Unbounded operators and their domains
Treat the domain as part of the operator and distinguish symmetric from self-adjoint realizations.
On this page
The question for today
What you’ll understand
- Check membership in a concrete unbounded operator domain.
- Define the adjoint of a densely defined operator.
- Explain why symmetry of a differential expression does not establish self-adjointness.
Differentiation and multiplication by an unbounded coordinate are indispensable operations, yet neither acts on every square-integrable function with a square-integrable result. Their domains specify which inputs are allowed. Two operators can use the same formula and still have different spectra or dynamics because their domains differ. This is why an integration-by-parts calculation is only the beginning of a self-adjointness argument: one must compare the full domains of the operator and its adjoint.
A formula together with an allowed input space
An operator A on H is a linear map from a linear subspace D(A) into H. It is densely defined when that subspace is dense in the Hilbert norm. On ℓ2, define (Ax)_n=n x_n with domain consisting of sequences for which the sum of n²|x_n|² is finite. Finite-support sequences lie in this domain, making it dense. The rule is unbounded because the unit input e_n has output norm n. Each permitted input nevertheless has a well-defined output in H.
The graph consists of pairs (x,Ax). A is closed when convergence of x_j to x and Ax_j to y forces x into D(A) with Ax=y. Closedness is a compatibility between input and output limits, not a norm bound. The diagonal example is closed and unbounded on a proper domain. Equipping D(A) with the graph norm makes it complete when A is closed. An everywhere-defined closed linear operator between Banach spaces would instead be bounded by the closed graph theorem.
The adjoint has a domain of its own
For densely defined A, a vector y belongs to D(A*) when the functional x↦inner-product(y,Ax), initially defined on D(A), is bounded with respect to the H norm of x. Riesz representation then determines A*y uniquely. Density ensures uniqueness of the representing vector. The adjoint's domain can be larger than D(A), and need not be all of H. This is the crucial difference from the adjoint of a bounded everywhere-defined operator.
A densely defined operator is symmetric if inner-product(y,Ax)=inner-product(Ay,x) for all x and y in D(A). Equivalently A is contained in A*: it has the same action on its domain, but the adjoint may admit more inputs. Self-adjoint means equality of both actions and domains. A symmetric operator has real expectation values on its domain, yet that alone does not grant the spectral theorem or a unique unitary evolution. Self-adjoint extension theory addresses which additional domains can supply those properties.
Boundary conditions change the adjoint calculation
On L2(0,1), take A=−i d/dx on the Sobolev domain H1_0(0,1), whose endpoint traces vanish. Integration by parts makes A symmetric because the boundary term vanishes. Its adjoint acts by the same differential expression on H1(0,1), with no endpoint condition. The domains differ, so this symmetric A is not self-adjoint. This example uses Sobolev weak derivatives and their endpoint traces, not an assumption that all L2 functions have ordinary endpoint values.
A different realization, with H1 functions satisfying f(1)=exp(iθ)f(0) for a fixed real θ, is self-adjoint. Matching the boundary term against every allowed test function imposes the same condition on the adjoint domain. Its allowed wave numbers are θ+2πk for integer k. The expression −i d/dx did not change, but the domain did. A boundary prescription therefore changes the mathematical observable, rather than merely changing how an existing operator is printed.
Work it through
An input can belong to the space but not the domain
For (Ax)_n=n x_n on ℓ2, compare x_n=1/n with y_n=1/n². Which sequence lies in D(A)?
Both sequences belong to ℓ2 because the sums of 1/n² and 1/n⁴ converge. Membership in H therefore does not distinguish them.
For x, the weighted sum is Σ n²/n²=Σ1, which diverges. Its formal image is the constant sequence one, so x is not in D(A).
For y, the weighted sum is Σ1/n², which converges. Thus y is in D(A), and Ay has coordinates 1/n. The latter image belongs to H but does not itself lie in D(A), so y is not in D(A²).
The domain of A is a genuine restriction, and composing A with itself introduces another restriction.
The living notebook
Explore the model
Interactive lab · spectral
Higher modes expose derivative sensitivity
Pause playback, raise N, and compare p=1 with a larger coefficient decay. The finite sine sum stays smooth for every finite N, but its high-frequency terms suggest why differentiation magnifies tails. The lab does not differentiate the sum, test a Sobolev domain, or decide self-adjointness.
- Sum at x = π/2
- 0.867
- Smallest wavelength
- 2π / 5
- Highest-mode coefficient
- 0.2
A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forEvery finite truncation may be admissible while its infinite limit fails the domain condition.
Pause & reason
Check your understanding
Your turn
Try it yourself
For the periodic derivative realization θ=0 on (0,1), check that f(x)=exp(2πix) is an eigenfunction and compute its eigenvalue. Would exp(iπx) satisfy the same boundary condition?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Check the endpoint traces before differentiating.
Compare with the solution
The first function has f(1)=f(0)=1 and lies in H1. Applying −i d/dx gives 2πf, so its eigenvalue is 2π. The second has values −1 and 1 at the endpoints, violating periodicity. It belongs to a different phase-twisted realization, so differentiating its formula alone cannot make it an eigenfunction of the periodic operator.
Keep the idea
Take it with you
- An unbounded operator consists of a formula and a specified domain.
- Adjoints of densely defined operators carry their own domain conditions.
- Symmetry does not imply self-adjointness; matching adjoint domains is essential.
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