An operator that satisfies some condition is a normal operator

Question

QUESTION:

Let $H$ be Hilbert space, $T$ is a bounded operator on $H$, $TT^{\ast }\geqslant T^{\ast }T$, proof $T$ is normal operator($TT^{\ast } =T^{\ast }T$).

I guess this question is missing condition (maybe $T$ is a compact operator?).

I don't know what "$TT^{\ast }\geqslant T^{\ast }T$" means, here's my guess:

1.$TT^{\ast }\geqslant T^{\ast }T\Leftrightarrow (TT^{\ast }x,x)\geqslant (T^{\ast }Tx,x)$

2.$TT^{\ast }\geqslant T^{\ast }T\Leftrightarrow ||TT^{\ast }||\geqslant ||T^{\ast }T||$

Answer 1

Although in the comments, @leslie townes has already answered this question(actually the result of a paper: Andô, T. On hyponormal operators. Proc. Amer. Math. Soc. 14, 290-291 (1963)), for the sake of the integrity of this question, I will paraphrase the paper's opinion as follows, which makes some sense to a newbie like me: Since the proof of the paper involves the use of some results of the exercises in Berberian's book(Berberian S K. Introduction to Hilbert space[M]. American Mathematical Soc., 1999.)(which is not difficult because of the excellent layout of this book), as for the results not proved in the following, they can be found in Berberian's book, or are well-know results(at least to me).

1.Problem review:

Let $H$ be a Hilbert space, $T$ is a bounded operator on $H$，$TT^{\ast }\leqslant T^{\ast }T$, proof $T$ is normal operator($TT^{\ast }=T^{\ast }T$).

As mentioned in the comments, $TT^{\ast }\leqslant T^{\ast }T\Leftrightarrow \left( TT^{\ast }x,x\right) \leqslant \left( T^{\ast }Tx,x\right) $, this operator is called hyponormal operator.

My initial intuition tells me that this problem seems to be missing conditions, that maybe compact operators work, but these are completely intuitive.

As @geetha290krm already mentioned, the left and right shifts on $l^2$ provide a counter-example: as is well-known, if $T=$ right shift on $l^2$, the $T^{\ast }=$ left shift on $l^2$, easy to verify $\left( TT^{\ast }x,x\right) < \left( T^{\ast }Tx,x\right)$.

@David Gao's comment made me realize that the dimension of Hilbert space is also a variable in this problem:

If Hilbert space is infinite dimensional, $T$ being a Hilbert-Schmidt operator is definitely enough to imply the result, the key here is that you can define traces on infinite-dimensional Hilbert spaces($tr(T^{\ast }T)<\infty $ and $tr(T^{\ast }T)=tr\left( TT^{\ast }\right) $), and that's no problem at all for finite dimensional Hilbert spaces, therefore, we can follow the proof of David Gao exactly and get the following result:

Let $H$ be a finite dimensional Hilbert space, $T$ is a hyponormal operator, then $T$ is normal operator.

Whether bounded operators on finite dimensional Hilbert spaces or Hilbert-Schmidt operator, they're all compact operator. Next, our goal is to prove:

Let $H$ be a Hilbert space(finite or infinite dimensional), $T$ is a hyponormal operator and compact, then $T$ is normal operator.

2.The proof method of Andô:

It must be mentioned, for general Banach spaces, compact operators must be completely continuous operators, and vice versa. In fact, for reflexive Banach spaces, completely continuous operators must also be compact operators. Therefore, for Hilbert spaces, completely continuous operator and compact operator are the same(Andô's paper uses the completely continuous operator to describe).

The proof method of Andô is a model of using spectrum as a tool to deal with functional analysis problems. The key to the method is proof:

Every nonzero hyponormal operator has a nonzero element in its spectrum.

Combine the following famous results:

Let $T$ be a compact operator in $B(X)$, in which $X$ is a Banach space. Then, the nonzero elements of the spectrum of $T$ are eigenvalues. There are only countably many eigenvalues, and, in the case of infinitely many, they form a sequence tending to $0$.

Let's repeat Andô's proof step by step.

• A family of closed linear subspaces is said to be total in case the only vector $x$ which is orthogonal to every subspace belonging to the family is $x=0$.
• An eigensubspace belonging to the eigenvalue $\mu$ ($\mu$-space of $T$), $H_{T}\left( \mu \right) =\left\{ x\in H\ |\ Tx=\mu x\right\} $.
• If the $\mu$-spaces of an operator $T$ are a total family, and $H_{T}\left( \mu \right) \subseteq H_{T^{\ast }}\left( \mu^{\ast } \right) $ for all $\mu$, then $T$ is normal.

Proof: Let $N$ be the null space of $TT^{\ast }-T^{\ast }T$, the problem is to show $N^{\bot }=\left\{ 0\right\} $.

One has $H_{T}\left( \mu \right) \subseteq N\ ,\ \forall \mu $: Since $x\in N_{T}\left( \mu \right) \subseteq N_{T^{\ast }}\left( \mu^{\ast } \right) $, $$TT^{\ast }x=T\left( \mu^{\ast } x\right) =\mu \mu^{\ast } x=T^{\ast }(\mu x)=T^{\ast }Tx.$$

It follows that if $x\bot N$, then $x\bot H_{T}\left( \mu \right) \ ,\ \forall \mu $, since the $\mu$-spaces of an operator $T$ are a total family, so $x=0$.

• Let $T$ be a hyponormal operator, then for every scalar $\mu$, $H_{T}\left( \mu \right) \subseteq H_{T^{\ast }}\left( \mu^{\ast } \right) $.

Proof: Since $T$ is hyponormal, $T-\mu I$ is also hyponormal, so $||\left( T-\mu I\right)^{\ast } x||\leqslant ||\left( T-\mu I\right) x||$.

For $x\in H_{T}(\mu )$, we can get $\left(T-\mu I\right)^{\ast } x=0$, this mean $x\in H_{T^{\ast }}(\mu^{\ast } )$.

• Let $T$ be a hyponormal operator, If the $\mu$-spaces of $T$ are a total family, then $T$ is normal.
• A closed linear subspace $N$ is said to reduce the operator $T$ in case both $N$ and $N^{\bot }$ are invariant under $T$.
• If T is an operator, and $N$ is a closed linear subspace, then $N$ reduce $T$ $\Leftrightarrow $ $N$ is invariant under both $T$ and $T^{\ast }$.
• Let $N$ be a closed linear subspace invariant under $T$, the restriction of $T$ to $N$ is the mapping $T/N:N\longrightarrow N$ deefined by $T/N\left( y\right) =Ty$.
• If $T$ is hyponormal, $N$ is invariant under $T$, and $T/N$ is normal, then $N$ reduces $T$.

Proof: we only need to show: $x\in N\ ,\ Tx\in N$.

Suppose $R=T/N$(normal), for any $y \in N$, $$\left( T^{\ast }x,y\right) =(x,Ty)=\left( x,Ry\right) =\left( R^{\ast }x,y\right) \Rightarrow \left( \left( T^{\ast }-R^{\ast }\right) x,y\right) =0.$$

So $\left( T^{\ast }-R^{\ast }\right) x\in N^{\bot }$, since $R^{\ast }x\in N$, one has $$\begin{gathered}||T^{\ast }x||^{2}=||\left( T^{\ast }x-R^{\ast }x\right) +R^{\ast }x||^{2}=||T^{\ast }x-R^{\ast }x||^{2}+||R^{\ast }x||^{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =||T^{\ast }x-R^{\ast }x||^{2}+||Rx||^{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =||T^{\ast }x-R^{\ast }x||^{2}+||Tx||^{2}\end{gathered} .$$

Notice that $||T^{\ast }x||\leqslant ||Tx||$, we can get $T^{\ast }x=R^{\ast }x\in N$.

• Let $T$ be a hyponormal operator, the $\mu-$space of $T$ reduce $T$.

Proof: we only need to show: $T/N_{T}\left( \mu \right) $ is normal.

For any $x\in N_{T}\left( \mu \right) \subseteq N_{T^{\ast }}\left( \mu^{\ast } \right) $, we have $$\left( Tx,Tx\right) =\mu \mu^{\ast } \left( x,x\right) =\left( T^{\ast }x,T^{\ast }x\right) .$$

• Let $H$ be a Hilbert space(finite or infinite dimensional), $T$ is a hyponormal operator and compact, then $T$ is normal operator.

Proof: It is sufficient to prove that the set of all eigenvectors for $T$ is total, in other words, the set $M$ of all vectors orthogonal to every eigenvector consists of only the null vector. Since $M$ is invariant under $T$, the restriction of $T$ to $M$, denoted by $T/M$, is also hyponormal.

The spectrum of $T/M$ consists of $0$ only, for $T/M$ is compact and has no eigenvalue by the definition of $M$. So $M={0}$.

Answer 2

[OP posted an answer as I was converting my own comments, which differ in substance from OP's exposition, to an answer] We can show that if $T$ is compact and satisfies your condition, then $T$ is normal.

Let us call a bounded operator $A$ on Hilbert space hyponormal if $AA^* \leq A^*A$; the hypothesis on $T$ is that the operator $T^*$ is hyponormal. Hyponormal operators generalize normal operators and there is quite a bit of general theory about them. One respect in which hyponormal operators are just like normal operators is the following (famous for normal operators, less well known for hyponormal operators).

Theorem. The eigenspaces of a hyponormal operator are reducing subspaces for that operator, and eigenspaces of a hyponormal operator corresponding to distinct eigenvalues are orthogonal to one another.

Proof. The eigenspaces of any operator (hyponormal or not) are invariant under that operator. If $A$ is hyponormal, then for any vector $\xi$ and $\lambda \in \mathbb{C}$ we also have \begin{align*} \|A^* \xi - \overline{\lambda} \xi\|^2 & = \langle AA^* \xi, \xi\rangle - \langle A^* \xi, \overline{\lambda} \xi\rangle - \langle \overline{\lambda} \xi, A^*\xi\rangle + \lambda \overline{\lambda} \langle \xi, \xi\rangle \\ & \leq \langle A^*A \xi, \xi\rangle - \langle A^* \xi, \overline{\lambda} \xi\rangle - \langle \overline{\lambda} \xi, A^* \xi\rangle + \lambda \overline{\lambda} \langle \xi, \xi\rangle \\ & = \langle A^*A \xi, \xi\rangle - \langle \lambda \xi, A \xi\rangle - \langle A \xi, \lambda \xi\rangle + \lambda \overline{\lambda} \langle \xi, \xi\rangle \\ & = \|A \xi - \lambda \xi\|^2, \end{align*} where we use $AA^* \leq A^* A$ in deducing the one inequality above. It follows that $\ker(A - \lambda I) \subseteq \ker(A^* - \overline{\lambda} I)$ for all $\lambda \in \mathbb{C}$, and thus each of the spaces $\ker(A - \lambda I)$ is invariant under $A^*$ and hence a reducing subspace for $A$.

If $\xi$ and $\eta$ are vectors with $A \xi = \lambda \xi$ and $A \eta = \mu \eta$ then (as just proved) $A^* \eta = \overline{\mu} \eta$, so that $\lambda \langle \xi, \eta\rangle = \langle A \xi, \eta\rangle = \langle \xi, A^* \eta\rangle = \langle \xi, \overline{\mu} \eta\rangle = \mu \langle \xi, \eta\rangle$, so that whenever $\lambda \neq \mu$ we deduce $\langle \xi, \eta\rangle = 0$. End of proof.

Because a direct sum of reducing subspaces for an operator is also reducing, it follows that for any hyponormal operator $A$ on a Hilbert space $H$, there is a direct sum decomposition $H = H_e \oplus H_0$ of the underlying Hilbert space into $A$-reducing subspaces, where $H_e = \bigoplus_{\lambda \in \sigma_p(A)} \ker(A - \lambda I)$ is the direct sum of the eigenspaces for $A$, if any (using the usual notation $\sigma_p(A) := \{\lambda \in \mathbb{C}: \ker(A - \lambda I) \neq 0\}$), and where $H_0$ is a subspace on which $A$ has no eigenvalues.

Note that the restriction of $A$ to $H_e$ is normal, so if the hyponormal operator $A$ is not normal, it is because the space $H_0$ is nonzero and the restriction $A_0$ of $A$ to $H_0$ is not normal. (In complete generality, the space $H_0$ can be nonzero even when $A$ is normal, and the operator $A_0$ can even itself be normal, e.g. if $A$ happens to be a normal operator without eigenvalues. So the decomposition just mentioned is not generally the decomposition of a hyponormal operator into a direct sum of a normal operator and a so-called "pure" hyponormal operator that one sometimes finds in the literature.)

If $A$ is assumed compact, it is possible to show that space $H_0$ must be trivial and the operator $A = A_e$ must be normal. When $A$ is compact, the operator $A_0$ is also compact; since it has no eigenvalues, by the well-known theory of compact operators (sometimes called the "Fredholm alternative"), one must have $\sigma(A_0) = \{0\}$. While it is generally possible for a nonzero compact operator to have only $\{0\}$ as its spectrum, this is not possible for a hyponormal operator, as shown (for example) by T. Ando or J. Stampfli (independently at about the same time) in the 1960s. I'll include a simplified version of their argument here.

First note that if $A$ is hyponormal we have $\|A^* \eta\| \leq \|A \eta\|$ for all $\eta$ (the left hand side squared is $\langle A A^* \eta, \eta \rangle$ while the right hand side is $\langle A^* A \eta, \eta \rangle$), and hence $$ \|A^* A^n \xi\| \leq \|A A^n \xi\| \leq \|A^{n+1}\| \|\xi\| $$ for all $\xi$ and $n \geq 1$, so that $\|A^* A^n\| \leq \|A^{n+1}\|$ for all $n \geq 1$.

Theorem (Andô or Stampfli). If $A$ is hyponormal then $\|A^n\| = \|A\|^n$ holds for all $n$.

Proof. The desired result clearly holds when $n=1$, and if it holds for all $1 \leq k \leq n$, we have \begin{align*} \|A^n\|^2 & = \|A^{*n} A^n\| \\ & = \|A^{*(n-1)} A^* A^n\| \\ & \leq \|A^{*(n-1)}\| \|A^* A^n\| \\ & \leq \|A^{*(n-1)}\| \|A^{n+1}\| \\ & = \|A^{n-1}\| \|A^{n+1}\|, \end{align*} using the observation made preceding the theorem for the second inequality. The inductive hypothesis implies that the extreme left and right hand sides of the above inequality are $\|A\|^{2n}$ and $\|A\|^{n-1} \|A^{n+1}\|$, respectively, and we deduce that $\|A\|^{n+1} \leq \|A^{n+1}\|$, as desired (the reverse inequality $\|A^{n+1}\| \leq \|A\|^{n+1}$ is obvious from sub-multiplicativity of the norm). End of proof.

Corollary. If $A$ is hyponormal then its norm is its spectral radius.

Proof. Apply the well known Gelfand formula for the the spectral radius in terms of the norm, and use $\|S^n\|^{1/n} = (\|S\|^n)^{1/n} = \|S\|$ for all $n$ by the lemma. End of proof.

Turning to the application: if $A$ is hyponormal and compact, the corollary implies that $A_0$ is the zero operator, which (since $A_0$ has no eigenvalues) implies that the space $H_0$ is trivial, so that $A = A_e$ is normal. If $T$ (in the original post) is compact and $T^*$ is hyponormal, then because $T^*$ is also compact we can apply the above to $A = T^*$ and deduce that $T^*$ (and hence also $T$) is normal.