# Continuous functions with fixed points

• Apr 29th 2010, 09:04 PM
serious331
Continuous functions with fixed points
1. Let $\displaystyle f : [0,1] \rightarrow [0,1]$ be continuous. Prove that f has a fixed point.

2. let $\displaystyle f: (0,1] \rightarrow (0,1]$ be continuous. Give a counterexample to show that f need not have a fixed point.

For both the questions, I supposed a sequence $\displaystyle {x_n} = \frac{1}{n}$, which converges to 0, but this does not look correct. Can anyone tell how this is proved?
• Apr 29th 2010, 09:10 PM
Drexel28
Quote:

Originally Posted by serious331
1. Let $\displaystyle f : [0,1] \rightarrow [0,1]$ be continuous. Prove that f has a fixed point.

Hint:

Spoiler:

The mapping $\displaystyle \varphi:[0,1]\to[0,1]:x\mapsto f(x)-x$ is continuous. Think IVT

Quote:

2. let $\displaystyle f: (0,1] \rightarrow (0,1]$ be continuous. Give a counterexample to show that f need not have a fixed point.
What do you think?
Quote:

For both the questions, I supposed a sequence $\displaystyle {x_n} = \frac{1}{n}$, which converges to 0, but this does not look correct. Can anyone tell how this is proved?
What does that sequence help with?
• Apr 29th 2010, 10:03 PM
serious331
Quote:

Originally Posted by Drexel28
Hint:

Spoiler:

The mapping $\displaystyle \varphi:[0,1]\to[0,1]:x\mapsto f(x)-x$ is continuous. Think IVT

What do you think?

What does that sequence help with?

I am sorry I am not good at writing proofs. but how can these questions be shown from the IVT?
• Apr 29th 2010, 10:29 PM
Drexel28
Quote:

Originally Posted by serious331
I am sorry I am not good at writing proofs. but how can these questions be shown from the IVT?

I'll give you one more hint and let you take one more stab.

Consider the $\displaystyle \varphi(x)$ above.

Now, $\displaystyle \varphi$ being the difference of two continuous real-valued functions is continuous. Also, $\displaystyle \varphi(0)=f(0)-0=f(0)$. Now, if $\displaystyle \varphi(0)=0$ we're done, right? So what can you assume?

Similarly $\displaystyle \varphi(1)=f(1)-1$. If it's zero we're done again, so we can assume not.

So now you should have two conditions which should immediately make you jump for joy at knowing the IVT, what are they?
• Apr 29th 2010, 11:01 PM
serious331
Quote:

Originally Posted by Drexel28
I'll give you one more hint and let you take one more stab.

Consider the $\displaystyle \varphi(x)$ above.

Now, $\displaystyle \varphi$ being the difference of two continuous real-valued functions is continuous. Also, $\displaystyle \varphi(0)=f(0)-0=f(0)$. Now, if $\displaystyle \varphi(0)=0$ we're done, right? So what can you assume?

Similarly $\displaystyle \varphi(1)=f(1)-1$. If it's zero we're done again, so we can assume not.

So now you should have two conditions which should immediately make you jump for joy at knowing the IVT, what are they?

So, do we have $\displaystyle \varphi(0) \geq 0 \geq \varphi(1)$, and from IVT there is a fixed point x in [0,1] such that $\displaystyle \varphi(x) = 0$?
Is that right?
• Apr 29th 2010, 11:09 PM
Drexel28
Quote:

Originally Posted by serious331
So, do we have $\displaystyle \varphi(0) \geq 0 \geq \varphi(1)$, and from IVT there is a fixed point x in [0,1] such that $\displaystyle \varphi(x) = 0$?
Is that right?

(Yes)
• Apr 29th 2010, 11:18 PM
serious331
Quote:

Originally Posted by Drexel28
(Yes)

Thank you so much Drexel28. Can you give me a hint on what counterexample I can present on the second question?
• Apr 29th 2010, 11:22 PM
Drexel28
Quote:

Originally Posted by serious331
Thank you so much Drexel28. Can you give me a hint on what counterexample I can present on the second question?

Honestly, it's not that bad to come up with one. It may sound stupid, but just draw a picture.