What is a contraction on a space $(X,d)$?

Question

I have been reading some proofs on the elementary theorems of differential equations. One such proof uses the concept of a "contraction". See the definition below.

Definition 4 Let $(X,d)$ be a space equipped with a distance function $d$. A function $\Phi:X\to X$ from $X$ to itself is a contraction if there is a number $k\in (0,1)$ such that for any pair of points $x,y\in X$ we have $$d(\Phi(x),\Phi(y))\leq kd(x,y).$$ It is important that the constant $k$ is strictly less than one.

Source. (Taken from this pdf.)

To put it into my own words, it seems that a contraction is simply a modified distance function that always yields a value less than the normal distance function for the same set of points. Am I understanding it correctly?

Would someone be able to offer an example of a contraction?

Thank you!

Answer 1

No, a contraction is not a metric(aka distance function). What a contraction does is bring every pair of points closer together(in the implicit metric). For instance, $f(x)=x/2$ is a contraction in the euclidean metric of $\mathbb{R}$.

Answer 2

Besides the fact that your understanding omits the mapping of points to other points, the existence of $k<1$ such that a certain thing is always less than $k$ is NOT the same as saying that that certain thing is always less than $1$. For example, consider the function $f(x) = \sqrt{1+x^2}$. Then one has $|f(x)-f(y)|<1\cdot|x-y|$, but there is no number $k<1$ such that for every two points $x,y$ one has $|f(x)-f(y)|\le k|x-y|$. The reason is that for any proposed value $k<1$, one can find very large values of $x,y$ for which $|f(x)-f(y)|>k|x-y|$ (although $|f(x)-f(y)|$ is still less than $|x-y|$).