Let f be a strictly monotonic positive valued continuous function defined on $[a,b]$ such that $f(a) < a$ and $f(b)>b$ where $b>a>0$ then prove that there exist some $c\in (a,b)$ such that $f’(c)>1$

My approach:

I know that if $f$ is strictly monotonous and continuous then $f$ is differentiable almost everywhere. As graph of $f$ crosses $y=x$, at some $c\in (a,b)$, $f’(c)>1.$ I am not fully convinced with my approach. Is this correct or I need differentiable condition in the problem

  • $\begingroup$ No. If $c$ is a point where the graph crosses the diagonal there's not reason to think that $f$ is differentiable at $c$, and if it is diifferentible at $c$ I don't see why it should follow that $f'(c)>1$. (It's not quite clear to me whether the claim is actually true...) $\endgroup$ – David C. Ullrich Sep 13 '18 at 16:38

If you mean to assert that there exists $c$ such that $f$ is differentiable at $c$ and $f'(c)>1$ then what you're trying to prove is false. (It's probaby true under some weaker notion of "$f'(c)>1$".)

Say $K\subset [0,1]$ is the Cantor set and $g:[0,1]\to[0,1]$ is the "Cantor-Lebesgue function". Then $g(0)=0$, $g(1)=1$, $g$ is continuous and non-decreasing, $g'(x)=0$ for every $x\in[0,1]\setminus K$, and if $x\in K$ it's not hard to see that $g$ is not differentiable at $x$ (because if $x\in K$ then for every $n\ge1$ we have $x\in[j/3^n,(j+1)/3^n]$ with $g((j+1)/3^n)-g(j/3^n)=1/2^n$).

So if you define $$f(t)=g(t)-\frac13+\frac23t$$then $f$ is continuous and strictly increasing, $f(0)<0$, $f(1)>1$, but $f'=2/3$ at every point where $f$ is differentiable.

  • $\begingroup$ Excellent !! Thanks a lot $\endgroup$ – Makar Sep 13 '18 at 17:53
  • 2
    $\begingroup$ (+1) It's actually the lack of absolute continuity that makes the claim false. If $f$ were absolutely continuous, then the statement can easily seen to be true: indeed, if one assumes $f'(x) \leq 1$ almost everywhere, then (in view of absolute continuity of $f$) we have $$ f(x) = f(a) + \int\limits_a^x f'(t) dt \leq f(a) + x - a < x, $$ where the last inequality is due to $f(a) < a$. But then setting $x=b$ we get $f(b) < b$ which is a contradiction. $\endgroup$ – Hayk Sep 13 '18 at 19:28
  • $\begingroup$ @Hayk Yes of course - in fact I'd typed up exactly that argument when I realized that for some reason I'd been assuming $f$ was Lipshcitz... $\endgroup$ – David C. Ullrich Sep 13 '18 at 20:27

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