Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $f$ is continuous on $[a,b]$ and $f\;'> 0$ on $(a,b)$, then $f$ is monotonically increasing on $[a,b]$ and proof follows from mean value theorem. Now suppose we have $f \in C[a,b]$ and $f\;'(x+)> 0$ for all $x \in (a,b)$. Does it follow that $f$ is monotonically increasing on $[a,b]$?

share|improve this question
add comment

1 Answer 1

Suppose that $a<u<v<b$ with $f(u)>f(v)$. Let $C=\{x\in[u,v]:f(x)=f(u)\}$; $f$ is continuous, so $C$ is a closed subset of $[u,v]$ and therefore has a maximum element. Without loss of generality we may assume that this maximum element is $u$. The intermediate value theorem then ensures that $f(x)<f(u)$ for all $x\in(u,v]$. But $$0<f\;'(u+)=\lim_{h\to 0^+}\frac{f(u+h)-f(u)}{h},$$ so there is a $y\in(u,b)$ such that $f(x)>f(u)$ for every $x\in(u,y)$. This contradiction shows that $f$ must be monotone increasing.

share|improve this answer
1  
@Tigran: I don’t think that you understand the argument. I’m not using the mean value theorem anywhere. Of course $f$ is constant on $C$; so what? $C$ is used merely to find its maximum, $u$, so that I have an interval $(u,v]$ on which $f(x)<f(u)$. –  Brian M. Scott Nov 2 '11 at 13:24
    
Now I got your argument. Thank you for your nice solution. –  Tigran Hakobyan Nov 2 '11 at 13:36
    
@Tigran: You’re very welcome. –  Brian M. Scott Nov 2 '11 at 13:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.