Suppose that is a continuously differentiable function such that at every point of the plane at least one of the partial derivatives is zero.
Prove that depends on just one variable (either or ). In other words, either or .
Proof (suggested by Lianxin He). Let be the (open) set where . We claim that is a union of vertical lines. Suppose not. Then there is an vertical open line segment such that (a) is not the entire line; (b) at least one of the endpoints of , say , is not contained in .
Since in , it follows that is constant on . On the other hand, at . We conclude that on , a contradiction. Thus, is a union of vertical lines.
The same argument shows that the set is a union of horizontal lines. Since by our assumption , it follows that one of the sets 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 regularity, which complicates the matter. Indeed, the example 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 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: 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 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 be a closed axes-aligned rectangle , and let be a vertex of . Then the connected component of in the set meets a side of not containing .
Assume Lemma 1 for now.
Lemma 2. In the notation of Lemma 1, for some vertex adjacent to .
Proof: We may assume is the bottom-left vertex and the component of in meets the right side of . Then separates the bottom-right vertex from the sides not containing it. This leads to a contradiction unless . Lemma 2 is proved.
Now the proof is finished as follows. Suppose attains distinct values at two points , lying on the same vertical line. For each point on this line pick such that . Applying Lemma 2 to all rectangles with and as vertices, we see that is constant on the horizontal line through . Since was arbitrary, 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 with . It also uses, perhaps surprisingly, the equality of components and quasicomponents in compact Hausdorff spaces (such as ).
We may assume is the bottom-left vertex of . Suppose the component of in does not meet either the top or the right side of . Then there exist disjoint sets which are open in and cover in such a way that contains and contains both the top and the right side of . There exists such that for all . The set is compact and nonempty. Let be its largest point in the lexicographic order. The points such that and satisfy . In particular, this holds when approaches from the right or from above (note that the construction of makes such approaches possible). But either or vanishes at — a contradiction.