I can't work out if this is true or not:
Let $f:\mathbb{R}^N\rightarrow \mathbb{R}$ be differentiable at zero with $\nabla f(0)=0$ and $f(0)=0$. Is it true that for any $\delta > 0$, there is a neighbourhood of $0$ over which $f$ is Lipschitz with constant $\delta$? i.e. does this hold:
$$\forall \delta > 0, \exists r > 0 \text{ such that } \forall x, y \in B_r(0), |f(x)-f(y)|<\delta |x-y|$$
where $B_r(0)$ is the $r-$ball around $0$.
I think it is true but I am struggling to prove it. I thought maybe I could use a proof by contradiction by assuming the opposite of the above and finding $x_n,y_n\rightarrow 0$ such that $|f(x_n)-f(y_n)|>\delta|x_n-y_n|$ to then somehow reach that $|\nabla f|>\delta$ but I can't get this to work. If it isn't true, I can't manage to make a counterexample. Note, I do not want to assume differentiability of $f$ for $x\neq 0$.
A counterexample is $f(x):=x^{3/2}\sin\frac1x$ if $x\ne0$, $f(0):=0.$
$f'(0)=\lim_{x\to0}\sqrt x\sin\frac1x=0$ but $f'(x)=\frac32\sqrt x\sin\frac1x-\frac1{\sqrt x}\cos\frac1x$ takes arbitrarily large values in any neighborhood of $0$.