Suppose $f$ is a continuous function on $(a,b)$ such that $$\lim_{h\to 0^+}\frac{f(x+h)-f(x)}{h}\geq 0$$ exists for all $x\in (a,b)$.

Prove that $f$ is an increasing function on $(a,b)$, i.e. $f(x_1)\geq f(x_0)$ for all $x_1\geq x_0$, $x_0,x_1\in (a,b)$.

What I tried is proof by contradiction: Suppose there exists $x_1> x_0$ but $f(x_1)<f(x_0)$.

By Intermediate Value Theorem, there exists $x_0<c_1<x_1$ such that $f(x_1)<f(c_1)<f(x_0)$.

By IVT again, there exists $x_0<c_2<c_1$ such that $f(c_1)<f(c_2)<f(x_0)$.

By repeated use of IVT, there exists $$x_0<\dots c_3<c_2<c_1$$ with $$f(c_1)<f(c_2)<f(c_3)<\dots<f(x_0)$$

Since $c_n$ is decreasing and bounded below, $L:=\lim c_n$ exists. Similarly, $\lim f(c_n)=f(L)$ exists.

Then $$\lim_{h\to 0^+}\frac{f(L+h)-f(L)}{h}=\lim_{n\to\infty}\frac{f(c_n)-f(L)}{c_n-L}\leq 0$$ since $f(c_n)-f(L)<0$ and $c_n-L>0$.

This is almost enough to be a contradiction had it been a strict $<0$. However at this point it seems not enough.

Thanks for any help. Note: I wouldn't rule out the possibility that the question itself is wrong..

Update: I think I got it. First prove the statement for a function $g$ with $\frac{g(x+h)-g(x)}{h}\geq\epsilon>0$ on $(a,b)$. The above proof will do the job since now a contradiction can be achieved.

Then apply it to the function $g(x)=f(x)+\epsilon x$. Since $\epsilon>0$ is arbitrary, it can be proved that $f$ is increasing.

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    $\begingroup$ This is not a simple problem. Any answer must make use of some form of completeness of real numbers. One approach is to use Dedekind's theorem. Also continuity of the function is necessary. $\endgroup$ – Paramanand Singh Dec 7 '16 at 7:46
  • $\begingroup$ @ParamanandSingh Do you happen to know a reference for this question? $\endgroup$ – yoyostein Dec 7 '16 at 7:51
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    $\begingroup$ It's available in Hardy's Pure Mathematics. I will give page no question no in a minute or so. $\endgroup$ – Paramanand Singh Dec 7 '16 at 7:53
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    $\begingroup$ It's page 208 question 19, 10th edition Hardy's A Course of Pure Mathematics. $\endgroup$ – Paramanand Singh Dec 7 '16 at 8:16
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    $\begingroup$ @Matteo: I am not aware of any results dealing with functions whose right derivatives are monotone. Perhaps you could ask this as a question and let's see what our community has to say on that. $\endgroup$ – Paramanand Singh Feb 3 at 14:07

See snapshot of Hardy's book question 19: enter image description here


As pointed out by Paramanand in the comments, without continuity no conclusion on monotonicity at right neighborhoods of points in the domain can be drawn, even though $$f'_+(x) = \lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h} \geq 0$$ $\forall x\in\Bbb R$. I show in this answer, too, for completeness, a counterexample that I already posted here. It's a function $f(x)$ with a well defined, non negative right derivative in $\Bbb R$, which however has $f(0) = 0$ and $f(x) <0$ for $x>0$.

enter image description here

The red lines are parabolic envelopes with equations $y=-x|x|$ and $y=-\frac{1}{2}x|x|$ and the function is $$f(x) = \begin{cases}\frac{\sqrt{2^{-k}}-\sqrt{2^{-k+1}}}{\sqrt[4]{2^{-k+1}}-\sqrt[4]{2^{-k}}}(x+\sqrt[4]{2^{-k+1}})- \sqrt{2^{-k+1}} & \left(-\sqrt[4]{2^{-k+1}}\leq x<-\sqrt[4]{2^{-k}}; \ k\in \Bbb Z\right)\\ 0 & (x=0)\\ \frac{\sqrt{2^{-k}}-\sqrt{2^{-k+1}}}{\sqrt[4]{2^{-k+1}}-\sqrt[4]{2^{-k}}}(x-\sqrt[4]{2^{-k+1}})+ \sqrt{2^{-k+1}}& \left(\sqrt[4]{2^{-k}}\leq x<\sqrt[4]{2^{-k+1}}; \ k\in \Bbb Z\right) .\end{cases}$$

Let us now demonstrate the following

Theorem Let $f(x)$ be a continuous function in $[a,b]$ such that \begin{equation}f'_+(x) = \lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h} \geq 0, \ \ \forall x\in[a,b).\tag{1}\label{eq:1}\end{equation} Then, for all $a\leq x \leq y\leq b$, $f(y)\geq f(x)$.

We will need the following variation of Rolle's Theorem.

Lemma If $f(x)$ a is continuous function in $[a,b]$ with well defined right derivative $f'_+(x)$, and such that $f(a) = f(b)$, then there are points $\alpha,\beta\in[a,b]$ for which $f'_+(\alpha)\geq 0$ and $f'_+(\beta)\leq 0$.

Since $f(x)$ is continuous in $[a,b]$, by Weierstrass Theorem it must reach its minimum and maximum in such interval. If $f(a) = f(b)$ is minimum and maximum then the function is constant and the Lemma is proved. If $f(a) = f(b)$ is a minimum, then the maximum must be $\beta \in (a,b)$, and we have $f(x) \leq f(\beta)$. Hence, if $x\in(\beta, b] $, then we obtain $\frac{f(x)-f(\beta)}{x-\beta}\leq 0$. Therefore $f'_+(\beta) \leq 0$. Also, by minimality of $a$, we have $f(x) \geq f(a)$ for $x\in [a,b]$ so that $f'_+(a) \geq 0$. Similary, we can prove the statement if $f(a) = f(b)$ is a maximum, and if both minimum and maximum are in the open interval $(a,b)$.$\square$

Now we are ready to prove the Theorem, by contradiction. Suppose there are $c_1, c_2\in[a,b]$ such that $c_1 < c_2$ and $f(c_1) > f(c_2)$.

As when proving the Mean Value Theorem, consider an auxiliary function $$F(x) = f(c_2)-f(x) + K(x-c_2),$$ with $$K = \frac{f(c_2)-f(c_1)}{c_2-c_1}< 0.$$ It is easy to verify that $F(x)$ satisfies the Lemma hypotheses in $[c_1,c_2]$, with $F(c_1) = F(c_2) = 0$, and \begin{equation}F'_+(x) = K-f'_+(x).\tag{2}\label{eq:2}\end{equation} By the Lemma there is a point $c\in[c_1,c_2]$ such that $$F'_+(c) \geq 0,$$ which, by \eqref{eq:2}, yields $$f'_+(c) \leq K < 0,$$ a contradiction. $\blacksquare$

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    $\begingroup$ The proof of your lemma requires slight modification. Suppose the maximum is attained at $\beta$ then we have $f(x) \leq f(\beta) $ and there is no need of strict inequality. $\endgroup$ – Paramanand Singh Feb 4 at 2:25
  • $\begingroup$ Nice proof altogether +1 $\endgroup$ – Paramanand Singh Feb 4 at 2:27
  • $\begingroup$ @ParamanandSingh thanks! I'll make the correction! $\endgroup$ – dfnu Feb 4 at 7:52

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