Criteria for Lipschitz continuity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:49:23Z http://mathoverflow.net/feeds/question/109613 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109613/criteria-for-lipschitz-continuity Criteria for Lipschitz continuity djoke 2012-10-14T13:54:57Z 2012-10-17T07:54:36Z <p>Is the following statement true. Assume that $f:[0,1]\to [0,1]$ is a continuous function such that $$\sup_t\lim\sup_{s\to t}\frac{|f(s)-f(t)|}{|t-s|}&lt;\infty,$$ then $f$ is Lipchitz continuous.</p> http://mathoverflow.net/questions/109613/criteria-for-lipschitz-continuity/109744#109744 Answer by Misha for Criteria for Lipschitz continuity Misha 2012-10-15T18:40:09Z 2012-10-15T18:40:09Z <p>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 <a href="http://www.ams.org/bookstore/pspdf/gsm-105-prev.pdf" rel="nofollow">http://www.ams.org/bookstore/pspdf/gsm-105-prev.pdf</a> ("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. </p> http://mathoverflow.net/questions/109613/criteria-for-lipschitz-continuity/109768#109768 Answer by Pietro Majer for Criteria for Lipschitz continuity Pietro Majer 2012-10-15T22:49:21Z 2012-10-17T07:54:36Z <p>Yes, it's not hard to show that $f$ is Lipschitz of constant $$k:=\sup_t\lim\sup_{s\to t}\frac{|f(s)-f(t)|}{|t-s|}\ .$$ </p> <p>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 $0:=t_0 &lt; t_1 &lt; t_2 &lt;\dots &lt; t_n:= 1$, and assume it is minimal. By minimality, it follows $J_{t_i}\cap J_{t_{i+1}}\neq\emptyset$ for all $0\le i &lt; n$. 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 $I$. $$*$$ <strong></strong> 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 <a href="http://en.wikipedia.org/wiki/Dini_derivative" rel="nofollow">Dini's derivatives</a> of $f$ . </p> <blockquote> <p><strong>Theorem.</strong> 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.</p> </blockquote> <p>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' <em>Real Analysis</em>, I think. </p> <p>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.</p> <p>Another well-known application of this theorem:</p> <blockquote> <p><strong>Theorem.</strong> 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.</p> </blockquote> <p>(as before life is simpler applying Zygmund's theorem first to $t\mapsto f+\epsilon t$ with $\epsilon > 0$).</p>