Show that a function $f(x)$ maps to a set of points.Fixed point theorem

Question

Show that the function $f(x)=\frac{1+x^2}{2}$ maps the set of points $0\leqslant x\leqslant 1$ into itself and has a fixed point in that interval even though there does not exists a positive constant $K<1$ such that $\vert f(a)-f(b) \vert \leqslant K \vert b-a\vert$.

Answer 1

If $0\leq x\leq 1$, then $0\leq x^2\leq 1$ as well, so it follows that $\frac{1+x^2}{2}$ lies in the interval $[\frac{1}{2},1]$. It isn't that hard to see that $f(1)=1$, so $f$ indeed has a fixed point in $[0,1]$.

The second bit I read as "Show that there exists no $k$ such that $|f(a)-f(b)|\leq k|b-a|$ holds for all $a,b\in[0,1]$. We will without loss of generality assume that $a>b$ and since $f$ is monotonically increasing, we can ignore the absolute values. We find that $\frac{f(a)-f(b)}{a-b} = \ldots = \frac{a+b}{2}$.

Suppose we have an arbitrary $k<1$ given. We then pick two arbitrary values $a,b\in(k,1]$ with $a\neq b$ to find that $\frac{f(a)-f(b)}{a-b}>\frac{k+k}{2}=k$. The last inequality therefore doesn't hold for these $a,b$. Since $k<1$, we can always find such $a,b\in(k,1]$ and we're done.

Answer 2

1. The function $f\colon [0,1]\to \mathbb{R}$, $\displaystyle x\mapsto \frac{1+x^2}{2}$ is increasing and continuous on its domain which is a closed and bounded interval. Therefore $\displaystyle \min_{x\in [0,1]}{\bigl(f(x)\bigr)} \text{ exists }\wedge\displaystyle \min_{x\in [0,1]}{\bigl(f(x)\bigr)}=f(0)=\frac{1}{2}$ and $\displaystyle \max_{x\in [0,1]}{\bigl(f(x)\bigr)}\text{ exists }\wedge \displaystyle \max_{x\in [0,1]}{\bigl(f(x)\bigr)}=f(1)=1$.
   The Intermediate Value Theorem guarantees that $\displaystyle f([0,1])\subseteq [\frac{1}{2},1\textbf]$ and it follows that $f$ maps $[0,1]$ into itself.

2. A fixed point of $f$ is an element of $[0,1]$, let's say $\alpha$, such that $f(\alpha)=a$. Well, for all $x\in [0,1]$ we have $$f(x)=x \iff 1+x^2=2x \iff (x-1)^2=0 $$ It follows that $f$ has one, and only one fixed point, namely at $\alpha=1$.

3. Finally, following David Mitra's suggestion, notice that for all $a,b\in[0,1]$, we have $\displaystyle \vert f(a)-f(b)\vert={1\over2}(a+b)\vert a-b\vert$, in particular for $a\neq b$ it follows that $$\vert f(a)-f(b)\vert\leq K\vert a-b\vert \iff\frac{a+b}{2}\leq K$$ and taking $a=1$ we have that if such $K$ exists, then $\displaystyle \frac{1+b}{2}\leq K$, for any $b\in [0,1]$. Since $K\in [0,1]$ it follows that $\displaystyle \frac{1+K}{2}\leq K$ and therefore $1\leq K$ which is a contradiction.