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QUestion:

let $f(x)$ is continuous everywhere in [a,b], differentiable except at a countable number of points in [a,b].and $f'(x)\ge 0$

show that $$f(x)\le f(b)$$

This problem is from this: [china BBS]http://www.duodaa.com/?qa=4999/一个证明题

My idea: if $f'(x)$ is integrable,Assmue that $f(x)$ is not derivative on $x_{i},x_{i}>x_{i-1},i=1,2,3,\cdots,n$,then we have $$0 \le \int_{x_{i-1}}^{x_{i}}f'(x)dx=f(x_{i})-f(x_{i-1}),i=1,2,\cdots, x_{0}=a,x_{n}=b$$ then have $$f(b)\ge f(x)$$

But I know this methods is not true,so How prove it? Thank you

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  • $\begingroup$ Please clarify what you mean by $f(x)$ is derivative. Do you mean it is the derivative of a differentiable function $F$? $\endgroup$
    – konewka
    Oct 14, 2014 at 11:25
  • $\begingroup$ With the assumptions, your method looks good, $f$ has to be an increasing function wherever differentiable, and continuity assures the value does not drop anywhere. $\endgroup$
    – Macavity
    Oct 14, 2014 at 11:39
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    $\begingroup$ @Macavity: if the points of non-differentiability are $\{0\} \cup \{1/n : n \in \mathbb{N}\}$, how does one form an open interval to the right of $0$ on which to integrate? $\endgroup$
    – Carl Mummert
    Oct 14, 2014 at 11:41
  • $\begingroup$ @Carl Mummert Youre right, I considered only finitely many discontinuities. However in your e.g. continuity at $x=0$ should assure a nbd to the right where the value is arbitrarily close to $f(0)$, and farther than that nbd the earlier argument holds. Not rigorous though :( $\endgroup$
    – Macavity
    Oct 14, 2014 at 11:52
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    $\begingroup$ @Macavity: the result would become false if we weaken the assumption to say the derivative is nonnegative except on a set of measure 0. We can make a counterexample from the Cantor staircase function. $\endgroup$
    – Carl Mummert
    Oct 14, 2014 at 15:21

1 Answer 1

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the function $f$ is bounded. let $x_n$ be the $n^{th}$ point of non-differentiability according to some pre-chosen numbering. for $\epsilon \gt 0$ cover each $x_n$ with an interval $I_n = (x-\frac{\epsilon}{2^n},x+\frac{\epsilon}{2^n})$.

let $S_{\epsilon} = \bigcup_n I_n$ the total measure of $S_{\epsilon}$ is no greater that $4\epsilon$. set $T_{\epsilon}= [a,b] \backslash S_{\epsilon}$

perhaps this construction can be used?

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