# Assume $f :[0,1] \rightarrow [0,1]$ is a continuous function. Show there exists a fixed point $x$, i.e. $f(x) = x$ [closed]

Assume $f :[0,1] \rightarrow [0,1]$ is a continuous function.

Show there is $x \in [0,1]$ with $f(x) = x.$

Is there an example of a continuous function $g:(0,1) \rightarrow (0,1)$ where this is false?

## closed as off-topic by Did, user186170, user223391, Joey Zou, Daniel W. FarlowAug 13 '16 at 22:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Marco Cantarini, Community, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

• Hint: $f(0)\geqslant0$ and $f(1)\leqslant1$. – Did Aug 13 '16 at 18:22
• The implication is pretty straightforward using the Intermediate Value Theorem. See e.g. Prove that a continuous function defined on an interval $[a,b]$ has a fixed point. – hardmath Aug 13 '16 at 18:44
• Well Shrink or stretch x for one part of the interval and let it "snap back" into place for the rest. i.e. f(x) = xw for x <= t and f(x) = (1 - tw)x + tw for x > t will do it. Or you could simply do $f(x) = x^2$. – fleablood Aug 13 '16 at 18:57

Let $h(x)=f(x)-x$. $$h(0)=f(0)-0=f(0)\ge0$$ $$h(1)=f(1)-1\le0$$ By the intermediate value theorem, $$h(x)=0\Rightarrow f(x)=x$$ for some $x\in[0,1]$.

This is not true in $(0,1)$. For example, consider $g(x)=x^2\lt x$ for all $x\in(0,1)$.

$f(x)=x+x(1-x)$ for example $f(x)$ is always greater then $x$ in $(0,1)$ but still doesn't leave the interval

• Even simpler is the example $g(x) = x^2$ given in the previous Answer. – hardmath Aug 13 '16 at 18:47

A demonstration without the intermediate value theorem.

If $f(0)\neq 0$ and $f(1)\neq 1$ then $f(0)-0>0$ and $f(1)-1<0$. Let $I_{n}=\{0=\frac{0}{n},\frac{1}{n},\frac{2}{n},\ldots, \frac{n-1}{n}, \frac{n}{n}=1\}$. For all $n$ set $k_n\in\{0,1,\ldots,n\}$ such that $$\begin{array}{cc} f\left(\frac{0}{n}\right)-\frac{0}{n}&\geq 0. \\ f\left(\frac{1}{n}\right)-\frac{1}{n}&\geq 0. \\ \vdots & \\ f\left(\frac{k_n-1}{n}\right)-\frac{k_n-1}{n}&\geq 0. \\ f\left(\frac{k_n}{n}\right)-\frac{k_n}{n}&\geq 0. \\ f\left(\frac{k_n+1}{n}\right)-\frac{k_n+1}{n}&\leq 0. \end{array}$$ Set $u_n=\frac{k_n}{n}$ and $v_n=u_n+\frac{1}{n}$. The sequences $u_n$ and $v_n$ are limited. By weierstrass theorem there is a convergent subsequence $u_{n_i}$ such that $\lim_{i\to \infty} u_{n_i}=u\in [0,1]$ and $\lim_{i\to \infty} v_{n_i}=v\in [0,1]$. By the continuity of $f$ we have $\lim_{i\to \infty}f(u_i)-u_i=f(u)-u\geq 0$ and $\lim_{i\to \infty}f(v_i)-v_i=f(v)-v\leq 0$.

Now note that $|u_{n_i}-v_{n_i}|=1/n$ implies $f(u)-u=\lim_{i\to \infty}f(u_i)-u_i= \lim_{i\to \infty}f(v_i)-v_i=f(v)-v$. By $0\leq f(u)-u=f(v)-v\leq 0$ we can only conclude $f(u)-u=0$, i.e. $$f(u)=u$$ Counter example. For $0<x<1$ we have $0<x^n<x$ for all $n\geq 2$. Then $f:(0,1)\to (0,1)$ definite by $f(x)=x^n$ is such that $f(x)<x$.

• Is it possible to use iterations $w_0=0$, $w_{k+1}=f(w_k)$ to prove existence? $\{w_k\}$ is compact, so there is a cluster point, which seems to be a fixed point. – Nyfiken Aug 13 '16 at 22:32
• @Nyfiken I think it is not possible. Because the starting $w_0$ point is not necessarily zero. – MathOverview Aug 13 '16 at 23:13