Let $T: \mathcal{H}\to\mathcal{H}’$ be a linear operator. $T$ is said to be bounded if $\left\lVert T(\xi)\right\rVert\leq C\left\lVert \xi \right\rVert$ for all $\xi \in\mathcal{H}$ where $C<\infty$. The smallest of such $C$ is called the operator norm of $T$, denoted $\left\lVert T \right\rVert$. $T$ is bounded iff $T$ is continuous.

The space of all the bounded linear operator that maps $\mathcal{H}$ to itself is denoted $B(\mathcal{H})$. It is a Banach space with operator norm.

The kernel $\text{ker}(T)$ of a bounded operator $T$ is a closed subspace of $\mathcal{H}$.

Riesz’s representation lemma. Let $T$ be a bounded (complex) functional of $\mathcal{H}$. Then $T$ can be represented by a state $\eta \in\mathcal{H}$ such that for all $\xi \in\mathcal{H}$, $T(\xi)=\left\langle \eta,\xi \right\rangle$.

$T\in B(\mathcal{H})$ is an isometry if $\left\langle T\xi,T \eta \right\rangle=\left\langle \xi,\eta \right\rangle$ for all $\xi,\eta \in\mathcal{H}$.

$U\in B(\mathcal{H})$ is unitary iff $U$ is surjective and isometry.

In general topology, a set $K$ is compact if every open cover of $K$ has a finite subcover.

Sequential compactness means that every sequence has a convergent subsequence whose limit lies in the set itself.

In metric spaces (like $\mathbb{R}^n$, Hilbert spaces, Banach spaces), these two definitions are equivalent. In general topological spaces, they are not equivalent.

set $K \subset H$ is relatively compact if its closure $\overline{K}$ is compact. Compact means: every sequence in $\overline{K}$ has a convergent subsequence with a limit in $\overline{K}$. So, relatively compact means: all sequences have convergent subsequences, possibly converging to points just outside $K$, but within its closure.

A linear operator $T: \mathcal{H} \to \mathcal{H}$ is compact if it maps bounded sets to relatively compact sets. In other words, take any bounded set $B \subset \mathcal{H}$, then $T(B)$ should have the property that every sequence in it has a convergent subsequence (in $\mathcal{H}$). This is stronger than just continuity. Compact operators “shrink” infinite-dimensional complexity down so that images behave almost like finite-dimensional sets.

A Hilbert space $\mathcal{H}$ is separable if there exists a countable dense subset $D\subset \mathcal{H}$.

The spectrum of $T: \mathcal{H}\to\mathcal{H}$ is like a generalization of the eigenvalues. It is defined as

\[\sigma(T) := \left\lbrace \lambda \in \mathcal{C} \,\middle\vert\, \text{ker}(T-\lambda \mathbb{1})\neq 0 \right\rbrace .\]

If $T:\mathcal{H}\to\mathcal{H}$ is compact and self-adjoint, then $T$ at least has one eigenvalue, $\pm\left\lVert T \right\rVert$.

Spectral theorem. Let $T: \mathcal{H}\to\mathcal{H}$ be compact and self-adjoint. We have the following:

  1. The spectrum is real.
  2. aa

If $T$ is a bounded operator and $[T,T^\ast]=0$ then $T$ is said to be normal.

The multiplication operator $M_ {g}$, $M_ {g} \psi := g\psi$, is the only normal operator up to unitary equivalence. If $T: \mathcal{H}\to\mathcal{H}$ is a bounded, normal operator, then there exists a bounded function $g$ such that $T = U^{\dagger} M_ {g} U$, where $U$ maps $\mathcal{H}$ to $L^{2}$ space.

Let $T: D(T)\to\mathcal{H}$ be an unbounded operator, $D(T)$ is the domain of $T$. We say $T$ is densely defined if $D(T)$ is dense in $\mathcal{H}$. That means, for each $x\in\mathcal{H}$, there exists a sequence $x_ {i}$ in $D(T)$ such that $x_ {i}\to x$ in $H$. Equivalently, $\overline{D}(T) = \mathcal{H}$.

If $T$ is densely defined, then one can define $T^\ast$.

If the graph $\text{Gr}(T):=\left\lbrace (x,Tx) \,\middle\vert\, x\in\mathcal{H} \right\rbrace$ is a closed subspace of $\mathcal{H}\otimes \mathcal{H}$, then $T$ is called closed.

When we write $E_ n \uparrow X$, it means$E_ 1 \subseteq E_ 2 \subseteq E_ 3 \subseteq \cdots$, an increasing sequence of sets, and the union of all the $E_ n$ is the whole set $X$.

The image of $T$ is sometimes written as $\text{ran}(T)$, the range of $T$.

$T$ is symmetric is $\left\langle T\eta,\xi \right\rangle=\left\langle \eta,T\xi \right\rangle$ on the domain of $T$. $T$ is said to be self-adjoint if it is symmetric on all of $\mathcal{H}$.

The space of pure states of quantum mechanics is the projective space $\mathcal{S}:=\mathbb{P}(\mathcal{H})$. The observables are (sometimes unbounded) self-adjoint operators defined on a dense subspace of $\mathcal{H}$.

The spectral theorem for self-adjoint operators says that, every self-adjoint operator $T$ on a Hilbert space is unitarily equivalent to a multiplication operator $M_ {\alpha}$ on some $L^2$-space. Note that, the theorem does not say that TTT is literally multiplication on your original Hilbert space. It only says that after a suitable change of coordinates (via a unitary transform), it becomes multiplication.

A bounded operator $T$ is said to be trace class (or of nuclear class) if

\[\left\lvert T \right\rvert := \sum_ {n=1}^\infty s_ n(T) < \infty,\]

where $s_ n(T)$ are the singular values of $T$, i.e. the eigenvalues of the positive operator

\[\left\lvert T \right\rvert := (T^\ast T)^{1/2}\]

arranged in decreasing order and counted with multiplicity. The number $\left\lvert T \right\rvert$​ is called the trace norm.