Assume $H$ is a Hilbert space and $V \in L_b(H)$.
I want to show, that the following propositions are equivalent:
- V is an isometry.
- For every orthonormal system $\{u_{\alpha}: \alpha \in A \}$ the set $\{Vu_{\alpha}: \alpha \in A \}$ is also an orthonormal system.
- There is an orthonormal basis $\{u_{\alpha}: \alpha \in A \}$ such that $\{Vu_{\alpha}: \alpha \in A \}$ is also an orthonormal system.
I have already proven that $(1) \Rightarrow (2)$:
$\forall x,y \in H:~~(x-y,x-y)=(x,x)-2(x,y)+(y,y)$
$\forall \alpha, \beta \in A:~~2(Vu_\alpha,Vu_\beta)=\Vert Vu_\alpha\Vert^2 -\Vert Vu_\alpha-Vu_\beta\Vert^2-\Vert Vu_\beta\Vert^2 = \Vert u_\alpha\Vert^2 -\Vert u_\alpha-u_\beta\Vert^2-\Vert u_\beta\Vert^2 = 2(u_\alpha,u_\beta)$
Therefore $V$ preserves the scalar product, which yields the claim.
I think $(2) \Rightarrow (3)$ is trivial because every orthonormal basis is an orthonormal system.
This exercise seems a bit odd to me because in my mind the third proposition should read: "There is an orthonormal basis $\{u_{\alpha}: \alpha \in A \}$ such that $\{Vu_{\alpha}: \alpha \in A \}$ is also an orthonormal basis."
In either way, can someone help me with this proof?
It is enough to assume that for some orthonormal basis $(u_{\alpha})$ the set $(Vu_{\alpha})$ is orthonormal. It is not necessary to assume that the latter is complete. Proof of 3) implies 1): any $v \in V$ can be expanded as $\sum c_{\alpha} u_{\alpha}$. Since $V$ is assumed to be a bounded operator we have $Vx=\sum c_{\alpha} Vv_{\alpha}$. Now I leave it to you to verify that $\|Vv\|^{2}=\sum |a_{\alpha}|^{2}$ which is also equal to $\|v\|^{2}$. Hence $V$ is an isometry.