Lipschitz does not bound increments by 1-variation

Question

Can you find a counterexample to the below claim?

Claim: Let $f:\mathbb{R}\to \mathbb{R}$ be Lipschitz, i.e.

$$|f(x)-f(y)| \leq K_1 | x - y |$$ for all $x,y$ .

Let $x, y :[0,T]\to \mathbb{R}$ be continuous of bounded variation (i.e. in $C_{BV}([0,T])$ ) and write $x_t:=x(t)$ and $y_t:=y(t)$. Then $f$ satisfies, for any such $ x, y $ :

$$ | f(x_t) - f(y_t) - (f(x_s)-f(y_s))| \\ \leq K _2 \sup_{t,s \in [0,T] } | ( x_t - y_t) - (x_s- y_s) | \\ (\leq K_2 \operatorname{Variation}{(x_{(\cdot)} - y _{(\cdot)})})$$

I know this is false from academic vox populi, and posted about this before but my previous posts went through too many edits

Some potentially illuminating workings:

First, note that if $f$ were linear (i.e. the $g(\cdot,\cdot)$ below is a constant) the claim would be true (it will become evident shortly).

In view of this, consider the non-linear part of $f$, $g(x,y) $, as defined through $f(x)-f(y)$$= g(x,y)(x-y)$. Since $|f(x)-f(y)|$ = $| g(x,y)(x-y )|$ $\leq K_1 | x - y |$, it follows that $g(x,y) \leq K_1$ for all $x,y$ in $C_{BV}([0,T])$. Then,

$$ | f(x_t) - f(y_t) - (f(x_s)-f(y_s))| \\ = | g(x_t,y_t)(x_t-y_t ) - g(x_s,y_s)(x_s-y_s)| \\ = | g(x_t,y_t) | \left|x_t-y_t - \frac{g(x_s,y_s)}{g(x_t,y_t)}(x_s-y_s)\right|\\ \leq K_2 \left|x_t-y_t - \frac{g(x_s,y_s)}{g(x_t,y_t)}(x_s-y_s)\right| \\ (\text{not true ?}) \leq K_2 \sup_{t,s \in [0,T] } | ( x_t - y_t) - (x_s- y_s) | $$

I think that if we choose a function such that its non-linear part can wiggle a lot such that $g(x_t,y_t)/g(x_s,y_s)$ can get very large, then this may work. For example, if $f(x_t) = 1/x_t$ were Lipschitz, this may work since $g(x_t,y_t)/g(x_s,y_s) = x_s y_s /(x_t y_t) $ can get as large as we need in the interval $[0,T]$.

Answer 1

Let $f: \mathbb{R} \to\mathbb{R}$ be $f(a)=a^2$ on $[-10,10]$, $f(a)=100$ elsewhere. Then $f$ is globally Lipschitz. Consider $x,y:[0,1]\to\mathbb{R}$ defines by $x_t = t$, $y_t = t-1$, then:

$$ \operatorname{Variation}( x_t - y_t ) \\ \operatorname{Variation}( 1 ) = 0 $$

but

$$ | f(x_t) - f(y_t) - ( f(x_s) - f(y_s) ) | \\ = | t^2- (t^2 + 1- 2 t) - ( s^2 - ( s^2 + 1 - 2 s) ) | \\ = 2 | t - s | \\ > 0 \text { whenever } t\neq s $$

as desired

Answer 2

Let $f$ to be a nonconstant $1$-Lipschitz function on $[-1.1]$ and extend to a $1$-Lipchitz function $\mathbb{R}\to\mathbb{R}$ by constant: $f(t)=f(-1)$ for all $t\leq -1$ and $f(t)=f(1)$ for all $t\geq 1$. Define $x,y\colon[0,2+\epsilon]\to\mathbb{R}$ by $x(t)=t-1$, $y(t)=t-1-\epsilon$. Then $x-y=\epsilon$ is a constant, hence $\operatorname{Variation}(x-y;[0,2+\epsilon])=0$, but obviously $f(x)-f(y)$ is nonconstant.