# Prove that if $f\in C([a,b])$, $f(a)=f(b)$ and $f_{-}'$ exists on $(a,b)$ then $\inf\{f_{-}'(x):x\in (a,b)\}\leq 0.$

Prove that if $f\in C([a,b])$, $f(a)=f(b)$ and $f_{-}'$ exists on $(a,b)$ then $\inf\{f_{-}'(x):x\in (a,b)\}\leq 0\leq \sup\{f_{-}'(x):x\in(a,b)\}.$

My Attempt: Since $f$ is a continuous function on a compact interval we must have $p,q\in [a,b]$ such that $f(p)=M$ and $f(q)=m$, where $M$ is the maximum value of $f$ and $m$ is the minimum value. Now we must have $f_{-}'(M)\leq 0$ and so the $\inf\{f_{-}'(x):x\in (a,b)\}\leq 0$. Similarily $f_{-}'(m)\geq 0$ and so $\sup\{f_{-}'(x):x\in(a,b)\}\geq 0.$

There is a futher generalization of this when $f(a)\not =f(b)$ then we have $$\inf\{f_{-}'(x):x\in (a,b)\}\leq \frac{f(b)-f(a)}{b-a}\leq \sup\{f_{-}'(x):x\in(a,b)\}.$$

Both these problems are very much related and I am not sure whether my attempted proof is correct or not. So it would be great if someone could give some feedback. Furthermore, if the proof works then I can't see a way to generalize this to the above-stated fact. So any hints in that direction would also be much appreciated.

• The general result is derived from the specific one by using function $g(x) =f(b) - f(x) - ((f(b) - f(a)) /(b-a)) (b-x)$. Dec 20 '17 at 8:59
• Your approach is essentially correct, but it appears you have considered right derivative in your mind. Dec 20 '17 at 9:42

If $\inf\{f_{-}'(x):x\in (a,b)\}> 0$ would hold then $f$ would grow strictly monotonious, so for every $c > a$ we would have $f(c) > f(a)$. This contradicts $f(a) = f(b)$.
• How can you say that the function will be strictly monotonous, I have only assumed that $f_'(x)$ exists.
• @SuperMario: the result is correct, but difficult to prove: if $f$ is increasing from the left at each point of an interval and is continuous on that interval then $f$ is increasing on that interval. Having a positive left derivative ensures that the function is increasing from left at the point under consideration. Dec 20 '17 at 8:55