Assume $f : \mathbb{R}^d \to \mathbb{R}$ is twice continuously-differentiable and let $U$ be a bounded convex set. I know that if $\sup_{x\in U}\lVert \nabla^2 f(x) \rVert \leq L$ then $\nabla f$ has Lipschitz constant $L$ in $U$. This can be proved by using the function $H(t) = \nabla f(x+t(y-x))$.
What if $U$ is not convex? Is there a counter-example where $\sup_{x\in U}\lVert \nabla^2 f(x) \rVert \leq L$ but $\nabla f|_U$ is not $L$-Lipschitz?
Edit: Perhaps the claim at least holds if $U$ is connected: for any $x,y$, pick a path $\gamma : [0,1] \to U$ such that $\gamma(0) = x, \gamma(1) = y$. Then \begin{align*} \lVert \nabla f(x)-\nabla f(y) \rVert &= \left\lVert \int_0^1 (\nabla f(\gamma(t)))' dt \right\rVert \\ &= \left\lVert \int_0^1 \nabla^2 f(\gamma(t))\gamma'(t) dt \right\rVert \\ &\leq L\int_0^1 \left\lVert\gamma'(t) \right\rVert dt \,. \end{align*} Can we show that $\int_0^1 \left\lVert \gamma'(t) \right\rVert dt \leq \lVert x-y\rVert$ for some appropriate choice of $\gamma$? I think we can always reparametrise such that $\left\lVert \gamma'(t) \right\rVert$ is constant, but can we show that this constant is bounded above by $\lVert x-y \rVert$? This holds for straight lines $\gamma(t) = x+t(y-x)$, but perhaps not in general.
In any case, this still wouldn't answer whether we can find a counterexample for bounded but non-connected $U$. You may assume for e.g. analyticity of $f$ if it helps, but ideally not.