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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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Banach closed-range theorem

Statement

Let K=R or C. Assume DC and let T:XY be bounded linear between Banach spaces. The following are equivalent: ranT is norm closed; ranT is norm closed; and there is C>0 such that dist(x,kerT)CTx for all xX. In that case ranT=(kerT),ranT=(kerT).

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From Closed range is equivalent to a quotient estimate, with its stated hypotheses: Let K=R or C. Assume DC. For a bounded linear map T:XY between Banach spaces, ranT is norm closedC>0 xX: dist(x,kerT)CTx.

[F2]

From Membership in the transpose range by an operator estimate, with its stated hypotheses: Let K=R or C. Let T:XY be bounded linear between normed spaces and fX. Then franTC0 xX: f(x)CTx. For any such C, a representing gY can be chosen with gC.

[F3]

From Surjectivity is equivalent to a lower bound for the transpose, with its stated hypotheses: Let K=R or C. Assume DC. For a bounded linear T:XY between Banach spaces, T is ontoC>0 gY: gCTg.

[F4]

From Elementary kernel and range annihilator identities, with its stated hypotheses: Let K=R or C. For a bounded linear T:XY between normed spaces, (ranT)=kerT,(ranT)=kerT,ranT=(kerT). The closure in the last identity is in Y.

[F5]

From The dual of a closed subspace is a dual quotient, with its stated hypotheses: Let K=R or C. Let X be normed and MX closed. Restriction R:XM induces a linear isometric bijection R~:X/MM,f+MfM. Also R1; its norm is 1 when M{0} and 0 when M={0}.

[F6]

From Distance to an annihilator is the restriction norm, with its stated hypotheses: Let K=R or C. If M is a closed linear subspace of a normed X and fX, then dist(f,M)=fM.

[F7]

From Annihilators and preannihilators are norm closed, with its stated hypotheses: Let K=R or C. For any normed X and arbitrary MX, NX, both MX and NX are norm-closed linear subspaces. Moreover N(N).

[F8]

From The transpose is bounded with the same norm, with its stated hypotheses: Let K=R or C. For a bounded linear T:XY between normed spaces, T:YX is bounded linear and T=T.

[F9]

From If (Y) is Banach then (\mathcal B(X,Y)) is Banach, with its stated hypotheses: Let X and Y be normed spaces over the same scalar field. If Y is Banach, then B(X,Y) is Banach for the operator norm.

[F10]

From A closed subspace of a Banach space is Banach, with its stated hypotheses: Let V be a Banach space and let WV be a closed linear subspace, equipped with the restricted norm. Then W is a Banach space.

Proof

1.1

The quotient-estimate lemma equates closedness of ranT with the stated estimate. Suppose these hold, and write N=kerT. If fN, then for every nN, f(x)=f(x+n)fx+n; infimizing gives f(x)CfTx.

F1
1.2

For the reverse implication, assume ranT is norm closed. Both duals are Banach because the scalar field is Banach, and T is bounded. The quotient-estimate lemma applied to T:YX therefore supplies C>0 with dist(g,kerT)CTg.

F1F8F9
1.3

Let Y0=ranTY; it is Banach as a closed subspace. The bounded map S:XY0, Sx=Tx, has dense range. The elementary identity and continuity give Y0=kerT.

F4F10
2.1

Domination yields franT. Conversely every Tg vanishes on N by evaluation. Thus ranT=N, which is norm closed.

F2F4F7step 1.1
2.2

For any hY0, the restriction quotient isometry supplies an extension gY with gY0=h. The distance formula gives h=dist(g,Y0). Direct evaluation gives Sh=Tg, so step 1.2 implies hCSh.

F5F6step 1.2step 1.3
3.1

The separately proved surjectivity criterion makes S onto Y0. Hence ranT=Y0 is norm closed, proving the reverse implication. The elementary primal closure identity now yields ranT=kerT; step 2.1 supplies the dual identity.

F3F4step 2.1step 2.2
4.1

If T=0, its two ranges are zero and the primal estimate is dist(x,X)=0. Both identities reduce to the same zero spaces by the elementary identities. The quotient and restriction steps allow Y0=0, so no nonzero-range assumption has entered either implication.

F4step 1.1step 3.1

Depends on

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Sources