# Intermediate value theorem proof

• Aug 13th 2011, 09:47 PM
whiteboard
Intermediate value theorem proof
have a continuous function f:[0,1] -> [0,1]

For some\$\displaystyle x1, f(x1)>x1 \$
For some\$\displaystyle x2, f(x2)>x2 \$

Using the Intermediate Value Theorem, show that there exists some x* such that f(x*)=x*

I have attempted the problem:

Make \$\displaystyle h(x)=f(x)-x\$. Then we are looking for the case when \$\displaystyle h(x)=0\$. So if f(0)=0 or f(1)=1, then we are done. So we assume neither case is true. Then we have that \$\displaystyle f(0)>0\$ because f must take on a value in the interval [0,1], and the same for \$\displaystyle f(1)<1\$. These two statements imply that \$\displaystyle h(0)>0 and h(1)<0\$. So then by the intermediate value theorem, we can say that \$\displaystyle h(x)=0\$ for some \$\displaystyle x [0,1] \$ which means that \$\displaystyle f(x)=x \$ for some \$\displaystyle x [0,1] \$

As I am pretty rusty on my proof skills I was wondering if somebody could check this for me.

Thanks
• Aug 13th 2011, 10:17 PM
CaptainBlack
Re: Intermediate value theorem proof
Quote:

Originally Posted by whiteboard
have a continuous function f:[0,1] -> [0,1]

For some\$\displaystyle x1, f(x1){\color{red}<}x1 \$
For some\$\displaystyle x2, f(x2)>x2 \$

Using the Intermediate Value Theorem, show that there exists some x* such that f(x*)=x*

I have attempted the problem:

Make \$\displaystyle h(x)=f(x)-x\$. Then we are looking for the case when \$\displaystyle h(x)=0\$. So if f(0)=0 or f(1)=1, then we are done. So we assume neither case is true. Then we have that \$\displaystyle f(0)>0\$ because f must take on a value in the interval [0,1], and the same for \$\displaystyle f(1)<1\$. These two statements imply that \$\displaystyle h(0)>0 and h(1)<0\$. So then by the intermediate value theorem, we can say that \$\displaystyle h(x)=0\$ for some \$\displaystyle x [0,1] \$ which means that \$\displaystyle f(x)=x \$ for some \$\displaystyle x [0,1] \$

As I am pretty rusty on my proof skills I was wondering if somebody could check this for me.

Thanks

The function \$\displaystyle h(x)=f(x)-x\$ is continuous on an interval containing both \$\displaystyle x_1\$ and \$\displaystyle x_2\$ , with \$\displaystyle h(x_1)<0\$ and \$\displaystyle h(x_2)>0\$, hence by theIntermediate value theorem there exists a \$\displaystyle c\$ between \$\displaystyle x_1\$ and \$\displaystyle x_2\$ such that \$\displaystyle h(c)=0\$ and so \$\displaystyle f(c)=c\$

CB