Rigidity of functions

Suppose that u\colon \mathbb R^2\to\mathbb R is a continuously differentiable function such that at every point of the plane at least one of the partial derivatives u_x, u_y is zero.

Prove that u depends on just one variable (either x or y). In other words, either u_x\equiv 0 or u_y\equiv 0.

Proof (suggested by Lianxin He). Let U be the (open) set where u_x\ne 0. We claim that U is a union of vertical lines. Suppose not. Then there is an vertical open line segment I\subset U such that (a) I is not the entire line; (b) at least one of the endpoints of I, say p, is not contained in U.

Since u_y=0 in U, it follows that u_x is constant on I. On the other hand, u_x=0 at p. We conclude that u_x= 0 on I, a contradiction. Thus, U is a union of vertical lines.

The same argument shows that the set V=\{u_y\ne 0\} is a union of horizontal lines. Since by our assumption U\cap V=\varnothing, it follows that one of the sets U,V is empty. QED

The above is a toy model of rigidity problems for maps on product spaces. For nonsmooth maps (bilipschitz or quasiconformal) one does not have the C^1 regularity, which complicates the matter. Indeed, the example u(x,y)=\max(x,y) shows that the above argument does not extend to Lipschitz functions.

However, one may want to weaken the assumption concerning the derivatives. Let us assume that u is a continuous function such that at every point at least one of the partial derivative vanishes. (The other derivative need not even exist). The conclusion remains the same: u depends on only one variable. The proof given below is due to A.M. Bruckner, G. Petruska, D. Preiss, and B.S. Thomson, who published it in “The equation u_xu_y=0 factors”, Acta Math. Hungar. 57 (1991), no. 3-4, 275-278. Actually, they prove a more general result, for separately continuous functions. The proof for continuous functions is simple (once you see it, of course!). The main idea is in the following lemma.

Lemma 1. Let R be a closed axes-aligned rectangle R, and let p be a vertex of R. Then the connected component of p in the set E=\{q\in R\colon u(q)=u(p)\} meets a side of R not containing p.

Assume Lemma 1 for now.

Lemma 2. In the notation of Lemma 1, u(r)=u(p) for some vertex r adjacent to p.

Proof: We may assume p is the bottom-left vertex and the component of p in E meets the right side of R. Then E separates the bottom-right vertex r from the sides not containing it. This leads to a contradiction unless u(r)=u(p). Lemma 2 is proved.

Now the proof is finished as follows. Suppose u attains distinct values at two points p_1, p_2 lying on the same vertical line. For each point p on this line pick i\in\{1,2\} such that u(p_i)\ne u(p). Applying Lemma 2 to all rectangles with p and p_i as vertices, we see that u is constant on the horizontal line through p. Since p was arbitrary, u is constant on all horizontal lines. QED

It remains to prove Lemma 1. The proof involves comparison with square cones, which are functions of the form f(x)=c\|x-a\|_1=c(|x_1-a_1|+|x_2-a_2|) with c>0. It also uses, perhaps surprisingly, the equality of components and quasicomponents in compact Hausdorff spaces (such as E).

We may assume p is the bottom-left vertex of R. Suppose the component of p in E does not meet either the top or the right side of R. Then there exist disjoint sets U,V which are open in R and cover E in such a way that U contains p and V contains both the top and the right side of R. There exists \epsilon>0 such that |u(q)-u(p)|\ge 2\epsilon \|q-p\|_1 for all q\in R\setminus (U\cup V). The set K=\{q\in U\colon |u(q)-u(p)|\le \epsilon \|q-p\|_1\} is compact and nonempty. Let r be its largest point in the lexicographic order. The points q\in U such that q_1\ge r_1 and q_2\ge r_2 satisfy |u(q)-u(r)|\ge \epsilon \|q-r\|_1. In particular, this holds when q approaches r from the right or from above (note that the construction of U makes such approaches possible). But either u_x or u_y vanishes at r — a contradiction.

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