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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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Closability is equivalent to density of the adjoint domain

Statement

Assume Countable Choice. Let T be a densely defined linear operator on H. Then T is closable if and only if D(T) is dense in H. In that case T=T and T=T; moreover if TT, then T is closable.

Facts & Assumptions

[A1]

For densely defined T the adjoint is defined by Tx,y=x,Ty for xD(T), yD(T), and Γ(T)={(y,w):Tx,y=x,w for all xD(T)}. Putting W(x,y):=(y,x) defines a bijective isometry of HH with W2=I, and W maps orthocomplements to orthocomplements: W(M)=(WM) (Adjoint of a densely defined operator, Hilbert space, Orthogonality and the orthogonal complement).

[A2]

If M is a linear subspace of the Hilbert space HH, then M=M (The double orthogonal complement of a subspace is its closure).

[A3]

For a densely defined T the operator T is closed. If in addition D(T) is dense, then T is defined, is closed, and contains T (The adjoint is well defined, closed, and reverses inclusions, Adjoint of a densely defined operator).

[A4]

T is closable when it has a closed extension; if T is closable then T is the least closed extension of T and TT (Closure of a closable operator, Densely defined, closed and closable operators, and cores).

Proof

technique · direct

Given: A densely defined linear operator T on H.

1.1

By [A1] we have Γ(T)=WΓ(T): indeed (y,w)Γ(T) means Tx,y=x,w for all xD(T), which is exactly (x,Tx),W(y,w)=0 for all xD(T), that is W(y,w)Γ(T).

A1
2.1

Assume in addition that D(T) is dense, so that T is defined. Replacing T by T in step 1.1 gives Γ(T)=WΓ(T), so by step 1.1 and unitarity of W, with W(M)=(WM) and W2=I, one has Γ(T)=W(WΓ(T))=WWΓ(T)=Γ(T)=Γ(T).

step 1.1A1A2
2.2

Conversely assume T is closable, and let T be its closure, a closed densely defined operator with Γ(T)Γ(T) and Γ(T)=Γ(T) by [A4]. Then D(T) is dense: if yD(T) then (y,0)Γ(T), so by step 1.1 applied to T and closedness of T we get (y,0)Γ(T)=WΓ(T)=WΓ(T), that is (y,0)=W(x,Tx)=(Tx,x) for some xD(T), forcing x=0 and y=0.

A1A2A4step 1.1
3.1

Under the hypothesis of step 2.1 the set Γ(T) is the graph of the operator T, which is closed by [A3]; hence T has a closed extension and is closable, and its closure is T by minimality in [A4].

step 2.1A3A4
3.2

With T closable as in step 2.2, apply step 1.1 to T and to T: using Γ(T)=Γ(T) and [A2] one gets Γ(T)=WΓ(T)=WΓ(T)=WΓ(T)=Γ(T), so T=T and D(T)=D(T) is dense by step 2.2.

A1A2step 1.1step 2.2
4.1

The two implications are steps 3.1 (density of D(T) gives closability, with closure T) and 3.2 (closability gives density of D(T), and T=T). If TT, then T is a closed extension of T by [A3], so T is closable.

step 3.1step 3.2

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