Let $H$ be a Hilbert space with orthonormal basis $(e_n)$ and let $T:H\to H$ be the (only) bounded operator that satisfies:
$T(e_n)=e_{n+1}$ for all $n\in \mathbb{N}$.
Show that $T$ is isometric but not unitary.
Let $H_1,H_2$ be Hilbert spaces and $T : H_1→ H_2$. Then
T is called unitary if:
$T$ is surjective, and
$T$ preserves the inner product.
Let $X,Y$ be normed spaces.
$T:X\to Y$.
Then $T$ is called isometric if it preserves the norm:
$||Tx||=||x||\quad \forall x\in X$.
I tried showing that T is isometric:
Since $(e_n)$ is an orthonormal basis of H:
Let $x\in H$. We want to show that $||Tx||=||x||$.
From a theorem in functional analysis, we have
$||x||^2=\sum_{n} |<x,e_n>|^2$ for every $x\in H$.So we can substitute $Tx\in H$ but i cannot see how to use then the given definition of the operator $T$?
From the same theorem, we get also:
$x=\sum_{n} <x,e_n>e_n$ for all $x\in H$ but then how to use this for showing that the operator is not unitary i.e. it does not preserve inner product, using the definition of $T$.
Best Answer
You mentioned all the data you need.
You have,using that $T$ is bounded for the first equality, and the continuity of the inner product in the fourth equality, \begin{align} \|Tx\|^2&=\Big\|\sum_n\langle x,e_n\rangle\,Te_n\Big\|^2 =\Big\|\sum_n\langle x,e_n\rangle\,e_{n+1}\Big\|^2\\[0.3cm] &=\sum_n|\langle x,e_n\rangle|^2=\|x\|^2. \end{align} So $T$ is an isometry. But it is not surjective, since $e_1$ is orthogonal to the range of $T$.