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Let $f(x)$ be a continuous function on $[a, b]$ (not necessarily derivable) and satisfy $f(a)<f(b)$. At the same time, the limit $g(x)=\lim_{t \to 0} \frac{f(x+t)-f(x-t)}{t}$ exists for all $x\in(a,b)$.

Prove that there is a $c\in(a,b)$ such that $g(c)\geqslant0$.

emacs drives me nuts
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  • What have you tried yet? Also, note that $g(x) = 2 × f'(x)$ –  Feb 13 '20 at 07:08
  • @Chief VS: $f'(x)$ can fail to exist at some points. For example, if $f(x) = |x|,$ then $f$ is continuous everywhere and $g(0) = 0,$ but $f'(0)$ does not exist. In fact, $f$ can be continuous everywhere and $g$ can exist everywhere, and $f'$ can fail to exist on a dense set of points (indeed, even sets having cardinality $c$ in every interval; even more, even sets having Hausdorff dimension $1$ in every interval). For more details, see my comments here. – Dave L. Renfro Feb 13 '20 at 07:19
  • I probably shouldn't do this, but your question is Lemma 1 on p. 708 of The first symmetric derivative by Charles Edward Aull [American Mathematical Monthly 74 #6 (June-July 1967), pp. 708-711]. – Dave L. Renfro Feb 13 '20 at 07:26
  • @DaveL.Renfro, Gotcha, i didn't knew that, thanks. –  Feb 13 '20 at 13:35

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