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

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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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