Let M be a connected complete Riemannian manifold and denote by $\nabla$ and $\triangle$ the Levi-Civita connection and the Laplace operator of M, respectively. A non-constant function f of class $C^{2}$ on $M$ is called transnormal if \begin{equation}\label{RK 1} |\nabla f|^{2}=b(f) \end{equation} for some real function b of class $C^{2}$ defined on the range of $f$, and $f$ is called isoparametric if, in addition to the first equation, it satisfies \begin{equation}\label{RK 2} \triangle(f) = a(f) \end{equation} for some continuous real function a defined on the range of $f$. The first equation implies that the level sets of $f$ are parallel to each other. The second equation implies that a level hypersurface has constant mean curvature. Qustion: If $f$ is transnormal function and the level sets of $f$ are parallel to each other, how to prove that $f$ is isoparametric function?

For spaces of constant curvature this has been done by Élie Cartan in Familles de surfaces isoparamétriques dans les espaces à courbure constante.

In general, however, your claim is not true. See Theorem 1.4 in Transnormal functions on Riemannian manifolds by Reiko Miyaoka.

`\Delta`

instead of`\triangle`

for the Laplacian operator. $\endgroup$ – Alex M. Mar 25 '19 at 14:22