2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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) $. $\endgroup$ Dec 20 '17 at 8:59
  • $\begingroup$ Your approach is essentially correct, but it appears you have considered right derivative in your mind. $\endgroup$ Dec 20 '17 at 9:42
0
$\begingroup$

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 $f(a) = f(b)$.

$\endgroup$
2
  • $\begingroup$ How can you say that the function will be strictly monotonous, I have only assumed that $f_'(x)$ exists. $\endgroup$
    – nls
    Dec 19 '17 at 13:07
  • $\begingroup$ @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. $\endgroup$ Dec 20 '17 at 8:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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