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Research article

Volterra integral operator and essential norm on Dirichlet type spaces

  • Received: 22 April 2021 Accepted: 06 July 2021 Published: 07 July 2021
  • MSC : 30D45, 30D50

  • In this paper, we study the boundedness and essential norm of Volterra integral operator Vg and integral operator Sg on Dirichlet type spaces DK,α.

    Citation: Liu Yang, Ruishen Qian. Volterra integral operator and essential norm on Dirichlet type spaces[J]. AIMS Mathematics, 2021, 6(9): 10092-10104. doi: 10.3934/math.2021586

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  • In this paper, we study the boundedness and essential norm of Volterra integral operator Vg and integral operator Sg on Dirichlet type spaces DK,α.



    First, let us introduce some necessary notation. Let D be the unit disk in the complex plane C, H(D) be the class of functions analytic in D and H be the class of bounded analytic functions on D. The Bloch space B ([34]) is the class of all fH(D) for which

    fB:=|f(0)|+supzD(1|z|2)|f(z)|<.

    The little Bloch space B0, consists of all fH(D) such that

    lim|z|1(1|z|2)|f(z)|=0.

    The Hardy space Hp(D) (0<p<) ([8,10]) is the set of fH(D) with

    fpHp=sup0<r<112π2π0|f(reiθ)|pdθ<.

    Suppose that 0<p<, α>1 and dAα(z)=(1|z|2)αdA(z)=1π(1|z|2)αdxdy. The weighted Bergman space Apα(D) ([34]) is the set of fH(D) with

    fpApα=D|f(z)|pdAα(z)<.

    Let α0. The Dirichlet type space Dα is the set of fH(D) with

    f2Dα=|f(0)|2+f2A2α<.

    If α=0, it gives classic Dirichlet space D. When α=1, it is Hardy space H2. When α>1, it turns into weighted Bergman spaces A2α2. Thus, the interesting scope is α(0,1). For more information relating to Dα, we refer to [23,25,26].

    In this paper, we use the weighted function in [9,30]. We always suppose that K:[0,)[0,) is a right-continuous and nondecreasing function. The weighted function K also satisfies

    10φK(s)sds< (A)

    and

    1φK(s)s2ds<, (B)

    where

    φK(s)=sup0t1K(st)/K(t),0<s<.

    Let α0 and Dirichlet type space DK,α denotes the spaces of function fH(D) satisfying

    f2DK,α=|f(0)|2+D|f(z)|2(1|z|2)K(1|z|2)dAα(z)<.

    When α>0, if the weighted function K satisfies (A) and (B), we easily to see that DαDK,αA2α1. By [9], there exist a small c>0, such that C1t1cK(t)C2tc, where 0<t<1, C1>0 and C2>0. Thus, when α1, we easily to see that A2α2+cDK,αA2α1c. Moreover, using high order characterization, it is not hard to check that DK,α turns into a Bergman type space, when α1. Thus, the interesting scope is α[0,1). For more results of DK,α spaces, we refer to [3,4,5,11,15,19,20].

    Let I be an arc of D and |I| be the normalized Lebesgue arc length of I. The Carleson square based on I, denoted by S(I), is defined by

    S(I):={z=reiθD:1|I|r<1,eiθI}.

    Let μ be a positive Borel measure on D. For 0<s<, μ is called an s-Carleson measure if

    supIDμ(S(I))|I|s<.

    We say that a function fH2(D) belongs to Morrey type space H2K if

    f2H2K=|f(0)|2+supID1K(|I|)I|f(ζ)fI|2dζ2π<,

    where

    fI=1|I|If(ζ)dζ2π, ID.

    This space was introduced by H. Wulan and J. Zhou in [29]. When K(t)=t, it gives the BMOA space, the space of those analytic functions f in the Hardy space Hp whose boundary functions having bounded mean oscillation on D. In the case K(t)=tλ, 0<λ<1, the space H2K gives classical Morrey space L2,λ. Morrey spaces L2,λ were introduced by Morrey in [16]. From [29], we know that fH2K if and only if

    supaD1|a|2K(1|a|2)D|f(z)|2(1|φa(z)|2)dA(z)<,

    where φa(z)=az1¯az.

    Let α0 and we say that a function fH(D) belongs to Morrey type space H2K,α if

    f2H2K,α=|f(0)|2+supaD1|a|2K(1|a|2)D|f(z)|2(1|φa(z)|2)dAα(z)<.

    It is easy to verify H2K,α is a Banach space under the above norm.

    Let f,gH(D). The Volterra integral operator Vg and the integral operator Sg are defined by

    Vgf(z):=z0g(w)f(w)dw,Sgf(z):=z0g(w)f(w)dw,zD,

    respectively. For gH(D), the multiplication operator Mg is defined by Mgf(z)=f(z)g(z). It is easy to see that Mg is related with Sg and Vg by

    Mgf(z)=f(0)g(0)+Sgf(z)+Vgf(z).

    It it well known that Vg is bounded on the Hardy space Hp (Bergman space Ap) if and only if gBMOA (gB). Vg is bounded on BMOA if and only if gBMOAlog (see [24]). For more information relating to Volterra integral operator Vg, we refer to [1,2,7,12,13,14,17,21,22,28,31,33].

    In this note, we study Volterra integral operator Vg acting from DK,α to H2K,α, that is, we prove that Vg:DK,αH2K,α is bounded if and only if gB, when 0<α<1. Meanwhile, the boundedness of Sg and the essential norm of Vg and Sg from DK,α to H2K,α are also studied.

    In this paper, the symbol fg means that f. We say that f\lesssim g if there exists a constant C such that f\leq Cg .

    In this section, we are going to give some auxiliary results.

    Lemma 1. Let (A) and (B) hold for K . Suppose that \alpha > 0 and f\in {\mathcal{D}_{K, \alpha}} , then

    |f(z)|\lesssim\|f\|_{{\mathcal{D}_{K, \alpha}}}\sqrt{\frac{K(1-|z|^2)}{(1-|z|^2)^{1+\alpha}}}, \ \ \ z\in{\mathbb{D}}.

    Proof. The proof is similar to [33], thus we omit it here. The proof is completed.

    Lemma 2. Let (B) hold for K . Suppose that \alpha > 0 . Then

    f_{a}(z) = \frac{(1-|a|^2)\sqrt{K(1-|a|^2)}}{(1-\overline{a}z)^{\frac{3+\alpha}{2}}}\in {\mathcal{D}_{K, \alpha}}

    and

    F_{a}(z) = \frac{(1-|a|^2)\sqrt{K(1-|a|^2)}}{\overline{a}(1-\overline{a}z)^{\frac{3+\alpha}{2}}}\in {\mathcal{D}_{K, \alpha}},

    where z, \ a\in{\mathbb{D}}.

    Proof. Since (B) holds, then from [9], there is some c\in (0, 1) , such that

    \varphi_{K}(t)\lesssim t^{1-c}, \ t\geq1. (1)

    Combining with K which is nondecreasing and Lemma 3.10 of [34], we obtain

    \begin{equation} \nonumber \begin{split} &\int_{{\mathbb{D}}}|f_{a}'(z)|^2\frac{1-|z|^2}{K(1-|z|^2)}dA_{\alpha}(z) = (1-|a|^2)^2\int_{{\mathbb{D}}}\left(\frac{(1-|z|^2)K(1-|a|^2)}{|1-\overline{a}z|^{5+\alpha}K(1-|z|^2)}\right)dA_{\alpha}(z) \\\lesssim&(1-|a|^2)^2\int_{{\mathbb{D}}}\left(\frac{(1-|z|^2)K(1-|a|)}{|1-\overline{a}z|^{5+\alpha}K(1-|z|)}\right)dA_{\alpha}(z) \\\lesssim&(1-|a|^2)^2\int_{{\mathbb{D}}}\left(\frac{(1-|z|^2)K(|1-\overline{a}z|)}{|1-\overline{a}z|^{5+\alpha}K(1-|z|)}\right)dA_{\alpha}(z) \\\lesssim&(1-|a|^2)^2\int_{{\mathbb{D}}}\left(\frac{(1-|z|^2)}{|1-\overline{a}z|^{5+\alpha}}\right)\varphi_{K}\left(\frac{|1-\overline{a}z|}{1-|z|}\right)dA_{\alpha}(z) \\\lesssim&(1-|a|^2)^2\int_{{\mathbb{D}}}\left(\frac{(1-|z|^2)^{1+\alpha}(|1-\overline{a}z|)^{1-c}}{|1-\overline{a}z|^{5+\alpha}(1-|z|)^{1-c}}\right)dA(z)\lesssim1. \end{split} \end{equation}

    where the third inequality is deduced by 1-|a|\leq |1-\overline{a}z| and K is nondecreasing, the last second inequality is deduced by 1-|z|\leq |1-\overline{a}z| and (1) . Thus, f_a\in {\mathcal{D}_{K, \alpha}} . Similar proof can be applied to F_a , thus we omit here. The proof is completed.

    Lemma 3. ([34]) Suppose that \alpha > -1 and \mu is a non-negative measure on {\mathbb{D}} . Then \mu is a (2+\alpha) -Carleson measure if and only if the following inequality

    \int_{{\mathbb{D}}}|f(z)|^2d\mu\lesssim\|f\|_{A_{\alpha}^2}^2

    holds for all f\in A_{\alpha}^2 .

    Lemma 4. ([32]) Let p > 1 and f\in H({\mathbb{D}}) . Then f\in {\mathcal{B}} if and only if the measure d\mu_f = |f'(z)|^2(1-|z|^2)^pdA(z) is a p -Carleson measure.

    Lemma 5. ([6]) Suppose that 1 < p < \infty , \alpha > -1 , \beta\geq 0 with \beta < 2+\alpha . Let f\in H({\mathbb{D}}) and z, w\in{\mathbb{D}} . Then

    \int_{{\mathbb{D}}}|f(z)-f(0)|^p\frac{(1-|z|^2)^\alpha}{|1-\overline{w}z|^\beta}dA(z)\lesssim\int_{{\mathbb{D}}}|f'(z)|^p\frac{(1-|z|^2)^{p+\alpha}}{|1-\overline{w}z|^\beta}dA(z).

    Lemma 6. Let (A) and (B) hold for K . Suppose that 0 < \alpha < 1 . Then f\in{\mathcal{H}_{K, \alpha}^2} if and only if

    \sup\limits_{I\subseteq{\partial {\mathbb{D}}}}\frac{1}{K(|I|)}\int_{S(I)}|f'(z)|^2(1-|z|^2)^{1+\alpha}dA(z) < \infty. (2)

    Proof. The proof is similar to Lemma 2.1 of [18]. Thus we omit here. The proof is complete.

    Theorem 1. Let (A) and (B) hold for K . Suppose that g\in H({\mathbb{D}}) and 0 < \alpha < 1 . Then V_g is bounded from {\mathcal{D}_{K, \alpha}} to {\mathcal{H}_{K, \alpha}^2} if and only if g\in {\mathcal{B}} . Moreover, the operator norm satisfies \|V_g\|\approx\|g\|_{{\mathcal{B}}} .

    Proof. For any I\in{\partial {\mathbb{D}}} , let a = (1-|I|)\zeta\in{\mathbb{D}} , where \zeta is the center of I . Then

    (1-|a|^2)\approx|1-\overline{a}z|, \ \ |K(1-|a|^2)\approx K(|I|), \ \ z\in S(I). (3)

    Let f_a be defined as in Lemma 2. Then

    |f_a(z)|^{2}\approx \frac{K(|I|)}{|I|^{1+\alpha}}, \ z\in S(I).

    Suppose that V_g is bounded from {\mathcal{D}_{K, \alpha}} to {\mathcal{H}_{K, \alpha}^2} . By Lemmas 4 and 6, we have

    \begin{equation} \nonumber \begin{split} &\frac{1}{|I|^{\alpha+1}}\int_{S(I)}|g'(z)|^2(1-|z|^2)^{\alpha+1}dA(z) \\\lesssim&\frac{1}{K(|I|)}\int_{S(I)}|f_a(z)|^2|g'(z)|^2(1-|z|^2)^{\alpha+1}dA(z) \\\lesssim&\frac{1}{K(|I|)}\int_{S(I)}|(V_gf_a)'(z)|^2(1-|z|^2)^{\alpha+1}dA(z) \\\lesssim&\|V_gf_a\|_{{\mathcal{H}_{K, \alpha}^2}}^2 < \infty. \end{split} \end{equation}

    Thus, g\in {\mathcal{B}} .

    On the other hand, suppose that g\in {\mathcal{B}} , by Lemma 4, we have d\mu_g = |g'(z)|^2(1-|z|^2)^{\alpha+1}dA(z) is a (\alpha+1) -Carleson measure. Let f\in {\mathcal{D}_{K, \alpha}} . From Lemma 6, we only need to prove that

    L = :\frac{1}{K(|I|)}\int_{S(I)}|(V_gf)'(z)|^{2}(1-|z|^2)^{1+\alpha}dA(z) < \infty.

    Since

    \begin{align*} L& = \frac{1}{K(|I|)}\int_{S(I)}|f(z)|^2|g'(z)|^{2}(1-|z|^2)^{1+\alpha}dA(z)\\ &\lesssim \frac{1}{K(|I|)}\int_{S(I)}|f(a)|^{2}|g'(z)|^{2}(1-|z|^2)^{1+\alpha}dA(z)\\ & \ \ \ \ \ \ \ +\frac{1}{K(|I|)}\int_{S(I)}|f(z)-f(a)|^{2}|g'(z)|^{2}(1-|z|^2)^{1+\alpha}dA(z)\\ & = M+N. \end{align*}

    Using Lemma 1 and (3) , we see that

    \begin{align*} M&\lesssim \|g\|_{{\mathcal{B}}}^2\|f\|^{2}_{{\mathcal{D}_{K, \alpha}}}. \end{align*}

    By Lemma 3, we have A_{\alpha-1}^2\subseteq L^2(d\mu_g) . Note that

    \|f\|_{A_{\alpha-1}^2}^2\approx\int_{{\mathbb{D}}}|f'(z)|^2(1-|z|^2)^{\alpha+1}dA(z)\leq\|f\|_{{\mathcal{D}_{K, \alpha}}}^2.

    Thus, {\mathcal{D}_{K, \alpha}}\subseteq A_{\alpha-1}^2 . Bearing in mind these facts, we are going to estimate N . Let z = \varphi_a(w) . Since |\varphi_a'(w)|(1-|w|^2) = 1-|\varphi_a(w)|^2, using Lemmas 3, 4, 5, we obtain

    \begin{align*} &N\approx \frac{(1-|a|^2)^{4}}{K(1-|a|^2)}\int_{S(I)}\left|\frac{f(z)-f(a)}{(1-\overline{a}z)^{2}}\right|^2d\mu_g(z)\\ &\leq \frac{(1-|a|^2)^{4}}{K(1-|a|^2)}\int_{{\mathbb{D}}}\left|\frac{f(z)-f(a)}{(1-\overline{a}z)^{2}}\right|^2d\mu_g(z)\\ &\lesssim\frac{(1-|a|^2)^{2+2}}{K(1-|a|^2)}\int_{{\mathbb{D}}}\left|\frac{f(z)-f(a)}{(1-\overline{a}z)^{2}}\right|^2(1-|z|^2)^{\alpha-1}dA(z)\\ &\leq\frac{(1-|a|^2)^{2}}{K(1-|a|^2)}\int_{{\mathbb{D}}}\frac{|f(z)-f(a)|^2(1-|a|^2)^{2}}{|1-\overline{a}z|^4}(1-|z|^2)^{\alpha-1}dA(z)\\ &\lesssim\frac{(1-|a|^2)^{2}}{K(1-|a|^2)}\int_{{\mathbb{D}}}|(f\circ\varphi_a)(w)-(f\circ\varphi_a)(0)|^2(1-|\varphi_a(w)|^2)^{\alpha-1}dA(w)\\ &\leq\frac{(1-|a|^2)^{1+\alpha}}{K(1-|a|^2)}\int_{{\mathbb{D}}}|(f\circ\varphi_a)(w)-(f\circ\varphi_a)(0)|^2(1-|w|^2)^{\alpha-1}dA(w)\\ &\leq\frac{(1-|a|^2)^{1+\alpha}}{K(1-|a|^2)}\int_{{\mathbb{D}}}|(f\circ\varphi_a)'(w)|^2(1-|w|^2)^{\alpha+1}dA(w)\\ &\leq\frac{(1-|a|^2)^{1+\alpha}}{K(1-|a|^2)}\int_{{\mathbb{D}}}|f'(\varphi_a(w)|^2(1-|\varphi_a(w)|^2)^2(1-|w|^2)^{\alpha-1}dA(w)\\ &\leq\frac{(1-|a|^2)^{1+\alpha}}{K(1-|a|^2)}\int_{{\mathbb{D}}}|f'(z)|^2(1-|z|^2)^2(1-|\varphi_a(z)|^2)^{\alpha-1}\frac{(1-|a|^2)^2}{|1-\overline{a}z|^4}dA(w)\\ & = \int_{{\mathbb{D}}}|f'(z)|^2\frac{(1-|z|^2)^{\alpha+1}}{K(1-|z|^2)}\frac{K(1-|z|^2)}{K(1-|a|^2)}\frac{(1-|a|^2)^{2(1+\alpha)}}{|1-\overline{a}z|^{2+2\alpha}}dA(w)\\ &\lesssim\int_{{\mathbb{D}}}|f'(z)|^2\frac{(1-|z|^2)^{\alpha+1}}{K(1-|z|^2)}\left(\frac{K(|1-\overline{a}z|)}{K(1-|a|)}\right)\frac{(1-|a|^2)^{2(1+\alpha)}}{|1-\overline{a}z|^{2+2\alpha}}dA(z)\\ &\lesssim\int_{{\mathbb{D}}}|f'(z)|^2\frac{(1-|z|^2)^{\alpha+1}}{K(1-|z|^2)}\varphi_K\left(\frac{|1-\overline{a}z|}{1-|a|}\right)\frac{(1-|a|^2)^{2(1+\alpha)}}{|1-\overline{a}z|^{2+2\alpha}}dA(z)\\ &\lesssim\int_{{\mathbb{D}}}|f'(z)|^2\frac{(1-|z|^2)^{\alpha+1}}{K(1-|z|^2)}\frac{(|1-\overline{a}z|)^{1-c}}{(1-|a|^2)^{1-c}}\frac{(1-|a|^2)^{2(1+\alpha)}}{|1-\overline{a}z|^{2+2\alpha}}dA(z)\lesssim\|f\|_{{\mathcal{D}_{K, \alpha}}}^2, \\ \end{align*}

    where the last second inequality is deduced by (1) . Combining the estimates M and N, we conclude that V_g:{\mathcal{D}_{K, \alpha}}\to {\mathcal{H}_{K, \alpha}^2} is bounded.

    Theorem 2. Let (A) and (B) hold for K . Suppose that g\in H({\mathbb{D}}) and 0 < \alpha < 1 . Then S_g is bounded from {\mathcal{D}_{K, \alpha}} to {\mathcal{H}_{K, \alpha}^2} if and only if g\in H^{\infty} . Moreover, the operator norm satisfies \|S_g\|\approx\sup_{z\in{\mathbb{D}}}|g(z)| .

    Proof. Suppose that S_g is bounded from {\mathcal{D}_{K, \alpha}} to {\mathcal{H}_{K, \alpha}^2} . Let a\in{\mathbb{D}} and

    F_a(z) = \frac{(1-|a|^2)\sqrt{K(1-|a|^2)}}{\overline{a}(1-\overline{a}z)^{\frac{3+\alpha}{2}}}.

    By Lemma 2, we have F_a\in{\mathcal{D}_{K, \alpha}} and \|F_a\|_{{\mathcal{D}_{K, \alpha}}}\lesssim 1 . For a\in {\mathbb{D}} and r > 0 , let D(a, r) denote the Bergman metric disk centered at a with radius r . From [34] we see that

    \frac{(1-|a|^2)^2}{|1-\overline{a} z|^4}\approx \frac{1}{(1-|z|^2)^2} \approx \frac{1}{(1-|a|^2)^2}

    when z\in D(a, r) . Using subharmonic property of |g|^2 , we have

    \begin{equation} \nonumber \begin{split} \infty > &\|S_gF_a\|_{{\mathcal{H}_{K, \alpha}^2}}^2 \\\gtrsim&\sup\limits_{b\in{\mathbb{D}}}\frac{1-|b|^2}{K(1-|b|^2)}\int_{{\mathbb{D}}}|F_{a}'(z)|^2|g(z)|^2\left(1-|\varphi_b(z)|^2\right)dA_{\alpha}(z) \\\gtrsim&\frac{1-|a|^2}{K(1-|a|^2)}\int_{{\mathbb{D}}}|F_{a}'(z)|^2|g(z)|^2\left(1-|\varphi_a(z)|^2\right)dA_{\alpha}(z) \\\gtrsim&\frac{1-|a|^2}{K(1-|a|^2)}\int_{D(a, r)}|F_{a}'(z)|^2|g(z)|^2\left(1-|\varphi_a(z)|^2\right)dA_{\alpha}(z) \\ = &\frac{1}{(1-|a|^2)^2}\int_{D(a, r)}|g(z)|^2dA(z)\gtrsim |g(a)|^2. \end{split} \end{equation}

    That is,

    \|S_g\|^2\gtrsim\|S_g\|^2\|F_a\|_{{\mathcal{D}_{K, \alpha}}}^2\gtrsim\|S_gF_a\|_{{\mathcal{H}_{K, \alpha}^2}}^2\gtrsim |g(a)|^2.

    Since a\in{\mathbb{D}} is arbitrary, we have

    \|g\|_{H^{\infty}}^2\lesssim\|S_g\|^2 < \infty.

    On the other hand. Let g\in H^{\infty} . Using (1), we can deduce that for f\in{\mathcal{D}_{K, \alpha}} ,

    \begin{equation} \nonumber \begin{split} &\frac{1-|a|^2}{K(1-|a|^2)}\int_{{\mathbb{D}}}|f'(z)|^2|g(z)|^2\left(1-|\varphi_a(z)|^2\right)dA_{\alpha}(z) \\\lesssim&\|g\|_{H^{\infty}}^2\int_{{\mathbb{D}}}|f'(z)|^2\frac{(1-|a|^2)^2K(1-|z|^2)}{|1-\overline{a}z|^2K(1-|a|^2)}\frac{(1-|z|^2)}{K(1-|z|^2)}dA_{\alpha}(z) \\\lesssim&\|g\|_{H^{\infty}}^2\int_{{\mathbb{D}}}|f'(z)|^2\frac{(1-|a|^2)^2K(|1-\overline{a}z|)}{|1-\overline{a}z|^2K(1-|a|)}\frac{(1-|z|^2)}{K(1-|z|^2)}dA_{\alpha}(z) \\\lesssim&\|g\|_{H^{\infty}}^2\int_{{\mathbb{D}}}|f'(z)|^2\frac{(1-|a|^2)^2\varphi_K\left(\frac{|1-\overline{a}z|}{1-|a|}\right)}{|1-\overline{a}z|^2}\frac{(1-|z|^2)}{K(1-|z|^2)}dA_{\alpha}(z) \\\lesssim&\|g\|_{H^{\infty}}^2\int_{{\mathbb{D}}}|f'(z)|^2\frac{(1-|a|^2)^2(|1-\overline{a}z|)^{1-c}}{|1-\overline{a}z|^2(1-|a|^2)^{1-c}}\frac{(1-|z|^2)}{K(1-|z|^2)}dA_{\alpha}(z) \\\lesssim&\|g\|_{H^{\infty}}^2\|f\|_{{\mathcal{D}_{K, \alpha}}}^2. \end{split} \end{equation}

    The proof is completed.

    Remark. Note that

    M_gf(z) = f(0)g(0)+S_gf(z)+V_gf(z).

    Hence, if (A) and (B) hold for K , then M_g is bounded from {\mathcal{D}_{K, \alpha}} to {\mathcal{H}_{K, \alpha}^2} if and only if g\in H^{\infty} .

    Let us recall the definition of essential norm. Suppose that X be a Banach space and T is a bounded linear operator on X . The essential norm of T is the distance of T to the closed ideals of compact operators, that is

    \|T\|_{e} = inf\{\|T-S\|:S \ is\ a \ compact \ operator\ on \ X\}.

    Note that T is compact if and only if \|T\|_{e} = 0 .

    Lemma 7. Suppose that 0 < \alpha < 1 and K satisfies the conditions (A) and (B). Let g\in {\mathcal{B}} . Then V_{g_r}:{\mathcal{D}_{K, \alpha}}\rightarrow {\mathcal{H}_{K, \alpha}^2} is compact. Here g_r(z) = g(rz), 0 < r < 1, z\in{\mathbb{D}}.

    Proof. Let \{f_n\} be any function sequence such that \|f_n\|_{{\mathcal{D}_{K, \alpha}}}\lesssim1 and f_n\rightarrow0 uniformly on compact subsets of {\mathbb{D}} as n\rightarrow \infty . We need only to show that

    \lim\limits_{n\rightarrow \infty}\|T_{g_r}f_n\|_{{\mathcal{H}_{K, \alpha}^2}} = 0.

    Since

    |g'_r(z)|\lesssim\frac{\|g\|_{{\mathcal{B}}}}{1-r^2}, \ z\in{\mathbb{D}}.

    Combining with (1), we have

    \begin{equation} \nonumber \begin{split} &\sup\limits_{a\in{\mathbb{D}}}\frac{1-|a|^2}{K(1-|a|^2)}\int_{{\mathbb{D}}}|f_{n}(z)|^2|g_r'(z)|^2\left(1-|\varphi_a(z)|^2\right)dA_{\alpha}(z) \\\lesssim&\frac{\|g\|_{{\mathcal{B}}}^2}{(1-r^2)^2}\sup\limits_{a\in{\mathbb{D}}}\frac{1-|a|^2}{K(1-|a|^2)}\int_{{\mathbb{D}}}|f_n(z)|^2\left(1-|\varphi_{a}(z)|^2\right)dA_{\alpha}(z) \\\lesssim&\frac{\|g\|_{{\mathcal{B}}}^2}{(1-r^2)^2}\int_{{\mathbb{D}}}|f_n(z)|^2\frac{1-|z|^2}{K(1-|z|^2)}\left(\frac{(1-|a|^2)^2K(1-|z|)}{K(1-|a|)|1-\overline{a}z|^2}\right)dA_{\alpha}(z) \\\lesssim&\frac{\|g\|_{{\mathcal{B}}}^2}{(1-r^2)^2}\int_{{\mathbb{D}}}|f_n(z)|^2\frac{1-|z|^2}{K(1-|z|^2)}\left(\frac{(1-|a|)^2\frac{K(|1-\overline{a}z|)}{K(1-|a|)}}{|1-\overline{a}z|^2}\right)dA_{\alpha}(z) \\\lesssim&\frac{\|g\|_{{\mathcal{B}}}^2}{(1-r^2)^2}\int_{{\mathbb{D}}}|f_n(z)|^2\frac{1-|z|^2}{K(1-|z|^2)}\left(\frac{(1-|a|)^2\varphi_K\left(\frac{|1-\overline{a}z|}{1-|a|}\right)}{|1-\overline{a}z|^2}\right)dA_{\alpha}(z) \\\lesssim&\frac{\|g\|_{{\mathcal{B}}}^2}{(1-r^2)^2}\int_{{\mathbb{D}}}|f_n(z)|^2\frac{1-|z|^2}{K(1-|z|^2)}\left(\frac{(1-|a|)^2\frac{|1-\overline{a}z|^{1-c}}{(1-|a|)^{1-c}}}{|1-\overline{a}z|^2}\right)dA_{\alpha}(z) \\\lesssim&\frac{\|g\|_{{\mathcal{B}}}^2}{(1-r^2)^2}\int_{{\mathbb{D}}}|f_n(z)|^2\frac{1-|z|^2}{K(1-|z|^2)}\left(\frac{(1-|a|)^{1+c}}{|1-\overline{a}z|^{1+c}}\right)dA_{\alpha}(z) \\\lesssim&\frac{\|g\|_{{\mathcal{B}}}^2}{(1-r^2)^2}\int_{{\mathbb{D}}}|f_n(z)|^2\frac{1-|z|^2}{K(1-|z|^2)}dA_{\alpha}(z). \end{split} \end{equation}

    Note that \|f_n\|_{{\mathcal{D}_{K, \alpha}}}\lesssim1 and Lemma 1, the argument is then finished by the Dominated Convergence Theorem.

    Let X and Y be two Banach spaces with X\subset Y . If f\in Y , then the distance from f to X is defined as

    dist_Y(f, X) = \inf\limits_{g\in X}\|f-g\|_Y.

    We also need the following lemma.

    Lemma 8. ([27]) If f\in{\mathcal{B}}, then

    \begin{align*} \limsup\limits_{|z|\to1^-}(1-|z|^2)|f'(z)|\approx {\rm dist}_{{\mathcal{B}}}(f, {\mathcal{B}}_0)\approx\limsup\limits_{r\to 1^-}\|f-f_r\|_{{\mathcal{B}}}. \end{align*}

    Theorem 3. Suppose 0 < \alpha < 1 , g\in{\mathcal{B}} and K satisfy the conditions (A) and (B). Then V_g : {\mathcal{D}_{K, \alpha}}\rightarrow {\mathcal{H}_{K, \alpha}^2} satisfies

    \|V_g\|_{e} \approx dist(g, {\mathcal{B}}_0)\approx\limsup\limits_{|z|\to1^-}(1-|z|^2)|g'(z)|.

    Proof. Let \{I_n\} be the subarc sequence of {\partial {\mathbb{D}}} , such that |I_n|\rightarrow0 as n\rightarrow \infty , w_n = (1-|I_n|)\zeta_n\in{\mathbb{D}} , where \zeta_n is the center of I_n . n = 1, 2, .... Then

    1-|w_n|\approx |1-\overline{w_n}z|\approx |I_n|, \ \ z\in S(I_n).

    Thus, by [9], we know that

    K(1-|w_n|)\approx K(|I_n|), \ \ z\in S(I_n).

    Take

    f_n(z) = \frac{(1-|a_n|^2)\sqrt{K(1-|a_n|^2)}}{(1-\overline{a_n}z)^{\frac{3+\alpha}{2}}}.

    Then f_{n}\rightarrow0 uniformly on the compact subsets of {\mathbb{D}} as n\rightarrow \infty and \|f_{n}\|_{{\mathcal{D}_{K, \alpha}}}\lesssim1 . Thus, for any compact operator S from {\mathcal{D}_{K, \alpha}} to {\mathcal{H}_{K, \alpha}^2} , we have

    \lim\limits_{n\rightarrow \infty}\|Sf_{n}\|_{{\mathcal{H}_{K, \alpha}^2}}\rightarrow0.

    Therefore

    \begin{equation} \nonumber \begin{split} \|V_g-S\|\gtrsim&\lim\limits_{n\rightarrow \infty}\sup \left(\|V_gf_{n}\|_{{\mathcal{H}_{K, \alpha}^2}}-\|Sf_{n}\|_{{\mathcal{H}_{K, \alpha}^2}}\right) \\ = &\limsup\limits_{n\rightarrow \infty}\|V_gf_{n}\|_{{\mathcal{H}_{K, \alpha}^2}} \\\approx&\limsup\limits_{n\rightarrow \infty}\left(\frac{1}{K(|I_n|)}\int_{S(I_n)}|(V_gf_n)'(z)|^2(1-|z|^2)^{1+\alpha}dA(z)\right)^{\frac{1}{2}} \\ = &\limsup\limits_{n\rightarrow \infty}\left(\frac{1}{K(|I_n|)}\int_{S(I_n)}|f_n(z)|^2|g'(z)|^2(1-|z|^2)^{1+\alpha}dA(z)\right)^{\frac{1}{2}} \\\approx&\limsup\limits_{n\rightarrow \infty}\left(\frac{1}{|I_n|^{1+\alpha}}\int_{S(I_n)}|g'(z)|^2(1-|z|^2)^{1+\alpha}dA(z)\right)^{\frac{1}{2}} \\\gtrsim&\limsup\limits_{n\rightarrow \infty}(1-|w_n|^2)|g'(w_n)|. \end{split} \end{equation}

    On the other hand, by Lemma 7, V_{g_r}: {\mathcal{D}_{K, \alpha}}\rightarrow {\mathcal{H}_{K, \alpha}^2} is compact operator. Combining this with Theorem 1 and the linearity of V_g respect to g implies

    \|V_g\|_{e}\leq\|V_g-V_{g_r}\| = \|V_{g-g_r}\|\approx\|g-g_r\|_{{\mathcal{B}}}.

    Together with Lemma 8, we have

    \|V_g\|_{e}\lesssim \limsup\limits_{|z|\to1^-}(1-|z|^2)|g'(z)|\approx dist(g, {\mathcal{B}}_0).

    The proof is completed.

    Corollary 1. Suppose 0 < \alpha < 1 and K satisfies the conditions (A) and (B). If g\in H({\mathbb{D}}), then V_g:{\mathcal{D}_{K, \alpha}}\to{\mathcal{H}_{K, \alpha}^2} is compact if and only if g\in{\mathcal{B}}_0.

    Theorem 4. Suppose 0 < \alpha < 1 and K satisfies the conditions (A) and (B). If g\in H({\mathbb{D}}) and S_g is bounded from {\mathcal{D}_{K, \alpha}} to {\mathcal{H}_{K, \alpha}^2} , then

    \|S_g\|_{e}\approx\sup\limits_{z\in{\mathbb{D}}}|g(z)|.

    Proof. For compact operators S , it follows that

    \|S_g\|_{e} = \inf\limits_{S}\|S_g-S\|\leq\|S_g\|\lesssim\sup\limits_{z\in{\mathbb{D}}}|g(z)|.

    On the other hand, we choose the sequence \{a_n\}\subset {\mathbb{D}} such that |a_n|\rightarrow 1 . We define

    f_{n}(z): = \frac{(1-|a_n|^2)\sqrt{K(1-|a_n|^2)}}{(1-\overline{a_n}z)^{\frac{3+\alpha}{2}}}, \; \; z\in{\mathbb{D}}.

    It follows from the proof of Lemma 2 that \|f_{n}\|_{{\mathcal{D}_{K, \alpha}}}\lesssim1 . It is easy to check that f_{n} converges to zero uniformly on any compact subsets of {\mathbb{D}} . Then \|Sf_{n}\|_{{\mathcal{H}_{K, \alpha}^2}}\rightarrow0 as n\rightarrow \infty for any compact operator S from {\mathcal{D}_{K, \alpha}} to {\mathcal{H}_{K, \alpha}^2} . So

    \begin{equation} \nonumber \begin{split} \|S_g-S\|\gtrsim&\limsup\limits_{n\rightarrow \infty}\|(S_g-S)f_{n}\|_{{\mathcal{H}_{K, \alpha}^2}} \\\geq&\limsup\limits_{n\rightarrow \infty}(\|S_gf_{n}\|_{{\mathcal{H}_{K, \alpha}^2}}-\|Sf_{n}\|_{{\mathcal{H}_{K, \alpha}^2}}) \\ = &\limsup\limits_{n\rightarrow \infty}\|S_gf_{n}\|_{{\mathcal{H}_{K, \alpha}^2}}. \end{split} \end{equation}

    From the proof of Theorem 2, we have

    \begin{equation} \nonumber \begin{split} \|S_gf_{n}\|_{{\mathcal{H}_{K, \alpha}^2}}\gtrsim&|g(a_n)|. \end{split} \end{equation}

    Since \{a_n\}\subseteq{\mathbb{D}} is arbitrary, we have

    \|S_g\|_{e} \gtrsim\sup\limits_{z\in {\mathbb{D}}} |g(z)|.

    The proof is completed.

    Corollary 2. Suppose 0 < \alpha < 1 and K satisfy the conditions (A) and (B). If g\in H({\mathbb{D}}), then S_g:{\mathcal{D}_{K, \alpha}}\to{\mathcal{H}_{K, \alpha}^2} is compact if and only if g = 0.

    In this paper, we give some equivalent characterizations of Volterra integral operator and essential norm from Dirichlet type spaces {\mathcal{D}_{K, \alpha}} to Morrey type spaces {\mathcal{H}_{K, \alpha}^2} .

    The authors thank the referee for useful remarks and comments that led to the improvement of this paper. This work was supported by NNSF of China (No. 11801250, No.11871257), Overseas Scholarship Program for Elite Young and Middle-aged Teachers of Lingnan Normal University, Yanling Youqing Program of Lingnan Normal University (No. YL20200202), the Key Program of Lingnan Normal University (No. LZ1905), The Innovation and developing School Project of Guangdong Province (No. 2019KZDXM032) and Education Department of Shaanxi Provincial Government (No.19JK0213).

    We declare that we have no conflict of interest.



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