Differential operators are usually unbounded, so their domains carry real mathematical content. Closedness and adjoints are the tools that keep such operators analytically controllable.
Operators with domains
For unbounded operators, the domain is part of the operator. Many pathologies come from changing the domain while keeping the same formula.
Definition. An unbounded operator \(T:X\supset D(T)\to Y\) is a linear map defined on a linear subspace \(D(T)\). The domain is part of the operator. Two formulas that agree pointwise but have different domains define different operators.
Definition. The graph norm is \(\|x\|_T=\|x\|_X+\|Tx\|_Y\) on \(D(T)\). The operator is closed if \(D(T)\) is complete in this graph norm, equivalently if its graph is closed in \(X\times Y\).
Closed operators and boundedness
A closed everywhere-defined operator on a Banach space is bounded. Thus genuine unbounded operators must have proper domains.
Theorem. If \(X,Y\) are Banach and \(T:X\to Y\) is a closed linear operator defined on all of \(X\), then \(T\) is bounded.
Proof. This is exactly the closed graph theorem: the domain is all of \(X\), the graph is closed by assumption, and both spaces are Banach. Therefore \(T\in\mathcal L(X,Y)\). Hence any genuinely unbounded closed operator must have a proper domain.
Adjoints of densely defined operators
Dense domain is needed so that the adjoint is well-defined. Without density, the functional defining the adjoint may not determine a unique vector.
Definition. Let \(T:D(T)\subset H\to K\) be densely defined. A vector \(y\in K\) belongs to \(D(T^*)\) if the functional \(x\mapsto\langle Tx,y\rangle_K\) is bounded on \(D(T)\) with respect to the \(H\)-norm. Then \(T^*y\in H\) is the unique vector satisfying \(\langle Tx,y\rangle=\langle x,T^*y\rangle\) for all \(x\in D(T)\).
Proposition. The adjoint \(T^*\) of a densely defined operator is closed.
Proof. Suppose \(y_n\in D(T^*)\), \(y_n\to y\) in \(K\), and \(T^*y_n\to z\) in \(H\). For every \(x\in D(T)\), \(\langle Tx,y\rangle=\lim_n\langle Tx,y_n\rangle=\lim_n\langle x,T^*y_n\rangle=\langle x,z\rangle\). Thus the functional \(x\mapsto\langle Tx,y\rangle\) is represented by \(z\), so \(y\in D(T^*)\) and \(T^*y=z\).
Symmetric and self-adjoint
Symmetric is a formal identity on a domain; self-adjoint is an equality of operators and domains. The distinction is central for differential operators.
Definition. A densely defined operator \(T:D(T)\subset H\to H\) is symmetric if \(T\subset T^*\), meaning \(D(T)\subset D(T^*)\) and \(T^*x=Tx\) for \(x\in D(T)\). It is self-adjoint if \(T=T^*\), including equality of domains.
Proposition. If \(T\) is self-adjoint, then \(\langle Tx,x\rangle\in\mathbb R\) for \(x\in D(T)\), and \(T\pm iI\) are injective with closed range.
Proof. Reality follows from \(\langle Tx,x\rangle=\langle x,Tx\rangle=\overline{\langle Tx,x\rangle}\). For \(x\in D(T)\), \(\|(T\pm iI)x\|^2=\|Tx\|^2+\|x\|^2\), because the cross term is \(\pm i\langle x,Tx\rangle\mp i\langle Tx,x\rangle=0\). Hence \(\|(T\pm iI)x\|\ge\|x\|\), which gives injectivity and closed range.
Multiplication operators
Multiplication operators show unboundedness in its simplest form. The formula is harmless, but the domain selects exactly where it is square-integrable.
Example. On \(L^2(\Omega,\mu)\), let \(m:\Omega\to\mathbb R\) be measurable and define \(Tf=mf\) on \(D(T)=\{f\in L^2:mf\in L^2\}\). Then \(T\) is self-adjoint. Closedness follows because if \(f_n\to f\) and \(mf_n\to g\) in \(L^2\), a subsequence converges a.e.; hence \(g=mf\) a.e. and \(f\in D(T)\). Symmetry is immediate from real-valued \(m\), and the adjoint domain calculation gives equality.