Let $V: [0, T] (\:\subset \mathbb{R}) \to \mathbb{R}$ be Lipschitz continuous, i.e. $$ | V_D(x,y) |:= |V(x)-V(y)| \leq K | x - y | \quad\forall x, y \in [0, T] , $$ where $V_D(\cdot,\cdot)$ is the difference function associated with $V$. Define the $p$-variation by: $$ \| V\|_p = \left(\sup_{D \subset [0,T]} \sum_{x_j \in D}\lvert V(x_j)-V(x_{j-1}) \rvert ^p\right)^{\frac{1}{p}}. $$ Here, $D = \{t_0, t_1, \ldots, t_n\}$, with $t_0=0$, $t_n=T$ and $t_j > t_{j-1}$ is referred to as a partition of the interval $[0,T]$.

Finally, let $X_s, Y_s: \; [0,T] \to \mathbb{R}$ be two continuous functions of finite $p$-variation. Then, we can consider the composition $ V_D(X,Y): [0,T] \to \mathbb{R}$, which, naturally, satisfies
$$
V_D(X_s,Y_s) = V(X_s) - V(Y_s).
$$
**Can you find a counterexample to the claim that $V$ being Lipschitz implies the $p$-variation of $V$ satisfies a Lipschitz-like condition, as stated below?**

$$| V_D(x,y) | \leq K | x - y | \implies \|V_D(X,Y)\|_p \leq \tilde{K} \| X - Y \|_p $$

Some thoughts:

Another (potential) way of phrasing this is to say: can you find an example of a function that is Lipschitz with respect to the Euclidean distance but not with respect to the $p$-variation distance?

Apparently, it’s a well-known fact that this is false. In fact, I have heard from experts that this has been disproven,apparently in this paper, I could not find where in this paper, however.

If $V $ were linear, this would be true, so from what I have been told, a possibly good idea would be to make the linear part of $V $ more explicit by writing:

$$V(x) - V (y) =: g(x,y) (x-y)$$