Let $f: X \to \mathbb{R}^n$ where $X$ is a Banach space. $f$ is called locally Lipschitz around $\overline{x}$ if there exist $L > 0$ and a neighborhood $U$ of $\overline{x}$ such that $$\|f(x) - f(x')\| \leqslant L\|x - x'\|, \quad \text{for all} ~ x,x' \in U.$$
As I mentioned on the title, assume that $f$ is locally Lipschitz around $\overline{x}$ and differentiable on a neighborhood of $\overline{x}$.
- Is $f'$ necessarily continuous at $\overline{x}$?
- Is $f'$ locally Lipschitz around $\overline{x}$?
Edit 1: $f$ is now differentiable on a neighborhood of $\overline{x}$. This is my proof for part 1.
Let $u \in X$, $x_n \to \overline{x}$ and $y_n = f'(x_n)u$. We will show that $y_n$ converges to $f'(\overline{x})u$. The Lipschitz property of $f$ ensures the boundedness of $(y_n)$, then by Bolzano-Weierstrass theorem, $(y_n)$ has a subsequence (denoted the same as $(y_n)$) converges to some $y \in \mathbb{R}^n$. Because $y_n = f'(x_n)u$, there exist $h_{n(k)} := \|x_n - \overline{x}\|^\frac{1}{2}$ and $u_k \in X$ such that, for all $k \in \mathbb{N}$, \begin{equation} \|u_k - u\| < \dfrac{1}{k}, \quad \|y_{n(k)} - y_n\| < \dfrac{1}{k}, \quad \text{where} ~ y_{n(k)} := \dfrac{f(x_n + h_{n(k)}u) - f(x_n)}{h_{n(k)}}. \end{equation} Setting \begin{equation} y'_{n(k)} := \dfrac{f(\overline{x} + h_{n(k)}u) - f(\overline{x})}{h_{n(k)}}, \end{equation} it is needed to verify that $y'_{n(k)} \to y_n$ (as $k \to \infty$) for all $n \in \mathbb{N}$. Due to the Lipschitz continuity of $f$, there exists $L > 0$ such that, for $k$ large enough, \begin{align} \|y'_{n(k)} - y_{n(k)}\| &= \dfrac{1}{h_{n(k)}} \|f(\overline{x} + h_{n(k)}u) - f(x_n + h_{n(k)}u) + f(x_n) - f(\overline{x})\| \\ &\leqslant \dfrac{1}{h_{n(k)}} \big(\|f(\overline{x} + h_{n(k)}u) - f(x_n + h_{n(k)}u)\| + \|f(x_n) - f(\overline{x})\|\big) \\ &\leqslant \dfrac{1}{h_{n(k)}} \cdot 2L \cdot \|x_n - \overline{x}\| = 2L \cdot \sqrt{h_{n(k)}}. \end{align} Hence, \begin{equation} \|y'_{n(k)} - y_n\| \leqslant \|y'_{n(k)} - y_{n(k)}\| + \|y_{n(k)} - y_n\| < 2L \cdot \sqrt{h_{n(k)}} + \dfrac{1}{k}. \end{equation} The right-hand side tends to $0$, which means $y'_{n(k)}$ tends to $y_n$ as $k \to \infty$. Letting $n \to \infty$, one gets $y'_{n(k)} \to y$. By the setting of $y'_{n(k)}$, it follows that $y = f'(\overline{x})u$.
This is not true. Here is a counterexample. Take $$ f(x) = \begin{cases}x^2 \sin(x^{-1}) & \text{ if } x\ne 0,\\ 0 & \text{ if } x=0. \end{cases} $$ then $$ f'(x) = \begin{cases} 2x \sin(x^{-1}) - \cos(x^{-1}) & \text{ if } x\ne 0,\\ 0 & \text{ if } x=0. \end{cases} $$ Then $f'$ is bounded on bounded intervals, hence $f$ is locally Lipschitz continuous (by mean-value theorem). However, $f'$ is not continuous at $x=0$.