Why do we only consider the spectral theorem for self-adjoint operators

Question

In our functional analysis lecture and also in books I've read about this theme (or on Wikipedia for example), the spectral theorem is only stated for (un)bounded self-adjoint operators. Sometimes this is extended to normal operators but never further.

Why is this the case? What pathological properties do non-self-adjoint / non-normal operators have that make defining the functional calculus or proving the spectral theorem impossible?

Answer 1

As already mentioned in the comments, if you want to define a functional calculus for non-normal with at least some desirable properties, you soon run into the problem that the complex functions $z\bar z$ and $\bar z z$ are identical, but $TT^\ast$ and $T^\ast T$ are not. There are some ways to deal with that problem though.

1. Holomorphic functional calculus: If $f$ is a holomorphic function on a neighborhood of $\sigma(T)$, then one can define $f(T)$ using a variant of the Cauchy integral formula, i.e. $$ f(T)=\frac 1{2\pi i}\int_\Gamma f(z)(z-T)^{-1}\,dz, $$ where $\Gamma$ is a contour enclosing $\sigma(T)$. This works even for bounded operators on Banach spaces and some classes of unbounded operators such as sectorial ones. The problem mentioned above is avoided since holomorphic functions don't depend on $\bar z$ (in the sense that $\partial_{\bar z}f=0$).
2. Dilations: For every contraction $T$ on a Hilbert space $H$ there exists a Hilbert space $K$, an isometry $V\colon H\to K$ and a unitary $U$ on $K$ such that $T^n=V^\ast U^n V$ for all $n\in\mathbb{N}$. Then one can define $f(T)=V^\ast f(U) V$ for let's say $f\colon \mathbb{D}\to \mathbb{C}$ continuous. This definition is consistent with the naive definition for polynomials in $z$ and $\bar f(T)=f(T)^\ast$, but in general, $f(T)g(T)=(fg)(T)$ does not hold.
3. Double operator integrals: Given a pair of noncommuting bounded self-adjoint operators $S$ and $T$ (which is equivalent to a non-normal operator by taking the real and imaginary part), you can always make them commute in the following way: Let $S_2(H)$ be the space of all Hilbert-Schmidt operators. If we let $S$ act on $S_2(H)$ be left multiplication and $T$ by right multiplication, then the resulting operators $L(S)$ and $R(T)$ (on $S_2(H)$!) commute. Hence we can define $f(L(S),R(T))A$ for any continuous $f$ on $\sigma(S)\times \sigma(T)$ and $A\in S_2(H)$. For some functions $f$ the operator $f(S,T)$ satisfies $\|f(S,T)A\|\leq C\|A\|$ for the operator norm of $A$. In this case, $f(L(S),R(T))$ can be extended to a bounded linear operator from $B(H)$ to $B(H)$, and we can take $A=1$. Then $f(S,T)=f(L(S),R(T))I$ is a bounded operator on $H$ (no longer $S_2(H)$). In the case of a polynomial, this corresponds to shifting all occurences of $S$ to the left and all occurences of $T$ to the right (e.g. if $p(x,y)=x^3y$, then $p(S,T)=S^3 T$).

There are many more functional calculi (see for example these lecture notes with a focus on functional calculus for generators of operator semigroups), so there is really no reason to stop at self-adjoint or normal operators, although self-adjointness or normality can make life considerably easier at times.