Throughout, G : U → R G:U\to\mathbb{R} G : U → R denotes the function given by
G ( z ) = φ ( z ) + μ 2 ∥ z ∥ 2 , G(z)=\varphi(z)+\frac{\mu}{2}\,\lVert z\rVert^{2}, G ( z ) = φ ( z ) + 2 μ ∥ z ∥ 2 ,
which is convex on U U U precisely because φ \varphi φ is semiconvex on U U U with constant μ \mu μ .
Proof of claim 1. Let x 0 ∈ U x_{0}\in U x 0 ∈ U . The set U U U belongs to the topology of Metric Open Sets Form a Topology and satisfies x 0 ∈ U x_{0}\in U x 0 ∈ U and U ⊆ U U\subseteq U U ⊆ U , so x 0 x_{0} x 0 is an interior point of U U U in R n \mathbb{R}^{n} R n by Interior of a Subset of a Topological Space . Since U U U is convex and G G G is convex on U U U , A Convex Function is Lipschitz on a Ball around an Interior Point supplies ρ , L 0 ∈ R \rho,L_{0}\in\mathbb{R} ρ , L 0 ∈ R with 0 < ρ 0<\rho 0 < ρ and 0 ≤ L 0 0\le L_{0} 0 ≤ L 0 such that B ˉ d E ( x 0 , ρ ) ⊆ U \bar{B}_{d_{E}}(x_{0},\rho)\subseteq U B ˉ d E ( x 0 , ρ ) ⊆ U and
∣ G ( y ) − G ( x ) ∣ ≤ L 0 ∥ y − x ∥ for all x , y ∈ B ˉ d E ( x 0 , 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). G ( y ) − G ( x ) ≤ L 0 ∥ y − x ∥ for all x , y ∈ B ˉ d E ( x 0 , ρ ) .
Put R = ∥ x 0 ∥ + ρ R=\lVert x_{0}\rVert+\rho R = ∥ x 0 ∥ + ρ , a nonnegative real number. If x ∈ B ˉ d E ( x 0 , ρ ) x\in\bar{B}_{d_{E}}(x_{0},\rho) x ∈ B ˉ d E ( x 0 , ρ ) then ∥ x − x 0 ∥ = d E ( x 0 , x ) ≤ ρ \lVert x-x_{0}\rVert=d_{E}(x_{0},x)\le\rho ∥ x − x 0 ∥ = d E ( x 0 , x ) ≤ ρ by Closed Ball in a Metric Space and claim 2 of Elementary Properties of the Euclidean Norm on R n \mathbb{R}^n R n , whence by the triangle inequality (claim 6 there)
∥ x ∥ = ∥ x 0 + ( x − x 0 ) ∥ ≤ ∥ x 0 ∥ + ∥ x − x 0 ∥ ≤ R . \lVert x\rVert=\lVert x_{0}+(x-x_{0})\rVert\le\lVert x_{0}\rVert+\lVert x-x_{0}\rVert\le R . ∥ x ∥ = ∥ x 0 + ( x − x 0 )∥ ≤ ∥ x 0 ∥ + ∥ x − x 0 ∥ ≤ R .
Next, for arbitrary x , y ∈ R n x,y\in\mathbb{R}^{n} x , y ∈ R n the triangle inequality gives ∥ y ∥ ≤ ∥ x ∥ + ∥ y − x ∥ \lVert y\rVert\le\lVert x\rVert+\lVert y-x\rVert ∥ y ∥ ≤ ∥ x ∥ + ∥ y − x ∥ and ∥ x ∥ ≤ ∥ y ∥ + ∥ x − y ∥ \lVert x\rVert\le\lVert y\rVert+\lVert x-y\rVert ∥ x ∥ ≤ ∥ y ∥ + ∥ x − y ∥ , while ∥ x − y ∥ = ∥ ( − 1 ) ( y − x ) ∥ = ∥ y − x ∥ \lVert x-y\rVert=\lVert(-1)(y-x)\rVert=\lVert y-x\rVert ∥ x − y ∥ = ∥( − 1 ) ( y − x )∥ = ∥ y − x ∥ by claim 5 there; hence
∣ ∥ y ∥ − ∥ x ∥ ∣ ≤ ∥ y − x ∥ . \bigl|\,\lVert y\rVert-\lVert x\rVert\,\bigr|\le\lVert y-x\rVert . ∥ y ∥ − ∥ x ∥ ≤ ∥ y − x ∥ .
Consequently, for x , y ∈ B ˉ d E ( x 0 , ρ ) x,y\in\bar{B}_{d_{E}}(x_{0},\rho) x , y ∈ B ˉ d E ( x 0 , ρ ) , using 0 ≤ ∥ x ∥ ≤ R 0\le\lVert x\rVert\le R 0 ≤ ∥ x ∥ ≤ R and 0 ≤ ∥ y ∥ ≤ R 0\le\lVert y\rVert\le R 0 ≤ ∥ y ∥ ≤ R ,
∣ ∥ y ∥ 2 − ∥ x ∥ 2 ∣ = ∣ ∥ y ∥ − ∥ x ∥ ∣ ⋅ ( ∥ y ∥ + ∥ x ∥ ) ≤ 2 R ∥ y − x ∥ . \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 . ∥ y ∥ 2 − ∥ x ∥ 2 = ∥ y ∥ − ∥ x ∥ ⋅ ( ∥ y ∥ + ∥ x ∥ ) ≤ 2 R ∥ y − x ∥ .
Since φ ( z ) = G ( z ) − μ 2 ∥ z ∥ 2 \varphi(z)=G(z)-\tfrac{\mu}{2}\lVert z\rVert^{2} φ ( z ) = G ( z ) − 2 μ ∥ z ∥ 2 for z ∈ U z\in U z ∈ U , the triangle inequality for the absolute value gives, for all x , y ∈ B ˉ d E ( x 0 , ρ ) x,y\in\bar{B}_{d_{E}}(x_{0},\rho) x , y ∈ B ˉ d E ( x 0 , ρ ) ,
∣ φ ( y ) − φ ( x ) ∣ ≤ ∣ G ( y ) − G ( x ) ∣ + μ 2 ∣ ∥ y ∥ 2 − ∥ x ∥ 2 ∣ ≤ ( L 0 + μ R ) ∥ y − x ∥ . \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 . φ ( y ) − φ ( x ) ≤ G ( y ) − G ( x ) + 2 μ ∥ y ∥ 2 − ∥ x ∥ 2 ≤ ( L 0 + μ R ) ∥ y − x ∥ .
So claim 1 holds with this ρ \rho ρ and with L = L 0 + μ R L=L_{0}+\mu R L = L 0 + μ R , which is nonnegative.
Proof of claim 2. Let S ⊆ U S\subseteq U S ⊆ U , let x ∈ S x\in S x ∈ S and let ε ∈ R \varepsilon\in\mathbb{R} ε ∈ R with 0 < ε 0<\varepsilon 0 < ε . Apply claim 1 at the point x 0 = x x_{0}=x x 0 = x of U U U to obtain ρ , L ∈ R \rho,L\in\mathbb{R} ρ , L ∈ R with 0 < ρ 0<\rho 0 < ρ , 0 ≤ L 0\le L 0 ≤ L , B ˉ d E ( x , ρ ) ⊆ U \bar{B}_{d_{E}}(x,\rho)\subseteq U B ˉ d E ( x , ρ ) ⊆ U and
∣ φ ( w ) − φ ( z ) ∣ ≤ L ∥ w − z ∥ for all z , w ∈ B ˉ d E ( 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). φ ( w ) − φ ( z ) ≤ L ∥ w − z ∥ for all z , w ∈ B ˉ d E ( x , ρ ) .
Since 0 ≤ L 0\le L 0 ≤ L we have 0 < L + 1 0<L+1 0 < L + 1 , so ε / ( L + 1 ) \varepsilon/(L+1) ε / ( L + 1 ) is a positive real number; let δ \delta δ be the smaller of ρ \rho ρ and ε / ( L + 1 ) \varepsilon/(L+1) ε / ( L + 1 ) , a positive real number.
Let y ∈ S y\in S y ∈ S satisfy ∥ y − x ∥ < δ \lVert y-x\rVert<\delta ∥ y − x ∥ < δ . Then d E ( x , y ) = ∥ y − x ∥ < δ ≤ ρ d_{E}(x,y)=\lVert y-x\rVert<\delta\le\rho d E ( x , y ) = ∥ y − x ∥ < δ ≤ ρ , so y ∈ B ˉ d E ( x , ρ ) y\in\bar{B}_{d_{E}}(x,\rho) y ∈ B ˉ d E ( x , ρ ) ; also x ∈ B ˉ d E ( x , ρ ) x\in\bar{B}_{d_{E}}(x,\rho) x ∈ B ˉ d E ( x , ρ ) by claim 1 of Elementary Properties of the Closed Ball in a Metric Space . Therefore, using 0 ≤ L 0\le L 0 ≤ L and ∥ y − x ∥ < δ \lVert y-x\rVert<\delta ∥ y − x ∥ < δ ,
∣ φ ( y ) − φ ( x ) ∣ ≤ L ∥ y − x ∥ ≤ L δ ≤ L L + 1 ε < ε , \bigl|\varphi(y)-\varphi(x)\bigr|\le L\,\lVert y-x\rVert\le L\,\delta\le\frac{L}{L+1}\,\varepsilon<\varepsilon, φ ( y ) − φ ( x ) ≤ L ∥ y − x ∥ ≤ L δ ≤ L + 1 L ε < ε ,
the last inequality because 0 ≤ L < L + 1 0\le L<L+1 0 ≤ L < L + 1 gives L / ( L + 1 ) < 1 L/(L+1)<1 L / ( L + 1 ) < 1 and 0 < ε 0<\varepsilon 0 < ε . This is the assertion of claim 2.