Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Pick out the true statements.
(a) Let $f : [0, 2] → [0, 1]$ be a continuous function. Then, there always exists $x ∈ [0, 1]$ such that $f(x) = x$.
(b) Let $f : [0, 1] → [0, 1]$ be a continuous function which is continuously differentiable in $(0, 1)$ and such that $|f’(x)| ≤ 1/2$ for all $x ∈(0, 1)$. Then, there exists a unique $x ∈ [0, 1]$ such that $f(x) = x$.
(c) Let $S$ = {$p = (x, y) ∈ \mathbb{R}^2 : x^2 + y^2 = 1$}. Let $f : S → S$ be a continuous function. Then, there always exists $p ∈ S$ such that $f(p) = p$.

by brouwer fixed point theorem we can say (a) is true but how can I verify the other options.

share|cite|improve this question
Part (a) follows from the intermediate value theorem applied to $f(x) - x$, so quoting the Brouwer fixed point theorem is a bit much. – Justin Young Jan 6 '13 at 6:15

You are right for a), as we can see $f$ as a self-map of $[0,2]$.

For b) show that $f$ is a contraction map: by the Mean value theorem we can see that $$|f(x) - f(y)| \le \frac{1}{2}|x-y|$$ for all $x,y \in [0,1]$. Now apply Banach's fixed point theorem.

Draw c), what is it? What kind of maps can you think of on this space?

share|cite|improve this answer
c) is false. For example take a rotation on the circle – user52188 Jan 6 '13 at 6:19

Here are some hints.

b): Suppose $f$ has two fixed points, $f(a) = a$ and $f(b) = b$, with $b > a$. What does the fundamental theorem of calculus have to say about the maximum possible value of $f(b)-f(a)$?

c): One special case of continuous functions on a set that immediately come to mind are its smooth symmetries. What are they for the circle?

share|cite|improve this answer
I wouldn't call the mean value theorem the "fundamental theorem of calculus"; this is normally reserved for the relation between integrals and anti-derivatives... – Henno Brandsma Jan 6 '13 at 6:23
Right... I'm suggesting to bound $f(b) - f(a) = \int_a^b f'(x) dx.$ – user7530 Jan 6 '13 at 6:24
Ok, that's a way to do it too. I still think the mean value theorem is more direct though. – Henno Brandsma Jan 6 '13 at 6:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.