Let $$ f:\mathbb{R}\rightarrow\mathbb{R}\\ f(x) = x-\frac{x^2}{2}. $$ Prove that $f$ is restricted to a function from $[0,1]$ to himself and prove that $f$ is a shrinking map but not a contraction.
For the first part, I know that $\frac{df}{dx} = 1 - x$ so the function is monotonically increasing in $[0,1]$, assuming the max value in $x=1$. Now, $f(1)=\frac{1}{2}$ so we have the first point. Moreover,
$$ \begin{align*} d(f(x),f(y))&=\left| x-\frac{x^{2}}{2} -y+\frac{y^{2}}{2}\right|\\ &= \left|(x-y) - \frac{(x^{2}-y^{2})}{2}\right| \\ &= \left| (x-y) (1-\frac{(x+y)}{2})\right|\\ &< \left|x-y\right|= d(x,y), \end{align*} $$ so $f$ it's a shrinking map. But I have no idea how to prove that $f$ is not a contraction. Can someone help me? Thanks before.
I suppose that, in this context, a contraction is a map $f$ such that there is some $K\in[0,1)$ for which you always have$$d\bigl(f(x),f(y)\bigr)<Kd(x,y).$$Note that, if $x\ne0$,$$\frac{d\bigl(f(x),f(0)\bigr)}{d(x,0)}=\left|\frac{f(x)-f(0)}{x-0}\right|=1-\frac x2$$and that$$\lim_{x\to0}1-\frac x2=1.$$Therefore, there is no such $K$.