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

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/gsm105prev.pdf ("A First Course in Sobolev Spaces" by G.Leoni). Here $mes$ is the (outer) Lebesgue measure and $$ D^+(f)(x)= \lim sup_{y\to x} \frac{f(x)f(y)}{xy} $$ 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 $$ M= \sup_{x} \lim sup_{y\to x} \frac{f(x)f(y)}{xy} $$ Then, you conclude that for every $s$ and $t$ in the interval $[0,1]$, $$ f(s)f(t)\le mes (f([s,t])) \le M st $$ and, hence, your function $f$ is $M$Lipschitz. 


Yes, it's not hard to show that $f$ is Lipschitz of constant $$k:=\sup_t\lim\sup_{s\to t}\frac{f(s)f(t)}{ts}\, .$$ 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 $f(s)f(t)\le lts$. 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 $$f(1)f(0)\le \sum_{i=0}^{2n1}\bigf(s_{i+1})  f(s_i)\big\le l\sum_{ i = 0 }^{2n1} (s_{i+1}  s_i)=l\, , $$ 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, $$f(b)f(a)\le k ba$$ 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 $f$ .
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 wellknown application of this theorem:
(as before life is simpler applying Zygmund's theorem first to $t\mapsto f+\epsilon t$ with $\epsilon > 0$). 

