Assume that $H_1,H_2$ are separable Hilbert spaces, $B$ is a separable Banach space and $H_1\subset B\subset H_2$. Assume further that the inclusion mappings are continuous and have dense images.

Given any closed, densely defined operator $T$ in $B$, is it then true that there exists a closed, densely defined operator $\tilde T$ in $H_2$ which extends $T$ (viewed as an operator in $H_2$)? In other words, is $T$ closable as an operator in $H_2$?

In a paper by Gill, Basu, Zachary and Steadman, it is pointed out that it is well known that the answer to my question is 'yes' (See Theorem 4). They argue that the adjoint of $T$, say $T':B'\rightarrow B'$, is a closed operator, which is densely defined if we restrict it to $H_2'\subset B'$. However, it is not explained why the restriction of $T'$ is densely defined, and it does not seem obvious to me that it is.

This question is related to another question of mine, Pointwise approximation of a closed operator.

Gill, Tepper L.; Basu, Sudeshna; Zachary, Woodford W.; Steadman, V., Adjoint for operators in Banach spaces, Proc. Am. Math. Soc. 132, No. 5, 1429-1434 (2004). ZBL1048.46014.


The answer seems to be no. A counter example follows.

Let $H_1$ be a separable Hilbert space with norm $\lVert \cdot \rVert_1$ and let $U$ be a bounded, injective, self-adjoint operator on $H_1$ with $0\in\sigma(U)$. Then we may consider the completion $H_2$ of $H_1$ with respect to the norm $\lVert \psi\rVert_2 := \lVert U \psi \rVert_1$. Since $U$ is not boundedly invertible, we have $H_2\supsetneq H_1$. On the other hand, $H_1$ is clearly dense in $H_2$. Furthermore, the extension $\bar{U}:H_2\rightarrow H_1$ of $U$ to $H_2$ is a unitary transformation.

For definiteness, let $H_1=\ell^2$, and let \begin{align} U e_n &= n^{-4}e_n + n^{-3}e_1, \mbox{ if }n\geqslant 2,\\ U e_1 &= a_1 e_1 + \sum_{n=2}^\infty n^{-3}e_n, \end{align} with $a_1=2\sum_{n=2}^\infty n^{-2}$. Furthermore, consider the closed, densely defined operator $T$ defined by $Te_n=n^4e_n$. Then $U$ is bounded and positive definite with $0\in\sigma(U)$, and $UT$ is not closable. Since $H_1\subseteq H_2$, we may consider $T$ as a densely defined operator in $H_2$, and we denote this operator by $T_2$.

Suppose now that $T_2$ is closable, with closure $\bar{T}_2$. Then $UT\subseteq \bar{U}\bar{T}_2\!\!\upharpoonright_{H_{1}\cap\mathrm{dom}(\bar{T}_2)}$. Since $\bar{U}$ is boundedly invertible, $\bar{U}\bar{T}_2$ is closed, which ensures that $\bar{U}\bar{T}_2\!\!\upharpoonright_{H_{1}\cap\mathrm{dom}(\bar{T}_2)}$ is closed. But then $UT$ is closable, and this is a contradiction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.