Unique Solution in a Banach Space

Question

This is a question from an exercise sheet that I am having trouble with.

Let X be a Banach Space.

Suppose there is a mapping $T:X$ $\mapsto$X $\ni$$\forall$ $x,y$$\in$X, $||Tx-Ty||$ $\leq$ $c||x-y||$, where $0\lt c\lt1$ .

Prove that there exist a unique solution where $x\in X$, $x \ne 0$ such that $Tx=x$.

The following hint was given:

Let $x_0$$\ne$$0$, and set $x_{n+1}=T^nx_n$

I understand the basics of Banach Spaces and theorems such as Holder's inequality, however I have no idea where to begin with questions such as this one.

Answer 1

First show that the orbit of $0$ is bounded by using the triangle inequality along with $\|T^n(0)-T^{n-1}(0)\|\leq c^{n-1}\|T(0)\|$. (Sum of geometric series is finite)

Now, consider the sequence $\{T^i(0)\}$. This is Cauchy because $\|T^n(0)-T^m(0)\|\leq c^m*(\text{something bounded})$. Here, ($n\geq m$)

Since the space is banach, consider the limit of the sequence, call it $l$. Since $T$ is continuous (Lipschitz), $T(l)=\lim_{i\to \infty}T(T^i(0))=l$. Hence $l$ is a fixed point.

As for uniqueness, if $x,y$ are fixed then $\|x-y\|=\|T(x)-T(y)\|\leq c\|x-y\|$ which gives that $x=y$. As shown by gerw, it is possible that $0$ is the only fixed point.

Answer 2

As stated, the problem is wrong. Here are some counterexamples:

• Let $X = \{0\}$. Then, there is no point $x \in X$, $x \ne 0$.
• Let $X$ be arbitrary and define $Tx = c \, x$ for some $c \in (0,1)$. Then, $T x = x$ implies $x = 0$.