# Question about normal operators

I have a question about definitions and theorems because I am a little bit confused.

By definition we say that a (possibly unbounded) operator $T$ on a Hilbert space $H$ is normal if $D(T)$ is dense in the Hilbert space $H$ and is closed. Moreover it must hold $TT^*=T^*T$.

A Theorem says that, if $T$ is normal, then $D(T)=D(T^*)$. But to prove that $TT^*=T^*T$ we have to know what $D(T^*)$ is, right?

To show what I mean here an example: if $T:l^2\rightarrow l^2$ is defined by $T((x_n)_n)=(\lambda_nx_n)$ for fixed $(\lambda_n)_n$ with $D(T)=\{x\in l^2:\sum_{n=1}^{\infty}{|\lambda_nx_n|}<\infty\}$. I know that the adjoint is given by $T^*((x_n)_n)=(\overline{\lambda_n}x_n)$, but the question is on which domain $D(T^*)$. My idea is that $D(T)=D(T^*)$ because $T$ is a normal operator (from my point of view) because $D(T)$ is dense and the graph of $T$ is closed (why?) but how to prove that? Because if I know that, I may conclude that $D(T)=D(T^*)$ by the theorem about normal operators. Any suggestion? Hints?Solutions?

Maybe I have to work this out in another way.

Thank you very much :)

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"graph of T is closed (why?)" Isn't the graph closed because you defined $T$ to be closed in the second sentence? (at least that's the definition of a closed operator) – Horstenson Jan 14 '14 at 23:13
I'm not so sure about the rest but there is for example the theorem that says $Graph(T^*)=orthogonal_complement(V(Graph(T)))$ (forgot latex symbol) where $V({a,b})={-b,a}$, so you "know" $D(T^*)$ – Horstenson Jan 14 '14 at 23:23
@Horstenson: yes i tried this also but got that G(T^*)=G (\overline{T}) but this is not what you want. If someone can tell more this would be nice. – UuPgol193 Jan 14 '14 at 23:28
@Horstenson: You can use \perp to get $\perp,$ if that's what you were looking for. – Cameron Buie Jan 15 '14 at 16:46
@Cameron Buie thanks, that was what I wanted – Horstenson Jan 15 '14 at 17:46

## 1 Answer

Let $T : \mathcal{D}(T)\subseteq l^{2}\rightarrow l^{2}$ be the linear operator you have defined. Then $T$ is densely defined because each standard basis element $e_{j}=\{\delta_{j,n}\}_{n=1}^{\infty}$ is in $\mathcal{D}(T)$. Define $L : \mathcal{D}(T)\subseteq l^{2}\rightarrow l^{2}$ by $L\{ x_{n}\}_{n=1}^{\infty} = \{\overline{\lambda_{n}}x_{n}\}_{n=1}^{\infty}$. I'll prove to you that $L=T^{\star}$ (equality by domain and action), and prove that $T$ is normal. I believe that will answer your question. I'll use the notation $A \preceq B$ for linear operators to mean that the graph of $A$ is a subset of the graph of $B$; equivalently $\mathcal{D}(A)\subseteq \mathcal{D}(B)$ and $Ax=Bx$ for all $x \in \mathcal{D}(A)$. Keep in mind that $T^{\star}$ is uniquely defined because $T$ is densely-defined. No a priori assumptions are made about the domain of $T^{\star}$. The Hilbert space adjoint may be defined in several different ways. The most useful definition I know is this: $y\in\mathcal{D}(T^{\star})$ iff there exists $z \in X$ such that $$(Tx,y)=(x,z),\;\;\; x \in \mathcal{D}(T).$$ If such a $z$ exists, then $y \in \mathcal{D}(T^{\star})$ and $T^{\star}y=z$. The reason such a definition is useful is that it gives you a nice equation to play with.

First notice that $$(Tx,y)=\sum_{n=0}^{\infty}\lambda_{n}x_{n}\overline{y_{n}}=\sum_{n=0}^{\infty}x_{n}\overline{\overline{\lambda_{n}}y_{n}}=(x,Ly),\;\;\; x,y\in\mathcal{D}(T)=\mathcal{D}(L).$$ Applying the above definition of adjoint with $z=Ly$, one finds that every $y\in \mathcal{D}(L)$ is also in $\mathcal{D}(T^{\star})$ and $T^{\star}y=z$ where $z=Ly$. In other words, $L\preceq T^{\star}$ because $\mathcal{D}(L)\subseteq \mathcal{D}(T^{\star})$ and $T^{\star}y=Ly$ for all $y \in \mathcal{D}(L)$.

Conversely, suppose $y \in \mathcal{D}(T^{\star})$. Equivalently, there exists $z$ such that $$(Tx,y)=(x,z),\;\;\; x \in \mathcal{D}(T).$$ In particular, the above holds for each standard basis element $x=e_{j}$, which implies $$\lambda_{j}\overline{y}_{j}=(Te_{j},y)=(e_{j},z)=\overline{z_{j}}.$$ Becuase $\{ z_{j}\} \in l^{2}$, then $\sum_{j}|\lambda_{j}y_{j}|^{2} < \infty$, which implies that $y \in \mathcal{D}(T)=\mathcal{D}(L)$; furthermore $T^{\star}y=z$ is shown by the above to be $Ly$. Therefore $T^{\star}\preceq L$, which completes the proof that $L=T^{\star}$. Hence, $L$ is closed because $T^{\star}$ is closed.

Finally, $x \in \mathcal{D}(T^{\star}T)$ iff $x \in \mathcal{D}(T)$ with $Tx\in\mathcal{D}(T^{\star})$, and, in that case, $T^{\star}Tx = T^{\star}(Tx)$. Equivalently, $$\sum_{j=0}^{\infty}|\lambda_{j}|^{4}|x_{j}|^{2} < \infty,$$ and, in that case, $$T^{\star}T \{ x_{n}\}_{n=1}^{\infty} = \{ |\lambda_{n}|^{2}x_{n}\}_{n=1}^{\infty}.$$ It's not hard to see why $T^{\star}T=TT^{\star}$.

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The corollary form this argumentation os $D(T)=D(T^*)$ – user121557 Jan 15 '14 at 15:59
@user121557: I'm not sure what you are asking or saying. Please explain. If you want to comment on my post, please add the comment under my post. I just happened to see this, but I didn't get a message. – TrialAndError Jan 15 '14 at 17:02