---
abstract: |
    We show that if the Kobayashi–Royden metric of a complex manifold is continuous and positive at a given point and any non-zero tangent vector, then the "derivatives" of the higher order Lempert functions exist and equal the respective Kobayashi metrics at the point. It is a generalization of a result by M. Kobaya-\
    shi for taut manifolds.
address:
- |
    Institute of Mathematics and Informatics\
    Bulgarian Academy of Sciences\
    Acad. G. Bonchev 8, 1113 Sofia, Bulgaria
- |
    Carl von Ossietzky Universität Oldenburg\
    Institut für Mathematik, Fakultät V\
    Postfach 2503\
    D-26111 Oldenburg, Germany
author:
- Nikolai Nikolov and Peter Pflug
title: On the derivatives of the Lempert functions
---

<span>This note was written during the stay of the first named author at the Universität Oldenburg supported by a grant from the DFG, Az. PF 227/8-2 (November – December 2006). He likes to thank both institutions for their support.</span>

Introduction and results
========================

Let $\Bbb D\subset\Bbb C$ be the unit disc. Let $M$ be an $n$-dimensional complex manifold. Recall first the definitions of the Lempert function $\tilde k_M$ and the Kobayashi–Royden pseudometric $\kappa_M$ of $M$: $$\aligned
\tilde k^\ast_M(z,w)&=\inf\{|\alpha|:\exists f\in\mathcal
O(\Bbb D,M):f(0)=z,f(\alpha)=w\},\\
\tilde k_M&=\tanh^{-1}\tilde k^\ast_M,\\
\kappa_M(z;X)&=\inf\{|\alpha|:\exists f\in\mathcal O(\Bbb D,M):
f(0)=z,\alpha f_\ast(d/d\zeta)=X\},
\endaligned$$ where $X$ is a complex tangent vector to $M$ at $z$. Note that such an $f$ always exists (cf. @Win; according to @Din, page 49, this was already known by J. Globevnik).

The Kobayashi pseudodistance $k_M$ can be defined as the largest pseudodistance bounded by $\tilde k_M$. Note that if $k_M^{(m)}$ denotes the $m$-th Lempert function of $M$, $m\in\Bbb N$, that is, $$k_M^{(m)}(z,w)=\inf\{\sum_{j=1}^m\tilde k_M(z_{j-1},z_{j}):z_0,\dots,z_{m}\in M,
z_0=z,z_m=w\},$$ then $$k_M(z,w)=k_M^{(\infty)}:=\inf_m k_M^{(m)}(z,w).$$

By a result of M.-Y. Pang (see @Pang), the Kobayashi–Royden metric is the "derivative" of the Lempert function for taut domains in $\Bbb C^n;$ more precisely, if $D\subset\Bbb C^n$ is a taut domain, then $$\kappa_D(z;X)=\lim_{t\nrightarrow 0}\frac{\tilde k_D(z,z+tX)}{t}.$$

In @KobS, S. Kobayashi introduces a new invariant pseudometric, called the Kobayashi–Buseman pseudometric in @Jar-Pfl. One of the equivalent ways to define the Kobayashi–Buseman pseudometric $\hat\kappa_M$ of $M$ is just to set $\hat\kappa_M(z;\cdot)$ to be largest pseudonorm bounded by $\kappa_M(z;\cdot)$. Recall that $$\hat\kappa_M(z;X)=\inf\{\sum_{j=1}^m\kappa_M(z;X_j):m\in\Bbb N,\
\sum_{j=1}^mX_j=X\}.$$ Thus it is natural to consider the new function $\kappa_M^{(m)}$, $m\in\Bbb N$, namely, $$\kappa_M^{(m)}(z;X)=\inf\{\sum_{j=1}^m\kappa_M(z;X_j):
\sum_{j=1}^mX_j=X\}.$$ We call $\kappa_M^{(m)}$ the <span>*$m$-th Kobayashi pseudometric*</span> of $D$. It is clear that $\kappa_M^{(m)}\ge\kappa_M^{(m+1)}$ and if $\kappa_M^{(m)}(z;\cdot)=\kappa_M^{(m+1)}(z;\cdot)$ for some $m$, then $\kappa_M^{(m)}(z;\cdot)\\=\kappa_D^{(j)}(z;\cdot)$ for any $j>m$. It is shown in @Nik-Pfl that $\kappa_M^{(2n-1)}=\kappa_M^{(\infty)}:=\hat\kappa_M$, and $2n-1$ is the optimal number, in general.

We point out that all the introduced objects are upper semicontinuous. Recall that this is true for $\kappa_M$ (cf. @KobS2). It remains to check this for $\tilde k_M$. We shall use a standard reasoning. Fix $r\in(0,1)$ and $z,w\in M$. Let $f\in\mathcal O(\Bbb D,M)$, $f(0)=z$ and $f(\alpha)=w$. Then $\tilde
f=(f,\hbox{id}):\Delta\to\tilde M=M\times\Delta$ is an embedding. Setting $\tilde f_r(\zeta)=\tilde f(r\zeta)$, by @Roy, Lemma 3, we may find a Stein neighborhood $S\subset\tilde M$ of $\tilde
f_r(\Bbb D)$. Embed $S$ as a closed complex manifold in some $\Bbb
C^N$ and denote by $\psi$ the respective embedding. Moreover, there is an open neighborhood $V\subset \Bbb C^N$ of $\psi(S)$ and a holomorphic retraction $\theta:V\to\psi(S)$. Then, for $z'$ near $z$ and $w'$ near $w$, we may find, as usual, $g\in\mathcal O(\Bbb D,V)$ such that $g(0)=\psi(z',0)$ and $g(\alpha/r)=\psi(w',\alpha)$. Denote by $\pi$ the natural projection of $\tilde M$ onto M. Then $h=\pi\circ\psi^{-1}\circ\theta\circ g\in\mathcal O(\Bbb D,M)$, $h(0)=z'$ and $h(\alpha/r)=w'$. So $r\tilde
k^\ast_M(z',w')\le\alpha$, which implies that $\ds\limsup_{z'\to
z,w'\to w}\tilde k_M(z',w')\le\tilde k_M(z,w)$.

To extend Pang's result on manifolds, we have to define the "derivatives" of $k^{(m)}_M$, $m\in\Bbb N^\ast=\Bbb
N\cup\{\infty\}$. Let $(U,\varphi)$ be a holomorphic chart near $z$. Set $$\mathcal D k^{(m)}_M(z;X)=\limsup_{t\nrightarrow 0,w\to
z,Y\to\varphi_\ast X}
\frac{k^{(m)}_M(w,\varphi^{-1}(\varphi(w)+tY))}{|t|}.$$ Note that this notion does not depend on the chart used in the definition and $$\mathcal D k^{(m)}_M(z;\lambda X)=|\lambda|\mathcal D
k^{(m)}_M(z;X),\quad \lambda\in\Bbb C.$$ Replacing $\limsup$ by $\liminf$, we define $\underline{\mathcal D} k_M^{(m)}$.

From M. Kobayashi's paper @KobM it follows that, if $M$ is a taut manifold, then $$\hat\kappa_M(z;X)=\mathcal D k_M(z;X)=\underline{\mathcal D}
k_M(z;X),$$ that is, the Kobayashi–Buseman metric is the "derivative" of the Kobayashi distance. The proof there also leads to $$\kappa_M^{(m)}(z;X)=\mathcal D
k^{(m)}_M(z;X)=\underline{\mathcal D} k_M^{(m)}(z;X),\quad m\in\Bbb
N^{\ast}.\leqno{(\ast)}$$

We say that a complex manifold $M$ is hyperbolic at $z$ if $k_M(z,w)>0$ for any $w\neq z$. We point out that the following conditions are equivalent:

\(i) M is hyperbolic at $z;$

\(ii) $\ds\liminf_{z'\to z,w\in M\setminus U}\tilde k_M(z',w)>0$ for any neighborhood $U$ of $z;$

\(iii) $\ds\underline{\kappa}_M(z;X):=\liminf_{z'\to z,X'\to
X}\kappa_M(z';X')>0$ for any $X\neq 0;$

The implication (i)$\Rightarrow$(ii) $\Rightarrow$(iii) are almost trivial (cf. @Jar-Pfl) and the implication (iii)$\Rightarrow$(i) is a consequence of the fact that $k_M$ is the integrated form of $\kappa_M$.

In particular, if $M$ is hyperbolic at $z$, then it is hyperbolic at any $z'$ near $z$.

Since if $M$ is taut, then it is k-hyperbolic and $\kappa_M$ is a continuous function, the following theorem is a generalization of $(\ast)$.

\[th1\] Let $M$ be a complex manifold and $z\in M$.

<span>(i)</span> If $M$ is hyperbolic at $z$ and $\kappa_M$ is continuous at $(z,X)$, then $$\kappa_M(z;X)=\mathcal D\tilde k_M(z;X)=\underline{\mathcal D}
\tilde k_M(z;X).$$

<span>(ii)</span> If $\kappa_M$ is continuous and positive at $(z,X)$ for any $X\neq 0$, then $$\kappa_M^{(m)}(z;\cdot)=\mathcal D k_M^{(m)}(z;\cdot)=\underline{\mathcal D}
k_M^{(m)}(z;\cdot),\quad m\in\Bbb N^\ast.$$

The first step in the proof of Theorem \[th1\] is the following

\[pr2\] For any complex manifold $M$ one has that $$\kappa_M^{(m)}\ge\mathcal D k_M^{(m)}, \quad m\in\Bbb N^\ast.$$

Note that when $M$ is a domain, a weaker version of Proposition \[pr2\] can be found in @Jar-Pfl, namely, $\hat\kappa_M\ge\mathcal D k_M$ (the proof is based on the fact that $\mathcal D k_M(z;\cdot)$ is a pseudonorm).

Examples
========

The following examples show that the assumption on continuity in Theorem \[th1\] is essential.

$\bullet$ Let $A$ be a countable dense subset of $\Bbb
C_\ast.$ In @Die-Sib (see also @Jar-Pfl), a pseudoconvex domain $D$ in $\Bbb C^2$ is constructed such that:

\(a) $(\Bbb C\times\{0\})\cup(A\times\Bbb C)\subset D;$

\(b) if $z_0=(0,t)\in D,$ $t\neq 0,$ then $\kappa_D(z_0;X)\ge C||X||$ for some $C=C_t>0$. (One can be shown that even $\mathcal D\tilde
k_D(z_0;X)\ge C||X||.$)

Then it is easy to see that $\underline{
\kappa}_D(\cdot;e_2)=\mathcal D k^{(3)}_D(\cdot;e_2)=k^{(5)}_D=0$ and $\hat\kappa_D(z_0;X)\ge c||X||,$ where $e_2=(0,1)$ and $c>0.$ Thus $$\hat\kappa_D(z_0;X)>\underline{ \kappa}_D(z_0;e_2)=\mathcal
D k^{(3)}_D(z_0;e_2)=\mathcal D k^{(5)}_D(z_0;X),\quad X\neq 0.$$

This phenomena obviously extends to $\Bbb C^n$, $n>2$ (by considering $D\times\Bbb D^{n-2}$). So the inequalities in Proposition \[pr2\] are strict in general.

$\bullet$ If $D$ is a pseudoconvex balanced domain with Minkowski function $h_D$, then (cf. @Jar-Pfl) $$h_D=\kappa_D(0;\cdot)=\mathcal D{\tilde k}_D(0;\cdot).$$ Therefore, $\mathcal D{\tilde k}_D(0;X)>\underline{\mathcal
D}\tilde k_D(0;X)$ if $\kappa_D(0;\cdot)$ is not continuous at $X$. On the other hand, if $\hat D$ denotes the convex hull of $D$, then $$h_{\hat D}=\hat\kappa_D(0;\cdot)=\mathcal
D{k}_D(0;\cdot)=\underline{\mathcal D}k_D(0;\cdot)=
\underline{\hat\kappa}_D(0;\cdot).$$

$\bullet$ Modifying the first example leads to a pseudoconvex domain $D\subset\Bbb C^2$ with $$L_{\mathcal
Dk_D}(\gamma)>0=L_{k_D}(\gamma)= L_{\underline{\mathcal
D}k_D}(\gamma),$$ where $\gamma:[0,1]\to\Bbb C^2$, $\gamma(t):=(ti/2,1/2)$, and $L_\bullet(\gamma)$ denotes the respective length.

Indeed, choose a dense sequence $(r_j)$ in $[0,i/2].$ Put $$u(\lambda)=\sum_{k=1}^\infty
\frac{1}{k^2}\log\tfrac{|\lambda-1/k|}{4},\quad
v(\lambda)=\sum_{j=1}^\infty\frac{u(\lambda/2-r_j)}{2j^2},\quad\lambda\in\Bbb
C,$$ and $$D=\{z\in\Bbb C^2:\psi(z)=|z_2|e^{\|z\|^2+v(z_1)}<1\}.$$ It is easy to see that $v$ is a subharmonic function on $\Bbb C$. Hence $D$ is a pseudoconvex domain with $(\Bbb C
\times\{0\})\cup(\bigcup_{j,k=1}^\infty \{r_j+1/k\}\times\Bbb
C)\subset D.$ Observe that $u|_{\Bbb D}<-1$ and so $D$ contains the unit ball $\Bbb B_2.$ Note also that $$k_D(a,b)=0,\quad a,b\in\gamma([0,1]).$$ Set $\widehat \psi(z)=\|z\|^2/2-\log\psi(z)$. Fix $z^0\in\Bbb B_2$ with $\Ree z^0_1\le 0,\;\Imm z^0_2\ge1/e$. Since $u(\lambda)\ge
u(0)$ for $\Ree \lambda\le 0,$ we have $$||z^0||/2<\widehat\psi(z^0)<1-u(0)=:8C.$$

Let $\varphi\in\mathcal O(\Bbb D,D)$, $\varphi(0)=z^0.$ Following the estimates in the proof of Example 3.5.10 in @Jar-Pfl, we see that $\|\varphi'(0)\|<C$. Hence, $\kappa_D(z^0;X)\geq C\|X\|$, $X\in\Bbb C^2$. Since $k_D$ is the integrated form of $\kappa_D,$ it follows that $$k_D(a,a-te_1)\geq Ct,\quad
a\in\gamma([0,1]),\;0\le t\le 1/2-1/e,\;e_1=(1,0).$$ Hence $\mathcal D k_D(a;e_1)\geq C$ and therefore, $L_{\mathcal
Dk_D}(\gamma)\ge C/2>0,$ which completes the proof of this example.

Note that it shows that, with respect to the lengths of curves, $\mathcal D k_D$ behaves different than the "real" derivative of $k_D$ (cf. @Ven or @Jar-Pfl2, page 12). Moreover, it implies that, in general, $\mathcal D k_D\neq\underline{\mathcal
D}k_D$.

<span>**Questions.**</span> It will be interesting to know examples showing that, in general, $\kappa_D\neq\mathcal D\tilde k_D$. It remains also unclear whether $\mathcal D k_D$ is holomorphically contractible (see @Jar-Pfl). Recall that $\int\mathcal D
k_D=k_D;$ but we do not know if $\int\underline{\mathcal D}k_D=k_D.$

Proofs
======

<span>*Proof of Proposition \[pr2\].*</span> First, we shall consider the case $m=1$. The key is the following

\[th4\][^1] @Roy Let $M$ be an $n$-dimensional complex manifold and $f\in\mathcal O(\Bbb D,M)$ regular at $0$. Let $r\in(0,1)$ and $D_r=r\Bbb D\times\Bbb D^{n-1}$. Then there exists $F\in\mathcal O(D_r,M)$, which is regular at $0$ and $F|_{r\Bbb D\times\{0\}}=f$.

Since $\kappa_M(z;0)=\mathcal D\tilde k_M(z;0)=0$, we may assume that $X\neq 0$. Let $\alpha>0$ and $f\in\mathcal O(\Bbb D,M)$ be such that $f(0)=z$ and $\alpha f_\ast(d/d\zeta)=X$. Let $r\in(0,1)$ and $F$ as in Theorem \[th4\]. Since $F$ is regular at $0$, there exist open neighborhoods $U=U(z)\subset M$ and $V=V(0)\subset D_r$ such that $F|_V:V\to U$ is biholomorphic. Hence $(U,\varphi)$ with $\varphi=(F|_V)^{-1}$, is a chart near $z$. Note that $\varphi_\ast(X)=\alpha e_1$, where $e_1=(1,0,\dots,0).$

If $w$ and $Y$ are sufficiently near $z$ and $\alpha e_1$, respectively, then $$g(\zeta):=F(\varphi(w)+\zeta Y/\alpha),\quad \zeta\in r^2\Bbb D,$$ belongs to $\mathcal O(r^2\Bbb D,M)$ with $g(0)=w$ and $g(t\alpha)=\varphi^{-1}(\varphi(w)+tY),$ $t<r^2/\alpha.$ Therefore, $r^2\widetilde k^\ast_M(w,\varphi^{-1}(\varphi(w)+tY))\leq t\alpha$. Hence $r^2\mathcal D\tilde k_M(z;X)\le\alpha$. Letting $r\to 1$ and $\alpha\to\kappa_M(z;X)$ we get that $\mathcal D\tilde
k_M(z;X)\le\kappa_M(z;X).$

Let now $m\in\Bbb N$. By definition, $\kappa_M^{(m)}(z;\cdot)$ is the largest function with the following property:

For any $X=\sum_{j=1}^mX_j$ one has that $\kappa_M^{(m)}(z;X)\le\sum_{j=1}^m\kappa_M(z;X_j)$.

To prove that $\kappa_M^{(m)}\ge\mathcal D k_M^{(m)}$ it suffices to check that $\mathcal D k^{(m)}_M(z;\cdot)$ has the same property. Following the above notation and choosing $Y_j\to\varphi_\ast X_j$ with $\sum_{j=1}^m Y_j=Y$, we set $w_0=w$ and $w_j=\varphi^{-1}(\varphi(w)+t\sum_{k=1}^{j}Y_j)$. Since $$k^{(m)}_M(w,w_q)\le\sum_{j=1}^m\tilde k_M(w_{j-1},w_j),$$ it follows by the case $m=1$ that $$\mathcal D
k^{(m)}_M(z;X)\le\sum_{j=1}^m\mathcal D
k_M(z;X_j)\le\sum_{j=1}^m\kappa_M(z;X_j).$$

Finally, let $m=\infty$ and $n=\dim M$. Since $\hat\kappa_M=\kappa_M^{(2n-1)}$ and $k_M\le k_M^{(2n-1)}$, we get that $\mathcal D k_M\le\hat\kappa_M$ using the case $m=2n-1$.

<span>*Proof of Theorem \[th1\].*</span> We may assume that $X\neq
0$. In virtue of Proposition \[pr2\], we have to show that $$\kappa_M^{(m)}(z;X)\le\underline{\mathcal D}k^{(m)}_M(z;X).$$ For simplicity we assume that $M$ is a domain in $\Bbb C^n$.

\(i) Fix a neighborhood $U=U(z)\Subset M.$ Applying the hyperbolicity of $M$ at $z$, there are a neighborhood $V=V(z)\subset U$ and a $\delta\in (0,1)$ such that, if $h\in\mathcal O(\Bbb D,M)$ with $h(0)\in V$, then $h(\delta\Bbb D)\subset U$. Hence, by the Cauchy inequalities, $||h^{(k)}(0)||\leq c/\delta^k$, $k\in\Bbb N.$

Now choose sequences $M\ni w_j\to z$, $\Bbb C_\ast\ni t_j\to 0$, and $\Bbb C^n\ni Y_j\to X$ such that $$\frac{\widetilde k_M(w_j,w_j+t_jY_j)}{|t_j|}\to\underline{\mathcal
D}\widetilde k_M(z;X).$$ There are holomorphic discs $g_j\in\mathcal O(\Bbb D,M)$ and $\beta_j\in (0,1)$ with $g_j(0)=w_j$, $g_j(\beta_j)=w_j+t_jY_j,$ and $\beta_j\le\widetilde k^\ast_M(w_j,w_j+t_jY_j)+|t_j|/j.$ Note that $\widetilde k^\ast_M(w_j,w_j+t_jY_j)\le c_1||t_jY_j||\le c_2 |t_j|.$

Write $$w_j+t_jY_j=g_j(\beta_j)=w_j+g_j'(0)\beta_j+ h_j(\beta_j).$$ Then $$||h_j(\beta_j)||\leq c\sum_{k=2}^\infty
\left(\tfrac{\beta_j}{\delta}\right)^k\leq c_3|\beta_j|^2\le
c_4|t_j|^2,\quad j\ge j_0.$$

Put $\widehat Y_j=Y_j-h_j(\beta_j)/t_j.$ We have that $g_j(0)=w_j$ and $\beta_jg_j'(0)/t_j=\widehat Y_j\to X$. Therefore, $$\kappa_M(w_j;\widehat
Y_j)\leq\frac{\beta_j}{|t_j|}\leq\frac{\widetilde
k^\ast_M(z_j,w_j+t_jY_j)}{|t_j|}+\frac{1}{j}.$$ Hence with $j\to\infty$, we get that $\kappa_M(z;X)=\underline{\kappa}_M(z;X)\leq \underline{\mathcal
D}\widetilde k_M(z;X)$.

\(ii) The proof of the case $m\in\Bbb N$ is similar to the next one and we omit it. Now, we shall consider the case $m=\infty$.

Note first that our assumption implies that $M$ is hyperbolic at $z$ and, by the contrary, $$\forall\varepsilon>0\
\exists\delta>0:||w-z||<\delta,||Y-X||<\delta||X||$$ $$\Rightarrow|\kappa_M(w;Y)-\kappa_M(z;X)|<\varepsilon\kappa_M(z;X).\leqno{(1)}$$ Moreover, the proof of (i) shows that $$\tilde k_M(a,b)\ge\kappa_M(a;b-a+o(a,b)),\hbox{ where }\lim_{a,b\to
z}\frac{o(a,b)}{||a-b||}=0.\leqno{(2)}$$

Choose now sequences $M\ni w_j\to z$, $\Bbb C_\ast\ni t_j\to 0$, and $\Bbb C^n\ni Y_j\to X$ such that $$\frac{k_M(w_j,w_j+t_jY_j)}{|t_j|}\to\underline{\mathcal
D}k_M(z;X).$$ There are points $w_{j,0}=w_j,\dots,w_{j,m_j}=w_j+t_jX_j$ in $M$ such that $$\sum_{k=1}^{m_j}\tilde k_M(w_{j,k-1},w_{j,k})\le
k_M(w_j,w_j+t_jY_j)+\frac{1}{j}.\leqno{(3)}$$ Set $w_{j,k}=w_j$ for $k>m_j$. Since $$k_M(w_j,w_{j,l})\le\sum_{j=1}^l\tilde k_M(w_{j,k-1},w_{j,k})\le
k_M(w_j,w_j+t_jY_j)+\frac{1}{j}\le c_2|t_j|+\frac{1}{j},$$ then $k_M(w_j,w_{j,l})\to 0$ uniformly in $l.$ Then the hyperbolicity of $M$ at $z$ implies that $w_{j,l}\to z$ uniformly in $l.$ Indeed, assuming the contrary and passing to a subsequence, we may suppose that $w_{j,l_j}\not\in U$ for some $U=U(z).$ Then $$0=\lim_{j\to\infty}k_M(w_j,w_{j,l})\ge\ds\liminf_{z'\to z,w\in
M\setminus U}\tilde k_M(z',w)>0,$$ a contradiction.

Fix now $R>1.$ Then (1) implies that $$\kappa_M(z;w_{j,k}-w_{j,k-1})\le R\kappa_M(w_{j,k};w_{j,k}-w_{j,k-1}
+o(w_{j,k},w_{j,k-1})),\ j\ge j(R).$$ It follows by this inequality, (2) and (3) that $$\sum_{k=1}^{m_j}\kappa_M(z;w_{j,k}-w_{j,k-1})
\le Rk_M(w_j,w_j+t_jYj)+\frac{R}{j}.$$ Since $\hat\kappa_M(z;t_jY_j)$ is bounded by the first sum, we obtain that $$\hat\kappa_M(z;Y_j)\le R\frac{k_M(w_j,w_j+t_jYj)+1/j}{|t_j|}.$$ Note that $\hat\kappa_M(z;\cdot)$ is a continuous function. Hence with $j\to\infty$ and $R\to 1,$ we get that $\hat\kappa_M(z;X)\le\underline{\mathcal D}k_M(z;X)$.

<span>**Remark.**</span> It follows by the above proofs and a standard diagonal process that $\underline{\kappa}_M(z;\cdot)=\underline{\mathcal D}\tilde
k(z;\cdot)$ if $M$ is hyperbolic at $z$.

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[^1]: We may replace Theorem \[th4\] by the approach used in the proof of the upper semicontinuity of $\tilde k_M$.

{'source': 'arXiv_out/0801'}