# a problem on fixed point on different sets under continuous mapping

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.

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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 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?

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c) is false. For example take a rotation on the circle –  user52188 Jan 6 at 6:19
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 $f(b)-f(a)$?
Right... I'm suggesting to bound $f(b) - f(a) = \int_a^b f'(x) dx.$ –  user7530 Jan 6 at 6:24