I have read in textbooks the following phrase, "assuming the gradient is locally Lipschitz continuous...", but how does one prove that the gradient is locally Lipschitz continuous?
If we have a continuously differentiable function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, from looking at previous stackexchange questions on this topic it seems we can show the gradient is Lipschitz continuous via Taylor's formula, but was wondering if anyone has insight or can point to papers/resources that clarify this further. Thank you.
It is entirely possible, even for twice differentiable functions, to have a gradient that isn't locally Lipschitz. Namely, the function $g(t)=\begin{cases}t\lvert t\rvert^{1/2}\cos\frac1t&\text{if }t\ne 0\\ 0&\text{if }t=0\end{cases}$ is differentiable with derivative $g'(t)=\begin{cases}\lvert t\rvert^{1/2}\cos \frac1t-\frac{\lvert t\rvert^{1/2}}{t}\sin\frac1t&\text{if }t\ne 0\\ 0&\text{if }t=0\end{cases}$, which is unbounded in all neighbourhoods of $0$. Since a differentiable function $g:I\to\Bbb R$ is Lipschitz if and only if its derivative is bounded, $g$ isn't locally Lipschitz. $g$ is itself the derivative of the function $G(x)=\int_0^x g(t)\,dt$, which ends up being a twice-differentiable counterexample to your claim for $n=1$. For $n>1$, consider $f(x)=G(x_1)$.