Why do we introduce the continuous functional calculus for self-adjoint operators?

Question

The question is, as the title already says:

Why do we introduce the continuous functional calculus for self-adjoint operators?

To be more specific, why self-adjoint operators?

It is introduced (in short) as follows:

Let $T$ be a linear bounded operator on the Hilbert space $H$, then we can define $T^0=I$, $T^n= T \circ \dots \circ T$ ($n$ times). Thus we have a "polynomial" functional calculus for $T$. Since $P(\sigma(T))$ is dense in $C(\sigma(T))$, we can approximate continuous functions to get a continuous functional calculus.

Question is, why only for self-adjoint operators? Shouldn't it be possible to do the same with any kind of operator?

Answer 1

Nice question. Let us see what's going on in this setting.

Stone-Weierstrass theorem says that if $X$ is a compact subset of $\mathbb{C}$ and $f:X\to\mathbb{C}$ is a continuous function then, there exists a sequence $p_n(z,\bar{z})$ of polynomials in $z$ and $\bar{z}$ (with $z\in X$) such that, $p_n\to f$ uniformly on $X$ : for any $\varepsilon>0$ there exists a $k_0\in\mathbb{N}$ such that, $|f(z)-p_n(z,\bar{z})|<\varepsilon$ for all $z\in X$, for all $n\geq k_0$.

Firstly if $T\in\mathcal{L}(H)$ to make sense of $p_n(T,T^*)$ in a meaningful way you want $T$ and $T^*$ to commute ie. $TT^*=T^*T$ so you want your operator to be a normal operator. Now self-adjoint operators are normal. So they fits the bill. But recall that continuous functional calculus with full generality is not only true for self-adjoint operators. They are also true for normal operators.

Furthermore, if you just consider polynomials in $z$ ie. elements in $P(X)$ then, it isn't dense in $C(X)$. One easy example to illustrate this :

Consider, $X=S^1$, the unit circle in $\mathbb{C}$. Then, it is a compact subset. But, $P(X)$ is not dense in $C(X)$ where, $P(X)$ is the space of polynomials in $z$ ie. one-variable polynomials. To see this define, $f:S^1\to\mathbb{C}$ by, $f(z)=\bar{z}$. Try to show that $f$ is not a uniform limit of polynomials (defined on $S^1$) in $z$.

Answer 2

Let's take $T$ to be the nilpotent matrix $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ acting on $\mathbb{C}^2$, which is of course not self-adjoint, or even normal. The polynomial functional calculus applied to $T$ gives that if $f(x) = \sum f_i x^i$ is a polynomial then

$$f(T) = f_0 + f_1 T$$

since $T^2 = 0$. Equivalently, we have

$$f(T) = f(0) + f'(0) T.$$

This already suggests what goes wrong if we try to generalize to the case that $f$ is continuous; this expression depends on the derivative of $f$! In this case $\sigma(T) = \{ 0 \}$ consists of a single point but it's certainly not the case that if $f_n$ is a sequence of polynomials converging to a continuous function $f$ in $C(\sigma(T))$ (which just means that $f_n(0) \to f(0)$) then $f_n(T)$ is even guaranteed to converge at all, e.g. we could take $f_n(x) = nx$, so that $f_n(T) = n T$ diverges. By generalizing $T$ to larger nilpotent matrices we can get $f(T)$ to be sensitive to higher derivatives of $f$ as well.

The significance of $C(\sigma(T))$ in the case that $T$ is self-adjoint is that it is also the $C^{\ast}$-algebra generated by $T$, and we can apply Gelfand-Naimark to this subalgebra. This also generalizes to the case that $T$ is normal. But in the non-normal case $T$ and $T^{\ast}$ don't commute so the $C^{\ast}$-algebra generated by $T$ is not commutative.

For $T$ an arbitrary bounded operator the best we can get, in a certain precise sense, is the holomorphic functional calculus.

Answer 3

Suppose for a bounded operator $T\in\mathcal B(H)$ there exists a continuous functional calculus in the strong sense of a unital $*$-homomorphism $\Phi_T: C(\sigma(T))\to \mathcal B(H)$ that depends only on the values of functions on $\sigma(T)$, is contractive, and extends the polynomial calculus by sending the coordinate function $z(\lambda)=\lambda$ on $\sigma(T)$ to $T$. Let $\bar z(\lambda)=\overline{\lambda}$ on $\sigma(T)$. Since $\Phi_T$ is a $*$-homomorphism, $\Phi_T(\bar z)=\Phi_T(z)^{*}=T^{*}$. In the commutative algebra $C(\sigma(T))$, we have $z\bar z=\bar z z$. Applying $\Phi_T$ and using multiplicativity gives $$ TT^{*}=\Phi_T(z)\Phi_T(\bar z)=\Phi_T(z\bar z)=\Phi_T(\bar z z)=\Phi_T(\bar z)\Phi_T(z)=T^{*}T, $$ so $T$ is normal. Thus, the existence of a continuous functional calculus with these natural properties already forces normality. There is no way to push a continuous $C(\sigma(T))$-calculus onto a non-normal operator without sacrificing multiplicativity, $*$-structure, or continuity.

Conversely, if $T$ is normal, the Gelfand–Naimark machinery (or the spectral theorem) produces the unique unital $*$-homomorphism $$\Phi_T: C(\sigma(T))\to C^{*}(T)\subset\mathcal B(H)$$ with $\Phi_T(z)=T$ and $\|\Phi_T(f)\|=\|f\|_{\infty,\sigma(T)}$. In the self-adjoint case, where $\sigma(T)\subset\mathbb R$, the calculus inherits order and positivity and extends uniquely to bounded Borel functions via the projection-valued spectral measure $E_T$ by defining $f(T)=\int f\,dE_T$. This is precisely the extra structure exploited in analysis and probability. For a non-normal operator $T$, we can retain a functional calculus only by shrinking the function class, for example to holomorphic functions via the Dunford–Riesz integral, which abandons the continuous $C(\sigma(T))$ domain for a different, resolvent-based theory.

Answer 4

There exist several functional calculi. Selfadjointness is not a requirement.

For instance, the Riesz-Dunford calculus $$ f(A) = \frac{1}{2\pi i} \oint_{\partial \sigma(A)} f(\lambda) (\lambda - A)^{-1} d\lambda $$ works for arbitrary continuous operators in a complex Banach space. The integration path is any simple counterclockwise Jordan curve in $\mathbb{C}$ which surround the spectrum $\sigma(A)$. This calculus is basically an extension of Cauchy's formula from Complex Analysis, $f(z) = \frac{1}{2\pi i} \oint \frac{f(\zeta)}{\zeta - z} d\zeta$, from numbers to operators.

The calculus for selfadjoint operators allows for an extension to densely defined unbounded (discontinuous) selfadjoint operators. It has a counterpart for normal operators by decomposing a normal operator A into its real part $(A+A^*)/2$ and its imaginary part $(A-A^*)/(2i)$.

Why selfadjoint operators?

The main reason is historical. At first, there wasn't a spectral decomposition of linear operators (in fact, Stefan Banach became active after WWI, whereas Hilbert was already active in the decades before Banach!), but a decomposition of hermitian (or symmetric) quadratic forms, analogous to the principal axis transform, and mainly for what we would call today a selfadjoint compact operator (or its "inverse").

Often, the method was even just for "infinite Hermitian matrices", where one restricted first to $n$-dimensional subspaces and $n \times n$-matrices, and then took a limit $n\to \infty$ for the corresponding quadratic form. "Completely continuous" operators then came out naturally. But besides the classical eigenvalues, a "continuous spectrum" occurred in the infinite dimensional case and this was new and required new methods for its treatment.

One can extend spectral calculus to $C^*$-algebras and to some generalizations of them. Moreover, $H^\infty$-calculus extends even in a completely different direction by starting with unitary operators and then extending to contractions.

But again: Why selfadjoint operators? Because they have a real spectrum, and one usually starts mathematics with real numbers and not with complex ones. :-) The spectral calculus for selfadjoint operators also had connections to solving moment problems, in particular the Hamburger moment problem, and this was important for probability, systems analysis, statistics or mechanics.

Moreover, many (stationary) problems from physics are associated to selfadjoint forms, e.g. complex extensions of a kinetic energy form, or the Dirichlet form for the Laplacian operator in potential theory. These quadratic forms allow for a variational formulation, and this was even one of the historical origins for spectral theory, namely via Lagrange multipliers $\lambda$ by introducing elements of unit norm via a side condition. Studying the Dirichlet problem for the Laplacian was one of the main topics in 19th century analysis.

I'd recommend to you to read something about the history of functional analysis. (If you understand German or can use an online translation - e.g. google translate for pictures or .pdf -, then I'd recommend the book "Heuser, Harro: Funktionalanalysis" in its last German edition. It has an emphatic appendix about the emergence of functional analysis, but this appendix has never been translated into English, and the former editions even don't have that appendix!)

In David Hilbert's papers, you'll most often read about quadratic forms and their spectra.