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I was thinking about the following problem:
Let $f:\mathbb R \rightarrow \mathbb R$ be continuous with $f(0)=f(1)=0.$Then which of the following is not possible?

(a) $f([0,1])=\{0\},$

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

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

(d) $f([0,1])=[-1/2,1/2].$

Can someone point me in the right direction? Thanks in advance for your time.

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    Right direction: (b)2012-12-17

3 Answers 3

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Hints:

$f(x)=0\;\;,\;\;f(x)=\sin \pi x\;....$

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    also thought of that2013-05-30
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By the Extreme Value Theorem, any function that is continuous on a closed and bounded interval (in this case, $[0, 1]$) has a global maximum and minimum.

The interval $[0, 1)$ in (b) has $1$ as a least upper bound, but not as a local maximum, which contradicts the Extreme Value Theorem.

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    Thanks a lot @Joe Zeng. +1 from me.2012-12-17
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Let $[a,b]$ be a closed interval containing 0. Consider the function

$f : x\mapsto\cases{ 3ax, & if $0\le x\le \frac13$ \\ a + (3x-1)(b-a) & if $ \frac13\le x \le \frac23$ \\ (3-3x)b & if $ \frac23\le x \le 1$ } $

The graph of $f$ is a piecewise-linear zigzag which starts at $\langle0,0\rangle$, zigs down to $\langle\frac13, a\rangle$, zags up to $\langle \frac23, b\rangle$, and then zigs back down to $\langle 1, 0\rangle$.

So $f$ is a continuous function on $[0,1]$ with range $[a,b]$, and $f(0)=f(1) = 0$.

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    @david Oh, I see your point. It can't have $f(0)=f(1)=0$ *and* a range of $[a,b]$ unless $0\in [a,b]$. Yes, quite so. I will edit the question appropriately. Thank you.2012-12-17