# Intuition behind Contraction Mapping Theorem

I understand that this theorem works for complete metric spaces, but I have been studying this purely for normed vector spaces. I want to check my heuristic understanding of the theorem and proof:

Statement of theorem:

Let $(V, ||.||)$ be a complete normed vector space and $U \subset V$ a closed subset of the space. Let $f: U \rightarrow U$ be a contraction mapping such that $\exists K \in (0,1)$ such that $\forall u, v \in V, ||f(u) - f(v)|| \leq K||u-v||$ .

Then there is a unique $w \in U$ such that $f(w) = w$

The proof constructs a sequence by taking any $u_0 \in U$ and doing $f(u_n) = u_{n+1}$.

My understanding:

Since $K <1$ and we are mapping each time between $U$, we are mapping a 'smaller' and 'smaller' such that eventually $f(u_n)$ will just map to itself thus reaching a fixed point.

And this limit exists because this sequence $u_n$ is Cauchy and in a Banach space all Cauchy sequences will converge to some limit.

My questions

Why are we free to take any $u_0 \in U$? Does our choice of $u_0$ affect the convergence rate?

Why is it necessarily the case that the limit is unique? And how is this affected by the map $f?$

Does the theorem still apply for $K =1?$

What applications are there to this theorem?

• To answer your first question, yes the choice of $u_0$ affects the convergence rate. If $u_0$ is the fixed point then the $f(u_0) -u_0=0$. – fred Apr 13 '16 at 18:17

The fixed point is unique because if $f(u)=u$ and $f(v)=v$ then $$||u-v||=||f(u)-f(v)||\leq K||u-v||$$ and since $K<1$ this is only possible if $u=v$.
If $K=1$ then either uniqueness or existence can fail. For instance, the map $f(u)=u$ satisfies the hypotheses with $K=1$, but every point is a fixed point. And if $V=\mathbb{R}$ and $f(x)=x+\frac{1}{1+e^x}$, then $0\leq f^{\prime}(x)<1$ for all $x$ (so the contraction property holds with $K=1$ by the mean value theorem), but $f$ has no fixed point.
• Thanks, is there any case where $K =1$ but both uniqueness and existence still hold? Or is there a proof of why this can not happen? – Kevin Apr 13 '16 at 18:29
• If $X$ is a compact metric space and $f:X\to X$ satisfies $d(f(x),f(y))<d(x,y)$ for all $x\neq y\in X$, then it's still possible to show that $f$ has a unique fixed point. My memory is that the proof is very different from the case $K<1$. – carmichael561 Apr 13 '16 at 18:32