Orthogonal Operator Infinite Dimensional Inner Product Space

Question

I know that on a finite dimensional inner product space, a unitary (or orthogonal) operator preserves the inner product. That is, having $\|T(x)\|=\|x\|$ for all $x\in V$ is equivalent to having $\langle T(x),T(y) \rangle =\langle x,y\rangle$ for all $x,y\in V$ which is equivalent to $TT^\ast=T^\ast T=I$. I am hoping someone can give me an example of this failing in infinite dimensions?

Answer 1

Consider $\Bbb V$ the real vector space formed by the quasi-null families of real numbers $(x_n)_{n \in \Bbb Z}$, "quasi-null" meaning that $\{ n \in \Bbb Z ~:~x_n \neq 0\}$ is finite. The operations are defined in the usual way. Define the inner product: $$\langle x, y \rangle = \sum_{n \in \Bbb Z}x_n y_n, \qquad x = (x_n)_{n \in \Bbb Z}, y = (y_n)_{n \in \Bbb Z} \in \Bbb V$$ Consider the right/left shift operators $R,L: \Bbb V \rightarrow \Bbb V$ such that $(Rx)_n = x_{n-1}$ and $(Lx)_n = x_{n+1}$. They are clearly isomorphisms and $R = L^{-1}$.

We have also that $R^\ast = L$, and hence $L^\ast = R$. And $R, L$ are unitaries and normals. If you let $\Bbb W = \{ x \in \Bbb V ~:~ x_n = 0,~\forall~n < 0 \} \leq \Bbb V$, then $\Bbb W$ is $R$-invariant, but not $L$-invariant. Having this in mind, we can consider the restriction $R_{|_\Bbb W}:\Bbb W \rightarrow \Bbb W$. Finally, we have that $(R_{|_\Bbb W})^\ast$ is the operator: $$L': \Bbb W \rightarrow \Bbb W \\ (L'x)_n = \begin{cases} 0 \mbox{, if } n < 0 \\ x_{n+1} \mbox{, if } n \geq 0\end{cases}$$ So, it's not hard to check that $R_{|_\Bbb W}$ is not normal.