Suppose I have $f: \mathbb{R}^n \rightarrow \mathbb{R}$ and $f$ is M-Lipschitz. Does this imply that its partial derivatives are bounded? I think yes, but I am not sure if my reasoning is valid:
Since $f$ is M-Lipschitz, by definition that means its gradient must be bounded by $M$ (here I use Euclidian norm):
$$
||\nabla f|| \leq M
$$
Clearly, each of its partial derivatives must be bounded by some constant $< M$. Assume this is not true and there exists some $i$-th partial derivative $\frac{\partial f}{\partial x_i}$ that can take values $\geq M+1$. Then:
$$
||\nabla f|| = \sqrt{… + \left(\frac{\partial f}{\partial x_i} \right)^2 + …} \geq \sqrt{(M+1)^2}
$$
which is a contradiction. Thus, each partial derivative must be bounded by some constant that is less than $M$.
Best Answer
$f$ can be Lipschitz but not differentiable everywhere : $f(x) = ||x||$. If $f$ is differentiable at $x$, then by continuity of the absolute value $$\left|\frac{\partial f}{\partial x_i}(x)\right| = \left|\lim_{h \rightarrow 0} \frac{f(x+he_i)-f(x)}{h} \right| = \lim_{h \rightarrow 0} \left| \frac{f(x+he_i)-f(x)}{h} \right| \leq M.$$ Actually this is how I prove $||\nabla f|| \leq M$ and I would be curious to see how to prove it without the preceeding computations.