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Let ${M}, \, {N}$ be two Riemannian manifolds, and let $u_n: {M} \to {N}$ be a sequence of harmonic maps.

Question. Suppose that $u_n$ converges uniformly to a (necessarily continuous) function $u$. Is $u$ a harmonic map?

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The answer to your question is no (i.e., if you do not impose any further assumptions). Consider the undoloidal Delaunay cylinder in $\mathbb R^3$, which come in a real 1-parameter family. These surfaces are of constant mean curvature $H=\tfrac{1}{2}$, and they are rotational surfaces. The family parameter $w$ is (equivalent) the ratio of the minimum and the maximum radius. Their conformal type is that of a 2-punctured sphere, on which we put any metric in that conformal class to obtain a fixed Riemannian manifold $M$. By the Lawson correspondence, there exist minimal surfaces $f=f_w$ in the 3-sphere $\mathrm{SU}(2)$ with the same induced metric, but possibly with periods. The Lawson correspondence is given as follows (in the reverse direction): for the Maurer Cartan form $\omega=f^{-1}df\in\Omega^1(M,\mathfrak{su}(2))$ of $f,$ $*\omega$ is a closed 1-form which integrates up (on the universal covering) to give a constant mean curvature surface in $\mathbb R^3=\mathfrak{su}(2).$ The minimal surfaces are again rotational surfaces, and they integrate up (with the same period), so we obtain conformally parameterised minimal surfaces which are actually well-defined as maps $f=f_w\colon M\to SU(2).$ Of course, conformally parameterised minimal surfaces are harmonic maps. If $w$ goes to 0, the Delaunay cylinder converge (uniformly) against a chain of spheres, and the same holds true for the surfaces $f_w$ in $\mathrm{SU}(2),$ i.e., $f_w$ converges against an infinite chain of minimal spheres. The limit is not smooth, but continuous, and is only smooth away from the touching points. In fact, the Gauss maps are harmonic maps to $\mathbb S^2$, but the limit is not even continuous anymore. For a nice animation of the family of Delaunay cylinders in $\mathbb R^3,$ see the video A flow through the one-parameter family of Delaunay unduloids and nodoids.

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When $N=\mathbb{R}$, this is a classical result by Friedrichs and Gårding :

Theorem. (1) If a sequence of real-valued harmonic functions converges uniformly on compact subsets of $M$, then the limit of the sequence is a harmonic function.

(2) Furthermore, the derivatives of the members of the sequence converge uniformly on compact sets to the derivatives of the limit of the sequence.

See p. 220 of

Greene, R. E.; Wu, H., Embedding of open Riemannian manifolds by harmonic functions, Ann. Inst. Fourier 25, No. 1, 215-235 (1975). ZBL0307.31003.

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    $\begingroup$ This case is too special. $\endgroup$
    – gaoqiang
    Commented Dec 12, 2022 at 13:33
  • $\begingroup$ I guess that a special case is better than nothing at all... $\endgroup$
    – Francesco Polizzi
    Commented Dec 12, 2022 at 13:45
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    $\begingroup$ At any rate, are you sure that the general case does not follow from this? By Nash Embedding Theorem, there is an isometrical embedding $\iota \colon N \to \mathbb{R}^n$ (where $\mathbb{R}^n$ has the standard Euclidean metric) for sufficiently big $n$. Then, composing any harmonic map $f \colon M \to N$ with $\iota$, we get an harmonic map $\tilde{f} \colon M \to \mathbb{R}^n$, with image $N$, and for such map the harmonicity is equivalent to the harmonicity of its $n$ components. Now just apply Friedrichs and Gårding result to each component of the sequence $\{\iota\circ u_n\}$. $\endgroup$
    – Francesco Polizzi
    Commented Dec 12, 2022 at 17:10
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    $\begingroup$ It is not true in general that the composition of the inclusion with a harmonic map is again harmonic. For example, consider an arc length parametrized great circle on a 2-sphere, or holomorphic maps from compact Riemann surfaces to the 2-sphere. The reason for not being harmonic again is that the tension field of the composition map has normal components. $\endgroup$
    – Sebastian
    Commented Dec 12, 2022 at 23:50
  • $\begingroup$ As noted by @Sebastian, the composition of the inclusion with a harmonic map is not necessarily harmonic. As far as I know, It's commonly required that the inclusion is totally geodesic, not only isometric. $\endgroup$
    – gaoqiang
    Commented Dec 13, 2022 at 3:37

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