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Proof of Local Lipschitz Bound and Continuity for a Semiconvex Function on an Open Convex Set

lemmalem:semiconvex-locally-lipschitz-2026a
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· 3,451 chars · 8 deps · depth 12 Reason: Proof of the local Lipschitz bound and continuity of a semiconvex function on an open convex set: convexify by adding the quadratic, apply the local Lipschitz theorem for convex functions at an interior point, and subtract the quadratic.

Adds the quadratic to obtain a convex function, applies the local Lipschitz theorem for convex functions at an interior point, and subtracts the quadratic, which is Lipschitz on a bounded set; the continuity estimate then follows by choosing the radius small.

Proof

Throughout, G:URG:U\to\mathbb{R} denotes the function given by

G(z)=φ(z)+μ2z2,G(z)=\varphi(z)+\frac{\mu}{2}\,\lVert z\rVert^{2},

which is convex on UU precisely because φ\varphi is semiconvex on UU with constant μ\mu.

Proof of claim 1. Let x0Ux_{0}\in U. The set UU belongs to the topology of Metric Open Sets Form a Topology and satisfies x0Ux_{0}\in U and UUU\subseteq U, so x0x_{0} is an interior point of UU in Rn\mathbb{R}^{n} by Interior of a Subset of a Topological Space. Since UU is convex and GG is convex on UU, A Convex Function is Lipschitz on a Ball around an Interior Point supplies ρ,L0R\rho,L_{0}\in\mathbb{R} with 0<ρ0<\rho and 0L00\le L_{0} such that BˉdE(x0,ρ)U\bar{B}_{d_{E}}(x_{0},\rho)\subseteq U and

G(y)G(x)L0yxfor all x,yBˉdE(x0,ρ).\bigl|G(y)-G(x)\bigr|\le L_{0}\,\lVert y-x\rVert\qquad\text{for all }x,y\in\bar{B}_{d_{E}}(x_{0},\rho).

Put R=x0+ρR=\lVert x_{0}\rVert+\rho, a nonnegative real number. If xBˉdE(x0,ρ)x\in\bar{B}_{d_{E}}(x_{0},\rho) then xx0=dE(x0,x)ρ\lVert x-x_{0}\rVert=d_{E}(x_{0},x)\le\rho by Closed Ball in a Metric Space and claim 2 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n, whence by the triangle inequality (claim 6 there)

x=x0+(xx0)x0+xx0R.\lVert x\rVert=\lVert x_{0}+(x-x_{0})\rVert\le\lVert x_{0}\rVert+\lVert x-x_{0}\rVert\le R .

Next, for arbitrary x,yRnx,y\in\mathbb{R}^{n} the triangle inequality gives yx+yx\lVert y\rVert\le\lVert x\rVert+\lVert y-x\rVert and xy+xy\lVert x\rVert\le\lVert y\rVert+\lVert x-y\rVert, while xy=(1)(yx)=yx\lVert x-y\rVert=\lVert(-1)(y-x)\rVert=\lVert y-x\rVert by claim 5 there; hence

yxyx.\bigl|\,\lVert y\rVert-\lVert x\rVert\,\bigr|\le\lVert y-x\rVert .

Consequently, for x,yBˉdE(x0,ρ)x,y\in\bar{B}_{d_{E}}(x_{0},\rho), using 0xR0\le\lVert x\rVert\le R and 0yR0\le\lVert y\rVert\le R,

y2x2=yx(y+x)2Ryx.\bigl|\,\lVert y\rVert^{2}-\lVert x\rVert^{2}\,\bigr|=\bigl|\,\lVert y\rVert-\lVert x\rVert\,\bigr|\cdot\bigl(\lVert y\rVert+\lVert x\rVert\bigr)\le 2R\,\lVert y-x\rVert .

Since φ(z)=G(z)μ2z2\varphi(z)=G(z)-\tfrac{\mu}{2}\lVert z\rVert^{2} for zUz\in U, the triangle inequality for the absolute value gives, for all x,yBˉdE(x0,ρ)x,y\in\bar{B}_{d_{E}}(x_{0},\rho),

φ(y)φ(x)G(y)G(x)+μ2y2x2(L0+μR)yx.\bigl|\varphi(y)-\varphi(x)\bigr|\le\bigl|G(y)-G(x)\bigr|+\frac{\mu}{2}\bigl|\,\lVert y\rVert^{2}-\lVert x\rVert^{2}\,\bigr|\le\bigl(L_{0}+\mu R\bigr)\lVert y-x\rVert .

So claim 1 holds with this ρ\rho and with L=L0+μRL=L_{0}+\mu R, which is nonnegative.

Proof of claim 2. Let SUS\subseteq U, let xSx\in S and let εR\varepsilon\in\mathbb{R} with 0<ε0<\varepsilon. Apply claim 1 at the point x0=xx_{0}=x of UU to obtain ρ,LR\rho,L\in\mathbb{R} with 0<ρ0<\rho, 0L0\le L, BˉdE(x,ρ)U\bar{B}_{d_{E}}(x,\rho)\subseteq U and

φ(w)φ(z)Lwzfor all z,wBˉdE(x,ρ).\bigl|\varphi(w)-\varphi(z)\bigr|\le L\,\lVert w-z\rVert\qquad\text{for all }z,w\in\bar{B}_{d_{E}}(x,\rho).

Since 0L0\le L we have 0<L+10<L+1, so ε/(L+1)\varepsilon/(L+1) is a positive real number; let δ\delta be the smaller of ρ\rho and ε/(L+1)\varepsilon/(L+1), a positive real number.

Let ySy\in S satisfy yx<δ\lVert y-x\rVert<\delta. Then dE(x,y)=yx<δρd_{E}(x,y)=\lVert y-x\rVert<\delta\le\rho, so yBˉdE(x,ρ)y\in\bar{B}_{d_{E}}(x,\rho); also xBˉdE(x,ρ)x\in\bar{B}_{d_{E}}(x,\rho) by claim 1 of Elementary Properties of the Closed Ball in a Metric Space. Therefore, using 0L0\le L and yx<δ\lVert y-x\rVert<\delta,

φ(y)φ(x)LyxLδLL+1ε<ε,\bigl|\varphi(y)-\varphi(x)\bigr|\le L\,\lVert y-x\rVert\le L\,\delta\le\frac{L}{L+1}\,\varepsilon<\varepsilon,

the last inequality because 0L<L+10\le L<L+1 gives L/(L+1)<1L/(L+1)<1 and 0<ε0<\varepsilon. This is the assertion of claim 2.

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