# Possible flaw in the proof of the Eells-Sampson theorem on harmonic maps in Nishikawa's book

Background:

I am reading the book Variational Problems in Geometry by Seiki Nishikawa. The main purpose of this book is to prove the existence of harmonic maps $$M\to N$$ between two compact Riemannian manifolds $$M$$ and $$N$$ with the target manifold being nonpositively curved. Since a harmonic map is defined as a minimizer for the energy functional $$E$$, the idea is to deform an existing map $$u:M\to N$$ along the gradient lines of $$E$$, which is equivalent to solving an equation: $$\frac{\partial u_t}{\partial t}=\tau(u_t)\quad\quad\text{Eq.(1)}$$ where $$u_t$$ is a smooth variation of some given $$u_0$$, and $$\tau$$ denotes the tension field (along this direction, $$E$$ decreases fastest). By embedding $$N$$ into $$\mathbb R^q$$ using the Nash embedding theorem, we can regard $$N$$ as a submanifold and $$u$$ as a vector-valued function to simplify certain statements. After this embedding, the above equation takes a new form: $$(\Delta-\partial_t)u_t=\Pi(u_t)\quad\quad\text{Eq.(2)}$$ I'll omit the definition of $$\Pi(u_t)$$. Just know that they are equivalent in the sense that their solutions coincide, if given identical initial conditions. The difference is:

• Eq.(1) is intrinsic and is an equation in $$\Gamma(u_t^{-1}TN)$$;
• Eq.(2) is an equation in $$\mathbb R^q$$.

Theorem:

One of the main theorems is the existence of a solution to any one of the equations above defined for $$t\in[0,+\infty)$$, given some $$u_0$$ as the initial condition. Suppose now we have already established:

• local (w.r.t time) existence of a solution
• certain estimates

The proof goes as follows:

Proof. (The part where there's no problem, so I'll give a sketch only) Since we have established the local existence, for a given $$u_0$$, we can find a smooth $$u:M\times[0,T)\to N$$ for some $$T>0$$ such that it solves (1) and (2) and $$u(\cdot,0)=u_0$$. Now suppose $$T_0$$ is the supremum of all such $$T$$. We wish to prove $$T_0=+\infty$$. If not, we take an increasing sequence $$t_i\to T_0$$. By certain estimates, we know that $$\{u_{t_i}\}$$ and $$\{\partial_tu_{t_i}\}$$ are uniformly bounded and equicontinuous, respectively in the Holder spaces $$C^{2+\alpha}(M,\mathbb R^q)$$ and $$C^{\alpha}(M,\mathbb R^q)$$, for some $$0<\alpha<1$$.

Proof. (The part where I am confused. I'll write exact words) By the Ascoli-Arzela theorem, there exists a subsequence $$\{t_{i_k}\}$$ of $$\{t_i\}$$ and functions$$^1$$ $$u(\cdot,T_0)\in C^{2+\alpha}(M,\mathbb R^q)\quad\text{and}\quad\partial_t u(\cdot,T_0)\in C^\alpha(M,\mathbb R^q)$$ such that the subsequences $$\{u(\cdot,t_{i_k})\}\quad\text{and}\quad\{\partial_tu(\cdot,t_{i_k})\}$$ respectively, converge uniformly to $$u(\cdot,T_0)$$ and $$\partial_tu(\cdot,T_0)$$, as $$t_{i_k}\to T_0$$. Since for each $$t_{i_k}$$, we have $$\partial_tu(\cdot,t_{i_k})=\tau(u(\cdot,t_{i_k}))\quad\quad\text{Eq.(3)}$$ we also get at$$^2$$ $$T_0$$ $$\partial_tu(\cdot,T_0)=\tau(u(\cdot,T_0))\quad\quad\text{Eq.(4)}$$ consequently$$^3$$, we see that (1) has a solution in $$M\times[0,T_0]$$. Using $$u(\cdot,T_0)$$ again as the initial condition to solve (1), we extend the solution to $$M\times[0,T_0+\epsilon)$$ for some $$\epsilon>0$$, contradicting $$T_0$$ being the supremum. Hence $$T_0=\infty$$.

Questions:

1. How do we know that the limits do not depend on the choice of $$t_i$$. I think it can be argued using the uniformly boundedness and equicontinuity. Am I right?

2. How can we go from (3) to (4)? Eq. (3) is clearly obtained by (1), which is an equation in $$\Gamma(u_t^{-1}TN)$$. However, (3)$$\implies$$(4) would require the convergence of $$\partial_tu(\cdot,t_{i_k})$$ to $$\partial_tu(\cdot,T_0)$$. But this convergence is only in $$C^{\alpha}(M,\mathbb R^q)$$, not in $$\Gamma(u_t^{-1}TN)$$. Although I think I can fix this by writing them in the form of (2) to start with.

3. How can we conclude from (4) that (1) has a solution in $$M\times[0,T_0]$$. Note that $$\partial_tu(\cdot,T_0)$$ is so far only a notation for the limit of $$\partial_tu(\cdot,t_{i_k})$$, we have not proved that it is actually the (one-sided) derivative of $$u$$ at $$t=T_0$$. How can I fix this?

$$\newcommand{\R}{\mathbb{R}}\newcommand{\pa}{\partial}$$Edit: The answer is now LaTeXified.

Below are my notes on this. I reworked the proof:

Proof. Let $$S:=\big\{T\in[0,\infty):$$ the equation has a solution in $$C^{2+\alpha,1+\alpha/2}(M\times[0,T],N)\big\}$$. Let $$T_0:=\sup S$$. By existence of local solution, $$T_0>0$$. We claim that $$T_0=\infty$$.

Suppose $$T_0<\infty$$. By uniqueness of solution and definition of $$T_0$$, we have a solution $$u\in C^{2,1}(M\times[0,T_0),N)$$. Take $$\alpha<\alpha'<1$$. By the a priori estimate above, $$u_t$$ is uniformly bounded in $$t$$ in $$C^{2+\alpha'}(M,\R^L)$$.

Define $$u(x,T_0):=\int_0^{T_0}\pa_tu(x,t)\,dt+u(x,0).$$ For any sequence $$t_k\nearrow T_0$$, $$(u_{t_k})_{k=1}^\infty$$ has a subsequence that converges in $$C^{2+\alpha}(M,N)$$ by Arzelà–Ascoli, and its limit is necessarily $$u_{T_0}$$. Thus $$u_{T_0}\in C^{2+\alpha}(M,N)$$, and any such sequence must in fact converge to $$u_{T_0}$$ in $$C^{2+\alpha}(M,N)$$. In other words, $$u\in C^{2+\alpha,0}(M\times[0,T_0],N)$$, or equivalently, $$t\mapsto u_t$$ is continuous as a map $$[0,T_0]\to C^{2+\alpha}(M,N)$$.

Since $$\pa_tu_t=\tau(u_t)$$, we see that $$t\mapsto\pa_tu_t$$ has a continuous extension $$[0,T_0]\to C^{\alpha/2}(M,\R^L)$$. So in fact $$u\in C^{2+\alpha,1+\alpha/2}(M\times[0,T_0],N)$$, i.e., $$T_0\in S$$. Now, existence of local solution implies that $$u$$ can be extended to a solution on $$[0,T_0+\varepsilon]$$, which is a contradiction. $$\square$$

The a priori estimate refers to the one as given in the book: For $$0<\alpha<1$$, $$\sup_{t\in[0,T)}\Big(\big\|u_t\big\|_{C^{2+\alpha}(M,\R^L)}+\big\|\pa_tu_t\big\|_{C^\alpha(M,\R^L)}\Big)\leq C(M,N,f,\alpha).$$

• Can you change the images into LaTeX? Doing so will enable search engines to index your answer properly. – Mateusz Kwaśnicki Feb 14 at 6:56
• @MateuszKwaśnicki The answer is now LaTeXified. – Colescu Feb 14 at 8:49