Does $C^1$ imply locally Lipschitz on Banach spaces?

Question

Let $X,Y$ be Banach spaces, $\Omega \subseteq X$ open and $f: \Omega \to Y$ continuously (Fréchet-) differentiable. Does this imply that $f$ is already locally Lipschitz? For finite dimensional spaces this is trivially true, but I am not certain if this holds for infinite dimensional Banach spaces.

Answer 1

Let $Df_x: X \to Y$ denote the Fréchet derivative of $f$ at $x$. Then $Df_x$ is a bounded linear operator (with the operator norm locally uniformly bounded in $x$) such that $$\lim_{\|h\| \to 0} \frac{\|f(x+h) - f(x) - Df_x(h)\|}{\|h\|} = 0.$$

In particular, there is a $\delta > 0$ such that $\|h\|< \delta$ implies that $\|f(x+h) - f(x) - Df_x(h)\| < \|h\|$. As a result, for $\|x-y\| < \delta$, $$\|f(y) - f(x)\| \leq \|f(y) - f(x) - Df_x(y-x)\| + \|Df_x(y-x)\| < (1+ \|Df_x\|)\|y-x\|$$ which implies that $f$ is locally Lipschitz.