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Let $f:[0,1]\to\mathbb{R}$ be a continuous function that does not take on any of its values twice and with $f(0)<f(1)$. Show that $f$ is strictly increasing on $[0,1]$.

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3 Answers 3

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For $x\in (0,1)$ if $f(x)<f(0)$ then $f(x)<f(0)<f(1)$ so by Intermediate Value Theorem there exist $x<x_0 <1$ s.t. $f(x_0)=f(0)$ a contradiction whit hypothesis. So for $x\in (0,1)$ we have $f(x)>f(0)$. Now consider $0<x<y<1$ and $f(x)>f(y)$. We have $f(0)<f(y)<f(x)$ so by Intermediate Value Theorem there exist $x_1$ s.t. $0<x_1 <x$ and $f(x_1)=f(y)$ a contradiction whit hypothesis. So $f(x)<f(y)$ which We want.

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Hint: the image of a closed, bounded interval under a continuous function is again closed, bounded interval, thus

$$f([0\,,\,1])=[f(0)\,,\,f(1)]$$

Suppose now there are $\,x,y\in[0,1]\;\;s.t.\;\;x<y\;,\;f(x)> f(y)\;$ .

Since $\,f(x),f(y)\in[f(0),f(1)]\,$ , show that there must be $\,w>y\;\;s.t.\;\;f(w)=f(x)\,$

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    $\begingroup$ $f([0,1])=[f(0),f(1)]$ is not necessarily true. The statement $f([0,1])\subseteq[m,M]$ where $m$ is the minimum of $f$ and $M$ is the maximum of $f$ is more accurate. However, I do see your point. Thanks. $\endgroup$
    – user72691
    Apr 18, 2013 at 1:33
  • $\begingroup$ You are right indeed. $\endgroup$
    – DonAntonio
    Apr 18, 2013 at 1:50
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If $f$ is not strictly increasing then there exist $0\leq x_1<x_2\leq 1$ such that $f(x_1)\geq f(x_2)$.
From $f(0)<f(1)$ we conclude that either $f(0)\leq f(x_1)$ or $f(x_2)\leq f(1)$.
Now, in any case use the IVT to derive a contradiction.

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  • $\begingroup$ what's the meaning of IVT please. $\endgroup$ Apr 17, 2013 at 20:15
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    $\begingroup$ @HoseynHeydari: I mean the Intermediate Value Theorem. $\endgroup$
    – P..
    Apr 17, 2013 at 20:30

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