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Proof of Jensen's Lemma: the Contact Set of a Semiconvex Function at a Strict Maximum has Positive Measure

lemmalem:jensen-maximum-2026a
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· 6,272 chars · 10 deps · depth 16 Reason: Proof of Jensen's lemma by the contact-set argument: the map sending a maximiser to its perturbing vector is single-valued, Lipschitz with constant twice the semiconvexity constant, and onto the closed ball of perturbations, so the Lipschitz image bound forces the contact set to have positive measure.

For small perturbation bounds the contact set is compact and lies in the half ball, so the map sending a maximiser to its perturbing vector is single-valued and Lipschitz with constant twice the semiconvexity constant, and it maps the contact set onto the closed ball of perturbations; the Lipschitz image bound for Lebesgue measure then forces the contact set to have measure at least a fixed multiple of that of a ball.

Proof

Step 0: the contact-set lemma applies. The ball Bˉ\bar{B} is contained in UU, and by the continuity estimate for semiconvex functions, applied with the subset S=BˉS=\bar{B} of the open convex set UU, the restriction of φ\varphi to Bˉ\bar{B} satisfies hypothesis (H1) of Maximisers of Linearly Perturbed Continuous Functions on a Closed Ball: Existence, Localisation, and Compactness of the Contact Set; hypothesis (H2) there is exactly the strict maximum assumption made here. Consequently Maximisers of Linearly Perturbed Continuous Functions on a Closed Ball: Existence, Localisation, and Compactness of the Contact Set applies to x^\hat{x}, rr and the restriction of φ\varphi to Bˉ\bar{B}, and the sets M(p)M(p) and KδK_{\delta} of that lemma are the ones written here.

Step 1: the choice of δ0\delta_{0}. Apply the localisation of maximisers with ρ=r/2\rho=r/2 to obtain a positive real δ\delta_{\ast} such that KδBdE(x^,r/2)K_{\delta}\subseteq B_{d_{E}}(\hat{x},r/2) whenever 0<δδ0<\delta\le\delta_{\ast}. Let δ0\delta_{0} be the smaller of δ\delta_{\ast} and λr/4\lambda r/4; since 0<λ0<\lambda and 0<r0<r, this is a positive real number. Fix from now on a real δ\delta with 0<δδ00<\delta\le\delta_{0} where claims 1 to 3 are concerned.

Proof of claim 1. By the existence of maximisers the set KδK_{\delta} is nonempty, and by the compactness of the contact set it is compact in Rn\mathbb{R}^{n}; hence KδB(Rn)K_{\delta}\in\mathcal{B}(\mathbb{R}^{n}) by claim 3 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure. Since δδ0δ\delta\le\delta_{0}\le\delta_{\ast}, step 1 gives KδBdE(x^,r/2)K_{\delta}\subseteq B_{d_{E}}(\hat{x},r/2), and BdE(x^,r/2)BˉdE(x^,r/2)B_{d_{E}}(\hat{x},r/2)\subseteq\bar{B}_{d_{E}}(\hat{x},r/2) by claim 1 of Elementary Properties of the Closed Ball in a Metric Space.

Proof of claim 2. For xKδx\in K_{\delta} put

P(x)={pRn:pδ  and  φ(y)+pyφ(x)+px  for every yBˉ},P(x)=\bigl\{p\in\mathbb{R}^{n}:\lVert p\rVert\le\delta\ \text{ and }\ \varphi(y)+p\cdot y\le\varphi(x)+p\cdot x\ \text{ for every }y\in\bar{B}\bigr\},

which is nonempty by the definition of KδK_{\delta}.

(a) P(x)P(x) has exactly one element. Let p,pP(x)p,p'\in P(x). By the triangle inequality and claim 5 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n,

ppp+p2δ2δ0λr2.\lVert p-p'\rVert\le\lVert p\rVert+\lVert p'\rVert\le 2\delta\le 2\delta_{0}\le\frac{\lambda r}{2}.

By claim 1 above, xBˉdE(x^,r/2)x\in\bar{B}_{d_{E}}(\hat{x},r/2). Applying the Lipschitz estimate between maximisers with the convex set UU, the constant λ\lambda, the ball Bˉ\bar{B}, the vectors p,pp,p' and the points x=xx'=x, we obtain p=pp=p'. Write T(x)T(x) for the unique element of P(x)P(x); this defines a map T:KδRnT:K_{\delta}\to\mathbb{R}^{n}.

(b) TT is Lipschitz with constant 2λ2\lambda. Let x,xKδx,x'\in K_{\delta} and put p=T(x)p=T(x), p=T(x)p'=T(x'). As in (a), pp2δλr/2\lVert p-p'\rVert\le 2\delta\le\lambda r/2, and by claim 1 both xx and xx' lie in BˉdE(x^,r/2)\bar{B}_{d_{E}}(\hat{x},r/2); each of x,xx,x' maximises the corresponding perturbation over Bˉ\bar{B}. The Lipschitz estimate between maximisers therefore gives

T(x)T(x)2λxx,\lVert T(x)-T(x')\rVert\le 2\lambda\,\lVert x-x'\rVert ,

so TT is Lipschitz with constant 2λ2\lambda for the Euclidean distances.

(c) T(Kδ)=BˉdE(0Rn,δ)T(K_{\delta})=\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta). Every xKδx\in K_{\delta} satisfies T(x)δ\lVert T(x)\rVert\le\delta, that is, dE(0Rn,T(x))δd_{E}(0_{\mathbb{R}^{n}},T(x))\le\delta, so T(Kδ)BˉdE(0Rn,δ)T(K_{\delta})\subseteq\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta). Conversely let pRnp\in\mathbb{R}^{n} with pδ\lVert p\rVert\le\delta. By the existence of maximisers there is xM(p)x\in M(p); then xKδx\in K_{\delta} and pP(x)p\in P(x), so p=T(x)p=T(x). Hence BˉdE(0Rn,δ)T(Kδ)\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta)\subseteq T(K_{\delta}).

(d) The measure estimate. The set KδK_{\delta} is nonempty and compact by claim 1, and TT is Lipschitz with constant 2λ2\lambda by (b). Applying The Lebesgue Measure of a Lipschitz Image of a Compact Subset of Rn\mathbb{R}^n and using (c),

λn(BˉdE(0Rn,δ))=λn(T(Kδ))(2σn2λ)nλn(Kδ)=(4σnλ)nλn(Kδ),\lambda_{n}\bigl(\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta)\bigr)=\lambda_{n}\bigl(T(K_{\delta})\bigr)\le\bigl(2\,\sigma_{n}\cdot 2\lambda\bigr)^{n}\lambda_{n}(K_{\delta})=\bigl(4\,\sigma_{n}\,\lambda\bigr)^{n}\lambda_{n}(K_{\delta}),

which is the displayed inequality of claim 2.

(e) Positivity. Since 0<δ0<\delta, claim 1 of Elementary Properties of the Closed Ball in a Metric Space gives BdE(0Rn,δ)BˉdE(0Rn,δ)B_{d_{E}}(0_{\mathbb{R}^{n}},\delta)\subseteq\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta), and by claim 1 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure the open ball is a Borel set of positive λn\lambda_{n}-measure. The closed ball is compact, hence Borel, by claim 2 of A Closed Euclidean Ball is Convex and Compact and claim 3 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure, so monotonicity (claim 2 of Basic Properties of a Measure) gives 0<λn(BˉdE(0Rn,δ))0<\lambda_{n}(\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta)). Combining with (d), 0<(4σnλ)nλn(Kδ)0<(4\sigma_{n}\lambda)^{n}\lambda_{n}(K_{\delta}); as 0<σn0<\sigma_{n} and 0<λ0<\lambda make (4σnλ)n(4\sigma_{n}\lambda)^{n} positive, it follows that 0<λn(Kδ)0<\lambda_{n}(K_{\delta}).

Proof of claim 3. Let ZB(Rn)Z\in\mathcal{B}(\mathbb{R}^{n}) with λn(Z)=0\lambda_{n}(Z)=0, and suppose that every xKδx\in K_{\delta} lay in ZZ, that is, KδZK_{\delta}\subseteq Z. Then monotonicity (claim 2 of Basic Properties of a Measure) would give λn(Kδ)λn(Z)=0\lambda_{n}(K_{\delta})\le\lambda_{n}(Z)=0, contradicting claim 2. Hence some xKδx\in K_{\delta} satisfies xZx\notin Z.

Proof of claim 4. Let ρR\rho\in\mathbb{R} with 0<ρ0<\rho. By the localisation of maximisers there is a positive real δρ\delta_{\rho} with KδBdE(x^,ρ)K_{\delta}\subseteq B_{d_{E}}(\hat{x},\rho) for every real δ\delta with 0<δδρ0<\delta\le\delta_{\rho}. Let δ1\delta_{1} be the smaller of δ0\delta_{0} and δρ\delta_{\rho}; it is positive and satisfies δ1δ0\delta_{1}\le\delta_{0}, and every real δ\delta with 0<δδ10<\delta\le\delta_{1} satisfies δδρ\delta\le\delta_{\rho}, so KδBdE(x^,ρ)K_{\delta}\subseteq B_{d_{E}}(\hat{x},\rho).

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