Describe the closure and adjont of operator T

Question

Let $\mathcal{H}_2$ be a separable Hilbert space and $(T,D(T))$ be an operator in $\mathcal{H}_2$ such that there exists a Hilbert basis $(e_n)_{n\in \mathbb{N}}$ of $\mathcal{H}_2$ with $e_n \in D(T)$ for any $n \in \mathbb{N}$ and $Te_n = \mu_n e_n$.

a) Describe the closure $\bar{T}$ of $T$.

b) Describe the adjoint $T^*$ of $T$.

c) Suppose that $\mu_n \in \mathbb{R}$ for any $n \in \mathbb{N}$. Prove that $\bar{T}$ is self-adjoint.

---

Since $\mathcal{H}_2$ is a Hilbert space, its norm must be continuous, bounded and the space must be complete. Therefore for each element $Tx \in \mathcal{H}_2$ with $x \in D(T)$:

\begin{align*} ||{Tx}||_{\mathcal{H}_2} = ||{\sum_{n=1}^\infty\,h_ne_n}||_{\mathcal{H}_2} \leq \sum_{n=1}^\infty\,||{h_ne_n}||_{\mathcal{H}_2} \leq c_{\mathcal{H}_2}\,\sup_{n\in\mathbb{N}}\,||{h_ne_n}|| < \infty \end{align*} Therefore $T$ is limited. We can also formulate $x \in D(T)$ using the base $(e_n)_n$: $x = \sum_{n=1}^\infty \, b_ne_n$. Therefore, the $T$ operator works as follows: \begin{align} Tx = \sum_{n=1}^\infty\, b_n \, Te_n = \sum_{n=1}^\infty\,b_n\mu_ne_n \end{align} Let us now consider the following: \begin{align*} ||{Tx}||_{\mathcal{H}_2} &= ||{\sum_{n=1}^\infty\,b_n\mu_ne_n}||_{\mathcal{H}_2} \leq \sum_{n=1}^\infty \, ||{b_n\mu_ne_n}||_{\mathcal{H}_2} = \sum_{n=1}^\infty |\mu_n|\,||{b_ne_n}||_{\mathcal{H}_2} \\ &\leq \sup_{n\in\mathbb{N}}\,|\mu_n|\cdot \sum_{n=1}^\infty\,||{b_ne_n}||_{\mathcal{H}_2} = ||{\mu_n}||_\infty\cdot\sum_{n=1}^\infty\,||{b_ne_n}||_{\mathcal{H}_2} \leq ||{\mu_n}||_\infty\cdot c_{\mathcal{H}_2}\,\sup_{n\in\mathbb{N}}\,||{b_ne_n}||_{\mathcal{H}_2} \\ & \leq ||{\mu_n}||_\infty\cdot c_{\mathcal{H}_2}\,\sup_{n\in\mathbb{N}}\,|b_n|\,||{e_n}||_{\mathcal{H}_2}\,. \end{align*} Since $||{Tx}||_{\mathcal{H}_2}$ must be $< \infty$, it follows that $||{\mu_n}||_\infty < \infty$. It is also linear: \begin{align} &T(\lambda x) = \sum_{n=1}^\infty\, \lambda b_n \, Te_n = \lambda\,Tx, \\ &T(x+y) = \sum_{n=1}^\infty\, \lambda (b_n+c_n) \, Te_n = \sum_{n=1}^\infty\, b_n\,Te_n + \sum_{n=1}^\infty\, c_n\,Te_n= Tx + Ty \end{align} Thus $T \in \mathcal{L}(D(T),\mathcal{H}_1)$, and therefore equally continuous in $D(T)$. After the theorem of closed graph, the corresponding graph $G_T$ is a closed subspace. Finally $T$ is closed. Thus $T = \bar{T}$ follows.

Is this so far correct? And what do I do now? How do I determine the adjoint $T^*$?

Answer 1

No, there are several things that are incorrect.

1. $Tx\in \mathcal{H_2}$ for $x\in D(T)$ is true by definition, there is nothing to check.
2. If the operator $T$ is not assumed to be linear, then just requiring $Te_n=\mu_n e_n$ does not determine it on any other element than the basis vectors. In particular, you cannot prove that $T$ is linear, it is rather an assumption.
3. It is not true that $\sup_n |\mu_n|$ is necessarily finite. This would be true if $D(T)=\mathcal{H}_2$, but this is not a given. That means also that the operator $T$ is not necessarily bounded.
4. The closed graph theorem does not assert that every bounded linear operator is closed. This is only true if the domain of definition is complete (i.e. closed in the case of Banach spaces). In general, the operator $T$ is neither closed nor bounded.

I would recommend you to try again with the following more concrete example: $\mathcal{H}_2=\ell^2=\{(x_n)\mid \sum_n |x_n|^2<\infty\}$, $D(T)=\{(x_n)\mid x_n=0\text{ for all but finitely many }n\}$, $T(x_n)=(n x_n)$. Once you can solve this, it should be quite easy to tackle the more general case.