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Proof of Fibre Suprema along a Coordinate Map and Their Semicontinuous Envelopes

lemmalem:fibre-supremum-envelope-hilbert-2026a
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· 10,392 chars · 17 deps · depth 23 Reason: Proof of the fibre-suprema lemma: the fibres are nonempty because the orthonormal tuple lies in the subspace, taking suprema and infima fibrewise preserves the penalised inequality, and the envelopes inherit it through the approximation property, which also forces equality at the base point.

The fibres are nonempty because the tuple lies in the subspace; taking suprema and infima fibrewise preserves the penalised inequality, and the envelopes inherit it along approximating sequences, which also forces equality at the base point.

Proof

Each result cited is universally quantified over the data in its own statement, and is applied here to the data named in the statement of the lemma. The properties of the coordinate map Λ\Lambda and of the associated map Λ\Lambda^{\sharp} recorded in Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions §coordinates are used throughout, as is Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions §stability. Put

M1=u^(xˉ)v^(yˉ)α2ζˉωˉ2,M_{1}=\hat{u}(\bar{x})-\hat{v}(\bar{y})-\tfrac{\alpha}{2}\lVert\bar{\zeta}-\bar{\omega}\rVert^{2},

so that the standing hypothesis reads

u^(x)v^(y)α2ΛxΛy2M1for all x,yA.\hat{u}(x)-\hat{v}(y)-\tfrac{\alpha}{2}\lVert\Lambda x-\Lambda y\rVert^{2}\le M_{1}\qquad\text{for all }x,y\in A .

Fix CuRC_{u}\in\mathbb{R} an upper bound for the set of values of u^\hat{u} and cvRc_{v}\in\mathbb{R} a lower bound for the set of values of v^\hat{v}.

Claim 1 (clause 1). Let ζRm\zeta\in\mathbb{R}^{m}. Every component eie_{i} of ee lies in the linear subspace AA, so the span of ee is contained in AA by claim 3 of The Span of a Finite Family is the Smallest Subspace Containing It; in particular ΛζA\Lambda^{\sharp}\zeta\in A, and ΛΛζ=ζ\Lambda\Lambda^{\sharp}\zeta=\zeta by Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions §coordinates. Hence ΛζAζ\Lambda^{\sharp}\zeta\in A_{\zeta} and AζA_{\zeta} is nonempty.

The set {u^(x):xAζ}\{\hat{u}(x):x\in A_{\zeta}\} is therefore nonempty and is bounded above by CuC_{u}, so it has a least upper bound by Least Upper Bound Property of the Real Numbers, unique by Uniqueness of the Supremum and of the Infimum; and {v^(y):yAζ}\{\hat{v}(y):y\in A_{\zeta}\} is nonempty and bounded below by cvc_{v}, so it has a greatest lower bound by Existence of the Infimum of a Nonempty Subset of R\mathbb{R} Bounded Below, again unique. Thus UU and V\mathcal{V} are defined. Since CuC_{u} is an upper bound for each of the sets whose suprema define UU, we get U(ζ)CuU(\zeta)\le C_{u} for every ζ\zeta; likewise cvV(ζ)c_{v}\le\mathcal{V}(\zeta) for every ζ\zeta. Consequently, for every ζRm\zeta\in\mathbb{R}^{m} the number CuC_{u} belongs to the set AU(ζ)A_{U}(\zeta) of Upper and Lower Semicontinuous Envelopes of a Real-Valued Function (take r=1r=1) and cvc_{v} belongs to BV(ζ)B_{\mathcal{V}}(\zeta), so UU is bounded above near each point of Rm\mathbb{R}^{m} and V\mathcal{V} is bounded below near each point. This proves Claim 1.

Claim 2 (the fibrewise inequality). For all ζ,ωRm\zeta,\omega\in\mathbb{R}^{m},

U(ζ)V(ω)α2ζω2M1.U(\zeta)-\mathcal{V}(\omega)-\tfrac{\alpha}{2}\lVert\zeta-\omega\rVert^{2}\le M_{1}.

Proof. Fix ζ,ω\zeta,\omega and write π=α2ζω2\pi=\tfrac{\alpha}{2}\lVert\zeta-\omega\rVert^{2}. Let yAωy\in A_{\omega}. For every xAζx\in A_{\zeta} we have Λx=ζ\Lambda x=\zeta and Λy=ω\Lambda y=\omega, so the standing hypothesis gives u^(x)v^(y)πM1\hat{u}(x)-\hat{v}(y)-\pi\le M_{1}, that is u^(x)M1+v^(y)+π\hat{u}(x)\le M_{1}+\hat{v}(y)+\pi by claim 3 of Elementary Arithmetic in an Ordered Field. Thus M1+v^(y)+πM_{1}+\hat{v}(y)+\pi is an upper bound for {u^(x):xAζ}\{\hat{u}(x):x\in A_{\zeta}\}, whence U(ζ)M1+v^(y)+πU(\zeta)\le M_{1}+\hat{v}(y)+\pi by Upper Bound and Least Upper Bound, that is U(ζ)M1πv^(y)U(\zeta)-M_{1}-\pi\le\hat{v}(y). As yAωy\in A_{\omega} was arbitrary, U(ζ)M1πU(\zeta)-M_{1}-\pi is a lower bound for {v^(y):yAω}\{\hat{v}(y):y\in A_{\omega}\}, so U(ζ)M1πV(ω)U(\zeta)-M_{1}-\pi\le\mathcal{V}(\omega) by Lower Bound and Greatest Lower Bound in a Totally Ordered Set, which rearranges to the claim. This proves Claim 2.

Claim 3 (the envelope inequality). UU^{*} is upper semicontinuous on Rm\mathbb{R}^{m}, V\mathcal{V}_{*} is lower semicontinuous on Rm\mathbb{R}^{m}, and for all ζ,ωRm\zeta,\omega\in\mathbb{R}^{m},

U(ζ)V(ω)α2ζω2M1.U^{*}(\zeta)-\mathcal{V}_{*}(\omega)-\tfrac{\alpha}{2}\lVert\zeta-\omega\rVert^{2}\le M_{1}.

Proof. The semicontinuity assertions are claim 2 of Properties of the Upper Semicontinuous Envelope and claim 3 of Properties of the Lower Semicontinuous Envelope, by Duality, whose hypotheses are supplied by Claim 1.

Fix ζ,ωRm\zeta,\omega\in\mathbb{R}^{m} and write D=ζωD=\lVert\zeta-\omega\rVert. Let εR\varepsilon\in\mathbb{R} be positive. Choose a positive δ1\delta\le1 with

2δ+2αδ(D+1)ε,2\delta+2|\alpha|\,\delta\,(D+1)\le\varepsilon ,

which is possible: the number 2+2α(D+1)2+2|\alpha|(D+1) is positive, so the quotient ε2+2α(D+1)\tfrac{\varepsilon}{2+2|\alpha|(D+1)} exists and is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field, and taking for δ\delta the lesser of it and 11 gives 2δ+2αδ(D+1)=δ(2+2α(D+1))ε2\delta+2|\alpha|\delta(D+1)=\delta\bigl(2+2|\alpha|(D+1)\bigr)\le\varepsilon.

By claim 5 of Properties of the Upper Semicontinuous Envelope, applied to UU at ζ\zeta with δ\delta, there is ζRm\zeta'\in\mathbb{R}^{m} with dE(ζ,ζ)δd_{E}(\zeta',\zeta)\le\delta and U(ζ)U(ζ)<δ|U(\zeta')-U^{*}(\zeta)|<\delta. By claim 1 of Properties of the Lower Semicontinuous Envelope, by Duality the function V-\mathcal{V}, whose value at ξ\xi is the additive inverse of V(ξ)\mathcal{V}(\xi), is bounded above near each point of Rm\mathbb{R}^{m} and satisfies (V)=V(-\mathcal{V})^{*}=-\mathcal{V}_{*}; so claim 5 of Properties of the Upper Semicontinuous Envelope, applied to V-\mathcal{V} at ω\omega with δ\delta, gives ωRm\omega'\in\mathbb{R}^{m} with dE(ω,ω)δd_{E}(\omega',\omega)\le\delta and V(ω)V(ω)<δ|\mathcal{V}(\omega')-\mathcal{V}_{*}(\omega)|<\delta. Since dE(ξ,ξ)=ξξd_{E}(\xi,\xi')=\lVert\xi-\xi'\rVert by Euclidean Distance is a Metric on Rn\mathbb{R}^n, we have ζζδ\lVert\zeta-\zeta'\rVert\le\delta and ωωδ\lVert\omega-\omega'\rVert\le\delta.

By Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions §stability, ζωD+2δD+2\lVert\zeta'-\omega'\rVert\le D+2\delta\le D+2 and

α2ζω2α2D2α2(D+(D+2δ))(2δ)α2(2D+2)(2δ)=2αδ(D+1),\Bigl|\tfrac{\alpha}{2}\lVert\zeta'-\omega'\rVert^{2}-\tfrac{\alpha}{2}D^{2}\Bigr| \le\tfrac{|\alpha|}{2}\bigl(D+(D+2\delta)\bigr)\,(2\delta) \le\tfrac{|\alpha|}{2}\,(2D+2)\,(2\delta)=2|\alpha|\,\delta\,(D+1),

using δ1\delta\le1, claim 4 of Properties of the Absolute Value in an Ordered Field and claim 5 of Elementary Arithmetic in an Ordered Field. Combining, and using claim 6 of Properties of the Absolute Value in an Ordered Field on the three absolute values,

U(ζ)V(ω)α2D2<(U(ζ)+δ)(V(ω)δ)α2ζω2+2αδ(D+1)M1+ε,U^{*}(\zeta)-\mathcal{V}_{*}(\omega)-\tfrac{\alpha}{2}D^{2} <\bigl(U(\zeta')+\delta\bigr)-\bigl(\mathcal{V}(\omega')-\delta\bigr)-\tfrac{\alpha}{2}\lVert\zeta'-\omega'\rVert^{2}+2|\alpha|\delta(D+1) \le M_{1}+\varepsilon,

the last step by Claim 2 applied at (ζ,ω)(\zeta',\omega') and by the choice of δ\delta. As ε\varepsilon was an arbitrary positive real number, Comparison of Real Numbers with Arbitrary Positive Slack gives U(ζ)V(ω)α2D2M1U^{*}(\zeta)-\mathcal{V}_{*}(\omega)-\tfrac{\alpha}{2}D^{2}\le M_{1}. This proves Claim 3.

Claim 4 (clause 3). Since xˉAζˉ\bar{x}\in A_{\bar{\zeta}} and yˉAωˉ\bar{y}\in A_{\bar{\omega}}, we have u^(xˉ)U(ζˉ)\hat{u}(\bar{x})\le U(\bar{\zeta}) and V(ωˉ)v^(yˉ)\mathcal{V}(\bar{\omega})\le\hat{v}(\bar{y}) by Upper Bound and Least Upper Bound and Lower Bound and Greatest Lower Bound in a Totally Ordered Set; and U(ζˉ)U(ζˉ)U(\bar{\zeta})\le U^{*}(\bar{\zeta}) and V(ωˉ)V(ωˉ)\mathcal{V}_{*}(\bar{\omega})\le\mathcal{V}(\bar{\omega}) by claim 1 of Properties of the Upper Semicontinuous Envelope and claim 2 of Properties of the Lower Semicontinuous Envelope, by Duality. Writing π=α2ζˉωˉ2\pi=\tfrac{\alpha}{2}\lVert\bar{\zeta}-\bar{\omega}\rVert^{2}, these inequalities give

M1=u^(xˉ)v^(yˉ)π  U(ζˉ)V(ωˉ)π  U(ζˉ)V(ωˉ)π  M1,M_{1}=\hat{u}(\bar{x})-\hat{v}(\bar{y})-\pi\ \le\ U(\bar{\zeta})-\mathcal{V}(\bar{\omega})-\pi\ \le\ U^{*}(\bar{\zeta})-\mathcal{V}_{*}(\bar{\omega})-\pi\ \le\ M_{1},

the last inequality by Claim 3. Hence all four quantities are equal. Put a=U(ζˉ)u^(xˉ)a=U^{*}(\bar{\zeta})-\hat{u}(\bar{x}) and b=v^(yˉ)V(ωˉ)b=\hat{v}(\bar{y})-\mathcal{V}_{*}(\bar{\omega}); both are nonnegative by the inequalities above and claim 3 of Elementary Arithmetic in an Ordered Field, and a+b=0a+b=0 because the first and third quantities in the display are equal. Then aa+b=0a\le a+b=0 and ba+b=0b\le a+b=0 by the compatibility of the order with addition (an axiom of Ordered Field), so a=0a=0 and b=0b=0, that is U(ζˉ)=u^(xˉ)U^{*}(\bar{\zeta})=\hat{u}(\bar{x}) and V(ωˉ)=v^(yˉ)\mathcal{V}_{*}(\bar{\omega})=\hat{v}(\bar{y}). The chain of inequalities u^(xˉ)U(ζˉ)U(ζˉ)=u^(xˉ)\hat{u}(\bar{x})\le U(\bar{\zeta})\le U^{*}(\bar{\zeta})=\hat{u}(\bar{x}) then gives U(ζˉ)=u^(xˉ)U(\bar{\zeta})=\hat{u}(\bar{x}), and likewise V(ωˉ)=v^(yˉ)\mathcal{V}(\bar{\omega})=\hat{v}(\bar{y}). This proves Claim 4, which is clause 3.

Clause 2. By Claim 4 the right-hand side of the display in clause 2 equals M1M_{1}, so the display is exactly the inequality of Claim 3. Together with the semicontinuity assertions of Claim 3 this proves clause 2.

Claim 5 (clause 4). Let ζ1Rm\zeta_{1}\in\mathbb{R}^{m} and let r,εRr,\varepsilon\in\mathbb{R} be positive; let σ\sigma be the lesser of rr and ε2\tfrac{\varepsilon}{2}, positive because it is one of them (claim 9 of Elementary Order Arithmetic in an Ordered Field and claim 8 of the same lemma for ε2\tfrac{\varepsilon}{2}).

By claim 5 of Properties of the Upper Semicontinuous Envelope applied to UU at ζ1\zeta_{1} with σ\sigma, there is ζRm\zeta'\in\mathbb{R}^{m} with dE(ζ,ζ1)σd_{E}(\zeta',\zeta_{1})\le\sigma and U(ζ)U(ζ1)<σ|U(\zeta')-U^{*}(\zeta_{1})|<\sigma; hence U(ζ1)ε2U(ζ1)σ<U(ζ)U^{*}(\zeta_{1})-\tfrac{\varepsilon}{2}\le U^{*}(\zeta_{1})-\sigma<U(\zeta') by claim 6 of Properties of the Absolute Value in an Ordered Field. Since U(ζ)U(\zeta') is the least upper bound of {u^(x):xAζ}\{\hat{u}(x):x\in A_{\zeta'}\} and U(ζ)ε2<U(ζ)U(\zeta')-\tfrac{\varepsilon}{2}<U(\zeta'), the number U(ζ)ε2U(\zeta')-\tfrac{\varepsilon}{2} is not an upper bound of that set, so there is xAζx\in A_{\zeta'} with U(ζ)ε2<u^(x)U(\zeta')-\tfrac{\varepsilon}{2}<\hat{u}(x). Then xAx\in A, Λxζ1=ζζ1=dE(ζ,ζ1)σr\lVert\Lambda x-\zeta_{1}\rVert=\lVert\zeta'-\zeta_{1}\rVert=d_{E}(\zeta',\zeta_{1})\le\sigma\le r, and

U(ζ1)ε=(U(ζ1)ε2)ε2<U(ζ)ε2<u^(x).U^{*}(\zeta_{1})-\varepsilon=\Bigl(U^{*}(\zeta_{1})-\tfrac{\varepsilon}{2}\Bigr)-\tfrac{\varepsilon}{2}<U(\zeta')-\tfrac{\varepsilon}{2}<\hat{u}(x).

For yy, apply the same argument to V-\mathcal{V}, using (V)=V(-\mathcal{V})^{*}=-\mathcal{V}_{*} from claim 1 of Properties of the Lower Semicontinuous Envelope, by Duality: there is ωRm\omega'\in\mathbb{R}^{m} with dE(ω,ζ1)σd_{E}(\omega',\zeta_{1})\le\sigma and V(ζ1)σ<V(ω)-\mathcal{V}_{*}(\zeta_{1})-\sigma<-\mathcal{V}(\omega'), and then yAωy\in A_{\omega'} with V(ω)ε2<v^(y)-\mathcal{V}(\omega')-\tfrac{\varepsilon}{2}<-\hat{v}(y), because V(ω)\mathcal{V}(\omega') is the greatest lower bound of {v^(y):yAω}\{\hat{v}(y):y\in A_{\omega'}\} and V(ω)+ε2\mathcal{V}(\omega')+\tfrac{\varepsilon}{2} is therefore not a lower bound of it. Combining as above and negating (claim 4 of Elementary Order Arithmetic in an Ordered Field) gives Λyζ1r\lVert\Lambda y-\zeta_{1}\rVert\le r and v^(y)<V(ζ1)+ε\hat{v}(y)<\mathcal{V}_{*}(\zeta_{1})+\varepsilon. This proves Claim 5, which is clause 4, and completes the proof of the lemma.

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