Injective

**OntarioStud** · November 23rd 2006, 10:49 AM

Happy Thanksgiving to those of you in the States.

I have a problem:

Suppose that a function f [a,b] to R is continuous such that f attains its minimum or maximum on [a,b] at a point c such that
a<c<b.
Prove that f cannot be injective.

How would you work through this proof?

**CaptainBlack** · November 23rd 2006, 11:00 AM

Can you tell us why you need three accounts?

RonL

**ThePerfectHacker** · November 23rd 2006, 11:09 AM

Originally Posted by OntarioStud

Happy Thanksgiving to those of you in the States.

I have a problem:

Suppose that a function f [a,b] to R is continuous such that f attains its minimum or maximum on [a,b] at a point c such that
a<c<b.
Prove that f cannot be injective.

How would you work through this proof?

I am going to make this proof as if it was a maximum point the minimum point is anagolous.

Assume,
$f:[a,b]\to \mathbb{R}$ is an injective map.
Also that the function has a maximum $c\in (a,b)$ .
Subdivide your interval into,
$[a,c],[c,b]$
Since the function is continous on the full interval it is continous on these.
Without lose of generality assume $f(a)\geq f(b)$ . Pick any number $d$ such that $f(b) \leq d< f(c)$
Then that number (by our conditions) also exists on,
$f(a)\leq d< f(c)$
By the intermediate value theorem there is a point $m_{1,2}$ such that $f(m_1)=d$ on $[a,c]$ and $f(m_2)=d$ on $[c,b]$ . Thus, $f(m_1)=f(m_2)$ , and if $m_1\not = m_2$ the function fails to be injective. The only way it can is when $m_1=m_2=c$ but that is not possible.

I hope you can understand it, mathematicians are by nature lazy.

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