Ahlfors–David Regular Sets, Point Spectrum and Dirichlet Spaces

Abstract

Let E be a closed subset of the unit circle $\mathbb {T}$, and let $\alpha \in (0,1)$. Nikolski’s result states that if the Hausdorff dimension of E is strictly greater than $\alpha$, then for any operator T on a separable Hilbert space such that the point spectrum $\sigma _p(T)$ of T contains E, the series $\sum _{n}n^{\alpha -1}\Vert T^n\Vert ^{-2}$ converges. A partial converse of this result has been obtained by El-Fallah and Ransford. Namely they constructed, for any $\alpha$ strictly greater than the upper box dimension of E, an operator T on a separable Hilbert space such that $\sigma _p(T)$ contains E and $\frac{1}{n} \sum _{k=0}^{n-1}\left\| T^k\right\| ^2\lesssim n^{\alpha }$. In this paper, we improve on this latter result for regular sets. Indeed, for any Ahlfors–David regular set E and for any $\alpha$ strictly greater than the Hausdorff dimension of E there exists an operator T on a separable Hilbert space such that $\sigma _p(T)$ contains E and $\Vert T^n\Vert ^2\asymp n^{\alpha }$.

1 Introduction

Let T be a bounded linear operator on a complex Banach space X. The unit circle is denoted by $\mathbb {T}$, while $\sigma _p(T):=\{\lambda \in \mathbb {C}:\,\, \ker (T-\lambda I)\not =\emptyset \}$ represents the point spectrum of T. Jamison [10] showed that if X is separable and T is power-bounded, then $\sigma _p(T)\cap \mathbb {T}$ is at most countable. Later on, several authors (see, e.g., [1, 2, 4, 15, 16] and the references therein) have shown interest in the study of the relationship between the size of the set $\sigma _p(T)\cap \mathbb {T}$ and the growth of $\Vert T^n\Vert$ as $n\rightarrow \infty$.

In the case of separable Hilbert spaces, Nikolski [13] proved that if $\sigma _p(T)\cap \mathbb {T}$ has a positive $\gamma$-capacity, where $\gamma : \mathbb {T}\rightarrow (0,\infty )$ is integrable with positive Fourier coefficients, then there exists $N\in \mathbb {N}$ such that $\sum _n\widehat{\gamma }(n+N)\Vert T^n\Vert ^{-2}$ converges (see [11, Chapter 3, p. 31] for the definition of $\gamma$-capacity). As a by-product of this, if $\sigma _p(T)\cap \mathbb {T}$ contains a subset E of $\mathbb {T}$ with $\dim _H(E)>\alpha >0$, where $\dim _H(E)$ is the Hausdorff dimension of E, then the series $\sum _nn^{\alpha -1}\Vert T^n\Vert ^{-2}$ converges. El-Fallah-Ransford [8] proved that, as a partial converse of the preceding result, if $\alpha$ is strictly greater than the upper box dimension of E, then there exists an operator T on a separable Hilbert space such that $\sigma _p(T)\cap \mathbb {T}$ contains E and the series $\sum _nn^{\alpha -1}\Vert T^n\Vert ^{-2}$ diverges. (See [9, p. 41] for more details about the box dimension). Precisely, they constructed an operator T such that $\sigma _p(T)$ contains E and

$$\begin{aligned} \frac{1}{n} \sum _{k=0}^{n-1}\left\| T^k\right\| ^2 \lesssim \omega (n)^2\left| E_{1 / n}\right| , \quad n \geqslant 1, \end{aligned}$$

where $\omega : \mathbb {Z} \rightarrow (0, \infty )$ is a regular weight function satisfying $\sum _n \frac{1}{\omega (n)^2}<\infty$, $E_{1 / n}:=\{\zeta \in \mathbb {T}: \,\, {{\,\textrm{dist}\,}}(E,\zeta )<\frac{1}{n}\}$ with ${{\,\textrm{dist}\,}}(.,.)$ being the arc-length distance, and $\left| E_{1 / n}\right|$ is its Lebesgue measure. In particular, one can obtain

$$\begin{aligned} \inf _{n\le k\le 2n}\Vert T^k\Vert ^2\lesssim \omega (n)^2|E_{1/n}|,\quad n\ge 1. \end{aligned}$$

The main result of this paper is the following theorem.

Theorem 1.1

Assume that $E\subset \mathbb {T}$ is a closed Ahlfors–David regular set. If $\alpha >\dim _H(E)$, then there exists an operator T on a separable Hilbert space such that $\sigma _p(T)\cap \mathbb {T}$ contains E and

$$\begin{aligned} \Vert T^n\Vert ^2\asymp n^{\alpha }. \end{aligned}$$

The definitions of Ahlfors–David regular sets and Hausdorff dimension are recalled in Sect. 4. Theorem 1.1 is a corollary from the following more general result.

Theorem 1.2

Let E be closed subset of $\mathbb {T}$, if there exist an increasing function $\Lambda : [0,1]\rightarrow [0,+\infty )$ such that $\frac{\Lambda (t)}{t^c}$ is decreasing for some $c>0$, and a positive finite Borel measure $\mu$ on $\mathbb {T}$ satisfying

$$\begin{aligned} \int _0\frac{dt}{\Lambda (t)\mu (\overline{\zeta },t)}<\infty ,\quad \zeta \in E, \end{aligned}$$

(1)

where $\mu (\overline{\zeta },t)=\mu (\overline{\zeta } e^{-it},\overline{\zeta } e^{it}),$ then there exists an operator T on a separable Hilbert space such that $E \subset \sigma _p(T)$ and

$$\begin{aligned} \Vert T^n\Vert ^2\asymp n\Lambda \left( \frac{1}{n}\right) , \quad n\ge 1. \end{aligned}$$

In Sect. 2, we study the point spectrum of adjoint of the shift operator acting on some weighted Dirichlet spaces. In Sect. 3, we determine the growth of power of the adjoint of this shift operator. In Sect. 4, we give the proofs of the main theorems.

2 Point Spectrum

Let $\Lambda : [0,1]\rightarrow [0,+\infty )$ be a positive function, let $\mathbb {D}$ be the unit disc, and let $\mu$ be a positive finite Borel measure on the unit circle $\mathbb {T}$. The weighted Dirichlet integral of $f\in \textrm{Hol}(\mathbb {D})$ associated with $\Lambda$ and $\mu$ is defined as follows:

$$\begin{aligned} {\mathcal {D}}_{\Lambda ,\mu }(f)=\int _{\mathbb {D}}|f'(z)|^2\Lambda (1-|z|^2)P_\mu (z)dA(z), \end{aligned}$$

where dA denotes the normalized area measure on $\mathbb {D}$, and $P_\mu$ is the Poisson integral of $\mu$ on $\mathbb {T}$ given by

$$\begin{aligned} P_\mu (z)=\int _{\mathbb {T}}\frac{1-|z|^2}{|\zeta -z|^2}d\mu (\zeta ), \quad z\in \mathbb {D}. \end{aligned}$$

The associated weighted Dirichlet spaces ${\mathcal {D}}_{\Lambda ,\mu }$ consist of all analytic functions on $\mathbb {D}$ with finite weighted Dirichlet integral, i.e.,

$$\begin{aligned} {\mathcal {D}}_{\Lambda , \mu }:=\{f\in \textrm{Hol}(\mathbb {D}): {\mathcal {D}}_{\Lambda ,\mu }(f)<\infty \}. \end{aligned}$$

We associate to ${\mathcal {D}}_{\Lambda ,\mu }$ the following inner product

$$\begin{aligned} \left<f,g\right>_{\Lambda ,\mu }:=f(0)\overline{g(0)}+{\mathcal {D}}_{\Lambda ,\mu }(f,g),\quad f,g\in \textrm{Hol}(\mathbb {D}), \end{aligned}$$

where

$$\begin{aligned} {\mathcal {D}}_{\Lambda ,\mu }(f,g)=\int _{\mathbb {D}}f'(z))\overline{g'(z)}\Lambda (1-|z|^2)P_{\mu }(z)dA(z). \end{aligned}$$

${\mathcal {D}}_{\Lambda ,\mu }$ is a reproducing kernel Hilbert space, and denote K (or $K_{\Lambda ,\mu }$ if necessary) its reproducing kernel. The standard weighted Dirichlet spaces on $\mathbb {D}$, denoted ${\mathcal {D}}_\alpha$, correspond to $\Lambda (t)=t^{\alpha }$ and $\mu =m$ the normalized arc measure on $\mathbb {T}$. If $\Lambda =1$, then ${\mathcal {D}}_{\Lambda ,\mu }$ is the harmonically weighted Dirichlet spaces. (See, e.g., [6, 7]). Note that, in general, ${\mathcal {D}}_{\Lambda ,\mu }$ is not contained in the Hardy space $\mathrm {H^2 }$. In the following proposition, using a reasoning similar to that in [3], we establish the density of polynomials in ${\mathcal {D}}_{\Lambda ,\mu },$ for some regular weights $\Lambda$.

Proposition 2.1

Let $\mu$ be a positive finite Borel measure on $\mathbb {T}$, and let $\Lambda : [0,1]\rightarrow [0,+\infty )$ be an increasing function such that $\frac{\Lambda (t)}{t^c}$ is decreasing for some $c>0$. Then the polynomials are dense in ${\mathcal {D}}_{\Lambda ,\mu }.$

Proof

The proof uses the fact that the dilations $f_r(z):=f(r z)$ for $r\in [0,1)$ tend to f in the norm, i.e.,

$$\begin{aligned} \lim _{r \rightarrow 1}\left\| f_r-f\right\| _{\Lambda ,\mu }=0, \,\, \quad f\in {\mathcal {D}}_{\Lambda ,\mu }. \end{aligned}$$

To this end, it is sufficient to show that

$$\begin{aligned} \Vert f_r\Vert _{\Lambda ,\mu }\lesssim \Vert f\Vert _{\Lambda ,\mu },\quad f\in {\mathcal {D}}_{\Lambda ,\mu },\quad r\in [0,1). \end{aligned}$$

Since $\Lambda (2t)\asymp \Lambda (t)$, we have

$$\begin{aligned} \Lambda (1-|z|^2)\asymp \Lambda (1-|w|^2),\quad z\in D(w,\rho (w)), \end{aligned}$$

where $\rho (w)=(1-|w|)/2.$ Hence,

$$\begin{aligned} |f'(w)|^2\Lambda (1-|w|^2)\lesssim \int _{z\in \mathbb {D}}|f'(z)|^2\Lambda (1-|z|^2)\frac{(1-|z|^2)^2}{|1-\overline{z}w|^{4}}dA(z),\quad w\in \mathbb {D}, \end{aligned}$$

Denote

$$\begin{aligned} J(z,\zeta ):=\int _{w\in \mathbb {D}}\frac{(1-|w|^2)}{|1-w\overline{z}|^{4}|1-w\overline{\zeta }|^2}dA(w), \quad z\in \mathbb {D}. \end{aligned}$$

Using the following inequality (cf. [14, Lemma 2.5]):

$$\begin{aligned} J(rz,\zeta )\lesssim \frac{1}{(1-r^2|z|^2)|1-r\overline{\zeta }z|^2},\quad z\in \mathbb {D},\,\, \zeta \in \mathbb {T}, \,\, r\in [0,1), \end{aligned}$$

we obtain

$$\begin{aligned} \Vert f_r\Vert ^2_{\Lambda ,\mu }= & {} |f_r(0)|^2+\int _{w\in \mathbb {D}}|r f'(rw)|^2\Lambda (1-|w|^2)P_\mu (w)dA(w)\\\le & {} |f(0)|^2+\int _{\zeta \in \mathbb {T}}\int _{w\in \mathbb {D}}|f'(rw)|^2\Lambda (1-|rw|^2)\frac{(1-|w|^2)}{|1-w\overline{\zeta }|^2}dA(w)d\mu (\zeta )\\\lesssim & {} |f(0)|^2+\int _{\zeta \in \mathbb {T}}\int _{w\in \mathbb {D}}\int _{z\in \mathbb {D}}|f'(z)|^2\Lambda (1-|z|^2)\\{} & {} \quad \frac{(1-|z|^2)^2(1-|w|^2)}{|1-rw\overline{z}|^{4}|1-w\overline{\zeta }|^2} dA(z)dA(w) d\mu (\zeta )\\= & {} |f(0)|^2+\int _{\zeta \in \mathbb {T}}\int _{z\in \mathbb {D}}|f'(z)|^2\Lambda (1-|z|^2)(1-|z|^2)^{2}J(rz,\zeta )dA(z)d\mu (\zeta )\\\lesssim & {} |f(0)|^2+\int _{\zeta \in \mathbb {T}}\int _{z\in \mathbb {D}}|f'(z)|^2\Lambda (1-|z|^2)\frac{(1-|z|^2)}{|1-\overline{\zeta }z|^2}dA(z)d\mu (\zeta )\\= & {} |f(0)|^2+{\mathcal {D}}_{\Lambda ,\mu }(f)\\= & {} \Vert f\Vert ^2_{\Lambda ,\mu }. \end{aligned}$$

$\square$

Let $f\in \textrm{Hol}(\mathbb {D})$, and let $\zeta \in \mathbb {T}$, we write $f^*(\zeta ):=\lim _{r\rightarrow 1}f(r\zeta )$ its radial limit (if it exists) at $\zeta$. Note that, under conditions of Proposition 2.1, if $\sup _{0\le r<1}K(r\zeta ,r\zeta )<\infty$, then $f^*(\zeta )$ exists for every $f\in {\mathcal {D}}_{\Lambda ,\mu }$.

We denote $S_{\Lambda ,\mu }$ (or simply S) the shift operator acting on ${\mathcal {D}}_{\Lambda ,\mu }$ defined as follows:

$$\begin{aligned} S: {\mathcal {D}}_{\Lambda ,\mu }\longrightarrow {\mathcal {D}}_{\Lambda ,\mu },\,f\mapsto S f(z)=zf(z), \end{aligned}$$

and $S^*$ is its adjoint operator.

Theorem 2.1

Let $\mu$ be a positive finite Borel measure on $\mathbb {T}$, and let $\Lambda : [0,1]\rightarrow [0,+\infty )$ be an increasing function such that $\frac{\Lambda (t)}{t^c}$ is decreasing for some $c>0$, then

$$\begin{aligned} \sigma _p(S^*)\supset \left\{ \zeta \in \mathbb {T}: \, \int _0^1\frac{dx}{\Lambda (x)\mu (\overline{\zeta }, x)}<\infty \right\} . \end{aligned}$$

To prove Theorem 2.1, we need the following lemma.

Lemma 2.1

Let $\mu$ be a positive finite Borel measure on $\mathbb {T}$, and let $\Lambda : [0,1]\rightarrow [0,+\infty )$ be an increasing function such that $\frac{\Lambda (t)}{t^c}$ is decreasing for some $c>0$, then

The proof of Lemma 2.1 is inspired from [5]. For the sake of completeness, we include the proof here.

Proof

We have

$$\begin{aligned} K(z,z)=\sup \{|f(z)|^2: \,\Vert f\Vert _{\Lambda ,\mu }\le 1\},\quad z\in \mathbb {D}. \end{aligned}$$

Let $f\in {\mathcal {D}}_{\Lambda ,\mu }$, and let $z=r\in [\frac{1}{2},1)$. Consider

$$\begin{aligned} \Delta _{r}:=\left\{ x+iy\in \mathbb {D}: 0\le x\le r, \frac{x-1}{2} \le y \le \frac{1-x}{2}\right\} . \end{aligned}$$

We have

$$\begin{aligned} \begin{aligned}&\frac{1}{1-r} \int _{-\frac{(1-r)}{2}}^{\frac{1-r}{2}}|f(r+i s)| ds \\&\quad =\int _{-\frac{1}{2}}^{\frac{1}{2}}|f(r+i(1-r) y)| d y \\&\quad =\int _{-\frac{1}{2}}^{\frac{1}{2}}\left| f(i y)+\int _{0}^{r} f^{\prime }(t+i(1-t) y) dt\right| dy\\&\quad \le \int _{-\frac{1}{2}}^{\frac{1}{2}}|\left<f,K_{iy}\right>_{\Lambda ,\mu }| dy+\int _{[0, r] \times \left[ -\frac{1}{2}; \frac{1}{2}\right] }\left| f^{\prime }(t+i(1-t) y)\right| dy dt \\&\quad \le \Vert f\Vert _{\Lambda ,\mu }\int _{-\frac{1}{2}}^{\frac{1}{2}}\Vert K_{iy}\Vert _{\Lambda ,\mu } dy+\int _{[0, r] \times \left[ -\frac{1}{2}; \frac{1}{2}\right] }\left| f^{\prime }(t+i(1-t) y)\right| d y d t \\&\quad \lesssim \Vert f\Vert _{\Lambda ,\mu }+\int _{[0, r] \times \left[ -\frac{1}{2}; \frac{1}{2}\right] }\left| f^{\prime }(t+i(1-t) y)\right| d y d t. \end{aligned} \end{aligned}$$

Using the following change of variables:

$$\begin{aligned} \begin{aligned} \phi :[0, r] \times \left[ -\frac{1}{2}, \frac{1}{2}\right]&\longrightarrow \Delta _{r} \\ (t, y)&\longmapsto u+i v,\quad \text {with}\,\, \left\{ \begin{array}{l}u=t \\ v=(1-t) y,\end{array}\right. \end{aligned} \end{aligned}$$

we obtain

$$\begin{aligned} \begin{aligned}&\frac{1}{1-r} \int _{-\frac{(1-r)}{2}}^{\frac{1-r}{2}}|f(r+i s)| ds \\&\quad \lesssim \Vert f\Vert _{\Lambda ,\mu }+\int _{\Delta _{r}}\left| f^{\prime }(u+i v)\right| \frac{d u d v}{1-| u| } \\&\quad \lesssim \Vert f\Vert _{\Lambda ,\mu }+\left( {\mathcal {D}}_{\Lambda ,\mu }(f)\right) ^{\frac{1}{2}}\\&\quad \left( \int _{\Delta _{r}} \frac{dudv}{(1-|u|)^{2}\Lambda (1-|u+iv|^2)P_\mu (u+i v)}\right) ^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

Now, for $u+iv \in \Delta _r$, we have $\begin{aligned} |u+i v|&\le u+| v | \le u+\frac{1-u}{2} =\frac{1+u}{2}. \end{aligned}$ Then $1-|u+i v| \geqslant 1-\frac{1+u}{2}=\frac{1-u}{2}.$ Hence, , and Therefore,

$$\begin{aligned} \begin{aligned}&\frac{1}{1-r} \int _{-\frac{(1-r)}{2}}^{\frac{1-r}{2}}|f(r+i s)| ds\\&\quad \lesssim \Vert f\Vert _{\Lambda ,\mu }+({\mathcal {D}}_{\Lambda ,\mu }(f))^{\frac{1}{2}} \left( \int _{\Delta _{r}} \frac{d u d v}{(1-u)^2\Lambda (1-u)P_\mu (u)}\right) ^{\frac{1}{2}} \\&\quad =\Vert f\Vert _{\Lambda ,\mu }+({\mathcal {D}}_{\Lambda ,\mu }(f))^{\frac{1}{2}}\left( \int _{u=0}^{r} \int _{|v|\le \frac{1-u}{2}} d v \frac{d u}{(1-u)^{2}\Lambda (1-u)P_\mu (u)}\right) ^{\frac{1}{2}}\\&\quad \lesssim \Vert f\Vert _{\Lambda ,\mu }+({\mathcal {D}}_{\Lambda ,\mu }(f))^{\frac{1}{2}}\left( \int _{0}^{r} \frac{d u}{(1-u)\Lambda (1-u)P_\mu (u)}\right) ^{\frac{1}{2}}.\end{aligned} \end{aligned}$$

Moreover, the disc $D\left( r, \frac{1-r}{4}\right)$ is included in $\left\{ z=x+i y,|x-r| \lesssim \frac{1-r}{4},|y| \lesssim \right. \left. \frac{1-x}{2}\right\} .$ Thus

$$\begin{aligned} \begin{aligned} |f(r)|&\lesssim \frac{1}{(1-r)^{2}} \int _{r-\frac{1-r}{4}}^{r+\frac{1-r}{4}}\left( \int _{y=-\frac{(1-x)}{2}}^{\frac{1-x}{2}} |f(x+i y)| d y\right) d x \\&\lesssim \Vert f\Vert _{\Lambda ,\mu }+({\mathcal {D}}_{\Lambda ,\mu }(f))^{\frac{1}{2}} \left( \int _{0}^{r} \frac{d x}{(1-x)\Lambda (1-x)P_\mu (x)}\right) ^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \displaystyle |f(r)|^{2} \lesssim \left( 1+\int _{0}^{r} \frac{d x}{(1-x)\Lambda (1-x)P_\mu (x)}\right) \Vert f\Vert _{\Lambda ,\mu }^{2}, \end{aligned}$$

and we get

$$\begin{aligned} K(z, z) \lesssim 1+\int _{0}^{r} \frac{d x}{(1-x)\Lambda (1-x)P_\mu \left( x z^{*}\right) }. \end{aligned}$$

$\square$

Proof of Theorem 2.1

We have for any $\zeta \in \mathbb {T}$, and $t\in (0,1)$. Then

(2)

Let $\zeta \in \mathbb {T}$ such that $\int _0^1\frac{dx}{\Lambda (x)\mu (\overline{\zeta },x)}<\infty$. Combining inequality (2) with Lemma 2.1, we obtain $\sup _{0\le r<1} K(r\overline{\zeta },r\overline{\zeta })<\infty$. Consider now $L_{\overline{\zeta }}: {\mathcal {D}}_{\Lambda ,\mu }\rightarrow \mathbb {C}, f\mapsto f^*(\overline{\zeta })$. Since $L_{\overline{\zeta }}$ is continuous, it follows from Riesz representation theorem that there exists $k_{\overline{\zeta }}\in {\mathcal {D}}_{\Lambda ,\mu }$ such that $f^*(\overline{\zeta })=\left<f,k_{\overline{\zeta }}\right>_{\Lambda ,\mu }.$ Hence, $S^*k_{\overline{\zeta }}=\zeta k_{\overline{\zeta }}$. Indeed, we have

$$\begin{aligned} \left<f,S^*k_{\overline{\zeta }}\right>_{\Lambda ,\mu }=\left<Sf,k_{\overline{\zeta }}\right>_{\Lambda ,\mu }=\overline{\zeta } f^*(\overline{\zeta })=\left<f,\zeta k_\zeta \right>_{\Lambda ,\mu }. \end{aligned}$$

$\square$

3 Growth of Power of Shift Operator

Let $\mu$ be a positive finite Borel measure on $\mathbb {T}$, and let S be the shift operator acting on the Dirichlet space ${\mathcal {D}}_{\Lambda ,\mu }$ associated with $\mu$ and a positive function $\Lambda$.

Theorem 3.1

Let $\mu$ be a positive finite Borel measure on $\mathbb {T}$, and let $\Lambda : [0,1]\rightarrow [0,+\infty )$ be an increasing function such that $\frac{\Lambda (t) }{t^c}$ is decreasing for some $c>0$, and

$$\begin{aligned} \int _0\frac{dt}{\Lambda (t)}<\infty . \end{aligned}$$

(3)

We have

$$\begin{aligned} ||S^{*n}||^2\asymp n\Lambda \left( \frac{1}{n}\right) . \end{aligned}$$

To prove Theorem 3.1, we require the following lemmas.

Lemma 3.1

Let $\mu$ be a positive finite Borel measure on $\mathbb {T}$, and let $\Lambda : [0,1]\rightarrow [0,+\infty )$ be an increasing function such that $\frac{\Lambda (t) }{t^c}$ is decreasing for some $c>0$. We have

$$\begin{aligned} \Vert z^n\Vert _{\Lambda ,\mu }^{2}\asymp n\Lambda \left( \frac{1}{n}\right) ,\quad n\ge 1. \end{aligned}$$

Proof

Let $n\ge 1$. We have

$$\begin{aligned} {\mathcal {D}}_{\Lambda ,\mu }(z^{n})= & {} \int _{z \in \mathbb {D}} n^{2}\left( |z|^{2}\right) ^{n-1}\Lambda \left( 1-|z|^{2}\right) P[\mu ](z) d A(z) \nonumber \\= & {} n^{2} \int _{r=0}^{1}r^{2(n-1)}\Lambda \left( 1-r^{2}\right) \int _{\theta =0}^{2 \pi } \int _{t=0}^{2 \pi } \frac{1-r^{2}}{\left| e^{i t}-r e^{i \theta }\right| ^{2}} \frac{dt}{2\pi }d\mu \left( e^{i \theta }\right) dr^2 \nonumber \\= & {} \mu (\mathbb {T})n^{2} \int _{0}^{1} s^{n-1}\Lambda (1-s)ds\\\ge & {} \mu (\mathbb {T})\left( 1-\frac{2}{n}\right) ^{n-1}n^2 \int _{\frac{1}{n}}^{\frac{2}{n}}\Lambda (t)dt\nonumber \\\ge & {} \mu (\mathbb {T})\left( 1-\frac{2}{n}\right) ^{n-1}n\Lambda \left( \frac{1}{n}\right) .\nonumber \end{aligned}$$

(4)

Now,

$$\begin{aligned} \begin{aligned} \int _0^1r^{n-1} \Lambda (1-r)dr&\le \int _0^{\frac{1}{n}}\Lambda (t)dt+\int _{\frac{1}{n}}^1(1-t)^{n-1} \Lambda (t)dt\\&\le \frac{1}{n}\Lambda \left( \frac{1}{n}\right) +n^c\Lambda \left( \frac{1}{n}\right) \int _0^1 t^{n-1} (1-t)^c dt\\&= \left[ 1+n^{c+1}\textbf{B}(n,c+1)\right] n\Lambda \left( \frac{1}{n}\right) , \end{aligned} \end{aligned}$$

where $\textbf{B}$ is Beta function. Using equality (4), we obtain

$$\begin{aligned} {\mathcal {D}}_{\Lambda ,\mu }(z^n)\le \left[ 1+n^{c+1}\textbf{B}(n,c+1)\right] n\Lambda \left( \frac{1}{n}\right) . \end{aligned}$$

$\square$

In the case of $\mu =\delta _\zeta$, the Dirac measure at $\zeta \in \mathbb {T}$, the local weighted Dirichlet integral of $f\in \textrm{Hol}(\mathbb {D})$ is given by

$$\begin{aligned} {\mathcal {D}}_{\Lambda ,\zeta }(f)=\int _{\mathbb {D}}|f'(z)|^2\Lambda (1-|z|^2)\frac{1-|z|^2}{|\zeta -z|^2}dA(z), \end{aligned}$$

Suppose that $\Lambda \equiv 1$. Let $f\in \textrm{Hol}(\mathbb {D})$, say $f(z)=\sum _{n\in \mathbb {N}}a_nz^n$, we have

$$\begin{aligned} {\mathcal {D}}_\zeta (f):={\mathcal {D}}_{1,\zeta }(f)=\sum _{n\ge 1}\frac{1}{n(n+1)}\left| \sum _{k=1}^nka_k\zeta ^k\right| ^2, \end{aligned}$$

(5)

see [7, Theorem 7.2.6].

For the rest of the paper, we suppose that $\int _0\frac{dt}{\Lambda (t)}<\infty .$ Therefore, according to Lemma 2.1, the reproducing kernel of ${\mathcal {D}}_{\Lambda ,\zeta }$ satisfies $\sup _{0\le r<1}K_{\Lambda ,\zeta }(r\zeta ,r\zeta )<\infty$. Then the following space

$$\begin{aligned} {\mathcal {D}}^0(\Lambda ,\zeta )=\left\{ f\in {\mathcal {D}}_{\Lambda ,\zeta }:\,\,\, f^*(\zeta )=0\right\} , \end{aligned}$$

is closed in ${\mathcal {D}}_{\Lambda ,\zeta }.$ In order to extend formula (5) to ${\mathcal {D}}_{\Lambda ,\zeta },$ we endow the space ${\mathcal {D}}^0(\Lambda ,\zeta )$ with the following norm:

$$\begin{aligned} ||f||_{ 0, \Lambda , \zeta }^2:={\mathcal {D}}_{\Lambda ,\zeta }(f),\quad f\in \textrm{Hol}(\mathbb {D}). \end{aligned}$$

Additionally, we consider the weighted Bergman space $\mathcal {A}^2_{(1-|z|^2)\Lambda (1-|z|^2)}$ equipped with the following norm:

$$\begin{aligned} ||f||^2=\int _{\mathbb {D}}|f(z)|^2(1-|z|^2)\Lambda (1-|z|^2)dA(z),\quad f\in \textrm{Hol}(\mathbb {D}). \end{aligned}$$

Lemma 3.2

Let $\mu$ be a positive finite Borel measure on $\mathbb {T}$, and let $\Lambda : [0,1]\rightarrow [0,+\infty )$ be an increasing function such that $\frac{\Lambda (t) }{t^c}$ is decreasing for some $c>0$, and

$$\begin{aligned} \int _0\frac{dt}{\Lambda (t)}<\infty .\nonumber \\ \end{aligned}$$

For $f\in {\mathcal {D}}_{\Lambda ,\zeta }$ write $f(z)=\sum _{n\in \mathbb {N}}a_nz^n$. We have

$$\begin{aligned} {\mathcal {D}}_{\Lambda ,\zeta }(f)=\sum _{n\ge 1}||z^{n-1}||^2\left| \sum _{k=1}^nka_k\zeta ^k\right| ^2. \end{aligned}$$

Proof

Without loss of generality, we may assume that $\zeta =1$. We make the following identification

$$\begin{aligned} \textbf{T}:&{\mathcal {D}}^0(\Lambda ,1)\longrightarrow \mathcal {A}^2_{(1-|z|^2)\Lambda (1-|z|^2)}, \,\,\,f\mapsto \textbf{T}f(z)=f'(z)/(z-1). \end{aligned}$$

The operator $\textbf{T}$ is a surjective isometry. Indeed, let $f\in {\mathcal {D}}^0(\Lambda ,1),$ and denote $||f||_0:=||f||_{ 0, \Lambda , 1}$, we have

$$\begin{aligned} ||\textbf{T }f||^2= & {} \int _{\mathbb {D}}\left| \frac{f'(z)}{z-1}\right| ^2(1-|z|^2)\Lambda (1-|z|^2)dA(z)\\= & {} \int _{\mathbb {D}}|f'(z)|^2\frac{1-|z|^2}{|1-z|^2}\Lambda (1-|z|^2)dA(z)\\= & {} ||f||^2_0, \end{aligned}$$

then $\textbf{T}$ is an isometry. To prove that $\textbf{T}$ is surjective, we consider

$$\begin{aligned} \textbf{V}:&\mathbb {C}[z]\longrightarrow {\mathcal {D}}^0(\Lambda ,1), \,\,\,p\mapsto \textbf{V}p(z)=\int _z^1(\lambda -1)p(\lambda )d\lambda ,\nonumber \\ \end{aligned}$$

where $\mathbb {C}[z]$ is the set of polynomials. Let $p\in \mathbb {C}[z]$, we have

$$\begin{aligned} ||\textbf{V}p||_0^2= & {} \int _{\mathbb {D}}|\left( \textbf{V}p\right) '(z)|^2\frac{1-|z|^2}{|1-z|^2}\Lambda (1-|z|^2)dA(z)\nonumber \\= & {} \int _{\mathbb {D}}|p(z)|^2(1-|z|^2)\Lambda (1-|z|^2)dA(z)\nonumber \\= & {} ||p||^2, \end{aligned}$$

and

$$\begin{aligned} \textbf{T}\textbf{V}p(z)=\frac{\left( \textbf{V}p\right) '(z)}{z-1}=p(z),\quad z\in \mathbb {D}. \end{aligned}$$

Since $(1-|z|^2)\Lambda (1-|z|^2)$ is a radial weight, we get that polynomials are dense in $\mathcal {A}^2_{(1-|z|^2)\Lambda (1-|z|^2)}.$ It follows from equality (6) that $\textbf{V}$ extends to an isometry $\mathbf {\tilde{V}}$ on $\mathcal {A}^2_{(1-|z|^2)\Lambda (1-|z|^2)}$. Using equality (6), we obtain

$$\begin{aligned} \textbf{T}\mathbf {\tilde{V}}f=f, \quad f\in \mathcal {A}^2_{(1-|z|^2)\Lambda (1-|z|^2)}. \end{aligned}$$

Thus, $\textbf{T}$ is surjective. Moreover, since $(1-|z|^2)\Lambda (1-|z|^2)$ is a radial weight, we get that $\left( e_n(z):=\frac{z^n}{||z^n||}\right)$ is an orthonormal basis of $\mathcal {A}^2_{(1-|z|^2)\Lambda (1-|z|^2)}$. Hence, the following sequence

$$\begin{aligned} u_{n+1}(z){} & {} =\textbf{V}e_n(z)=\frac{1}{||z^n||}\left( \frac{1}{n+2}(z^{n+2}-1)-\frac{1}{n+1}(z^{n+1}-1)\right) ,\quad z\in \mathbb {D}\\ {}{} & {} \qquad \text {and}\,\, n\in \mathbb {N}, \end{aligned}$$

is an orthonormal basis of ${\mathcal {D}}^0(\Lambda ,1).$ Therefore, there exists a sequence $(c_n)_{n\ge 1}$ of complex numbers such that

$$\begin{aligned} f(z)=\sum _{n\ge 1} c_nu_n(z),\quad z\in \mathbb {D}. \end{aligned}$$

Thus,

$$\begin{aligned} {\mathcal {D}}_{\Lambda ,1}(f)=\sum _{n\ge 1}|c_n|^2. \end{aligned}$$

According to identification (6), we obtain

$$\begin{aligned} \left\{ \begin{array}{l} a_0=\sum _{n\ge 1}\frac{c_n}{||z^{n-1}||n(n+1)}, \\ \\ a_1=\frac{-c_1}{\Vert z^0\Vert }, \\ \\ a_n=\frac{1}{n}\left( \frac{c_{n-1}}{||z^{n-2}||}-\frac{c_{n}}{||z^{n-1}||}\right) ,\quad n\ge 2. \end{array}\right. \end{aligned}$$

Then

$$\begin{aligned} c_n=-||z^{n-1}||\sum _{k=1}^{n}ka_k,\,\,n\ge 1. \end{aligned}$$

$\square$

Lemma 3.3

Let $\mu$ be a positive finite Borel measure on $\mathbb {T}$, and let $\Lambda : [0,1]\rightarrow [0,+\infty )$ be an increasing function such that $\frac{\Lambda (t) }{t^c}$ is decreasing for some $c>0$, and

$$\begin{aligned} \int _0\frac{dt}{\Lambda (t)}<\infty . \end{aligned}$$

For $f\in {\mathcal {D}}_{\Lambda ,\zeta }$ write $f(z)=\sum _{n\in \mathbb {N}}a_nz^n$. Then

Proof

We have

$$\begin{aligned} z^pf(z)=\sum _{n\ge 0}a_{n-p}z^n \quad \text {with}\,\,a_{-1}=a_{-2}=...=a_{-p}=0,\,\, \text {and}\,z\in \mathbb {D}. \end{aligned}$$

By Theorem 3.2, we obtain

$$\begin{aligned} {\mathcal {D}}_{\Lambda ,\zeta }(z^pf)= & {} \sum _{n\ge 1}||z^{n-1}||^2\left| \sum _{k=1}^n ka_{k-p}\right| ^2\\ {}= & {} \sum _{n\ge p}||z^{n-1}||^2\left| \sum _{i=0}^{n-p}(i+p)a_i\right| ^2\\\le & {} 2\underbrace{\sum _{n\ge p}||z^{n-1}||^2\left| \sum _{k=0}^{n-p}k a_k\right| ^2}_{=I_1}+2\underbrace{\sum _{n\ge p}||z^{n-1}||^2\left| \sum _{k=0}^{n-p}pa_k\right| ^2}_{I_2}. \end{aligned}$$

Since $||z^{n-1}||\le ||z^{n-1-p}||$, we get

$$\begin{aligned} I_1\le {\mathcal {D}}_{\Lambda ,\zeta }(f). \end{aligned}$$

Let $A_n=\sum _{k=0}^nk a_k$, we have

$$\begin{aligned} a_n=\frac{1}{n}\left( A_n-A_{n-1}\right) ,\quad n\ge 1, \end{aligned}$$

and

$$\begin{aligned} \left| \sum _{k=0}^na_k\right| ^2= & {} \left| \sum _{k=1}^n\frac{1}{k}\left( A_k-A_{k-1}\right) +a_0\right| ^2\\= & {} \left| \sum _{k=1}^n\frac{1}{k}A_k-\sum _{k=0}^{n-1}\frac{1}{k+1}A_k+a_0\right| ^2\\= & {} \left| \sum _{k=1}^{n-1}\frac{1}{k(k+1)}A_k+\frac{1}{n}A_n+a_0\right| ^2\\\le & {} 4\left| \sum _{k=1}^{n-1}\frac{1}{k(k+1)}A_k\right| ^2+\frac{4}{n^2}|A_n|^2+2|a_0|^2. \end{aligned}$$

Then

$$\begin{aligned} I_2{} & {} \lesssim \underbrace{\sum _{n\ge p}p^2||z^{n-1}||^2\left| \sum _{k=1}^{n-p-1}\frac{1}{k(k+1)}A_k\right| ^2}_{Q_1}+\underbrace{\sum _{n\ge p}p^2\frac{||z^{n-1}||^2}{n^2}|A_{n-p}|^2}_{Q_2}\\{} & {} \qquad +\underbrace{|a_0|^2\sum _{n\ge p}p^2||z^{n-1}||^2}_{Q_3}. \end{aligned}$$

It follows from the inequality $|a_0|\le ||f||_{\Lambda ,\delta _{\zeta }}$ that

$$\begin{aligned} Q_3\le ||f||_{\Lambda ,\delta _{\zeta }}\times \sum _{n\ge p}p^2||z^{n-1}||^2. \end{aligned}$$

In addition,

$$\begin{aligned} Q_2= & {} \sum _{n\ge p}\frac{p^2}{n^2}||z^{n-1}||^2|A_{n-p}|^2\\\le & {} \sum _{n\ge p}||z^{n-1-p}||^2|A_{n-p}|^2\\= & {} {\mathcal {D}}_{\Lambda ,\zeta }(f). \end{aligned}$$

Moreover,

$$\begin{aligned} \left| \sum _{k=1}^{n}\frac{1}{k(k+1)}A_k\right| ^2\lesssim & {} \left( \sum _{k=1}^n\frac{1}{||z^{k-1}||k^2}\times ||z^{k-1}|||A_k|\right) ^2\\\le & {} \sum _{k=1}^n\frac{1}{k^4||z^{k-1}||^2}\times \sum _{k=1}^n||z^{k-1}||^2|A_k|^2\\\lesssim & {} \sum _{n\ge 1}\frac{1}{n^4||z^{n-1}||^2}\times {\mathcal {D}}_{\Lambda ,\zeta }(f). \end{aligned}$$

Then

$$\begin{aligned} Q_1= & {} \sum _{n\ge p}p^2||z^{n-1}||^2\left| \sum _{k=1}^{n-p}\frac{A_k}{k(k+1)}\right| ^2\\\lesssim & {} {\mathcal {D}}_{\Lambda ,\zeta }(f) \sum _{n\ge 1}\frac{1}{n^4||z^{n-1}||^2}\times \sum _{n\ge p}p^2||z^{n-1}||^2. \end{aligned}$$

$\square$

We are now in a position to prove Theorem 3.1.

Proof of Theorem 3.1

Let $p\in \mathbb {N}$. By definition, $\Vert S^{*p}\Vert =\sup _{\Vert f\Vert _{\Lambda ,\mu }=1}\Vert S^{*p}f\Vert _{\Lambda ,\mu }$. Then we get $\Vert S^{*p}\Vert ^2\ge \Vert z^p\Vert ^2_{\Lambda ,\mu }\asymp p\Lambda \left( \frac{1}{p}\right)$. Now, let $n\in \mathbb {N}$. Similar to the proof of Lemma 3.1, we have

$$\begin{aligned} \Vert z^{n-1}\Vert ^2=\int _0^1r^{n-1}(1-r)\Lambda (1-r)dr\asymp \frac{1}{n^2}\Lambda \left( \frac{1}{n}\right) . \end{aligned}$$

Then

$$\begin{aligned} \sum _{n}\frac{1}{n^4\Vert z^{n-1}\Vert ^2}&\asymp&\sum _n\frac{1}{n^2\Lambda \left( \frac{1}{n}\right) }\nonumber \\\le & {} 2 \sum _n\int _{1/(n+1)}^{1/n}\frac{1}{\Lambda (t)}dt\nonumber \\&\asymp&\int _0\frac{dt}{\Lambda (t)}<\infty . \end{aligned}$$

(6)

Furthermore,

$$\begin{aligned} \sum _{n\ge p}p^2\Vert z^{n-1}\Vert ^2&\asymp&\sum _{n\ge p}\frac{p^2}{n^2}\Lambda \left( \frac{1}{n}\right) \nonumber \\\le & {} p^2\Lambda \left( \frac{1}{p}\right) \sum _{n\ge p}\frac{1}{n^2}\nonumber \\&\asymp&p\Lambda \left( \frac{1}{p}\right) . \end{aligned}$$

(7)

Combining inequalities (6), (7), and Lemma 3.3, we obtain

$$\begin{aligned} \begin{aligned} \Vert S^{*p}\Vert ^2&=\sup _{\Vert f\Vert _{\Lambda ,\mu }=1}\left\{ |f(0)|^2+\int _\mathbb {T}{\mathcal {D}}_{\Lambda ,\zeta }(z^p f)d\mu (\zeta )\right\} \\&\lesssim \sup _{\Vert f\Vert _{\Lambda ,\mu }=1}\left\{ |f(0)|^2+ p\Lambda \left( \frac{1}{p}\right) {\mathcal {D}}_{\Lambda ,\mu }(f)\right\} \\&\lesssim p\Lambda \left( \frac{1}{p}\right) . \end{aligned} \end{aligned}$$

4 Proofs of the Main Theorems

We recall some definitions which will be used in what follows. Let $E\subset \mathbb {T}$, let $\delta >0$, and let $d\in [0,\infty )$. Consider

$$\begin{aligned} H^d_{\delta }(E)=\inf \left\{ \sum _{i=1}^\infty |I_i|^d:\,\, E\subset \bigcup _{i=1}^\infty I_i\,\, \text {and} \,\, |I_i|<\delta \right\} , \end{aligned}$$

where the infimum is taken over all countable intervals covering E. The Hausdorff outer measure of dimension d is given by $H^d(E)=\lim _{\delta \rightarrow 0}H^d_{\delta }(E)$, and the Hausdorff dimension of E is defined by

$$\begin{aligned} \dim _H(E)=\inf \{d\ge 0: \,\, H^d(E)=0\}. \end{aligned}$$

A closed subset E of $\mathbb {T}$ is an Ahlfors-David regular set if there exists a measure $\mu$ supported on E such that

$$\begin{aligned} C^{-1}t^d\le \mu (\zeta ,t)\le C t^d \end{aligned}$$

for all $\zeta \in {{\,\textrm{supp}\,}}\mu$ and $t\in [0,1]$. In this case, we have $\dim _H(E)=d$ (see [12] for more details).

Proof Theorem 1.2

Assume that there exist $\mu$ and $\Lambda$ satisfying the conditions of Theorem 1.2. Consider the operator $T=S^*$ on ${\mathcal {D}}_{\Lambda ,\mu }$. We have $\sigma _p(T)\supset E$. Indeed, for $\zeta \in E$, we have $\int _0\frac{dt}{\Lambda (t)\mu (\overline{\zeta },t)}<\infty ,$ it follows from Theorem 2.1 that $\zeta \in \sigma _p(T).$ The first assertion is proved. Since $\mu (\overline{\zeta },t)\Lambda (t)\lesssim \Lambda (t)$, condition (3) holds and the last assertion comes from Theorem 3.1.

We are now ready to prove Theorem 1.1.

Proof of Theorem 1.1

Since E is an Ahlfors–David set, $E^*:=\{\zeta \in \mathbb {T}\,: \overline{\zeta }\in E\}$ is too. Thus, there exists a measure $\mu$ supported on $E^*$ such that $\mu (\zeta ,t)\asymp t^d$ for any $\zeta \in E^*$, and $d=\dim _H(E)$. Let $\alpha >d$ and consider the function $\Lambda (t)=t^{1-\alpha }$, $t\in (0,1)$. We have

$$\begin{aligned} \int _0\frac{dt}{\Lambda (t)\mu (\overline{\zeta },t)}\asymp \int _0\frac{dt}{t^{1-(\alpha -d)}}<\infty ,\quad \zeta \in E. \end{aligned}$$

According to Theorem 1.2, we deduce the result.