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Is the following statement true. Assume that $f:[0,1]\to [0,1]$ is a continuous function such that $$\sup_t\lim\sup_{s\to t}\frac{|f(s)-f(t)|}{|t-s|}<\infty,$$ then $f$ is Lipchitz continuous.

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Yes, it is true, but could you provide some motivation for the question? –  Misha Oct 14 '12 at 21:21
I am to doing some research, I need to do this. Previously I was not suspicious that this is true, but I didn't find any simple proof. –  djoke Oct 15 '12 at 6:44
@Misha You told that it is true, but no additional comments. Maybe it is not true –  djoke Oct 15 '12 at 12:17

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up vote 4 down vote accepted

Even more is true: Suppose that $E$ is a measurable subset of an interval, $f$ is a function on $E$ so that for every $x\in E$, $|D^+(f)(x)|\le M$ (the quantity $|D^+(f)|$ is defined below). Then $$ mes(f(E))\le M \cdot mes(E), $$ see e.g. the proof of Lemma 3.13 in http://www.ams.org/bookstore/pspdf/gsm-105-prev.pdf ("A First Course in Sobolev Spaces" by G.Leoni). Here $mes$ is the (outer) Lebesgue measure and $$ |D^+(f)(x)|= \lim sup_{y\to x} \frac{|f(x)-f(y)|}{|x-y|} $$ Note that Lemma 3.13 assumes that $f$ is differentiable, but the proof uses only the upper bound $|D^+(f)(x)|\le M$ for all $x$. In your case, $E$ is an interval $[s,t]$ in $[0,1]$ and $$ M= \sup_{x} \lim sup_{y\to x} \frac{|f(x)-f(y)|}{|x-y|} $$ Then, you conclude that for every $s$ and $t$ in the interval $[0,1]$, $$ |f(s)-f(t)|\le mes (f([s,t])) \le M |s-t| $$ and, hence, your function $f$ is $M$-Lipschitz.

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@Misha. Thank you for your very interesting solution. –  djoke Oct 15 '12 at 19:57

Yes, it's not hard to show that $f$ is Lipschitz of constant $$k:=\sup_t\lim\sup_{s\to t}\frac{|f(s)-f(t)|}{|t-s|}\, .$$

Indeed, let's take any $l > k$. By assumption, any $t\in I:=[0,1]$ has a nbd $J_t:=]t',t''[\cap I$ such that for any $s\in J_t$ one has $|f(s)-f(t)|\le l|t-s|$. Extract a finite covering of $I$ by these nbd's, say $J_{t_0}\dots J_{t_{n }}$ with $0:=t_0 < t_1 < t_2 <\dots < t_n:= 1$, and assume it is minimal. By minimality, it follows $J_{t_i}\cap J_{t_{i+1}}\neq\emptyset$ for all $0\le i < n$. As a consequence, there is a sequence $0:=s_0 \le s_1 \le \dots \le s_{2n}:=1$ such that $s_{2i}=t_i$ and $s_{2i+1}\in J_{t_i}\cap J_{t_{i+1}}$, Therefore, since any two consecutive $s_j$'s are (in some order) a $t_i$ and an element of $J_{t_i}$, we have $$|f(1)-f(0)|\le \sum_{i=0}^{2n-1}\big|f(s_{i+1}) - f(s_i)\big|\le l\sum_{ i = 0 }^{2n-1} (s_{i+1} - s_i)=l\, , $$ and in fact $|f(1)-f(0)|\le k$ because the above inequality holds for any number $l > k$. This also implies, by rescaling and localizing, $$|f(b)-f(a)|\le k |b-a|$$ for all $a$ and $b$ in $I$. $$*$$ [edit] I realize now that there is another proof (which is, in a sense, the true proof) that follows from a beautiful theorem by Antoni Zygmund. Let $D f$ represent any of the Dini's derivatives of $f$ .

Theorem. Let $f:[a,b]\rightarrow\mathbb{R}$ be continuous, and assume that the set $f(\{Df \le 0\} )$ is nowhere dense. Then $f$ is increasing.

What is great of this theorem is the statement, natural and powerful; once you know the statement, the proof is not difficult by a nice argument by contradiction (assume that e.g. $f(a) > f(b)$ and consider a value $\lambda\in \big( f(b), f(a) \big)\setminus f(\{Df \le 0\} )$; then argue on the larger $t$ such that $f(t)=\lambda$...). There should be a proof in Mc Shane and Botts' Real Analysis, I think.

In your case, apply Zygmund's theorem to the functions $t\mapsto lt\pm f(t)$, with any $l > k$, as above and conclude that $f$ is $k$ Lipschitz.

Another well-known application of this theorem:

Theorem. Let $f:[a,b]\rightarrow\mathbb{R}$ be continuous and admit $f'(t) \ge0$ up to countably many points; or either, absolutely continuous and $f'(t)\ge 0$ a.e. Then $f$ is increasing.

(as before life is simpler applying Zygmund's theorem first to $t\mapsto f+\epsilon t$ with $\epsilon > 0$).

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This is very elegant proof. –  djoke Oct 16 '12 at 15:25
have another one ;) –  Pietro Majer Oct 16 '12 at 21:18

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