Calculate the original function of a Lipschitz function 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?
 A: 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.
A: 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.
A: 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.
A: Hint: look at $$\frac{|f(y)-f(x)|}{|y-x|} \le |y-x|$$
