8
$\begingroup$

We know that normal operators are "nice". In the finite dimensional case, the spectral theorem tells us everything we need to know. In the infinite dimensional case, we can define a continuous functional calculus over the (bounded) normal operators (admittedly I don't know much about the unbounded case). However, is there a context in which we should "expect" an operator to be normal? That is:

Is there a common application of linear algebra (or operator theory) in which we would expect the operator we're working with to be an arbitrary normal operator?

There are many contexts in which we might expect the operator in question to be self-adjoint (the Hessian matrix for instance), and there are certainly some in which we might expect the operator to be unitary/orthogonal (the orthogonal Procrustes problem comes to mind). However, I'm looking for a "natural" application in which we might expect a normal operator of any sort, but not expect a non-normal operator.

I found this MO thread about applications of the spectral theorem, but that hasn't led to any satisfying leads, unfortunately.

Thanks for any feedback.

$\endgroup$
2
$\begingroup$

Let $T$ be a bounded operator on a Hilbert space $H$. For any such thing we have a holomorphic functional calculus (by Cauchy residue formula), which is a continuous $\mathbb{C}$-algebra homomorphism:

$$Hol(\sigma(T)) \to \mathcal{B}(H)$$

Where we denote by $Hol(\sigma(T))$ the algebra of germs of holomorphic functions defined in some neighborhood $\sigma(T) \subset U\subset \mathbb{C}$ (endowed with the topology of uniform convergence on compact subsets). For some purposes it is desirable to upgrade this calculus to a larger class of functions.

Recall that the functional calculus extends the polynomial functional calculus which itself is a restriction of the natural "evaluation" $*$-homomorphism:

$$\mathbb{C}<z,\bar z> \to \mathcal{B}(H)$$

Where the LHS is the free $*$-algebra generated by $z$ and $\bar z$ with involution $z \mapsto \bar z$ and the homomorphism takes $z \mapsto T$ and $\bar z \mapsto T^*$

Consider now the embedding $\mathbb{C}[z,\bar z] \to C_0(\mathbb{C})$ of polynomials into complex valued continuous functions. This is naturally a homormophism of $*$-algebras (the involution on the LHS being $z \mapsto \bar z$). The RHS is moreover naturally a $C^*$-algebra and the embedding is dense in the topology determined by the $C^*$-norm (Stone-Weirstrass theorem).

In order for the evaluation calculus $\mathbb{C}<z,\bar z> \to \mathcal{B}(H)$ to upgrade to a continuous functional calculus it must first factor through $\mathbb{C}[z,\bar z]$ (this happens presicely when $T$ is normal). Suppose this happens, then we could say by the above paragraph that the polynomial functional calculus upgrades to a continuous functional calculus iff its continuous w.r.t. the topology on $\mathbb{C}[z,\bar z]$ inherited from the enbedding into $C_0(\mathbb{C})$. But since the spectral radius determines the norm of a $C^*$-algebra (which determines its topology) any homomorphism of $C^*$-algebras is automatically continuous.

We conclude that $T$ has a continuous functional calculus iff its evaluation homomorphism $\mathbb{C}<z,\bar z> \to \mathcal{B}(H)$ factors through $\mathbb{C}[z,\bar z]$. This is true iff $T$ is a normal operator. We conclude the following:

A bounded operator $T$ on a Hilbert space has a continuous functional calculus iff it is normal.

$\endgroup$

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.