Densely-defined unbounded operators in Hilbert spaces

hilbert-spacesoperator-theory

I've found the following exercise about densely-defined unbounded operators in Hilbert spaces quite challenging. Here's the text:

Let T : D(T) $\subset$ H $\rightarrow$ $\mathbb{C}$ be a linear operator such that its closure $\overline T$ is densely defined in H. Show if T is densely defined in H.

I tried to prove this by absurd but I'm not sure my thinking is correct.

Let's assume that T and its closure are densely defined in H (and therefore T is closable). Then, if $T \subset$ $\overline T$ it follows that $(\overline T)^{*} \subset T^{*}$. By hypothesis, T is dense in H so I can use the following theorem which states that:

If T : D(T) $\subset$ H $\rightarrow$ H is densely defined in H then: 1) $T^{*} = \overline{T^*}$ 2) T is closable if and only if $D(T^{*})$ is dense in H and $T \subset \overline T = T^{**}$

to prove that $T^{*}$ is dense in H. Furthermore, being $\overline T$ densely defined in H, we can conclude that $T^{**}$ is densely defined in H as well.

But then by a corollary of this theorem we have that if $T, T^*$ and $T^{**}$ are densely defined in H, then $T^* = {\overline{T}}^{*} = \overline{T^*} = T^{***}$.

However, this implies that $(\overline T)^{*} = T^{*}$.
I reasoned as follows: if T is extended (but not equal, by hypothesis we do not know if T is closed) by its closure, then $(\overline T)^*$ is extended by $T^*$ which equals $\overline {T^*}$. However, what I conclude is that: $(\overline T)^* = \overline{T^*}$ which means that simultaneously $(\overline T)^* \subset \overline{T^*}$ and $\overline {T^*} \subset( \overline T)^*$. But then, $\overline T \subset T$ which implies T and its closure must be equal which is absurd since by hypothesis $T \subset \overline T$.

Note: the notation I used $T \subset \overline T$ means that $\overline T$ extends T. $T^*$ does mean the adjoint of T.

I believe that the question here is subtle: I've just proved that the if and only if implication doesn't hold in this case. It is clear to me that T being densely defined in H does not imply that its extension $\overline T$ is dense in H (roughly saying: I don't know if the closure behaves well). In spite of this, I've encountered some difficulties in proving (or disproving) the vice-versa. Thus, if anyone could tell me whether the reasoning above is correct (or at least whether I'm on the right path towards the solution) it would be extremely appreciated.

Best Answer

  1. $\overline{D(T)}$ is dense in $H$

Let $x \in D(\overline{T}) \subset H$. Since, by hypothesis, $T$ is closable $\overline{G(T)}$ is the graph of a closed operator ($\overline T$): $\overline{G(T)} = G(\overline{T})$.

It follows that: $\exists$ {${x_n}$} $_{n \in \mathbb{N}}$ such that ${x}_n \in D(T)$ and $\lim_{n \rightarrow \infty}$ $(x_n, Tx_n) = (x,y) \in \overline{G(T)} = G(\overline{T})$. Therefore, $y = \overline{T}x$.

As a consequence, $\lim_{n \rightarrow \infty}$ $x_n = x \in D(\overline{T})$. Since - by definition of closure - $\overline{D(T)}$ contains the limit points of all succession in $D(T)$ we can conclude that: $x \in \overline{D(T)}$. Thus, $D(\overline{T}) \subset \overline{D(T)}$ and being $\overline{T}$ densely defined in $H$, it follows that $\overline{D(T)}$ is dense in $H$ as well.

  1. $D(T)$ is dense in $H$

$D(T)$ dense in $H$ means that: $\forall \phi \in H$ and $\forall \epsilon > 0$ $\exists \tilde{\phi} \in D(T)$ such that: $|\phi -\tilde{\phi}| < \epsilon$

  • $\overline{D(T)}$ dense in $H$: $\forall \phi \in H$ $\forall \epsilon > 0$ $\exists x \in \overline{D(T)}$ such that: $|\phi -x| < \epsilon /2$
  • $\overline{D(T)}$ is the closure of $D(T)$: $x \in \overline{D(T)} \subset H$ then $\forall \epsilon > 0$ $\exists \tilde{\phi} \in D(T)$ such that: $|x - \tilde{\phi}| < \epsilon /2$

Therefore: $|\phi - \tilde{\phi}| = |\phi - x + x - \tilde{\phi}| \le |\phi - x| + |x - \tilde{\phi}|< \epsilon$ which proves the thesis.

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