The (un)boundedness of an involutive operator

Question

This is a question I've been thinking about for a while, for which I do not have a satisfactory answer. Suppose that $T$ is a densely-defined operator on a Hilbert space $\mathcal{H}$ such that $R(T)\subseteq D(T)$ and $T^2x = x$ for all $x\in D(T)$. Is it possible for $T$ to be unbounded? Or must all densely-defined involutions be bounded? I would think that it must be bounded since coming up with an explicit counter-example is pretty challenging, but that may just be a testament to my lack of creativity with unbounded operators.

Answer 1

Well I feel a little silly for asking since I just came up with a really nice counter-example after thinking about $2\times 2$ matrices and what the involutions are in that setting. Let $(e_i)$ be an orthonormal basis for $\ell^2$. Define $D(T) = \operatorname{span}\{e_i: i\ge 0\}\subseteq \ell^2$ (which is clearly dense) and define $T$ via its matrix representation by

$$T=\left(\begin{array}{cccc} \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right) & & & \\ & \left(\begin{array}{cc} 0 & \frac{1}{2} \\ 2 & 0 \end{array}\right) & & \\ & & \left(\begin{array}{cc} 0 & \frac{1}{3} \\ 3 & 0 \end{array}\right) \\ & & & \ddots\end{array}\right).$$

$R(T)\subseteq D(T)$ since finite linear combinations of the $e_i$ get mapped to finite linear combinations of the $e_i$. Moreover $T^2x = x$ for any $x\in D(T)$ since $T$ is a direct sum of involutions.

Consider then the sequence $x_n = (0,\ldots,0,1,0,\ldots)$, where $1$ occurs in the $2n^{\text{th}}$ spot. Additionally, each $x_n$ has norm one. However $\|Tx_n\|= n\|x_n\|$ and so the sequence of $\|Tx_n\|$ diverges. Therefore $T$ is unbounded.