Unitary operators in Hilbert spaces

Question

Let $T$ be a bounded linear operator on a Hilbert space $H$. I have to show that the following are equivalent:

(i) $T$ is unitary

(ii) For every orthonormal basis $\{u_{\alpha}:\alpha\in \Lambda\}$, $\{T(u_{\alpha}):\alpha\in \Lambda\}$ is an orthonormal basis.

(iii) For some orthonormal basis $\{u_{\alpha}:\alpha\in \Lambda\}$, $\{T(u_{\alpha}):\alpha\in \Lambda\}$ is an orthonormal basis.

I have proved that (i)$\implies$ (ii). Also (ii)$\implies$ (iii) is obvious.

How to show that (iii)$\implies$ (i)? Please suggest anything?

Answer 1

For $(iii) \Rightarrow (i)$, you want to show that $$ \langle Tx,Ty\rangle = \langle x,y\rangle \qquad (\ast) $$ for all $x,y\in H$. First note that $(\ast)$ is true if $x,y\in S := \{u_{\alpha}\}$. By sesqui-linearity, $(\ast)$ is true if $x,y\in \text{span}(S)$. However, $\text{span}(S)$ is dense in $H$, so if $x\in H,v\in \text{span}(S)$ and $\epsilon > 0$, then $\exists u\in \text{span}(S)$ such that $\|x-u\| < \epsilon$. Hence, $$ |\langle Tx,Tv\rangle - \langle x,v\rangle| \leq |\langle Tx,Tv\rangle - \langle Tu,Tv\rangle| + |\langle Tu,Tv\rangle - \langle u,v\rangle| + |\langle u,v\rangle - \langle x,v\rangle| $$ Since $T$ is bounded, you can make the right hand side as small as you want, proving that $$ \langle Tx,Tv\rangle = \langle x,v\rangle $$ Similarly, one can replace $v$ by an arbitrary $y\in H$, proving $(\ast)$ for all $x,y\in H$.

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Edit: My apologies, in addition to proving $(\ast)$, one also needs to prove that $T$ is surjective. To see this, one simply notes that $R(T)$, the range of $T$ contains an orthonormal basis for $H$, so if $x\in H$, then write $$ x = \sum a_{\alpha} T(u_{\alpha}) \qquad (\dagger) $$ so that the series $y = \sum a_{\alpha} u_{\alpha}$ converges because $(\dagger)$ converges. Now $y\in H$ and $x = T(y)$ since $T$ is continuous. Hence, $R(T) = H$.

Answer 2

It is enough to compute $T^*T(u_\alpha)$ for all $\alpha$, since the operator is determined by its values at a basis.

We can compute the coordinates of this vector in the basis $u_\alpha$. So, we do

$$T^*T(u_\alpha)\cdot u_{\beta}=T(u_\alpha)\cdot T(u_\beta)=\delta_{\alpha,\beta}$$

Therefore $T^*T(u_\alpha)=\sum_\beta(T^*T(u_\alpha)\cdot u_{\beta})u_{\beta}= u_{\alpha}$, from where we get that $T^*T$ is the identity.

This is the same computation we did above, but hiding things a little.