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Function $f$ is continuous on $[a, b]$ and has a $f’_{-}$ on $(a, b)$. $f(a) = f(b)$, then show that $$\sup_{(a, b)} f’_{-}(x) \geq 0 \geq \inf_{(a, b)} f’_{-}(x) $$

Here’s my attempt. Consider 2 cases:

  1. $f$ is constant, then the statement is obvious.
  2. $f$ is not a constant. Since $f$ is continuous, there is a minimum $m$ and maximum $M$ on $[a, b]$. Then $m = f(x_1)$, $M = f(x_2)$. If both $x$s are in $(a, b)$, the statement is proven, because $f’_{-}(x_1) \leq 0, f’_{-}(x_2) \geq 0$. I don’t know what to do if only one of the $x$s lies in $(a, b)$. Do you have any ideas?
Gary
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ABlack
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  • It is shown here that if the one-sided derivative is strictly positive on the open interval then the function is strictly increasing. – Martin R Mar 17 '22 at 10:06

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