all isometry in space of sequence

Question

in my book definition of isometry is: Let $X$ and $Y$ be normed spaces with norms $\left\|\cdot\right\|_1$ and $\left\|\cdot\right\|_2$. A map $f : X \to Y$ is called an isometry if for any $a,b \in X$ one has:

$$\left\|f(a)-f(b)\right\|_1=\|a-b\|_2.$$

(functional analysis by Peter D.Lax page47-48)

Now in following Theorem 10 :

let X be the set of all convergence complex sequence to 0 with max norm . Let ${b_n}$ be an arbitrary sequence of complex numbers of absolute value 1: $|b_n|= 1$. Define the mapping U by : $Ux={b_n a_n } $ (43)

and $Px = \{ a^* _n \} (44) $ that $ { a^* _n } = a_{p(n)}$ and p(n) is a permutation from $\Bbb{N} $ itself.

theorem 10: Every linear isometry of the Banach space X is the composite of an isometry of type (43) and (44).

proof: exercise!

I can't prove this

Answer 1

I think its not true the way you wrote it. Counterexample: The map $$(a_1,a_2,a_3,a_4,\dots)\mapsto(a_1,a_2,a_2,a_3,a_3,a_3,a_4,a_4,a_4,a_4,\dots)$$ is a linear isometry, but I don't think you can decompose it into "complex rotation" $U$ and permutation $P$. Not even if you allow for linear combinations of multiple $P$ and $Q$.