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Suppose $ f$ is a continuous function on interval $I$ and that the derivative $ f'$ of $f$ exists and is nonnegative everywhere on the interior of $I$. Also suppose that there is no sub-interval of $I$ such that $f'(x)=0$ for all $x$ in the subinterval. Prove that $f$ is a strictly increasing function.

Attempt. I stepped on this when proving that $x-\sin x$ is strictly increasing. One could use the above result, or easier, the MVT on a positive $x>0$ and the nearest $n\pi+\pi/2$ from the left. However i could not transfer this idea to the general case, as stated above, in order to provide a proof of this general result.

Thanks for the help.

Nikolaos Skout
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  • Assume that there are $x<y$ in $I$ such that $f(x)=f(y)$. Let $z$ be the point where the maximum of $|f(r)-f(x)|$ is attained for $r\in [x,y]$. If $f(r)>f(x)$. Then by Lagrange's mean value theorem, there is a point in $(r,y)$ with negative derivative. If $f(r)<f(x)$, then there is a point in $(x,r)$ with negative derivative. Therefore, $f(r)=f(x)$, which implies that $f$ is constant on $(x,y)$. This contradicts that $f'$ cannot be zero on the interval $(x,y)$. – MoonLightSyzygy Dec 27 '19 at 22:24
  • I see, very nice. Just a typo: i guess you mean "if $f(z)>f(x)$ " etc – Nikolaos Skout Dec 27 '19 at 22:33
  • There are two parts. In one the assumption is $f(r)>f(x)$ in the other the assumption is $f(r)<f(x)$. – MoonLightSyzygy Dec 27 '19 at 22:34
  • Of course there are two parts. Point $z$ is where max is attained. Don't you take the cases: $f(z)>f(x), f(z)<f(x)$, which lead to contradiction, and so $f(z)=f(x)$, giving $f$ constant on $ [x,y]$? – Nikolaos Skout Dec 27 '19 at 22:41
  • See related https://math.stackexchange.com/q/1845927/72031 – Paramanand Singh Dec 28 '19 at 14:55

1 Answers1

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$f'(x) \ge 0$ implies that $f$ is (weakly) increasing: $$ a < b \implies f(a) \le f(b) \, , $$ this follows from the mean value theorem.

It remains to exclude that $f(a) = f(b)$ for some $a < b$: Then $$ a \le x \le b \implies f(a) \le f(x) \le f(b) = f(a) \, , $$ i.e. $f$ is constant on $[a, b]$, and therefore $f'(x) = 0$ on $[a, b]$, contradicting the assumption that there is no subinterval on which the derivative is zero.

Martin R
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