For a Lipschitz function, is the gradient (when defined) orthogonal to level set?

Question

For a differentiable function f, we know that the gradient is orthogonal to the level sets of f.

What if f is Lipschitz continuous? we know that its gradient exists almost everywhere. Is the gradient (when it exists) orthogonal to the level sets?

Answer 1

For a Lipschitz function, the normal of the level set may not be defined because there is no implicit function theorem for Lipschitz functions; for example, consider $$f(x) = d_K(x) = \inf_{y \in K} |x - y|$$ for any nonempty closed set $K \subseteq \mathbb{R}^n$. Then $d_K$ is Lipschitz with constant 1 and has $K$ for its zero level set.

However, if for a given $k\in\mathbb{R}$ the level set $K = \{f(x) = k\}$ is already known to have an almost everywhere well-defined normal $n(x)$, then by definition you can say that this normal is parallel to the gradient of $f$.

To illustrate this fact, consider the case of a Lipschitz level curve $K$. Then up to a rotation of the domain, $K$ can locally be written as $(y,\phi(y))$, where $y\in V\subseteq \mathbb{R}^{n-1}$. The normal vector for $K$ on the section $V \times \phi(V)$ is then defined to be $$n(y,\phi(y)) = \frac{\left<-\frac{\partial \phi}{\partial y_1}(y),-\frac{\partial \phi}{\partial y_2}(y),...,-\frac{\partial \phi}{\partial y_{n-1}}(y),1\right>}{\sqrt{\phi_{y_1}^2(y) +\phi_{y_2}^2(y) + ... + \phi_{y_{n-1}}^2(y) + 1}}.$$ Then, $f(y,\phi(y)) = k$ for all $y\in V$ and for any $i\in 1,2,...,n-1$, $$0 = \frac{\partial}{\partial y_i} f(y,\phi(y)) = \frac{\partial f}{\partial x_i}(y,\phi(y)) + \frac{\partial f}{\partial x_n}(y,\phi(y))\frac{\partial \phi}{\partial y_i}(y)$$ $$\implies -\frac{\partial f}{\partial x_i}(y,\phi(y)) = \frac{\partial f}{\partial x_n}(y,\phi(y))\frac{\partial \phi}{\partial y_i}(y).$$ Thus, after combining the above equations with the definition of $n(y,\phi(y)$ and writing the result in vector form we have $$\frac{\partial f}{\partial x_n}(y,\phi(y)) n(y,\phi(y)) = -\frac{\nabla f(y,\phi(y))}{\sqrt{\phi_{y_1}^2(y) +\phi_{y_2}^2(y) + ... + \phi_{y_{n-1}}^2(y) + 1}}.$$