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Suppose $f(x)$ is $L^1$ such that $$|f(x)-f(y)|\leq |x-y|^2$$ almost everywhere, does it imply $f$ is a constant function almost everywhere?

vonbrand
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Lucy
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4 Answers4

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Since $f$ is a lipschitz function then $f$ is continuous and so the inequality holds everywhere. Then notice that:

$$0\le\lvert\frac{f(x+h)-f(x)}{h}\rvert=\frac{\lvert f(x+h)-f(x)\rvert}{\lvert h\rvert}\le\lvert h\rvert$$

now let $h$ tend to $0$. So the derivative is $0$ everywhere and hence $f$ is constant.

user71352
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  • Thank you @user71352, I'm sorry that I modified the assumptions on $f$. It's not known if $f$ is continuous. – Lucy Jun 19 '14 at 18:21
  • @Lucy: user71352 has explained that the continuity follows from the Lipschitz condition – Jlamprong Jun 19 '14 at 18:36
  • @Jlamprong But the Lipschitz condition only holds almost every where. Does this still imply $f$ is continuous? – Lucy Jun 19 '14 at 18:37
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    @Lucy Choose $\tilde f$ such that $f = \tilde f$ a. e. and the equality holds everywhere for $f$. Then $\tilde f$ is constant, hence $f$ is constant a. e. – martini Jun 19 '14 at 18:48
  • @martini: how do we know that such an $\tilde{f}$ exists? – robjohn Jun 19 '14 at 19:41
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Yes, it means that $f$ is constant. In fact, $f$ is said to satisfy a Hölder condition if there exists $C, \alpha > 0$ such that: $$\| f(x) - f(y) \| \leq C \|x - y \|^\alpha$$ for every $x,y$ in $f$'s domain. Lipschitz functions are a particular case of Hölder functions with $\alpha = 1$. And if $\alpha > 1$, then $f$ must be constant, because $$0 \leq \frac{\|f(x) - f(y) \|}{\|x - y \|} \leq C \| x - y \|^{\alpha - 1}$$ If $\alpha > 1$, then $\alpha - 1 > 0$, and making $(x - y) \to 0$, we squeeze the incremental quotient between zeros. If the derivative is zero, then the function is constant. Once, I made a question that ended up being related to this subject, and I got a nice answer there. Maybe you'll like reading it.

Community
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Ivo Terek
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Define $$ m(x)=\mu\left\{y:|f(x)-f(y)|\gt|x-y|^2\right\} $$ Since $|f(x)-f(y)|\le|x-y|^2$ a.e., we have that $m(x)=0$ except on a set of measure $0$.

Claim: $f$ is constant on $E=\{\,x:m(x)=0\,\}$.

Proof: Choose any $a\lt b\in E$ and any $\epsilon\gt0$. Set $\delta=\min\left(\frac{\epsilon/3}{b-a},b-a\right)$ and $m=\left\lceil\frac{b-a}{\delta}\right\rceil$.

Then $m\delta\ge b-a$, yet $$ \begin{align} m\delta^2 &\le\left\lceil\frac{b-a}{\delta}\right\rceil\delta\frac{\epsilon/3}{b-a}\\ &\le\frac23\epsilon \end{align} $$ Since the complement of $E$ has measure $0$, $E$ is dense, therefore, we can choose $\{x_k\}_{k=0}^m\subset E$ so that $x_0=a$ and $x_m=b$ and $0\lt x_k-x_{k-1}\le\sqrt{\frac32}\,\delta$. Therefore, $$ \begin{align} |f(b)-f(a)| &=\left|\sum_{k=1}^mf(x_k)-f(x_{k-1})\right|\\ &\le\sum_{k=1}^m|x_k-x_{k-1}|^2\\[3pt] &\le\frac32m\delta^2\\[9pt] &\le\epsilon \end{align} $$ Since $\epsilon\gt0$ was arbitrary, we have that $f(b)=f(a)$ for $a,b\in E$. Thus, $f$ is constant on $E$.
QED

Since the complement of $E$ has measure $0$, $f$ is constant a.e.

robjohn
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Hint: look at $$\frac{|f(y)-f(x)|}{|y-x|} \le |y-x|$$

angryavian
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