Is every function Lipschitz in some Riemannian metric?

Question

Let $M =\mathbb{R}^{n}$ and let $f : M \to \mathbb{R}$ be a smooth function. Define the Riemannian metric $$ \|\delta x\|_{x} = (1+ \|\tfrac{\partial f}{\partial x}(x)\|)\|\delta x\| $$ on $M$ (here $\|\cdot\|$ without a subscript means the Euclidean norm). Put the usual norm on $\mathbb{R}$. Then it seems $f$ is Lipschitz continuous since $$ \sup_{x}\sup_{\delta x\neq 0} \frac{ |\frac{\partial f}{\partial x}(x)\delta x| } { \|\delta x\|_{x} } \leq \sup_{x}\sup_{\delta x\neq 0} \frac{ \|\frac{\partial f}{\partial x}(x)\|\|\delta x\| } { (1 + \|\frac{\partial f}{\partial x}(x)\|)\|\delta x\|} \leq \sup_{x}\sup_{\delta x\neq 0} 1 = 1 $$ Is my thinking right on this one? Thanks in advance.

Answer 1

Yes, as far as bounding the derivative, your reasoning is correct.

However, you have to be careful regarding the smoothness of the metric you define. When $\nabla f(x)$ approaches $0$, then $\| \nabla f(x)\|$ ceases to be smooth in $x$. So the Riemannian metric as you have defined it, is not necessarily smooth. This can be compensated by smoothing it and being a bit more careful with the inequalities, showing that the derivative of $f$ remains bounded.

Note, however, that your proof so far has only showed that every smooth function is Lipschitz in some Riemannian metric. For Lischitz functions you have to add another argument, that allows you to replace your Lipschitz function with a nearby smooth function, for which you construct your Riemannian metric.