A real function $f$ is continuous on $[0,2]$ and $f(0)=f(2).$

Question

I came across a problem in my analysis book that says:

A real function $f$ is continuous on $[0,2]$ and $f(0)=f(2).$ Then I have to prove that there exists a point $c$ in $[0,1]$ such that $f(c)=f(c+1).$ A hint is given as follows:
HINT: if $f(0)=f(1)$ then $c=0,1.$ If $f(0) \neq f(1)$, consider $g$ on $[0,1]$ defined by $g(x)=f(x)-f(x+1).$

But still I could not progress further.Could someone point me in the right direction? Thanks in advance for your time.

@Waqas: $$g(1)-g(0)=\bigl(f(1)-f(2)\bigr)-\bigl(f(0)-f(1)\bigr)=2f(1)-2f(0).$$ It really isn't all that clear how (or even if) that helps. — Cameron Buie, Feb 19 '13 at 04:34

score 2 · Accepted Answer · answered Feb 18 '13 at 12:53

Let $f(0)=f(2)=a$ and $f(1)=b$. Then $g(0)=a-b$ and $g(1)=b-a=-g(0)$. So because $a\neq b$ we know $g$ changes sign over the interval and $g$ is continuous because $f$ is continous. Now by the intermediate value theorem there exists a number $c\in [0,1]$ such that $g(c)=0$, so $f(c)-f(c+1)=0$ and we have $f(c)=f(c+1)$

score 0 · Answer 2 · answered Feb 18 '13 at 12:52

0

Do you know the intermediate value theorem? Prove that there is a $c$ so that $g(c)=0$

answered Feb 18 '13 at 12:52

Dominic Michaelis

19,935

Thanks a lot. I have got it. – Feb 18 '13 at 12:54

score 0 · Answer 3 · answered Feb 18 '13 at 12:53

0

If $f(0)<f(1)$, what is the sign of $g(0)$? of $g(1)$? What if $f(1)<f(0)$? Since $f$ is continuous, what can you say about $g$?

answered Feb 18 '13 at 12:53

Cameron Buie

102,994

score 0 · Answer 4 · answered Feb 18 '13 at 12:56

0

Use the intermediate value theorem for $g(x)$, and the fact that $g(0) = - g(1)$.

answered Feb 18 '13 at 12:56

Mattias Andersson

11

A real function $f$ is continuous on $[0,2]$ and $f(0)=f(2).$

4 Answers4