1 Introduction and preliminaries

It is well known that the Schwarz lemma serves as a very powerful tool to study several research fields in complex analysis. For example, almost all results in the geometric function theory have the Schwarz lemma lurking in the background [2–6].

On the other hand, Schwarz lemma at the boundary is also an active topic in complex analysis, various interesting results have been obtained [7–14]. Before summarizing these results, it is necessary to give some elementary contents on the boundary fixed points [\(\mathbb {D}\), \(\mathbb{N}\) denote the set of all positive integers. The boundary point \(\xi\in\partial\mathbb{D}\) is called a fixed point of \(f\in H(\mathbb{D,\mathbb{D}})\) if

$$\begin{aligned} f(\xi)=\lim_{r\rightarrow1^{-}}f(r\xi)=\xi. \end{aligned}$$

The classification of the boundary fixed points of \(f\in H(\mathbb {D,\mathbb{D}})\) can be performed via the value of the angular derivative

$$\begin{aligned} f'(\xi)=\angle\lim_{z\rightarrow\xi}\frac{f(z)-\xi}{z-\xi}, \end{aligned}$$

which belongs to \((0,\infty]\) due to the celebrated Julia-Carathédory theorem [13]. This theorem also asserts that the finite angular derivative at the boundary fixed point ξ exists if and only if the holomorphic function \(f'(z)\) has the finite angular limit \(\angle\lim_{z\rightarrow\xi}f'(z)\). For a boundary fixed point ξ of f, if

$$f'(\xi)\in(0,\infty), $$

then ξ is called a regular boundary fixed point. The regular fixed points can be attractive if \(f'(\xi)\in(0,1)\), neutral if \(f'(\xi)=1\), or repulsive if \(f'(\xi)\in(1,\infty)\).

By the Julia-Carathédory theorem [13] (see also [7]) and the Wolff lemma [11], if \(f\in H(\mathbb{D,\mathbb{D}})\) with no interior fixed point, then there exists a unique regular boundary fixed point ξ such that \(f'(\xi)\in(0,1]\); and if \(f\in H(\mathbb {D,\mathbb{D}})\) with an interior fixed point, then \(f'(\xi)>1\) for any boundary fixed point \(\xi\in\partial\mathbb{D}\).

In particular, Unkelbach [16] (see also [17]) obtain the following boundary Schwarz lemma.

Theorem A

If \(f\in H(\mathbb{D}, \mathbb{D})\) has a regular boundary fixed point 1, and \(f(0)=0\), then

$$\begin{aligned} f'(1)\geq\frac{2}{1+\vert f'(0)\vert }. \end{aligned}$$
(1)

Moreover, equality in (1) holds if and only if f is of the form

$$f(z)=-z\frac{a-z}{1-az},\quad \forall z\in\mathbb{D}, $$

for some constant \(a\in(-1,0]\).

Theorem A was improved 60 years later by Osserman [18] by removing the assumption \(f(0)=0\).

Theorem B

([18])

If \(f\in H(\mathbb{D}, \mathbb{D})\) with \(\xi=1\) as its regular boundary fixed point. Then

$$\begin{aligned} f'(1)\geq\frac{2(1-\vert f(0)\vert )^{2}}{1-\vert f(0)\vert ^{2}+\vert f'(0)\vert }. \end{aligned}$$
(2)

In [1], Frolova et al. proved the following theorem, which is an improvement of Theorem B.

Theorem C

([1])

If \(f\in H(\mathbb{D}, \mathbb{D})\) with \(\xi=1\) as its regular boundary fixed point. Then

$$\begin{aligned} f'(1)\geq\frac{2}{\operatorname{\Re e}(\frac{1-f(0)^{2}+f'(0)}{(1-f(0))^{2}} )}. \end{aligned}$$
(3)

Recently, Ren and Wang [15] offered an alternative and elementary proof of Theorem C and studied the extremal functions of the inequality (3). Their method of proof is quite different from that which Frolova et al. have used in [1].

In this paper, stimulated by the above-cited work (especially [15]), considering the zero of order, we obtain a version of boundary Schwarz lemma. This result is a generalization of the boundary Schwarz-Pick lemma obtained by Frolova et al. [1].

In order to prove the desired results, we first recall the classical Julia lemma [3] and the Julia-Carathéodory theorem [19].

Lemma 1

([3])

Let \(f\in H(\mathbb{D,\mathbb{D}})\) and let \(\xi\in\partial\mathbb{D}\). Suppose that there exists a sequence \({\{z_{n}\}}_{n\in\mathbb {N}}\subset\mathbb{D}\) converging to ξ as n tends to ∞, such that the limits

$$\begin{aligned} \alpha=\lim_{n\rightarrow\infty}\frac{1-\vert f(z_{n})\vert }{1-\vert z_{n}\vert } \end{aligned}$$

and

$$\begin{aligned} \eta=\lim_{n\rightarrow\infty}f(z_{n}) \end{aligned}$$

exist (finitely). Then \(\alpha>0\) and the inequality

$$\begin{aligned} \frac{\vert f(z)-\eta \vert ^{2}}{1-\vert f(z)\vert ^{2}}\leq\alpha\frac{\vert z-\xi \vert ^{2}}{1-\vert z\vert ^{2}} \end{aligned}$$
(4)

holds throughout the open unit disk \(\mathbb{D}\) and is strict except for Möbius transformations of  \(\mathbb{D}\).

Lemma 2

([19])

Let \(f\in H(\mathbb{D,\mathbb{D}})\) and let \(\xi\in\partial\mathbb{D}\). Then the following conditions are equivalent:

  1. (i)

    The lower limit

    $$\begin{aligned} \alpha=\liminf_{z\rightarrow\xi}\frac{1-\vert f(z)\vert }{1- \vert z\vert } \end{aligned}$$
    (5)

    is finite, where the limit is taken as z approaches ξ unrestrictedly in \(\mathbb{D}\);

  2. (ii)

    f has a non-tangential limit, say \(f(\xi)\), at the point ξ, and the difference quotient

    $$\begin{aligned} \frac{f(z)-f(\xi)}{z-\xi} \end{aligned}$$

    has a non-tangential limit, say \(f(\xi)\), at the point ξ;

  3. (iii)

    the derivative \(f'\) has a non-tangential limit, say \(f'(\xi)\), at the point ξ. Moreover, under the above conditions we have:

    1. (a)

      α in (i);

    2. (b)

      the derivatives \(f'(\xi)\) in (ii) and (iii) are the same;

    3. (c)

      \(f'(\xi)=\alpha\overline{\xi}f(\xi)\);

    4. (d)

      the quotient \(\frac{1-\vert f(z)\vert }{1- \vert z\vert }\) has the non-tangential limit α, at the point ξ.

Lemma 3

([17], p.35)

Let \(\varphi\in H(\mathbb{D}, \mathbb{D})\), and \(\varphi(z)=\sum_{n=0}^{\infty}b_{n} z^{\nu}\). Then

$$\begin{aligned} \vert b_{n}\vert \leq1-\vert b_{0}\vert ^{2},\quad n\geq1. \end{aligned}$$

2 Main results and their proofs

We now state and prove each of our main results given by Theorems 1 and 2 below.

Theorem 1

Let \(f\in H(\mathbb{D},\mathbb{D})\) with \(\xi=1\) as its regular boundary fixed point and suppose \(f(0)=f'(0)=\cdots=f^{(k-1)}(0)=0, a_{k}=\frac{f^{(k)}(0)}{k!}\neq0, k\in\mathbb{N}\), we can obtain:

  1. (I)

    if \(0<\vert a_{k}\vert <1\), then

    $$\begin{aligned} f'(1) \geq k+ \frac{\vert 1- a_{k} \vert ^{2}}{1- \vert a_{k} \vert ^{2}}\frac{2}{1+ \operatorname{\Re e}\frac{1-\overline{a_{k}}}{1-a_{k}}\frac{a_{k+1}}{1- \vert a_{k}\vert ^{2}}}, \end{aligned}$$
    (6)

    where \(a_{k+1}=\frac{f^{(k+1)}(0)}{(k+1)!}\). Equality holds in the inequality if and only if f is of the form

    $$\begin{aligned} f(z)=z^{k}\frac{a_{k}-z\frac{z-a}{1-az}\frac{a_{k}-1}{1-\overline {a_{k}}}}{1-z\frac{z-a}{1-az}\frac{a_{k}-1}{1-\overline{a_{k}}}\overline{a_{k}} },\quad \forall z\in \mathbb{D}, \end{aligned}$$
    (7)

    for some constant \(a\in[-1,1)\).

  2. (II)

    If \(\vert a_{k}\vert =1\), then \(f(z)=z^{k}\).

Proof

In view of Lemma 3, we consider the following two cases.

Case I If \(0< \vert a_{k} \vert <1\), let

$$\begin{aligned} g(z)= \textstyle\begin{cases}\frac{1-\overline{a_{k}}}{a_{k}-1}\frac{a_{k}-\frac {f(z)}{z^{k}}}{1-\overline{a_{k}} \frac{f(z)}{z^{k}}},&{0< \vert z \vert < 1},\\ 0,&{z=0}. \end{cases}\displaystyle \end{aligned}$$

It is elementary to see that \(g\in H(\mathbb{D,D})\), and \(\xi=1\) is its regular boundary fixed point. A straightforward computation shows that

$$\begin{aligned} f'(1)=k+\frac{\vert 1- a_{k}\vert ^{2}}{1- \vert a_{k}\vert ^{2}}g'(1) \end{aligned}$$
(8)

and

$$\begin{aligned} g'(0)=\frac{1-\overline{a_{k}}}{1-a_{k}} \cdot\frac{a_{k+1}}{1- \vert a_{k}\vert ^{2}}, \end{aligned}$$
(9)

which is no larger than 1 in modulus. Applying Lemmas 1 and 2 to the holomorphic function \(h:\mathbb{D\rightarrow\overline{\mathbb{D}}}\) defined by

$$\begin{aligned} h(z)=\frac{g(z)}{z},\quad \forall z\in\mathbb{D}, \end{aligned}$$

we obtain

$$\begin{aligned} g'(1)=1+h'(1) \geq1+ \frac{\vert 1- g'(0)\vert ^{2}}{1- \vert g'(0)\vert ^{2}} = \frac{2(1-\operatorname{\Re e}g'(0))}{1- \vert g'(0)\vert ^{2}}. \end{aligned}$$
(10)

In particular,

$$\begin{aligned} g'(1) \geq\frac{2}{1+ \operatorname{\Re e}g'(0)}. \end{aligned}$$
(11)

By combining (8), (9), and (11), we get the estimate in (6).

Furthermore, this bound in (6) is sharp. Indeed, if equality holds in (6) for \(z\in\mathbb{D}\), then we must have equalities in the corresponding inequalities in (4) and (11). Thus, we can obtain

$$\begin{aligned} g(z)=z\frac{z-a}{1-\overline{a}z}\frac{1-\overline{a}}{1-a} \end{aligned}$$
(12)

for some constant \(a\in\overline{\mathbb{D}}\), and \(g'(0)\in(-1,1]\), which is possible only if \(a\in[-1,1)\).

Consequently, f must be of the form

$$\begin{aligned} f(z)=z^{k}\frac{a_{k}-z\frac{z-a}{1-az}\frac{a_{k}-1}{1-\overline {a_{k}}}}{1-z\frac{z-a}{1-az}\frac{a_{k}-1}{1-\overline{a_{k}}}\overline{a_{k}} },\quad \forall z\in \mathbb{D}, \end{aligned}$$
(13)

for some constant \(a\in[-1,1)\).

Case II If \(\vert a_{k}\vert =1\), set

$$\begin{aligned} g(z)= \textstyle\begin{cases}\frac{f(z)}{z^{k}},&{0< \vert z \vert < 1},\\ a_{k},&{z=0}. \end{cases}\displaystyle \end{aligned}$$

It is clear that \(g\in H(\mathbb{D},\mathbb{D}), \vert g(0)\vert =\vert a_{k}\vert =1 \). Thus by the principle of the maximum modulus, g is a constant function, and \(g(z)=a_{k}=g(1)=1\), and hence \(f(z) \equiv z^{k}\). This completes the proof. □

Taking into account the relation \(\vert \frac{1-\overline {a_{k}}}{1-a_{k}} \cdot\frac{a_{k+1}}{1- \vert a_{k}\vert ^{2}}\vert \leq1\) and using (6) in Theorem 1, we can readily deduce the following corollary (the proof is omitted here).

Corollary 1

Let \(f\in H(\mathbb{D},\mathbb{D})\) with \(\xi=1\) as its regular boundary fixed point and suppose \(f(0)=f'(0)=\cdots=f^{(k-1)}(0)=0, a_{k}=\frac {f^{(k)}(0)}{k!}\neq0, k\in\mathbb{N}\); we have the following.

If \(0< \vert a_{k} \vert <1\), then

$$\begin{aligned} f'(1) \geq k+ \frac{\vert 1- a_{k} \vert ^{2}}{1- \vert a_{k} \vert ^{2}}. \end{aligned}$$
(14)

In particular,

$$\begin{aligned} f'(1) \geq k-1+ \frac{2}{1+\operatorname{\Re e}a_{k}}. \end{aligned}$$
(15)

Remark 1

When \(n=1\), it follows from (15) that

$$\begin{aligned} f'(1) \geq\frac{2}{1+\operatorname{\Re e}a_{1}}=\frac{2}{1+\operatorname{\Re e}f'(0)}. \end{aligned}$$

Note that

$$\begin{aligned} \frac{2}{1+\operatorname{\Re e}f'(0)}\geq\frac{2}{1+\vert f'(0)\vert }. \end{aligned}$$

Therefore, Theorem 1 (or Corollary 1) generalizes and improves Theorem A.

Theorem 2

Let \(f\in H(\mathbb{D},\mathbb{D})\) with \(\xi=1\) as its regular boundary fixed point and suppose \(f'(0)=\cdots=f^{(k-1)}(0)=0, a_{k}=\frac {f^{(k)}(0)}{k!}\neq0, k\in\mathbb{N}\), we can obtain:

  1. (I)

    If \(0<\vert a_{k}\vert <1-\vert f(0)\vert ^{2}\), then

    $$\begin{aligned} f'(1)\geq(k-1)\frac{\vert 1-f(0)\vert ^{2}}{1-\vert f(0)\vert ^{2}}+\frac{2}{\operatorname{\Re e}(\frac{1-f^{2}(0)+a_{k}}{(1-f(0))^{2}} )}. \end{aligned}$$
    (16)

    Equality holds in the inequality if and only if f is of the form

    $$\begin{aligned} f(z)=\frac{f(0)-z^{k}\frac{a-z}{1-az}\frac{1-f(0)}{1-\overline {f(0)}}}{1-z^{k}\frac{a-z}{1-az}\frac{1-f(0)}{1-\overline{f(0)}}\overline{f(0)}}. \end{aligned}$$
  2. (II)

    If \(\vert a_{k}\vert =1-\vert f(0)\vert ^{2}\), then

    $$\begin{aligned} f(z)=\frac{\frac{1-f(0)}{1-\overline{f(0)}}z^{k}+f(0)}{1+\overline {f(0)}\frac{1-f(0)}{1-\overline{f(0)}}z^{k}}. \end{aligned}$$
    (17)

Proof

Set

$$\begin{aligned} g(z)=\frac{f(z)-f(0)}{1-\overline{f(0)}f(z)}\frac{1-\overline{f(0)}}{1-f(0)}. \end{aligned}$$

It is not difficult to verify that \(g\in H(\mathbb{D,D})\), and \(\xi=1\) is its regular boundary fixed point. Elementary computations yield

$$\begin{aligned} f'(1)=\frac{\vert 1-f(0)\vert ^{2}}{1-\vert f(0)\vert ^{2}}g'(1) \end{aligned}$$
(18)

and

$$\begin{aligned} \frac{g^{(k)}(0)}{k!}=\frac{a_{k}}{1-\vert f(0)\vert ^{2}}\frac {1-\overline{f(0)}}{1-f(0)}. \end{aligned}$$
(19)

On the other hand, let

$$\begin{aligned} h(z)= \textstyle\begin{cases}\frac{g(z)}{z^{k}},&{0< \vert z \vert < 1},\\ \frac {g^{(k)}(0)}{k!},&{z=0}, \end{cases}\displaystyle \end{aligned}$$
(20)

which is in \(H(\mathbb{D}, \mathbb{D})\). By Lemma 3, we obtain the following results:

(I) If \(0< \vert a_{k}\vert <1-\vert f(0)\vert ^{2}\), then it follows from (19) that \(\vert \frac{g^{(k)}(0)}{k!}\vert <1\). By using Lemmas 1 and 2, we have

$$\begin{aligned} g'(1)=k+h'(1)\geq k+\frac{\vert 1-\frac{g^{(k)}(0)}{k!}\vert ^{2}}{1-\vert \frac{g^{(k)}(0)}{k!}\vert ^{2}}=k-1+ \frac{2 (1-\operatorname{\Re e}\frac{g^{(k)}(0)}{k!} )}{1-\vert \frac {g^{(k)}(0)}{k!}\vert ^{2}}. \end{aligned}$$

In particular,

$$\begin{aligned} g'(1)\geq k-1+\frac{2}{1+\operatorname{\Re e}\frac{g^{(k)}(0)}{k!}}. \end{aligned}$$

From the above relation and (18), we deduce that

$$\begin{aligned} f'(1)&\geq\frac{\vert 1-f(0)\vert ^{2}}{1-\vert f(0)\vert ^{2}} \biggl( k-1+\frac{2}{1+\operatorname{\Re e}\frac{g^{(k)}(0)}{k!}} \biggr) \\ &=\frac{\vert 1-f(0)\vert ^{2}}{1-\vert f(0)\vert ^{2}} \biggl( k-1+\frac{2}{1+\operatorname{\Re e}(\frac{a_{k}}{1-\vert f(0)\vert ^{2}}\frac {1-\overline{f(0)}}{1-f(0)} )} \biggr) \\ &=(k-1)\frac{\vert 1-f(0)\vert ^{2}}{1-\vert f(0)\vert ^{2}}+\frac {2}{\operatorname{\Re e}(\frac{1-f^{2}(0)+a_{k}}{(1-f(0))^{2}} )}. \end{aligned}$$

Applying a similar argument to Theorem 1, we deduce that equality holds in inequality (16) if and only if f is of the form

$$\begin{aligned} f(z)=\frac{f(0)-z^{k}\frac{a-z}{1-az}\frac{1-f(0)}{1-\overline {f(0)}}}{1-z^{k}\frac{a-z}{1-az}\frac{1-f(0)}{1-\overline{f(0)}}\overline{f(0)}}. \end{aligned}$$

(II) If \(\vert a_{k}\vert =1-\vert f(0)\vert ^{2}\), then we find from (19) and (20) that \(\vert h(0)\vert =\vert \frac {g^{(k)}(0)}{k!}\vert =1\). By the principle of the maximum modulus, h is a constant function, and \(h(z)=g(1)=1\), and hence \(g(z) \equiv z^{k}\), which yields the assertion (17). This completes the proof. □

Remark 2

By setting \(k=1\) in (16) of Theorem 2, we get the following estimate:

$$\begin{aligned} f'(1)\geq\frac{2}{\operatorname{\Re e}(\frac{1-f^{2}(0)+a_{1}}{(1-f(0))^{2}} )}=\frac {2}{\operatorname{\Re e}(\frac{1-f(0)^{2}+f'(0)}{(1-f(0))^{2}} )}, \end{aligned}$$

which is Theorem C obtained by Frolova et al. [1]. Thus, Theorem 2 is a generalization of Theorem C.