Papapetros Vaggelis wrote:

Let \(\displaystyle{\left(X,d\right)}\) be a compact metric space and \(\displaystyle{f:X\longrightarrow X}\) an isometry.

Prove that the function \(\displaystyle{f}\) is onto \(\displaystyle{X}\) .

Use this result in order to prove that the \(\displaystyle{\mathbb{R}}\) - normed space \(\displaystyle{\left(\ell^2,||\cdot||_{2}\right)}\)

is not compact.

Hello Vaggelis,

Suppose that $f$ is not onto. Then there exists an $x \in X$ such that $x \notin f(X)$. Since $f$ is an isometry this means that it will be continuous as a function and since $X$ is compact then $f(X)$ will also be a compact subset of $X$. So, $a= d(x, f(X))>0$.

We pick the sequence $x_0=x, \; x_1=f(x_0), \; \dots, x_n=f(x_{n-1}), \dots$. We note that for this the sequence holds $x_n \in f(X)$ for all $n \geq 1$. Since $f(X)$ is compact, as well as sequently compact, this means that the sequence has a convergent subsequence. Therefore, there exist $m,n $ such that $m>n\geq 1$ and $\rho(x_m, x_n)<a$.

But, $f$ is an isometry meaning that:

$$ \rho\left ( x_m, x_n \right )= \rho \left ( f(x_{m-1}), f\left ( x_{n-1} \right ) \right )=\cdots=\rho\left ( x_{m-n}, x \right )$$

On the other hand $x_{m-n} \in f(X)$ hence

$$d\left ( x, f(X) \right )\leq \rho \left ( x, x_{m-n} \right )=\rho(x_m, x_n)<a$$

a contradiction. Hence $f$ is onto.

In $\ell^2$ we consider the forward shift operator $S_r:\ell^2 \rightarrow \ell^2 $ given by:

$$S_r\left ( x_1, x_2, x_3, \dots \right )= \left ( 0, x_1, x_2, \dots \right )$$

We easily note that it is an isometry but not onto.