An upper bound for $\|2 \nabla f(x) - \nabla f(y)\|$ in terms of $\|x-y\|$ if the gradient is $L$-Lipschitz

Question

Suppose we have a function $f \colon \mathbb R^n \to \mathbb R$ with $L$-Lipschitz gradient in Euclidean $2$ norm, i.e., $\| \nabla f(x) - \nabla f(y)\| \le L\|x-y\|$. Is it possible to find a constant $C$ such that \begin{align*} \| 2\nabla f(x) - \nabla f(y)\| \le C\|x-y\|. \end{align*}

Answer 1

Impossible unless $f$ is constant. Your condition implies that $\nabla f$ is continuous. If your inequality is true, then let $y\to x$. Since $\nabla f(y) \to \nabla f(x)$, we have

$$ \nabla f(x) = 0.$$

This implies that $f$ is constant.

Answer 2

\begin{align*} \|2\nabla f(x) - \nabla f(y)\| &= \|\nabla f(x) + \nabla f(x) - \nabla f(y)\|\\& \leq \|\nabla f(x) \| + \|\nabla f(x) - \nabla f(y)\| \\ & \leq \|\nabla f(x) \| + L\|x-y\| \end{align*}

You need to find some kind of condition $\|\nabla f(x)\| \leq K \|x-y\|, \forall x,y \in \mathbb{R}^n$ in order to say something about the Lipschitz constant of the original expression. Actually, this condition implies that $\nabla f(x) = 0$ (take $y = x$) and hence $f(x)$ is a constant.