Harmonic mappings into Riemannian manifolds with non-positive sectional curvature.

Lemma 1. Let $U \in C^2(\Omega, \mathcal{M})$ be such that in some normal coordinate system on $\mathcal{M}$ (2) $$\Delta_{\alpha r} u^l + t \gamma^{\alpha \beta} \Gamma^l_{ik}(u) D_{\alpha} u^i D_{\beta} u^k = 0, \quad 1 \leq l \leq N.$$ Then if $t \in [0,1]$ we have $$\Delta_{\mathscr{X}}|u|^2 = \Delta_{\mathscr{X}} \sum_{1}^{N} (u^j)^2 \ge 2te$$ . In particular, if $U \in C^2(\Omega, \mathcal{M})$ is a harmonic map then for every $V \in \mathcal{M}$ $$\Delta_{\mathscr{X}}d^{V} \geq 2e$$ . PROOF OF LEMMA 1. $$\begin{split} \varDelta_{\mathscr{X}}|u|^2 &= 2u^j\varDelta_{\mathscr{X}}u^j + 2\gamma^{\alpha\beta}D_{\alpha}u^jD_{\beta}u^j \\ &= 2t\{\delta_{ik} - \Gamma^l_{ik}(u)u^l\}\gamma^{\alpha\beta}D_{\alpha}u^iD_{\beta}u^k + 2(1-t)\gamma^{\alpha\beta}D_{\alpha}u^jD_{\beta}u^j. \end{split}$$ Obviously, the last term here is non-negative. Since the $\Gamma^l_{ik}$ are calculated with respect to normal coordinates on $\mathcal{M}$ , we have (cf. [6, p. 211, formulas (50)–(52)]) $$\begin{split} g_{ik}(u)u^k &= g^{ik}(u)u^k = u^i\,,\\ \Gamma_{iik}(u)u^j + \Gamma_{ikl}(u)u^j &= \delta_{ik} - g_{ik}\,. \end{split}$$ The first of these formulas implies $$\Gamma^l_{ik}(u)u^l = \Gamma_{ijk}(u)u^j$$ . Hence, the second one is equivalent to $$\delta_{ik} - \varGamma^l_{ik}(u)u^l \,=\, g_{ik}(u) + \varGamma_{iki}(u)u^j \;.$$ Therefore we obtain $$\Delta_{\mathscr{X}}|u|^2 \ge 2t\{g_{ik}(u) + \Gamma_{iki}(u)u^j\}\gamma^{\alpha\beta}D_{\alpha}u^iD_{\beta}u^k.$$ Using Rauch's comparison theorem and the fact that $\mathcal{M}$ has non-positive sectional curvature one derives the inequality $$0 \le \Gamma_{iki}(u)u^j\xi^i\xi^k$$ for all $\xi \in \mathbb{R}^N$ , cf. [6, Lemma 6]. This concludes the proof of the first part of the lemma if we note that the trace $A_{\alpha\beta}B^{\alpha\beta}$ of the product of two positive semi-definite matrices $(A_{\alpha\beta})$ , $(B^{\alpha\beta})$ is non-negative provided that one of them is symmetric. The second part of the Lemma is immediate since a harmonic mapping satisfies (2) with t=1 and normal coordinates around any point $V \in \mathcal{M}$ , remembering that then $d^{V} = |u|^{2}$ .

Lemma 2. For given $\mathscr{X}$ and $\mathscr{M}$ let $U \in C^3(\Omega, \mathscr{M})$ be harmonic. Then there exists a constant $\tau \geq 0$ , depending only on $\mathscr{X}$ , such that $$\Delta_{\mathscr{X}}\{e+\tau d^0\} \geq 0 \quad on \ \Omega$$ . PROOF. We shall make use of the following differential inequality derived by Bochner [1] in the special case of a flat manifold $\mathcal{X}$ , and, in general, by Eells and Sampson [2, p. 123]: $$\frac{1}{2}\Delta_{\mathscr{X}}e \geq f_1 + f_2$$ where $$f_1 = -\gamma^{\alpha\beta}\gamma^{\mu\nu}R_{ijkl}(u)D_{\alpha}u^iD_{\mu}u^jD_{\beta}u^kD_{\nu}u^l$$ and $$f_2 = g_{ik}(u) P^{\alpha\beta} D_{\alpha} u^i D_{\beta} u^k .$$ Here $R_{ijkl}$ stands for the Riemannian curvature tensor of $\mathcal{M}$ : $$R_{ijkl} = g_{lh}R^h_{ijk}, \quad R^h_{ijk} = \frac{\partial \Gamma^h_{jk}}{\partial u^i} - \frac{\partial \Gamma^h_{ik}}{\partial u^j} + \Gamma^h_{il}\Gamma^l_{jk} - \Gamma^h_{jl}\Gamma^l_{ik}$$ and $P^{\alpha\beta}$ denotes the Ricci curvature tensor on $\mathscr{X}$ : $$P^{\alpha\beta}=\gamma^{\alpha\sigma}\gamma^{\mu\nu}P^{\beta}_{\sigma\mu\nu}$$ where $P^{\beta}_{\sigma\mu\nu}$ is the Riemann curvature tensor on $\mathscr{X}$ . Since the sectional curvature of $\mathcal{M}$ is assumed non-positive, we have $f_1 \ge 0$ . On the other hand, using a compactness argument one sees that there is a number $\tau$ such that $|f_2| \le \tau e$ . In view of Lemma 1 the proof is complete.

Lemma 3. For given $\mathscr X$ and $\mathscr M$ there is a function $k(\alpha,M)$ defined for $\alpha \in (0,1)$ and M>0 such that if $U \in C^3(\Omega,\mathscr M) \cap C^1(\mathscr X,\mathscr M)$ is a harmonic map with boundary values $U|_{\Sigma} = \Phi$ such that $\operatorname{dist}(\Phi(\Sigma),0) \leq M$ then $$\sup_{\mathscr{X}} d^0 \leq M^2$$ and $$\sup_{\mathscr{X}} e \leq k(\alpha, M)(1 + |\varphi|_{C^{1+\alpha}(\Sigma, \mathbb{R}^N)}^2)$$ where $\varphi$ is the standard representation for $\Phi$ . PROOF. The first inequality follows immediately from the maximum principle since, by Lemma 1, $d^0$ is a subharmonic function and, by assumption, $\sup_{\Sigma} d^0 \leq M^2$ . To prove the second inequality we notice that by Lemma 2 and the maximum principle $$\sup_{\mathscr{X}} e \leq \sup_{\mathscr{X}} \{e + \tau d^0\} \leq \sup_{\Sigma} \{e + \tau d^0\} \leq \sup_{\Sigma} e + \tau M^2,$$ whence we see that it is enough to estimate $\sup_{\Sigma} e$ . Let $x_0 \in \Sigma$ and $l \in T_{x_0} \mathcal{X}$ be a unit vector and pointing outwards such that $$\sup_{\mathcal{L}} e \, = \, e(x_0) \, \leqq \, (N+1) \, \left\| \frac{\partial U}{\partial l} \, (x_0) \, \right\|_{\mathcal{M}}^2,$$ where $\| \cdot \|_{\mathscr{M}}$ is the norm in $T\mathscr{M}$ induced by the metric in $\mathscr{M}$ . Excluding the trivial case when $\partial U(x_0)/\partial l$ is zero we walk a distance of $\frac{1}{2}$ on the geodesic ray from $U_0 = U(x_0)$ in the direction of $\partial U(x_0)/\partial l$ , to arrive at $V_0$ . An easy calculation then shows that $$\frac{\partial d^{V_0}}{\partial l}(x_0) = -\left\|\frac{\partial U}{\partial l}(x_0)\right\|_{\mathcal{M}}.$$ On the other hand, consider the Riesz decomposition for the subharmonic function $d^{V_0}$ , $$d^{V_0} = h + s$$ where $\Delta_{\mathcal{X}}h=0$ with $h=d^{V_0}$ on $\Sigma$ and where $\Delta_{\mathcal{X}}s\geq 0$ with s=0 on $\Sigma$ . Since l is pointing outwards, the maximum principle implies that $-\partial s/\partial l \leq 0$ and hence $$-\frac{\partial d^{V_0}}{\partial l}(x_0) \leq -\frac{\partial h}{\partial l}(x_0).$$ However, in view of the well-known Schauder estimates one easily realizes that $|h|_{C^1(\mathcal{X},\,\mathsf{R})} \leq k'(\alpha,M)|h|_{C^{1+\alpha}(\mathcal{Z},\,\mathsf{R})} \leq (N+1)^{-1}k(\alpha,M)(1+|\varphi|_{C^{1+\alpha}(\mathcal{Z},\,\mathsf{R}^N)}) \ ,$ and the statement of the lemma is proved.