Let $(X, \rho)$ be a compact metric space and let $f:X \rightarrow X$ be an isometry. Prove that $f$ is onto. Based on the above fact , prove that the $\ell^2$ space (that is the space of all real sequences $x_n$ such that $\sum \limits_{n=1}^{\infty} x_n^2$ converges) is not compact under the metric $ \rho \left ( x_n, y_n \right )= \sqrt{\sum \limits_{n=1}^{\infty}\left ( x_n-y_n \right )^2}$
Solution
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.
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In $\ell^2$ we consider the forward shift $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.
Solution
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.
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In $\ell^2$ we consider the forward shift $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.
We can reach the same result under the weakest assumption
ReplyDelete$$d\left ( f(x), f(y) \right )\geq d\left ( x, y \right )$$
Again $f$ is an isometry. In general , all functions $f$ that obey the above fact and that are defined on compact space $X$ are in fact isometries.
This result is left as an interesting and difficult at the same time exercise to the reader.
Also, about the forward shift operator one can see the topic in: mathimatikoi.org forum
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