A function with rectifiable graph satisfies Lipschitz condition on a large set

Question

Let $f$ be real-valued on $[0,1]$. Let $G$ be the graph of $f$, and suppose it is rectifiable. Let the length of $G$ be $L$. Put $\epsilon > 0$. Then there exists a positive constant $c$ and a measurable set $E \subset [0,1]$ such that $|E| > 1 - \epsilon$ and so that for $x,y\in E$, $f$ satisfies $|f(x) - f(y)| < C \dfrac{L}{\epsilon}|x - y|.$

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I know that if $f'(x)$ exists and is continuous, then $f$ satisfies the Lipschitz condition $|f(x)-f(y)|<M|x-y|$ for some constant $M$, but I'm not sure how to apply this for a rectifiable curve.

Answer 1

I will assume that $f$ is of bounded variation (this contains the case of continuous $f$). This argument can easily be generalized to the case where the set of discontinuities of $f$ has measure zero.

Since $f$ is a BV measurable function, there exists a (unique) Borel measure $\mu_f$ on $[0,1]$ such that $\mu_f(a,b]=f(b)-f(a)$ for all $a<b \in [0,1]$. By the Radon-Nikodym theorem, we have a decomposition $$d\mu_f=g\;dm+d\lambda$$ where $m$ is the Lebesgue measure, $\lambda$ is a measure singular to $m$, and $g$ is some measurable function. Note that $\int_0^1|g|dm \leq |\mu_f|\big([0,1]\big) \leq L$. Similarly $|\lambda| \big([0,1]\big) \leq L$.

From now on fix some $\epsilon>0$.

Define $K$ to be the support of the measure $\lambda$ so $m(K)=0$. Let $U_1$ be some finite union of open intervals containing $K$ such that $m(U_1)<\epsilon/2$ (such a $U_1$ exists by outer regularity of Lebesgue measure). Let $U_2$ be a finite union of open intervals such that $m(U_2)<\epsilon/2$ and $|g(x)|<2L/\epsilon$ for $x\not\in U_2$ (such a $U_2$ exists since $\int_0^1 |g|dm \leq L$). Define $E=[0,1] \backslash (U_1 \cup U_2)$ so $m(E)>1-\epsilon$, and moreover if $x,y$ lie in the same connected component of $E$ with $x<y$ then $$|f(x)-f(y)| \leq |\mu_f| (x,y] = \int_x^y |g| dm \leq \frac{2L}{\epsilon}|x-y|$$

In order to consider the case where $x,y$ lie in different components of $E$, note that $U_1 \cup U_2 = \bigcup_1^k I_j $ is a finite union of intervals of nonzero length, hence $ \alpha:= \min \{ m(I_j): 1 \leq j \leq k\}>0$. Thus if $x,y$ are both endpoints of one of the $I_j$ then we see that $$|\lambda|(x,y] = \frac{|\lambda|(I_j)}{m(I_j)}m(I_j) \leq \frac{L}{\alpha} |x-y|$$

Putting together the results of the previous two equations, we see that if $x,y \in E$ then $$|f(x)-f(y)| \leq \big(\frac{2}{\epsilon}+\frac{1}{\alpha}\big) L |x-y|$$ and we can define $C=2+\epsilon/\alpha$ so this last expression becomes $|f(x)-f(y)| \leq (CL/\epsilon) |x-y|$.

Answer 2

This answer is for functions $f$ of bounded variation. A continuous function is BV if and only if the graph is rectifiable, so this covers the continuous case. See the remarks below for the discontinuous case.

Since $f$ is BV, its distributional derivative $f'$ is a finite signed measure (we don't need to assume $f$ continuous for this). Let $g = M(|f'|)$ be the non-centered Hardy-Littlewood maximal function of the measure $|f'|$. The weak type (1,1) inequality applies to finite measures, and yields $$m(\{x: g(x)>\lambda \})\le \frac{C}{\lambda } |f'|([0,1]) = \frac{CL}{\lambda }$$ Choose $\lambda = CL/\epsilon $ and let $E = \{x: g(x)\le \lambda \}$. Then $m(E)\le \epsilon$ and for $x,y\in E$ we have $$ |f(x)-f(y)| \le \int_{[x,y]} d|f'| \le g(x) |x-y|\le \frac{CL}{\epsilon}|x-y| $$ as required.

Remarks

1. This type of argument is common in analysis on metric spaces, e.g., equation (1.6) here and references therein.

2. If we merely assume that the graph is a rectifiable set, then there is a counterexample. Let $E$ be a fat Cantor set and $f=\chi_E$. The graph is contained in the union of two lines, and the Lipschitz condition fails on any set of sufficiently large measure.