

The entries of the Hessian matrix $A({\mathbf {x}})=[a_{jk}({\mathbf {x}})], {\mathbf {x}}:=(x_1,x_2,\ldots ,x_n)$, for the function Q are given by

$$\begin{aligned}&a_{jk}({{\mathbf {x}}})=\frac{2Q({{\mathbf {x}}})^3}{x_j^2x_k^2}b_{jk}({{\mathbf {x}}}),\\&\quad \text {where }b_{jk}({{\mathbf {x}}}):=1-\delta _{jk}\frac{x_j}{Q({{\mathbf {x}}})} ~\text { and } \delta _{jk} \text { is the Kronecker symbol.} \end{aligned}$$

First we show that $\det A({{\mathbf {x}}})=0$. Clearly, the latter is equivalent to $\det B({{\mathbf {x}}})=0$, where $B({{\mathbf {x}}}):=[b_{jk}({{\mathbf {x}}})]$.

Subtract the last row of $B({{\mathbf {x}}})$ from each of the other rows. In the matrix we obtain, add to the last row the linear combination of all the other rows in which, for every $j=1,2,\ldots ,n-1$, the coefficient of the j-th row is equal to $Q({{\mathbf {x}}})/x_j$. The resulting matrix is upper-triangular, with the last diagonal entry equal to

$$\begin{aligned} 1-\frac{x_n}{Q({{\mathbf {x}}})}+\frac{x_n}{Q({{\mathbf {x}}})}\sum _{j=1}^{n-1}\frac{Q({{\mathbf {x}}})}{x_j}=1-\sum _{j=1}^n \frac{x_n}{x_j}+\sum _{j=1}^{n-1} \frac{x_n}{x_j}=0. \end{aligned}$$

Therefore, the determinant equals zero.

This argument, with an obvious modification, can be used to show that the determinants of all symmetric minors of $A({{\mathbf {x}}})$, i.e. minors of the form $[a_{jk}({{\mathbf {x}}})]_{j,k\in J}$, $J\subset \{1,2,\ldots ,n\}$, vanish, except for the symmetric minors of order one. They are simply diagonal entries of A, which are all negative. Therefore, according to Sylvester’s well-known criterion, the matrix $A({{\mathbf {x}}})$ is negative semi-definite for any ${{\mathbf {x}}}\in (0,+\infty )^n$, which was to be proved. $\square$

Lemma 8.2

Let Q be defined as in Lemma 8.1. Let ${{\mathbf {x}}},{\mathbf {y}}\in (0,+\infty )^n$, ${{\mathbf {x}}}\ne {\mathbf {y}}$. If

$$\begin{aligned} \lambda Q({{\mathbf {x}}})+(1-\lambda )Q({\mathbf {y}})=Q\big (\lambda {{\mathbf {x}}} + (1-\lambda ){\mathbf {y}}\big ) \end{aligned}$$

(8.3)

for some $\lambda \in (0,1)$, then ${{\mathbf {x}}}=\mu {\mathbf {y}}$ for some $\mu \in \mathbb {R}$.

Proof

Since by Lemma 8.1, Q is concave on $(0,+\infty )$, equality (8.3) for some $\lambda \in (0,1)$ implies the same equality for all $\lambda \in [0,1]$. Notice that the r.h.s. of (8.3), $f(\lambda ):=Q\big (\lambda {{\mathbf {x}}} + (1-\lambda ){\mathbf {y}}\big )$ is a rational function of $\lambda$. Therefore, extending f, as usual, to its removable singularities by continuity, we may conclude that $f(\lambda )=a\lambda +b$ for some $a,b\in \mathbb {R}$ and all $\lambda \in \mathbb {R}$. On the one hand, a function of this form has at most one zero. On the other hand, taking into account that for $\lambda =0$ all the components of the vector ${{\mathbf {x}}}_\lambda :=\lambda {{\mathbf {x}}} + (1-\lambda ) {\mathbf {y}}$ are positive, it is easy to see that $f(\lambda )=0$ for each $\lambda \in \mathbb {R}$ such that at least one component of ${{\mathbf {x}}}_\lambda$ vanishes.

All non-vanishing components of ${\mathbb {R}\ni \lambda \mapsto {{\mathbf {x}}}_\lambda }$ are positive constants. Therefore, if such components exist, then $a=0$ and hence $f(\lambda )\equiv Q({\mathbf {y}})>0$. It follows that if at least one of the components is non-vanishing, then $f(\lambda )$ does not vanish, which means that, in fact, all the components of ${\mathbb {R}\ni \lambda \mapsto {{\mathbf {x}}}_\lambda }$ are non-vanishing. This in turn would imply that ${{\mathbf {x}}}={\mathbf {y}}$ in contradiction to the hypothesis.

Thus we may conclude that all the components of ${{\mathbf {x}}}_\lambda$ vanish for the same value of $\lambda$ and the desired conclusion follows immediately. $\square$