# Does Lipschitz implies Lipschitz-like in $p$-variation?

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:

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