# two problems involving continuous functions

1. Let $f:[0,1]\to [0,1]$ be such that $|f(x)-f(y)|\leq \frac{1}{2}|x-y|$ $\forall x,y\in[0,1]$. Let $$A=\{x\in[0,1]: f(x)=x\}.$$ Then number of elements in $A$ is what?
2. Let $f:[0,1]\to[0,1]$ be continuous and such that $f(0)=f(1)$. Let $$A=\{(t,s) \in [0,1]\times [0,1]: t\neq s\text{ and }f(t)=f(s)\}.$$ Then the number of elements in $A$ is what?

My thoughts: For (1) from the given condition we get $f'(x) \leq 1/2$ if the derivative exist and $f$ is non-constant. So if $f(x)=x$ then $f'(x)=1$. So not possible. But if $f$ becomes a constant function then it has a fixed point. So the answer will be 1. Is my thinking correct?

For (2) I am taking $\sin x$ from $[0,n\pi]$ to $[0,1]$ then it has so many points satisfying the given condition. Now $[0,n\pi]$ is homeomorphic to $[0,1]$. So $A$ has infinitely many points. Is my thinking correct?

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You can find some good starting points on how to format mathematics on the site here and here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. –  Zev Chonoles Jan 2 '13 at 6:47
Please ask only one problem per question. Also, in question $2$, are you asking for the measure of the set $A$? –  Eric Naslund Jan 2 '13 at 7:07

2. Since $f$ is continuous, it attains its maximum $a:=\max_{0\le x\le 1} f(x)$ and minimum $b:=\min_{0\le x\le 1} f(x)$ in $[0,1]$. Assume $f(x_1)=a$ and $f(x_2)=b$. If $f(0)=f(1)<a$, then for every $y \in (f(0),a)$, there exists an $s\in(0,x_1)$ with $f(s)=y$ and there also exists $t\in (x_1,1)$ with $f(t)=y$. So in this case $A$ is uncountable. Similarly, if $f(0)=f(1)>b$, we find for evarey $y\in (b,f(0))$ an $s\in (0,x_2)$ and $t\in(x_2,1)$ with $f(s)=f(t)=y$. So again $A$ is uncountable. Remains the case that $f(0)=f(1)=a=b$. But then $f$ is constant, $A=[0,1]\times[0,1]$ uncountable.

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For 1. if $|A|>1$ then we can pick $t,s\in A$. Since they are in A we get $f(t)=t$ AND $f(s)=s$. Since $A$ is a subset of $[0,1]$ we get $|f(t)-f(s)|\leq\frac{1}{2}|t-s|<\frac{3}{4}|t-s|$. Notice that $|f(t)-f(s)|=|t-s|$ since $f(t)=t$ AND $f(s)=s$. This tells us that $|t-s|<\frac{3}{4}|t-s|$, clearly a contradiction. Suppose now that $|A|=0$. Pick $t\in[0,1]$ arbitrarily. Let $a_1=t$, $a_2=f(t)$, $a_3=f(f(t))$ and so on. This defines a sequence $\{a_n\}_{n=1}^\infty \subseteq [0,1]$. Now, let $m,n\in\mathbb{N}$ be arbitrary and WLOG assume $m\geq n$. The Triangle Inequality tells us that $|a_m -a_n|\leq |a_m-a_{m-1}|+|a_{m-1}-a_{m-2}|+\cdots+|a_{n+1}-a_n|$. Notice though that we can write the absolute value of the difference of two consecutive terms in the sequence as $|a_{k+1}-a_k|=|f^k(t)-f^{k-1}(t)|\leq\frac{1}{2}|f^{k-1}(t)-f^{k-2}(t)|<\frac{3}{4}|f^{k-1}(t)-f^{k-2}(t)|=\frac{3}{4}|a_{k}-a_{k-1}|$. By induction, $|a_{k+1}-a_k|<(\frac{3}{4})^{k-1}|a_2-a_1|=(\frac{3}{4})^{k-1}|f(t)-t|$. Therefore $|a_m -a_n|\leq |a_m-a_{m-1}|+|a_{m-1}-a_{m-2}|+\cdots+|a_{n+1}-a_n|<|f(t)-t|((\frac{3}{4})^{m-2}+(\frac{3}{4})^{m-3}+\cdots+(\frac{3}{4})^{n-1})<|f(t)-t|(\frac{3}{4})^{n-1}\frac{1}{1-\frac{3}{4}}$

It follows by definition that $a_n$ is a Cauchy sequence. Since $\mathbb{R}$ is a complete metric space with the Euclidean metric, it follows that every Cauchy Sequence converges. This means $a_n$ converges. Let $a$ be the limit of $a_n$. Since $[0,1]$ is compact, $a\in [0,1]$. We show $f(a)=a$. By Triangle Ineq. $|f(a)-a|\leq|f(a)-f(a_n)|+|f(a_n)-a|$. Since by definition $f(a_n)=a_{n+1}$ we get that $|f(a)-a|\leq|f(a)-f(a_n)|+|a_{n+1}-a|$. We have $|f(a)-f(a_n)|<\frac{3}{4}|a-a_n|$. Therefore $|f(a)-a|<\frac{3}{4}|a-a_n|+|a_{n+1}-a|$. By making both terms in the RHS of the inequality less than $\frac{\epsilon}{2}$ for arbitrary $\epsilon>0$, we show that $f(a)=a$. This tells us there exists an element in $[0,1]$ s.t. it is a fixed point. Therefore it follows that $|A|=1$.

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1>

$g: [0,1] \rightarrow \mathbb R$ defined by $g(x) =|f(x) - x|$ then g is continuous function.

$g$ attains its minimun at $a \in [0,1]$ if $f(a) \neq a$ then $g(f(a)) =|f(f(a)) - f(a)| \leq \frac{1}{2}|f(a)- a| \leq |f(a)- a|=g(a)$ contradicting the fact g attains minimum at $a$ hence $f(a) = a$.

then $f$ can have atmat one fixed point as if $f(x) = x$ and $f(y)= y$ then $|f(x)- f(y)| = |x-y|\leq \frac{1}{2}|x- y|$ (a contradiction)

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For problem $1$, $f$ is a contraction mapping on a complete metric space, so we see that it has a unique fixed point by the Banach fixed-point theorem.

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thanks for answering the first question.can you help me for the second question please. –  poton Jan 2 '13 at 7:27