Classification of Lipschitz Derivatives in Terms of Semicontinuity and the Baire Limit Functions

Let X and Y be metric spaces, f : X → Y be a function, x ∈ X. Denote

• ‖f‖lip= supu≠v∈X 1u−vX|fu−fv|Y \left\| f \right\|_{{\text{lip}}} = \mathop {{\text{sup}}}\limits_{u \ne v \in X} \frac{1} {{\left| {u - v} \right|_X }}\left| {f\left( u \right) - f\left( v \right)} \right|_Y ,

• 𝕃ipfx=limsupu,v→x,xu≠v 1u−vXfu−fvY {\rm{ {\mathbb L}ip }}f\left( x \right) = \mathop {\mathop {\lim {\rm{sup}}}\limits_{\left( {u,v} \right) \to \left( {x,x} \right)} }\limits_{u \ne v} {\rm{\;}}{1 \over {{{\left| {u - v} \right|}_X}}}{\left| {f\left( u \right) - f\left( v \right)} \right|_{_Y}} ,

• Lipfx=limsupu→x 1u−xX|fu−fx|Y {\rm{Lip }}f\left( x \right) = \mathop {\lim \sup }\limits_{u \to x} {\rm{\;\;}}{1 \over {{{\left| {u - x} \right|}_X}}}|f\left( u \right) - f\left( x \right){|_Y} ,

• lip fx=liminfr→0+supu∈Bx,r 1r|fu−fx|Y {\rm{lip}}\,f\left( x \right) = \mathop {\lim \inf }\limits_{r \to {0^ + }} \mathop {{\rm{\;sup}}}\limits_{u \in B\left( {x,r} \right)} {\rm{\;}}{1 \over r}|f\left( u \right) - f\left( x \right){|_Y} ;

We denote by X^d the set of all non-isolated points of X. Throughout the paper, we assume that sup ∅ = 0. As a consequence of this assumption we have 𝕃ip f(x) = Lip f(x) = lip f(x) = 0 for any x ∈ X \ X^d.

Obviously, if Y is a normed space then ∥ · ∥_lip is an extended seminorm on Y^X in the sense [13]. Moreover, ∥ · ∥_lip is a norm on the space Lip_a(X, Y ) of all Lipschitz functions f : X → Y vanishing at some fixed point a ∈ X.

We introduce some auxiliary notations:

• 𝕃ipr fx=f|B(x,r)lip=supu≠v∈B(x,r) 1u−vX|fu−fv|Y {\rm{\mathbb L}{ip}}{^r}{\rm{ }}f\left( x \right) = {\left\| {f|{B_{(x,r)}}} \right\|_{{\rm{lip}}}} = \mathop {\sup }\limits_{u \ne v \in B(x,r)} {\rm{\;\;}}{1 \over {{{\left| {u - v} \right|}_X}}}|f\left( u \right) - f\left( v \right){|_Y} ;

• Liprfx=supu∈Bx,r 1r|fu−fx|Y,Lip+rfx=supu∈Bx,r 1r|fu−fx|Y {\rm{Li}}{{\rm{p}}^r}f\left( x \right) = \mathop {{\rm{sup}}}\limits_{u \in B\left( {x,r} \right)} {\rm{\;}}{1 \over r}|f\left( u \right) - f\left( x \right){|_Y},{\rm{Lip}}_ + ^rf\left( x \right) = \mathop {{\rm{\;sup}}}\limits_{u \in B\left[ {x,r} \right]} {\rm{\;}}{1 \over r}|f\left( u \right) - f\left( x \right){|_Y} ;

• Lipr fx=sup0<ϱ<rLipϱfx,1ipr fx= inf0<ϱ<rLipϱfx {\rm{Li}}{{\rm{p}}_{\rm{r}}}\,f\left( x \right) = \mathop {{\rm{\;sup}}}\limits_{{\rm{0 < }}\varrho {\rm{ < }}r} {\rm{\;Li}}{{\rm{p}}^\varrho }f\left( x \right),1{\rm{i}}{{\rm{p}}_r}\,f\left( x \right) = {\rm{\;}}\mathop {{\rm{inf\;}}}\limits_{{\rm{0 < }}\varrho {\rm{ < }}r} {\rm{Li}}{{\rm{p}}^\varrho }f\left( x \right) .

𝕃ip fx= infr>0𝕃ipr fx, lip fx=liminfr→0+Liprfx. {\rm{{\mathbb L}ip}}\,f\left( x \right) = {\rm{\;}}\mathop {{\rm{inf}}}\limits_{r{\rm{ > 0}}} {\rm{\mathbb L\rm{ip}}^r}{\rm{\;}}f\left( x \right){\rm{,}}\,\,\,\,\,{\rm{lip}}\,f\left( x \right) = \mathop {\lim {\rm{inf}}}\limits_{r \to {0^ + }} {\rm{\;Li}}{{\rm{p}}^r}f\left( x \right).

Some authors (see, for example, [6, 2, 3]) define Lip f and lip f using the function Lip+rf {\rm{Lip}}_ + ^rf instead of Lip^r f. In the case where X is a normed space, we have Bx, r=Bx,r¯ B\left[ {x,\;r} \right] = \overline {B\left( {x,r} \right)} . Therefore, Liprfx=Lip+rfx {\rm{Li}}{{\rm{p}}^r}f\left( x \right) = {\rm{Lip}}_ + ^rf\left( x \right) for any continuous function f. But the previous equality does not hold for the discrete metric on X, nonconstant f and r = 1. However, we have the following

Let X and Y be metric spaces and f : X → Y be a function. Then, for any non-isolated point x ∈ X, the following equalities hold Lip f(x)=limsupr→0+ Liprf(x)=limsupr→0+Lip+rf(x), lip f(x)=lim infr→0+ Lip+rf(x). \eqalign{ & {\rm{Lip }}f\left( x \right) = \mathop {\lim \sup }\limits_{r \to {0^ + }} {\rm{\;Li}}{{\rm{p}}^r}{\rm{ }}f\left( x \right) = \mathop {\lim {\rm{sup}}}\limits_{r \to {0^ + }} {\rm{\;\;Lip}}_ + ^rf\left( x \right), \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{lip}}\,f\left( x \right) = \mathop {\lim {\rm{inf}}}\limits_{r \to {0^ + }} {\rm{\;Lip}}_ + ^rf\left( x \right). \cr}

Denote A=limsupr→0+ φr, B=limsupr→0+ ψr,a=liminfr→0+ φr, b=limr→0+ inf ψr. \eqalign{ & A = \mathop {\lim {\rm{sup}}}\limits_{r \to {0^ + }} {\rm{\;\;}}\varphi \left( r \right),\,\,\,\,\,\,\,B = \mathop {\lim {\rm{sup}}}\limits_{r \to {0^ + }} {\rm{\;\;}}\psi \left( r \right), \cr & a = \mathop {\lim {\rm{inf}}}\limits_{r \to {0^ + }} {\rm{\;\;}}\varphi \left( r \right),\,\,\,\,\,\,\,\,\,\,b = \mathop {\lim }\limits_{r \to {0^ + }} {\rm{\;inf\;}}\psi \left( r \right). \cr} Obviously, A ≤ B and a ≤ b. Let us show, that A ≥ B. Choose ϱ_n → 0⁺ such that ψ(ϱ_n) → B. Put rn=ϱn+ϱn2 {r_n} = {\varrho _n} + \varrho _n^2 . Since ψϱn≤rnϱnφrn \psi \left( {{\varrho _n}} \right) \le {{{r_n}} \over {{\varrho _n}}}\varphi \left( {{r_n}} \right) . Therefore, A≥limsupn→∞ φrn≥limn→∞ϱnrnψϱn=B. A \ge \mathop {\lim \sup }\limits_{n \to \infty } {\rm{\;}}\varphi \left( {{r_n}} \right) \ge \mathop {\lim }\limits_{n \to \infty } {{{\varrho _n}} \over {{r_n}}}\psi \left( {{\varrho _n}} \right) = B.

Now, we will show that a ≥ b. Choose r_n → 0⁺ such that 0 < r_n < 1 and φ(r_n) → a. Let ϱn=rn−rn2 {\varrho _n} = {r_n} - r_n^2 . Since 0 < ϱ_n < r_n, ψϱn≤rnϱnφrn \psi \left( {{\varrho _n}} \right) \le {{{r_n}} \over {{\varrho _n}}}\varphi \left( {{r_n}} \right) , we have b≤liminfn→∞ ψϱn≤limn→∞rnϱnφrn=a. b \le \mathop {\lim \inf }\limits_{n \to \infty } {\rm{\;}}\psi \left( {{\varrho _n}} \right) \le \mathop {\lim }\limits_{n \to \infty } {{{r_n}} \over {{\varrho _n}}}\varphi \left( {{r_n}} \right) = a.

Fix a non-isolated point x ∈ X. Denote φ(r) = Lip^r f(x), ψr=Lip+r fx \psi \left( r \right) = {\rm{Lip}}_ + ^r\,f\left( x \right) . Therefore, rφr= supu∈Bx,r |fu−fx|Y r\varphi \left( r \right) = {\rm{\;}}\mathop {{\rm{sup}}}\limits_{u \in B\left( {x,r} \right)} {\rm{\;}}|f\left( u \right) - f\left( x \right){|_Y} and rψr= supu∈Bx,r |fu−fx|Y r\psi \left( r \right) = {\rm{\;}}\mathop {{\rm{sup}}}\limits_{u \in B\left( {x,r} \right)} {\rm{\;}}|f\left( u \right) - f\left( x \right){|_Y} . Let 0 < ϱ < r. Then B (x, ϱ) ⊆ B [x, ϱ] ⊆ B (x, r), so ϱφϱ≤ϱψϱ≤rφr. \varrho \varphi \left( \varrho \right) \le \varrho \psi \left( \varrho \right) \le r\varphi \left( r \right). Hence, φ and ψ satisfy the condition from Lemma 2.2. Thus, by Lemma 2.2 we conclude that limsupr→0+ φr=limsupr→0+ ψr and liminfr→0+ φr=liminfr→0+ ψr. \mathop {\lim {\rm{sup}}}\limits_{r \to {0^ + }} {\rm{\;}}\varphi \left( r \right) = \mathop {\lim {\rm{sup}}}\limits_{r \to {0^ + }} {\rm{\;\;}}\psi \left( r \right){\rm{ and }}\mathop {\lim {\rm{inf}}}\limits_{r \to {0^ + }} {\rm{\;\;}}\varphi \left( r \right) = \mathop {\lim {\rm{inf}}}\limits_{r \to {0^ + }} {\rm{\;}}\psi \left( r \right). By the definition, we have fx=liminfr→0+ φr f\left( x \right) = \mathop {\lim {\rm{inf}}}\limits_{r \to {0^ + }} {\rm{\;\;}}\varphi \left( r \right) .

It remains to show that Lip fx=limsupr→0+ ψr {\rm{Lip}}\,f\left( x \right) = \mathop {\lim {\rm{sup}}}\limits_{r \to {0^ + }} {\rm{\;\;}}\psi \left( r \right) . Denote for any r > 0 αr= sup0<ϱ<r ψϱ and βr=sup |fu−fx|Y|u−x|X:0<|u−x|X<r. \alpha \left( r \right) = {\rm{\;}}\mathop {{\rm{sup}}}\limits_{{\rm{0 < }}\varrho < r} {\rm{\;}}\psi \left( \varrho \right)\,\,\,\,{\rm{and}}\,\,\,\beta \left( r \right) = {\rm{\;sup}}\left\{ {{\rm{\;}}{{|f\left( u \right) - f\left( x \right){|_Y}} \over {|u - x{|_X}}}:0 < |u - x{|_X} < r} \right\}. Therefore, αr= sup0<ρ<r sup0<u−xX≤ρ 1ρ|fu−fx|Y ≤ sup0<ρ<r sup0<u−xX≤ρ 1u−xX|fu−fx|Y=βr. \eqalign{ & \alpha \left( r \right) = \;\mathop {{\rm{\;sup}}}\limits_{0 < \rho < r} {\rm{\;}}\;{\rm{\;}}\mathop {{\rm{sup}}}\limits_{0 < {{\left| {u - x} \right|}_X} \le \rho } {\rm{\;}}\;{1 \over \rho }|f\left( u \right) - f\left( x \right){|_Y} \cr & \,\,\,\,\,\,\,\,\,\,\, \le \;{\rm{\;}}\mathop {{\rm{sup}}}\limits_{0 < \rho < r} {\rm{\;}}\;{\rm{\;}}\mathop {{\rm{sup}}}\limits_{0 < {{\left| {u - x} \right|}_X} \le \rho } {\rm{\;}}\;{1 \over {{{\left| {u - x} \right|}_X}}}|f\left( u \right) - f\left( x \right){|_Y} = \beta \left( r \right). \cr} On the other hand, we have βr= sup0<ρ<r sup|u−x|X=ρ 1ρ|fu−fx|Y ≤ sup0<ρ<r sup0<u−xX≤ρ 1ρ|fu−fx|Y = sup0<ρ<r Lip+ρfx=αr. \eqalign{ & \beta \left( r \right) = \;{\rm{\;}}\mathop {{\rm{sup}}}\limits_{0 < \rho < r} {\rm{\;}}\;{\rm{\;}}\mathop {{\rm{sup}}}\limits_{|u - x{|_X} = \rho } {\rm{\;}}\;{1 \over \rho }|f\left( u \right) - f\left( x \right){|_Y} \cr & \,\,\,\,\,\,\,\,\,\,\,\, \le \;{\rm{\;}}\mathop {{\rm{sup}}}\limits_{0 < \rho < r} {\rm{\;}}\;{\rm{\;}}\mathop {{\rm{sup}}}\limits_{0 < {{\left| {u - x} \right|}_X} \le \rho } {\rm{\;}}\;{1 \over \rho }|f\left( u \right) - f\left( x \right){|_Y} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {\rm{\;}}\mathop {{\rm{sup}}}\limits_{0 < \rho < r} {\rm{\;Lip}}_ + ^\rho f\left( x \right) = \alpha \left( r \right). \cr} We have shown, that α(r) = β(r) for r > 0. Hence, Lip fx= infr>0 βr= infr>0 αr=limsupr→0+ ψr. {\rm{Lip}}\,f\left( x \right) = {\rm{\;}}\mathop {{\rm{inf}}}\limits_{r > 0} {\rm{\;}}\beta \left( r \right) = {\rm{\;}}\mathop {{\rm{inf}}}\limits_{r > 0} {\rm{\;}}\alpha \left( r \right) = \mathop {\lim {\rm{sup}}}\limits_{r \to {0^ + }} {\rm{\;\;}}\psi \left( r \right).

Let X and Y be metric spaces and γ ≥ 0. A function f : X → Y is called

• γ-Lipschitz if ∥f∥_lip ≤ γ;

• Lipschitz if ∥f∥_lip < ∞;

• locally Lipschitz if 𝕃ip f < ∞;

• pointwise Lipschitz if Lip f < ∞;

• weakly pointwise Lipschitz if lip f < ∞.

Denote

• 𝕃(f) = {x ∈ X : 𝕃ip f(x) < ∞};

• 𝕃^∞(f) ={x ∈ X : 𝕃ip f(x) = ∞} = X \ 𝕃(f);

• L(f) = {x ∈ X : Lip f(x) < ∞};

• L^∞(f) = {x ∈ X : Lip f(x) = ∞} = X \ L(f);

• ℓ(f) = {x ∈ X : lip f(x) < ∞};

• ℓ^∞(f) = {x ∈ X : lip f(x) < ∞}; = X \ ℓ(f).

Inequalities ( 2.4) yield the next assertion.

Let X and Y be metric spaces, and f : X → Y be a function. Then 𝕃(f) ⊆ L(f) ⊆ ℓ(f) and ℓ^∞(f) ⊆ L^∞(f) ⊆ 𝕃^∞(f).

Let X and Y be normed spaces, G be an open subset of X, and f : G → Y have a locally bounded Gateaux derivative f′. Then, 𝕃ip fx=limsupu→x ‖f′x‖ {\rm{{\mathbb L}ip}}\,f\left( x \right) = \mathop {\lim {\text{sup}}}\limits_{u \to x} \left\| {f'\left( x \right)} \right\| for x ∈ X and so, f is locally Lipschitz. Moreover, if f is C¹ function, then 𝕃ip f(x) = ∥f′(x)∥, x ∈ X.

The following result was also stated in [7], but the proof contains a small blunder. Here, we provide the correct proof.

Let f : X → Y, where X and Y are normed spaces and assume that there exists the Fréchet derivative df (x₀) of f at a point x₀ ∈ X. Then lip f(x₀) = Lip f(x₀) = ∥df (x₀)∥.

It is enough to consider the case X ≠ {0}. Denote by A = df (x₀) the Fréchet derivative of f at the point x₀. We have (3.1) fx−fx0=Ax−x0+αx for x∈X, f\left( x \right) - f\left( {{x_0}} \right) = A\left( {x - {x_0}} \right) + \alpha \left( x \right)\,\,\, for x \in X, where α is a function, such that limx→x0αx‖x−x0‖=0 \mathop {\lim }\limits_{x \to x_0 } \frac{{\alpha \left( x \right)}}{{\left\| {x - x_0 } \right\|}} = 0 . By (3.1) we have ‖fx−fx0‖≤‖A‖‖x−x0‖+‖αx‖, \left\| {f\left( x \right) - f\left( {x_0 } \right)} \right\| \leq \left\| A \right\|\left\| {x - x_0 } \right\| + \left\| {\alpha \left( x \right)} \right\|, hence ‖fx−fx0‖||x−x0‖≤‖A‖+‖αx‖x−x0‖‖. \frac{{\left\| {f\left( x \right) - f\left( {x_0 } \right)} \right\|}}{{\left\| {x - x_0 } \right\|}} \leq \left\| A \right\| + \left\| {\frac{{\alpha \left( x \right)}} {{\left\| {x - x_0 } \right\|}}} \right\|. Thus, lipfx0≤Lip fx0=limsupx→x0 ‖fx−fx0‖||x−x0‖ ≤limx→x0‖A‖+αx‖x−x0‖ =‖A‖. \eqalign{ & {\text{lip }}f\left( {x_0 } \right) \leq {\text{Lip}}\,f\left( {x_0 } \right) = \mathop {\lim {\text{sup}}}\limits_{x \to x_0 } {\text{}}\frac{{\left\| {f\left( x \right) - f\left( {x_0 } \right)} \right\|}} {{\left\| {x - x_0 } \right\|}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \leq \mathop {\lim }\limits_{x \to x_0 } \left( {\left\| A \right\| + \left\| {\frac{{\alpha \left( x \right)}} {{\left\| {x - x_0 } \right\|}}} \right\|} \right) = \left\| A \right\|. \cr}

We want to prove that lip f(x₀) ≥ ∥A∥. Fix ε > 0. Then, there exists e ∈ X such that ∥e∥ = 1 and Ae ≥ A−ε. \left\| {Ae} \right\|\, \ge \,\left\| A \right\| - \varepsilon . For any r > 0, denote x_r = x₀ + re. Note that r = ∥x_t − x₀∥, and x_r → x₀ as r → 0. So, αxrr→0 {{\alpha \left( {{x_r}} \right)} \over r} \to 0 as r → 0. Therefore, by (3.1) fxr−fx0=Axr−x0+αxr=rAe+αxr, f\left( {{x_r}} \right) - f\left( {{x_0}} \right) = A\left( {{x_r} - {x_0}} \right) + \alpha \left( {{x_r}} \right) = rAe + \alpha \left( {{x_r}} \right), so for any r > 0 Lip+rfx0=1r sup‖u−x‖≤r ‖fx−fx0‖≥1r‖fxr−fx0‖ =Ae+αxrr≥‖Ae‖−αxrr≥‖A‖−ε−αxrr. \eqalign{ {\text{Lip}}_ + ^r f\left( {x_0 } \right) &= \frac{1} {r}\mathop {{\text{sup}}}\limits_{\left\| {u - x} \right\| \leq r} \left\| {f\left( x \right) - f\left( {x_0 } \right)} \right\| \geq \frac{1} {r}\left\| {f\left( {x_r } \right) - f\left( {x_0 } \right)} \right\| \cr & = \left\| {Ae + \frac{{\alpha \left( {x_r } \right)}} {r}} \right\| \geq \left\| {Ae} \right\| - \left\| {\frac{{\alpha \left( {x_r } \right)}} {r}} \right\| \geq \left\| A \right\| - \varepsilon - \left\| {\frac{{\alpha \left( {x_r } \right)}} {r}} \right\|. \cr} Thus, by Proposition 2.1 we conclude that lip fx0=liminfr→0+ Lip+rfx0≥limr→0+‖A‖−ε−αxrr=‖A‖−ε. {\text{lip}}\,f\left( {x_0 } \right) = \mathop {\lim {\text{inf}}}\limits_{r \to 0^ + } {\text{Lip}}_ + ^r f\left( {x_0 } \right) \geq \mathop {\lim }\limits_{r \to 0^ + } \left( {\left\| A \right\| - \varepsilon - \left\| {\frac{{\alpha \left( {x_r } \right)}} {r}} \right\|} \right) = \left\| A \right\| - \varepsilon . Since the ε was chosen arbitrarily, the proof is finished.

Let X be a topological space, 𝒜 ⊆ 2^X and f:X→𝕉¯ f:X \to \overline {{\rm{\mathbb R}}} be an A-upper semicontinuous function. Then

• (i)

  f⁻¹ ([γ, +∞]) ∈ 𝒜_c for any γ ∈ ℝ;

• (ii)

  f⁻¹ [[−∞, +∞)] ∈ 𝒜_σ and, so, f⁻¹ [{+∞}] ∈𝒜_σc;

• (iii)

  f⁻¹ [{−∞}] ∈ 𝒜_δ and, so, f⁻¹ ((−∞, +∞]) ∈𝒜_δc;

• (iv)

  f is 𝒜_cσ-lower semicontinuous.

• (i)

  For any γ ∈ ℝ we have that f⁻¹ ([−∞, γ)) ∈ 𝒜 and then f−1γ, +∞=X\f−1−∞, γ∈Ac. {f^{ - 1}}\left( {\left[ {\gamma ,\; + \infty } \right]} \right) = X\backslash {f^{ - 1}}\left( {\left[ { - \infty ,\;\gamma } \right)} \right) \in {{\cal A}_c}.

• (ii)

  Since f⁻¹ [[−∞, n)] ∈ 𝒜 for any n ∈ ℕ, we conclude that f−1−∞, +∞=∪n=1∞f−1−∞, n∈Aσ, {f^{ - 1}}\left[ {\left[ { - \infty ,\; + \infty } \right)} \right] = \mathop \cup \limits_{n = 1}^\infty {f^{ - 1}}\left[ {\left[ { - \infty ,\;n} \right)} \right] \in {{\cal A}_\sigma }, and so, f⁻¹ [{+∞}] = X \ f⁻¹ [[−∞, +∞)] ∈ 𝒜_{σ c}.

• (iii)

  Since f⁻¹ [[−∞, −n)] ∈ 𝒜 for any n ∈ ℕ, we have that f−1−∞=∩n=1∞f−1−∞, −n∈Aδ, {f^{ - 1}}\left[ {\left\{ { - \infty } \right\}} \right] = \mathop \cap \limits_{n = 1}^\infty {f^{ - 1}}\left[ {\left[ { - \infty ,\; - n} \right)} \right] \in {{\cal A}_\delta }, and so, f⁻¹ [(−∞, +∞]] = X \ f⁻¹ [{−∞}]∈ 𝒜_δc.

• (iv)

  Let γ ∈ ℝ and γ_n ↓ γ. Since f⁻¹ [[γ_n, +∞]] ∈ 𝒜_c by (i), we conclude that f−1γ, +∞=∪n=1∞f−1γn, +∞∈Acσ, {f^{ - 1}}\left[ {\left( {\gamma ,\; + \infty } \right]} \right] = \mathop \cup \limits_{n = 1}^\infty {f^{ - 1}}\left[ {\left[ {{\gamma _n},\; + \infty } \right]} \right] \in {{\cal A}_{c\sigma }}, i.e., f is 𝒜_cσ-lower semicontinuous.

Let X be a topological space, 𝒜 ⊆ 2^X, fx:X→𝕉¯ f_n:X \to \overline {{\rm{\mathbb R}}} , be an 𝒜-upper semicontinuous function for any n ∈ ℕ and f:X→𝕉¯ f:X \to \overline {{\rm{\mathbb R}}} be a function such that f(x) = sup_n∈ℕ f_n(x) for any x ∈ X. Then f is an 𝒜_cσ-lower semicontinuous function.

Consider γ ∈ ℝ. By Proposition 4.1(iv) the functions f_n are 𝒜_cσ-lower semicontinuous. So, fn−1γ, +∞∈Acσ f_n^{ - 1}\left[ {\left( {\gamma ,\; + \infty } \right]} \right] \in {{\cal A}_{c\sigma }} for any n ∈ ℕ. Consequently, f−1γ, +∞=∪n=1∞fn−1γ, +∞∈Acσ. {f^{ - 1}}\left[ {\left( {\gamma ,\; + \infty } \right]} \right] = \mathop \cup \limits_{n = 1}^\infty f_n^{ - 1}\left[ {\left( {\gamma ,\; + \infty } \right]} \right] \in {{\cal A}_{c\sigma }}. Thus, f is an 𝒜_cσ-lower semicontinuous function.

Observe that f is an 𝒜-upper semicontinuous function if and only if −f is 𝒜-lower semicontinuous. Therefore, using Proposition 4.1 and 4.2 with g = −f we obtain the following two propositions.

Let X be a topological space, 𝒜 ⊆ 2^X and f:X→𝕉¯ f:X \to \overline {{\rm{\mathbb R}}} be an 𝒜-lower semicontinuous function. Then

• (i)

  f⁻¹ [[−∞, γ]] ∈𝒜_c for any γ ∈ ℝ;

• (ii)

  f⁻¹ [(−∞, +∞]] ∈ 𝒜_σ and, so, f⁻¹ [{−∞}] ∈ 𝒜_{σ c};

• (iii)

  f⁻¹ [{+∞}] ∈ 𝒜_δ and, so, f⁻¹ [[−∞, +∞)] ∈ 𝒜_δc;

• (iv)

  f is 𝒜_cσ-upper semicontinuous.

Let X be a topological space, 𝒜 ⊆ 2^X, fn:X→𝕉¯ {f_n}:X \to \overline {{\rm{\mathbb R}}} be an 𝒜-lower semicontinuous function for any n ∈ℕ and f:X→𝕉¯ f:X \to \overline {{\rm{\mathbb R}}} be a function such that f(x) = inf_n∈ℕf_n(x) for any x ∈ X. Then f is an 𝒜_cσ-upper semicontinuous function.

Let X and Y be metric spaces, f : X → Y be a continuous function and r > 0. Then lip_r f : X → [0, +∞] is an upper semicontinuous function.

Let x₀ ∈ X and γ > lip_r f(x₀). Then infϱ<r Lipϱfx0=liprfx0<γ. {\rm{\;}}\mathop {{\rm{inf}}}\limits_{\varrho < r} {\rm{\;Li}}{{\rm{p}}^\varrho }f\left( {{x_0}} \right) = {\rm{1i}}{{\rm{p}}_r}f\left( {{x_0}} \right) < \gamma . So, there is positive ϱ < r such that Lip^ϱ f(x₀) < γ. Pick γ₁ such that Lip^ϱ f(x₀) < γ₁ < γ. Then we choose ϱ₁ such that γ1γϱ<ϱ1<ϱ {{{\gamma _1}} \over \gamma }\varrho < {\varrho _1} < \varrho . So, γϱ₁ > γ₁ϱ. Therefore, supu∈Bx0,ϱ |fu−fx0|Y=ϱLipϱfx0<γ1ϱ. {\rm{\;}}\mathop {{\rm{sup\;}}}\limits_{u \in B\left( {{x_0},\varrho } \right)} \;|f\left( u \right) - f\left( {{x_0}} \right){|_Y} = \varrho {\rm{Li}}{{\rm{p}}^\varrho }f\left( {{x_0}} \right) < {\gamma _1}\varrho . Then |fu−fx0|Y<γ1ϱ for any u∈Bx0, ϱ. |f\left( u \right) - f\left( {{x_0}} \right){|_Y} < {\gamma _1}\varrho \,\,\,{\rm{for\,\,any}}\,\,\,u \in B\left( {{x_0},\;\varrho } \right). By the continuity of f at x₀ there exists δ> 0 such that ϱ₁ + δ < ϱ and |fx−fx0|Y<γϱ1−γ1ϱ for any x∈U=Bx0, δ. |f\left( x \right) - f\left( {{x_0}} \right){|_Y} < \gamma {\varrho _1} - {\gamma _1}\varrho \,\,\,{\rm{for\,\,any }}x \in U = B\left( {{x_0},\;\delta } \right). Consider x ∈ U and u ∈ B(x, ϱ₁). Then u−x0|X≤u−x|X+|x−x0|X<ϱ1+δ<ϱ, \left| {u - {x_0}{|_X} \le } \right|u - x{|_X} + |x - {x_0}{|_X} < {\varrho _1} + \delta < \varrho , and so, u ∈ B(x₀, ϱ). Therefore, fu−fx|Y≤fu−fx0|Y+|fx0−fx|Y<γ1ϱ+γϱ1−γ1ϱ=γϱ1. \left| {f\left( u \right) - f\left( x \right){|_Y} \le } \right|f\left( u \right) - f\left( {{x_0}} \right){|_Y} + |f\left( {{x_0}} \right) - f\left( x \right){|_Y} < {\gamma _1}\varrho + \left( {\gamma {\varrho _1} - {\gamma _1}\varrho } \right) = \gamma {\varrho _1}. Thus, 1ϱ1|fu−fx|Y≤γ {1 \over {{\varrho _1}}}|f\left( u \right) - f\left( x \right){|_Y} \le \gamma for any u ∈ B(x, ϱ₁). Hence, Lip^ϱ1 f(x) ≤ γ. But 0 < ϱ₁ < r. Therefore, lip_r f(x) ≤ γ for any x ∈ U. Thus, lip_r f is upper semicontinuous at x₀.

Let X and Y be metric spaces and f : X → Y be a continuous function. Then lip f : X → [0, +∞] is a ℱ_σ-lower semicontinuous function.

By (2.1) and (2.3) we conclude that lip fx=supn∈𝕅 lip1nfx {\rm{lip}}\,f\left( x \right) = {\rm{su}}{{\rm{p}}_{n \in {\rm{\mathbb N}}}}\,{\rm{li}}{{\rm{p}}_{{1 \over n}}}f\left( x \right) for any x ∈ X. By Lemma 5.1, the functions lip1nf {\rm{li}}{{\rm{p}}_{{1 \over n}}}f are 𝒯-upper semicontinuous, where 𝒯 is the topology of X. Therefore, by Proposition 4.2 lip f is 𝒯_cσ-lower semicontinuous. This means that lip f is ℱ_σ-lower semicontinuous.

Let X and Y be metric spaces, f : X → Y be a continuous function and r > 0. Then Lip_r f : X → [0, +∞] is a lower semicontinuous function.

Fix r > 0. Let x₀ ∈ X and γ < Lip_r f(x₀). Then supϱ<r Lipϱfx0=Liprfx0>γ. {\rm{\;}}\mathop {{\rm{sup}}}\limits_{\varrho < r} {\rm{\;Li}}{{\rm{p}}^\varrho }f\left( {{x_0}} \right) = {\rm{Li}}{{\rm{p}}_r}f\left( {{x_0}} \right) > \gamma . So, there is ϱ ∈ (0, r) such that Lip^ϱ f(x₀) > γ. Pick γ₁ such that γ<γ1<Lipϱfx0. \gamma < {\gamma _1} < {\rm{Li}}{{\rm{p}}^\varrho }f\left( {{x_0}} \right). Therefore, supu∈Bx0,ϱ fu−fx0Y=ϱ Lipϱfx0>γ1ϱ. {\rm{\;}}\mathop {{\rm{sup\;}}}\limits_{u \in B\left( {{x_0},\varrho } \right)} \;{\left| {f\left( u \right) - f\left( {{x_0}} \right)} \right|_Y} = \varrho {\rm{Li}}{{\rm{p}}^\varrho }f\left( {{x_0}} \right) > {\gamma _1}\varrho . Thus, there is u ∈ B(x₀, ϱ) with fu−fx0Y>γ1ϱ. {\left| {f\left( u \right) - f\left( {{x_0}} \right)} \right|_Y} > {\gamma _1}\varrho . Then we choose ϱ₁ such that ϱ<ϱ1<minr, γ1γϱ \varrho < {\varrho _1} < {\rm{\;min\;}}\left\{ {r,\;{{{\gamma _1}} \over \gamma }\varrho } \right\} . Consequently, γϱ₁ < γ₁ϱ. By the continuity of f at x₀ there exists δ > 0 such that ϱ + δ < ϱ₁ and |fx−fx0|Y<γ1ϱ−γϱ1 for any x∈U:=Bx0, δ. |f\left( x \right) - f\left( {{x_0}} \right){|_Y} < {\gamma _1}\varrho - \gamma {\varrho _1}\,\,\,\,\,\,{\rm{for\,\,any }}x \in U: = B\left( {{x_0},\;\delta } \right). Consider x ∈ U. Then u−x|X≤u−x0|X+|x0−x|X<ϱ+δ<ϱ1, \left| {u - x{|_X} \le } \right|u - {x_0}{|_X} + |{x_0} - x{|_X} < \varrho + \delta < {\varrho _1}, and, so, u ∈ B(x, ϱ₁). Consequently, fu−fxY≥fu−fx0Y−fx−fx0Y>γ1ϱ−γ1ϱ−γϱ1=γϱ1. \left| {f\left( u \right) - f{{\left( x \right)}_Y}} \right| \ge {\left| {f\left( u \right) - f\left( {{x_0}} \right)} \right|_Y} - {\left| {f\left( x \right) - f\left( {{x_0}} \right)} \right|_Y} > {\gamma _1}\varrho - \left( {{\gamma _1}\varrho - \gamma {\varrho _1}} \right) = \gamma {\varrho _1}. Hence, Lip^ϱ1 f(x) > γ. But 0 < ϱ₁ < r. Therefore, Lip_r f(x) > γ for any x ∈ U. Thus, Lip_r f is lower semicontinuous at x₀.

Let X and Y be metric spaces and f : X → Y be a continuous function. Then Lip f : X → [0, +∞] is a ℱ_σ-upper semicontinuous function.

By (2.1) and (2.2) we conclude that Lip fx=infn∈𝕅 Lip1nfx {\rm{Lip}}\,f\left( x \right) = {\rm{in}}{{\rm{f}}_{n \in {\rm{\mathbb N}}}}\,{\rm{Li}}{{\rm{p}}_{{1 \over n}}}f\left( x \right) for any x ∈ X. By Lemma 5.3, the functions Lip1nf {\rm{Li}}{{\rm{p}}_{{1 \over n}}}f are 𝒯-lower semicontinuous where 𝒯 is the topology of X. Therefore, by Proposition 4.4 Lip f is 𝒯_cσ-upper semicontinuous. This means that Lip f is ℱ_σ-upper semicontinuous.

Let X and Y be metric spaces and f : X → Y be a function. Then 𝕃ip f : X → [0, +∞] is an upper semicontinuous function.

Fix x₀ ∈ X and γ > 𝕃ip f(x₀). Since 𝕃ip fx0= inf r>0𝕃iprfx0 {\rm{{\mathbb L}ip}}\,f\left( {{x_0}} \right) = {\rm{\;}}\mathop {{\rm{inf}}}\limits_{r > 0} {\rm{\mathbb Lip}}{^{\rm{r}}}f\left( {{x_0}} \right) , there exists r > 0 such that 𝕃ip^r f(x₀) < γ. Set ϱ=r2 \varrho = {r \over 2} and consider x ∈ B(x₀, ϱ). Then B(x, ϱ) ⊆ B(x₀, r). Consequently, 𝕃ip fx≤𝕃ipϱfx= supu≠v∈Bx,ϱ 1u−vX|fu−fv|Y ≤ supu≠v∈Bx0,r 1u−vX|fu−fv|Y=𝕃iprfx0<γ \eqalign{ & {\rm{{\mathbb L}ip}}\,f\left( x \right) \le {\rm{\mathbb L}}{{\rm{ip}}^\varrho }f\left( x \right) = {\rm{\;}}\mathop {{\rm{sup}}}\limits_{u \ne v \in B\left( {x,\varrho } \right)} {\rm{\;}}{1 \over {{{\left| {u - v} \right|}_X}}}|f\left( u \right) - f\left( v \right){|_Y} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \le \;{\rm{\;}}\mathop {{\rm{sup}}}\limits_{u \ne v \in B\left( {{x_0},r} \right)} {\rm{\;}}\;{1 \over {{{\left| {u - v} \right|}_X}}}|f\left( u \right) - f\left( v \right){|_Y} = {\rm{\mathbb L}}{{\rm{ip}}^r}f\left( {{x_0}} \right) < \gamma \cr} and, hence, 𝕃ip f is upper semicontinuous.

Theorems 5.2, 5.4, 5.5, and Propositions 4.1, 4.3 yield the following assertions.

Let X and Y be metric spaces and f : X → Y be a continuous function. Then

• (i)

  ℓ(f) is a G_δσ-set;

• (ii)

  ℓ^∞(f) is an F_σδ-set;

• (iii)

  L(f) is an F_σ-set;

• (iv)

  L^∞(f) is a G_δ-set.

• (i)

  We have 𝓁(f)={x∈X:lip f(x)<∞}=(lip f)-1[[-∞, +∞)]=X\(lip f)-1[{+∞}] \eqalign{ & {\rm{\cal l}}\left( f \right) = \{ x \in X:{\rm{lip}}\,f\left( x \right) < \infty \} = {({\rm{lip}}\,f)^{ - 1}}[[ - \infty ,\; + \infty )] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = X\backslash {({\rm{lip}}\,f)^{ - 1}}\left[ {\left\{ { + \infty } \right\}} \right] \cr} and, since lip f is ℱ_σ-lower semicontinuous, by Proposition 4.3(iii) (lip f)−1+∞∈Fσδ, {({\rm{lip}}\,f)^{ - 1}}\left[ {\left\{ { + \infty } \right\}} \right] \in {{\cal F}_{\sigma \delta }}, so ℓ(f) = X \ (lip f)⁻¹ [{+∞}] ∈ 𝒢_{δ σ}.

• (ii)

  It follows immediately from (i).

• (iii)

  It is easy to see, that Lf=∪k=1∞(Lip f)-1[[0,k)] L\left( f \right) = \bigcup\nolimits_{k = 1}^\infty {{{({\rm{Lip}}\,f)}^{ - 1}}[[0,k)]} . Since Lip f is an ℱ_σ-upper semicontinuous function, each set (Lip f)⁻¹[[0, k)] is of F_σ type, hence L(f) is an F_σ-set as a countable sum of F_σ-sets.

• (iv)

  It follows from (iii).

Let X and Y be metric spaces and f : X → Y be an arbitrary function. Then

• (i)

  𝕃(f) is an open set;

• (ii)

  𝕃^∞(f) is a closed set.

If E ⊆ ℝ and f : ℝ → ℝ such that lip f ≤ χ_E then |f(a) − f(b)| ≤ µ([a, b] ∩ E) for every a, b ∈ ℝ (where a < b) so, f is Lipschitz and hence absolutely continuous.

In the above, µ denotes the Lebesgue measure. We will state the following

Let γ > 0 and f : [0, 1] → ℝ be a function such that lip f(x) ≤ γ for any x ∈ [0, 1]. Then f is γ-Lipschitz.

Extend f to f˜:𝕉→𝕉 \tilde f:{\rm{\mathbb R}} \to {\rm{\mathbb R}} by f˜x=f0 \tilde f\left( x \right) = f\left( 0 \right) if x < 0 and f˜x=f1 \tilde f\left( x \right) = f\left( 1 \right) if x > 1. Let g=1γf˜ g = {1 \over \gamma }\tilde f and E = [0, 1]. Then by Lemma 6.1 we conclude that 1γfx−fy=gx−gy≤μx, y∩E=x−y {1 \over \gamma }\left| {f\left( x \right) - f\left( y \right)} \right| = \left| {g\left( x \right) - g\left( y \right)} \right| \le \mu \left( {\left[ {x,\;y} \right] \cap E} \right) = \left| {x - y} \right| for any x, y ∈ [0, 1].

Let A be a convex subset of the normed space X, f : X → ℝ be a function, and a, b ∈ A. Moreover, let T : [0, 1] → A be an affine function given by T (u) = a + u(b − a) for 0 ≤ u ≤ 1 and g = f ◦ T : [0, 1] → ℝ. Then, (6.1) lip g≤ ‖b−a‖lip f∘T. {\text{lip}}\,g \leq \,\left\| {b - a} \right\|\left( {\left( {{\text{lip}}\,f} \right) \circ T} \right).

It is enough to consider the case where a ≠ b. Fix u₀∈ [0, 1] and observe, that (6.2) ‖Tu−Tu0‖=‖u−u0b−a‖=u−u0‖b−a‖, 0≤u≤1. \left\| {T\left( u \right) - T\left( {u_0 } \right)} \right\| = \left\| {\left( {u - u_0 } \right)\left( {b - a} \right)} \right\| = \left| {u - u_0 } \right|\left\| {b - a} \right\|,\,\,\,\,\,0 \leq u \leq 1. We have (6.3) lip gu0=liminfr→0+sup0<u−u0<rfTu−fTu0r. {\rm{lip}}\,g\left( {{u_0}} \right) = \mathop {\lim {\rm{inf}}}\limits_{r \to {0^ + }} \mathop {{\rm{\;sup}}}\limits_{0 < \left| {u - {u_0}} \right| < r} {{\left| {f\left( {T\left( u \right)} \right) - f\left( {T\left( {{u_0}} \right)} \right)} \right|} \over r}. Put x₀ = T (u₀). Substituting ϱ = r ∥b − a∥ and x = T (u) in (6.3) and taking into account (6.2) we obtain that lip gu0=liminfϱ→0+ sup0<u−u0<ϱ‖b−a‖ fTu−fTu0ϱ/‖b−a‖ =‖b−a‖liminfϱ→0+ sup0<‖x−x0‖<ϱx∈T0,1 fx−fx0ϱ ≤‖b−a‖liminfϱ→0+sup0<‖x−x0‖<ϱfx−fx0ϱ =‖b−a‖lip fx0. \eqalign{ & {\text{lip}}\,g\left( {u_0 } \right) = \mathop {\lim {\text{inf}}}\limits_{\varrho \to 0^ + } \mathop {{\text{sup}}}\limits_{0 < \left| {u - u_0 } \right| < \frac{\varrho } {{\left\| {b - a} \right\|}}} {\text{}}\frac{{\left| {f\left( {T\left( u \right)} \right) - f\left( {T\left( {u_0 } \right)} \right)} \right|}} {{\varrho /\left\| {b - a} \right\|}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\| {b - a} \right\|\mathop {\lim {\text{inf}}}\limits_{\varrho \to 0^ + } {\text{}}\mathop {{\text{}}\mathop {{\text{sup}}}\limits_{0 < \left\| {x - x_0 } \right\| < \varrho } }\limits_{x \in T\left[ {\left[ {0,1} \right]} \right]} {\text{}}\frac{{\left| {f\left( x \right) - f\left( {x_0 } \right)} \right|}} {\varrho } \cr &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \leq \left\| {b - a} \right\|\mathop {\lim {\text{inf}}}\limits_{\varrho \to 0 + } \mathop {{\text{sup}}}\limits_{0 < \left\| {x - x_0 } \right\| < \varrho } \frac{{\left| {f\left( x \right) - f\left( {x_0 } \right)} \right|}} {\varrho } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = b - a{\text{lip}}\,f\left( {x_0 } \right). \cr} As lip f∘T‖b−a‖=lip f∘T⋅dTdu \left( {\left( {{\text{lip}}\,f} \right) \circ T} \right)b - a = \left( {\left( {{\text{lip}}\,f} \right) \circ T} \right) \cdot \left\| {\frac{{dT}} {{du}}} \right\| , the right side of inequality (6.1) is reminiscent of the “chain rule” for the usual derivative. Nevertheless, the inequality can be strict. To see that, take a function f : ℝ² → ℝ defined as f(x, y) = y, (x, y) ∈ ℝ² and consider the usual distance on ℝ². By Theorem 3.2 we have, lipfx, y=‖dfx, y‖=∂f∂xx,y2+∂f∂yx,y2=1 {\text{lip }}f\left( {x,y} \right) = \left\| {df\left( {x,y} \right)} \right\| = \sqrt {\left( {\frac{{\partial f}} {{\partial x}}\left( {x,y} \right)} \right)^2 + \left( {\frac{{\partial f}} {{\partial y}}\left( {x,y} \right)} \right)^2 } = 1 . Let a = (0, 0), b = (1, 0), T (u) = a + (b − a)u = (u, 0), u ∈ [0, 1] and g = f ◦ T. Therefore, g(u) = f(u, 0) = 0 and so, lip g(u) = 0 < 1 = ∥b − a∥ lip f(T (u)) for any u.

Let D be a convex subset of a normed space X, f : D → ℝ be a function and γ ≥ 0. Then f is γ-Lipschitz if and only if lip f(x) ≤ γ for any x ∈ D.

Fix a, b ∈ D. Define T : [0, 1] → D as Tu=a+ub−a for u∈0,1 T\left( u \right) = a + u\left( {b - a} \right)\,\,\,{\rm{for}}\,\,\,\,u \in \left[ {0,1} \right] and put g = f ◦ T : [0, 1] → ℝ. Applying Lemma 6.3 we get lip gu≤ ‖b−a‖lip fTu≤ ‖b−a‖γ, {\text{lip}}\,g\left( u \right) \leqslant \,\left\| {b - a} \right\|{\text{lip}}\,f\left( {T\left( u \right)} \right) \leq \,\left\| {b - a} \right\|\gamma , for any u ∈ [0, 1]. Therefore, Corollary 6.2 implies that g is Lipschitz with the constant γ₁ = ∥b − a∥ γ. Thus, fa−fb=g0−g1≤γ10−1=γ‖a−b‖. \left| {f\left( a \right) - f\left( b \right)} \right| = \left| {g\left( 0 \right) - g\left( 1 \right)} \right| \leq \gamma _1 \left| {0 - 1} \right| = \gamma \left\| {a - b} \right\|. So, f is γ-Lipschitz on D.

By ∥·∥_∞ we denote standard norm on space of bounded real functions B(D) defined on a set D, i.e., ‖h‖∞= supx∈D hx \left\| h \right\|_\infty = \mathop {{\text{sup}}}\limits_{x \in D} {\text{}}\left| {h\left( x \right)} \right| for any h: D → ℝ.

Let f : D → ℝ be a continuous function, where D is a convex subset of some normed space X. Then, ∥f∥_lip = ∥lip f∥_∞.

We simply check, that if ∥f∥_lip < ∞ or ∥lip f∥_∞ < ∞, then flip=inf γ>0:f is γ-Lipschitz =inf γ>0:lip fx≤γ for any x∈D = sup x∈Dlip fx=‖lip f‖∞, \eqalign{ & \left\| f \right\|_{{\text{lip}}} = {\text{inf}}\,\left\{ {\gamma > 0:f\,{\text{is}}\,\,\gamma {\text{ - Lipschitz}}} \right\} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{inf}}\,\left\{ {\gamma > 0:{\text{lip}}\,f\left( x \right) \leq \gamma \,\,\,{\text{for}}\,{\text{any}}\,\,\,x \in D} \right\} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{}}\mathop {{\text{sup}}}\limits_{x \in D} {\text{lip}}\,f\left( x \right) = \left\| {{\text{lip}}\,f} \right\|_\infty , \cr} where the second equality follows from Theorem 6.4.

Therefore, lip is an isometric injection of the normed space Lip_a(D, ℝ) with some a ∈ X, into the space B(D).

Let D be a locally convex subset of a normed space X and let f : D → ℝ be a function. Then (lip f)∨=(Lip f)∨=𝕃ip f. {({\rm{lip}}\,f)^ \vee } = {({\rm{Lip}}\,f)^ \vee } = {\rm{{\mathbb L}ip}}\,f.

Since lip f ≤ Lip f ≤ 𝕃ip f and 𝕃ip f is upper semicontinuous by Theorem 5.5, we have (lip f)∨≤ (Lip f)∨≤(𝕃ipf)∨=𝕃ip f. {({\rm{lip}}\,f)^ \vee } \le \,{({\rm{Lip}}\,f)^ \vee } \le {({\rm{{\mathbb L}ip }}f)^ \vee } = {\rm{{\mathbb L}ip}}\,f. Therefore, it is enough to prove that 𝕃ip f ≤ (lip f)^∨. Fix x₀ ∈ D. The case where (lip f)^∨(x₀) = ∞ is obvious. So, we suppose that (lip f)^∨(x₀) < ∞. Let γ > (lip f)^∨(x₀). Then, there exists a convex neighborhood U of x₀, such that lip fx<γ for any x∈U. {\rm{lip}}\,f\left( x \right) < \gamma \,\,\,{\rm{for\,\,any}}\,x \in U. By Theorem 6.4, the function f is γ-Lipschitz on U. Hence, 𝕃ipfx0= infr>0 f|Bx0,rlip ≤‖f|U‖lip≤γ, \eqalign{ & {\rm{ {\mathbb L}ip }}f\left( {x_0 } \right) = {\text{}}\mathop {{\text{inf}}}\limits_{r > 0} \left\| {{\text{}}f|_{B\left( {x_0 ,r} \right)} } \right\|_{{\text{lip}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \leq \left\| {f|_U } \right\|_{{\text{lip}}} \leq \gamma , \cr} where B (x₀, r) means the ball in the metric subspace D. Passing to the limit with γ → (lip f)^∨(x₀) we obtain the desired inequality.