Closure of an unbounded symmetric operator

Question

The following theorems are from "Introductory functional analysis with applications" by Kreyszig.

My question reagards an apparent contradiction in the theorems:

1-st (p.525),

Theorem 1: If a linear operator $T$ is defined on all of a complex Hilbert space $H$ and satisfies \begin{equation} \langle Tx,y \rangle=\langle x,Ty \rangle , \end{equation} then $T$ is bounded.

(p.533),

Definition 1: Let $T: D(T) \rightarrow H$ be a linear operator which is densely defined on $H$, that is, $\overline{D(T)}=H$. Then $T$ is called symmetric if \begin{equation} \forall x,y \in D(T):\langle Tx,y\rangle=\langle x,Ty \rangle . \end{equation}

(p.534),

Definition 2: Let $T: D(T) \rightarrow H$ be a linear operator which is densely defined on $H$, that is, $\overline{D(T)}=H$. Then $T$ is called self-adjoint if \begin{equation} T=T^* . \end{equation}

$T^*$ is the Hilbert-adjoint operator defined on p.527. After which we have a definition of a closed linear operator on p.535 and a closure of a linear operator on p.537. The following theorem on p.537 is,

Theorem 2: Let $T: D(T) \rightarrow H$ be a linear operator which is densely defined on $H$, that is, $\overline{D(T)}=H$. Then if $T$ is symmetric, its closure $\overline{T}$ exists and is unique.

So here is my concern. If $T$ is symmetric on the whole space $H$ (hence self-adjoint), it must be bounded. If it's unbounded it can be at most symmetric on a dense domain $D(T)$ in $H$. However if such an operator has a symmetric closure $\overline{T}$, then its domain $D(\overline{T})=\overline{D(T)}=H$ (even if you look at the proof of Theorem 2 in the book, the way $\overline{T}$ is defined, I can't see how it's domain is anything different than $H$), but it still continues to be unbounded at some point in $D(T)$ since it's an extension. Then how doesn't that violate Theorem 1?

Answer 1

As was stated in the comments, the problem is where you write $D(\overline{T}) = \overline{D(T)}$. The closure of $T$, if it exists, is the linear operator whose graph is equal to the closure of the graph of $T$ in $H\times H$. But this does not imply that the domain of $T$ is closed in $H$. If $G(T)$ is the graph of $T$, it is worth noting that $D(T)$ is the projection of $G(T)$ onto the first product space of $H\times H$. However, that a set is closed in $H\times H$ does not imply that the projection onto the first product space of $H\times H$ is closed in $H$, as can be seen here.

A counterexample which is a generalisation of one mentioned in the comments is to take any unbounded real-valued sequence $a = (a(n))_{n\in\mathbb{N}}$ and then to define $A: D(A) \to \ell_{2}$ on the domain \begin{equation} D(A) := \{ x \in \ell_{2} : ax = (a(n)x(n))_{n\in\mathbb{N}} \in \ell_{2} \} \end{equation} by $Ax := ax$.

Note that $D(A)$ contains the sequences with finite support so $A$ is densely defined. Since $a$ is a real-valued sequence it is straightforward to check that $A$ is symmetric. To show that $A$ is closed, suppose $(x_{n})_{n\in\mathbb{N}}$ is a sequence in $D(A)$ and $x,y\in \ell_{2}$ such that $\lim_{n\to\infty} x_{n} = x$ and $\lim_{n\to\infty} Ax_{n} = y$. Since convergence in $\ell_{2}$ implies pointwise convergence it follows that \begin{equation} y(k) = \lim_{n\to\infty} (Ax_{n})(k) = \lim_{n\to\infty} a(k) x_{n}(k) = a(k) x(k) \end{equation} for each $k\in\mathbb{N}$. Consequently, $ax = y\in \ell_{2}$, so $x\in D(A)$ and $y = ax = Ax$. This shows that $G(A)$ is closed and hence that $A$ is closed.

On the other hand, if it is supposed for a contradiction that $D(A) = \ell_{2}$, then by the closed graph theorem $A$ is bounded and it can further be shown that $\|A\| = \sup\{|a(n)| : n\in\mathbb{N} \}$. But this contradicts that $a$ is an unbounded sequence. Therefore $D(A) \neq H$.