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I have a question which came up studying a certain (system of) PDE:

If I can write a function $f(x,y,z)$ as $g(x-y,z)$ as well as $h(x^2-z, y)$ with $g$, $h$ continuous and differentiable, does it follow that $f$ is constant? I can't imagine it does but I can't find any counterexample..

Thanks

1 Answers 1

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Yes, $f$ is constant.

For any $x_1$ and $x_2$, the form of $g$ implies that $f(x_1,y,z) = f(x_2,y+x_2-x_1,z)$, while that of $h$ implies that $f(x_1,y,z) = f(x_2,y,z+x_2^2-x_1^2)$. That is, $f$ is constant along certain parallel lines in the $xy$ plane, and along parallel parabolas in the $xz$ plane. We will show that you can reach any point $(x,y,z)$ from $(0,0,0)$ by travelling only along these curves along which $f$ is constant.

Let $G_s(x,y,z) = (x+s,y+s,z)$ and $H_t(x,y,z) = (x+t,y,z+(x+t)^2-x^2)$, so that $G_s$ and $H_t$ preserve $f$ for all $s$ and $t$. Starting from $(0,0,0)$, you can reach any point $(0,0,2s^2)$ on the positive $z$-axis via $G_s \circ H_{-s} \circ G_{-s} \circ H_s$, and any point $(0,0,-2s^2)$ on the negative $z$-axis via the reverse, $H_s \circ G_{-s} \circ H_{-s} \circ G_s$. From the $z$-axis, you can reach any point on the $xz$-plane using $H$, and from there to any point in $\mathbb R^3$ using $G$.