The Dual Potential, the Involution Kernel and Transport in Ergodic Optimization

Abstract

Consider the shift σ acting on the Bernoulli space $\varSigma =\{ 1,2,\ldots,n\}^{\mathbb{N}}$. We denote $\hat{\varSigma }=\{ 1,2,\ldots,n\}^{\mathbb{Z}} =\varSigma \times \varSigma$. We analyze several properties of the maximizing probability μ _∞, A of a Hölder potential $A:\varSigma \rightarrow \mathbb{R}$. Associated to A(x), via the involution kernel, W(x, y), $W:\hat{\varSigma }\rightarrow \mathbb{R}$, one can get the dual potential A ^∗(y), where $(x,y) \in \hat{\varSigma }$. We denote $\mu _{\infty,A^{{\ast}}}$ the maximizing probability for A ^∗. We would like to consider the transport problem from μ _∞, A to $\mu _{\infty,A^{{\ast}}}$. In this case, it is natural to consider the cost function c(x, y) = I(x) − W(x, y) +γ, where I is the deviation function for μ _∞, A , as the limit of Gibbs probabilities μ _{β A} for the potential β A when β → ∞. The value γ is a constant which depends on A. We could also take c = −W above. We denote by $\mathcal{K} = \mathcal{K}(\mu _{\infty,A},\mu _{\infty,A^{{\ast}}})$ the set of probabilities $\hat{\eta }(x,y)$ on $\hat{\varSigma }$, such that $\pi _{x}^{{\ast}}(\hat{\eta }) =\mu _{\infty,A},\,\,\text{and}\,\,\pi _{y}^{{\ast}}(\hat{\eta }) =\mu _{\infty,A^{{\ast}}}\,.$ We describe the minimal solution $\hat{\mu }$ (which is invariant by the shift on $\hat{\varSigma }$) of the Transport Problem, that is, the solution of

$$\displaystyle{\inf _{\hat{\eta }\in \mathcal{K}}\int \int c(x,y)\,d\,\hat{\eta } =\, -\,\max _{\hat{\eta }\in \mathcal{K}}\int \int (W(x,y)-\gamma )\,d\,\hat{\eta }.\,}$$

The optimal pair of functions for the Kantorovich Transport dual Problem is (−V, −V ^∗), where we denote the two calibrated sub-actions by V and V ^∗, respectively, for A and A ^∗. We show that the involution kernel W is cyclically monotone. In other words, satisfies a twist condition in the support of $\hat{\mu }$. We analyze the question: is the support of $\hat{\mu }$ a graph? We also investigate the question of finding an explicit expression for the function $f:\varSigma \rightarrow \mathbb{R}$ whose c-subderivative determines the graph. We also analyze the same kind of problem for expanding transformations on the circle.

Appendix

Here we consider first the shift $\varSigma =\{ 0,1\}^{\mathbb{N}}$, and Σ as a metric space with the usual distance:

$$\displaystyle{d(x,y) = \left \{\begin{array}{cc} 0,\,\,\text{if}\, & \quad x = y\\ (1/2)^{n },\, &\,\,\text{if} \,\quad n =\min \{ i\,\,\vert \,\,x_{ i}\neq y_{i}\}. \end{array} \right.}$$

Additionally, we suppose that Σ is ordered by x < y, if x _i  = y _i for i = 1. . n − 1, and x _n  = 0 and y _n  = 1.

As the usual, we consider the dynamical system (Σ, σ) where σ: Σ → Σ is given by σ(x) = σ(x ₁, x ₂, x ₃, …) = (x ₂, x ₃, x ₄, …).

• (a) Potentials and the Involution Kernel

  As usual we denote

  $$\displaystyle{\tau _{x}^{{\ast}}(y) = (x_{ 1},y_{1},y_{2},y_{3},\ldots )\text{ and }\tau _{y}(x) = (y_{1},x_{1},x_{2},x_{3},\ldots ),}$$

  and

  $$\displaystyle{\hat{\sigma }(x,y) = (\sigma (x),\tau _{x}^{{\ast}}(y))\text{ and }\hat{\sigma }^{-1}(x,y) = (\tau _{ y}x,\sigma ^{{\ast}}(y)),}$$

  the skew product map, where σ ^∗(y = (y ₁, y ₂, y ₃, …)) = (y ₂, y ₃, y ₄, …).

We also define τ _k, y x = (y _k , y _k−1, … y ₂, y ₁, x ₀, x ₁, x ₂, …), where x = (x ₀, x ₁, x ₂, …),  y = (y ₁, y ₂, y ₃, …). In a similar way we define τ _k, y ^∗ x. 

Given a continuous function $A:\varSigma \rightarrow \mathbb{R}$, remember that a continuous function $W:\varSigma \times \varSigma \rightarrow \mathbb{R}$ is an involution kernel for A if $(W \circ \hat{\sigma }^{-1} - W + A \circ \hat{\sigma }^{-1})(x,y)$ does not depends on x; In this case the continuous function $A^{{\ast}}(y) = (W \circ \hat{\sigma }^{-1} - W + A \circ \hat{\sigma }^{-1})(x,y)$ is called the W-dual potential of A.

As in [2] we define the cocycle Δ _A (x, x′, y), where

$$\displaystyle{\varDelta _{A}(x,x',y) =\sum _{n\geq 1}A \circ \hat{\sigma }^{-n}(x,y) - A \circ \hat{\sigma }^{-n}(x',y) =\sum _{ n\geq 1}A \circ \tau _{n,y}(x) - A \circ \tau _{n,y}(x'),}$$

and its dual version $\varDelta _{A^{{\ast}}}(x,y,y')$, where

$$\displaystyle{\varDelta _{A^{{\ast}}}(x,y,y') =\sum _{n\geq 1}A^{{\ast}}\circ \hat{\sigma }^{n}(x,y) - A^{{\ast}}\circ \hat{\sigma }^{n}(x,y') =\sum _{ n\geq 1}A^{{\ast}}\circ \tau _{ n,x}^{{\ast}}(y) - A^{{\ast}}\circ \tau _{ n,x}^{{\ast}}(y').}$$

Note that:

• (i) Δ _A (x, x′, y) = −Δ _A (x′, x, y), in particular Δ _A (x, x, y) = 0,

• (ii) Δ _A (x, x′, y) +Δ _A (x′, x″, y) = Δ _A (x, x″, y),

• (iii) Δ _A (x, x′, y) = Δ _A (τ _y x, τ _y x′, σ ^∗(y)) + [A ∘τ _y x − A ∘τ _y x′], 

and the same relations are true for $\varDelta _{A^{{\ast}}}(x,y,y')$.

Using this properties one can prove that, for any involution kernel we have $W(x,y) - W(x',y) =\varDelta _{A}(x,x',y)\text{ and }W(x,y) - W(x,y') =\varDelta _{A^{{\ast}}}(x,y,y').$

From this fact, we get that the difference between two involution kernels for A is a continuous function of y: {Involution kernels for A}∕C⁰(Σ) = W ⁰, where W ⁰(x, y) = Δ _A (x, x′, y) for a fix x′ ∈ Σ is called a fundamental involution kernel of A. Indeed, the property (iii) shows that W ⁰ is an involution kernel for A.

On the other hand, given another involution kernel, W we have W(x, y) − W(x′, y) = Δ _A (x, x′, y), thus

$$\displaystyle{W(x,y) = W(x',y) +\varDelta _{A}(x,x',y) = W(x',y) + W^{0}(x,y) = g(y) + W^{0}(x,y),}$$

where g(y) = W(x′, y) ∈ C ⁰(Σ).

As an example we compute the general dual potential. First for W ⁰(x, y) = Δ _A (x, x′, y) we get:

$$\displaystyle{\begin{array}{ccc} A_{0}^{{\ast}}(y)& =&(W^{0}(\tau _{y}x,\sigma ^{{\ast}}(y))) - W^{0}(x,y) + A(\tau _{y}x) \\ & =& \varDelta _{A}(\tau _{y}x,x',\sigma ^{{\ast}}(y)) -\varDelta _{A}(x,x',y) + A(\tau _{y}x) \\ & =& A(\tau _{y}x') +\varDelta _{A}(\tau _{y}x',x',\sigma ^{{\ast}}(y)).\\ \end{array} }$$

Given another involution kernel, W we have W(x, y) = W(x′, y) + W ⁰(x, y) thus

$$\displaystyle{A^{{\ast}}(y) = (W \circ \hat{\sigma }^{-1} - W + A \circ \hat{\sigma }^{-1})(x,y) = W(x',\sigma ^{{\ast}}(y)) - W(x',y) + A_{ 0}^{{\ast}}(y).}$$

• (b) The Twist Property of an Involution Kernel

If $A:\varSigma \rightarrow \mathbb{R}$ is a potential and W an arbitrary involution kernel for A, as we said before, W has the twist property, if for any, a, b, a′, b′ ∈ Σ

$$\displaystyle{W(a,b) + W(a',b') < W(a,b') + W(a',b),}$$

provided that a < a′ and b < b′.

If we rewrite this inequality as,

$$\displaystyle{\begin{array}{ccc} W(a,b) + W(a',b')& <&W(a,b') + W(a',b) \\ W(a,b) - W(a',b)& <&W(a,b') - W(a',b') \\ \varDelta _{A}(a,a',b) & <& \varDelta _{A}(a,a',b'), \end{array} }$$

we get an alternative criteria for the twist property, that is, W has the twist property, if for any, a, a′ ∈ Σ the function y → Δ _A (a, a′, y), is strictly increasing, provided that a < a′.

Remark 5

This characterization shows a very important fact. The twist property is a property of A, so we can said that A is a twist potential or equivalently A has a twist involution kernel (as, obviously other involution kernel is also twist).

Remark 6

As an initial approximation we can consider a different setting of dynamics. Let T(x) = −2x ( mod 1 ), and

$$\displaystyle{\tau _{0}x = -\frac{1} {2}x + \frac{1} {2},\text{ and }\tau _{1}x = -\frac{1} {2}x + 1,}$$

the inverse branches that defines the skew maps (that are not the actual natural extension of T):

$$\displaystyle{\hat{T}(x,y) = (T(x),\tau _{x}^{{\ast}}(y))\text{ and }\hat{T}^{-1}(x,y) = (\tau _{ y}x,T^{{\ast}}(y)).}$$

So, one can compute an involutive (that is, A ^∗(y) = A(y)) smooth kernel for A ₁(x) = x and A ₂(x) = x ² given by

$$\displaystyle{W_{1}(x,y) = -\frac{1} {3}(x + y)\text{ and }W_{2}(x,y) = \frac{1} {3}(x^{2} + y^{2}) -\frac{4} {3}xy.}$$

As a corollary we get that any potential A(x) = a + bx + cx ² has a smooth involution kernel given by W(x, y) = a + bW ₁(x, y) + cW ₂(x, y). 

Here and in the next paragraphs, we will denote

$$\displaystyle{W_{A}(x,y):= a + bW_{1}(x,y) + cW_{2}(x,y),}$$

where A(x) = a + bx + cx ² is a polynomial of degree 2.

We observe that the twist property can be derived from the positivity of the second mix derivative of the involution kernel when it is smooth. Note that,

$$\displaystyle{\frac{\partial ^{2}W_{1}} {\partial x\partial y} = 0,\text{ and }\frac{\partial ^{2}W_{2}} {\partial x\partial y} = -\frac{4} {3},}$$

thus W ₁ is not twist and W ₂ is. Actually any potential A(x) = a + bx + cx ² where c > 0 is twist.

Remark 7

In this remark we are going to consider the case of A(x) = a + bx + cx ² where c < 0 (not twist). In this case we will be able to compute the calibrated subaction explicitly, which, we believe, it is interesting in itself.

As a first example consider A(x) = −(x − 1)² which is a convex potential.

From [30, 31] we get that the unique maximizing measure for this potential is μ _∞  = δ _2∕3, so the critical value is m = A(2∕3). Using the fact that m = A(2∕3) one can show that there is a unique (up to constants) calibrated subaction ϕ given by:

$$\displaystyle{\phi (x) = W(x,2/3) - W(2/3,2/3) = -\frac{1} {3}x^{2} + \frac{2} {9}x}$$

where the kernel is given by

$$\displaystyle{W(x,y) = -(1/3)x^{2} - (1/3)y^{2} + (4/3)xy - (2/3)x - (2/3)y.}$$

As a second example consider $A(x) = -(x -\frac{1} {2})^{2}$ which it is also a concave potential.

The general arguments in [31] shown that any maximizing measure for this potential is μ _∞  = (1 − t)δ _1∕3 + t δ _2∕3, where t ∈ [0, 1], so the critical value is m = A(1∕3) = A(2∕3). In this case the involutive smooth involution kernel is:

$$\displaystyle{W(x,y) = -(1/3)x^{2} - (1/3)y^{2} + (4/3)xy - (2/3)x - (1/3)y.}$$

It is easy to verify that,

$$\displaystyle{\phi (x) = V _{1}(x)\chi _{_{[(0,1/2)]}}(x) + V _{2}(x)\chi _{_{[1/2,1]}}(x) =\max \{ V _{1}(x),V _{2}(x)\},}$$

is indeed a calibrated subaction for A, where

V ₁(x) = W(x, 1∕3) − W(1∕3, 1∕3) = Δ(x, 1∕3, 1∕3) = −(1∕3)x ² + (1∕9)x,

V ₂(x) = W(x, 2∕3)−W(2∕3, 2∕3) = Δ(x, 2∕3, 2∕3) = −(1∕3)x ²+(5∕9)x−2∕9,

Note that,

$$\displaystyle{\begin{array}{ccc} \phi (\tau _{0}x)& =& V _{1}(\tau _{0}x)\chi _{_{[(0,1/2)]}}(\tau _{0}x) + V _{2}(\tau _{0}x)\chi _{_{[1/2,1]}}(\tau _{0}x) \\ & =& V _{1}(\tau _{0}x) =\varDelta (\tau _{0}x,1/3,1/3) \\ & =& \varDelta (\tau _{1/3}x,\tau _{1/3}1/3,T^{{\ast}}1/3) \\ & =&\varDelta (x,1/3,1/3) - [A(\tau _{1/3}x) - A(\tau _{1/3}1/3)] \\ & =& V _{1}(x) - [A(\tau _{0}x) - m].\\ \end{array} }$$

Thus ϕ(τ ₀ x) + A(τ ₀ x) − m = V ₁(x). Analogously, ϕ(τ ₁ x) + A(τ ₁ x) − m = V ₂(x) so

$$\displaystyle{\begin{array}{ccc} \phi (x)& =& \max \{V _{1}(x),V _{2}(x)\} \\ & =&\max \{\phi (\tau _{0}x) + A(\tau _{0}x) - m,\phi (\tau _{1}x) + A(\tau _{1}x) - m\} \\ & =& \max _{y\in \varSigma }\{\phi (\tau _{y}x) + A(\tau _{y}x) - m\}.\\ \end{array} }$$

• (c) Twist Criteria

Is natural to consider a criteria for the twist property for a class of functions that has a small dependence on the cubic (or higher order) terms. Let P ₂ ⁺ = { p(x) = a + bx + cx ² | c > 0} be the set of strictly convex polynomial. Consider p ∈ P ₂ ⁺, and define

$$\displaystyle{\mathcal{C}_{\varepsilon }(p) =\{ A \in \text{C}^{3}([0,1])\vert A(x) = p(x) +\varepsilon R(x),\,\text{where}\,\frac{\partial R} {\partial x} \in \text{C}^{3}([0,1])\}}$$

Theorem 8

For any p ∈ P ₂ ⁺ , there exists $\varepsilon > 0$ such that all $A \in \mathcal{C}_{\varepsilon }(p)$ is twist.

Proof

Consider p ∈ P ₂ ⁺ fixed. So, p has a smooth and involutive involution kernel given by

$$\displaystyle{W_{p}(x,y) = (a + bW_{1} + cW_{2})(x,y),}$$

that is, p ^∗(y) = p(y), where $W_{1}(x,y) = -\frac{1} {3}(x + y)$ and $W_{2}(x,y) = \frac{1} {3}(x^{2} + y^{2}) -\frac{4} {3}xy$, are the involution kernel associated to x and x ² respectively. Let, $A = p +\varepsilon R \in \mathcal{C}_{\varepsilon }(p)$, and W _R be the involution kernel for R. Since R is C³ we get that, its corresponding involution kernel W _R is C² in the variable x. Using the linearity of the cohomological equation, we get $W_{A}(x,y) = p(W)(x,y) +\varepsilon W_{R}(x,y)$, and differentiating with respect to x, we have

$$\displaystyle{\begin{array}{ccc} \frac{\partial } {\partial x}W_{A}(x,y)& =&(b \frac{\partial } {\partial x}W_{1} + c \frac{\partial } {\partial x}W_{2})(x,y) +\varepsilon \frac{\partial } {\partial x}W_{R}(x,y)\, = \\ & & -\frac{1} {3}b + \frac{2} {3}cx -\frac{4} {3}cy +\varepsilon \frac{\partial } {\partial x}W_{R}(x,y)\\ \end{array} }$$

Since $-\frac{4} {3}c < 0$, and $\frac{\partial } {\partial x}W_{R}(x,y) \in \text{C}^{0}([0,1]^{2})$ the compactness of [0, 1]² implies that $\frac{\partial } {\partial x}W_{A}(x,\cdot )$ is a strictly decreasing function for any $\varepsilon$ small enough, which is sufficient to ensure the twist property.

Remark 8

If, A ∈ C^∞([0, 1]) is strongly convex, we can consider a perturbation of A of order 2 given by

$$\displaystyle{B_{\varepsilon }(x) = A(0) - A'(0)x + \frac{A''(0)} {2} x^{2} +\varepsilon \sum _{ n\geq 3}\frac{A^{(n)}(0)} {n!} x^{n} \in \mathcal{C}_{\varepsilon }(p_{ A}),}$$

where $p_{A} = A(0) - A'(0)x + \frac{A''(0)} {2} x^{2} \in P_{ 2}^{+}$. Thus, we can find $\varepsilon _{0} > 0$ such that $B_{\varepsilon }$ is twist for any $0 <\varepsilon <\varepsilon _{0}$.

• (d) The Involution Kernel is Bi-Hölder

We consider now T(x) = 2x (mod 1) on the interval [0, 1] and the shift σ on $\varOmega =\{ 0,1\}^{\mathbb{N}}$. A natural question is the regularity of the involution kernel W. We denote τ _j , j = 0, 1 the two inverse branches of T. Given $w = (w_{1},w_{2},\ldots ) \in \{ 0,1\}^{\mathbb{N}}$ we denote by τ _k, w the transformation in [0, 1] given by $\tau _{k,w}(x) = (\tau _{w_{k}} \circ \tau _{w_{k-1}} \circ \,\ldots \,\circ \tau _{w_{1}})\,(x).$ We have that, for a fixed x ₀

$$\displaystyle{\varDelta (x,x_{0},w) =\sum _{ k=1}^{\infty }A(\tau _{ k,w}(x)) - A(\tau _{k,w}(x_{0}))}$$

and, the involution kernel W can be described as: for any (x, w) we have W(x, w) = Δ(x, x ₀, w). It is easy to see that W is Hölder on the variable x. Consider a, b ∈ Ω and suppose that d(a, b) = 2⁻ⁿ. In this way a _j  = b _j , j = 1, 2…, n − 1, n. We denote $\bar{a} =\sigma ^{n}(a)$ and $\bar{b} =\sigma ^{n}(b)$.

Proposition 7

Suppose A is α−Hölder. Consider a,b ∈Ω such that d(a,b) = 2 ⁻ⁿ . For a fixed x ∈ [0,1] we have | W(x,a) − W(x,b) |≤ C (2 ⁻ⁿ ) ^α .

Proof

Note that for z = τ _n, a (x) = τ _n, b (x) and z ₀ = τ _n, a (x ₀) = τ _n, b (x ₀) we have

$$\displaystyle\begin{array}{rcl} & W(x,a) - W(x,b) =\sum _{ k=1}^{\infty }A(\tau _{k,a}(x)) - A(\tau _{k,a}(x_{0})) - A(\tau _{k,b}(x)) + A(\tau _{k,b}(x_{0})) =& {}\\ & \sum _{k=1}^{\infty }[\,A(\tau _{k,a}(x)) - A(\tau _{k,b}(x))\,] - [\,A(\tau _{k,a}(x_{0})) - A(\tau _{k,b}(x_{0}))\,] = & {}\\ & \sum _{k=1}^{\infty }[\,A(\tau _{k,\bar{a}}(z)) - A(\tau _{k,\bar{b}}(z))\,] - [\,A(\tau _{k,\bar{a}}(z_{0})) - A(\tau _{k,\bar{b}}(z_{0}))\,]. & {}\\ \end{array}$$

Note also that | z − z ₀ | ≤ d(a, b) = 2⁻ⁿ. Consider z = z ₀ + h, then

$$\displaystyle\begin{array}{rcl} & \,A(\tau _{k,\bar{a}}(z_{0} + h)) - A(\tau _{k,\bar{a}}(z_{0})) \leq C_{A}\,d(\tau _{k,\bar{a}}(z_{0} + h),\tau _{k,\bar{a}}(z_{0}))^{\alpha } \leq & {}\\ & C_{A}\,(\,2^{-k}\,h\,)^{\alpha } = C_{A}(\,2^{-k}\,)^{\alpha }\,h^{\alpha }. & {}\\ \end{array}$$

Then,

$$\displaystyle\begin{array}{rcl} & \sum _{k=1}^{\infty }[\,A(\tau _{k,\bar{a}}(z)) - A(\tau _{k,\bar{a}}(z_{0}))\,] - [\,A(\tau _{k,\bar{b}}(z)) - A(\tau _{k,\bar{b}}(z_{0}))\,]& {}\\ & \leq C_{A}\sum _{k=1}^{\infty }2\,(\,2^{-k}\,)^{\alpha }\,h^{\alpha } \leq C_{A}\sum _{k=1}^{\infty }2\,(2^{\alpha })^{-k}\,\,h^{\alpha } \leq C\,d(a,b)^{\alpha }. & {}\\ \end{array}$$

From the above we get:

Theorem 9

If $A: S^{1} \rightarrow \mathbb{R}$ is Hölder then $W: S^{1} \times \{ 0,1\}^{\mathbb{N}} \rightarrow \mathbb{R}$ is bi-Hölder.

• (e) The Fenchel-Rockafellar Theorem Given $f: \mathbb{R} \rightarrow \mathbb{R}$ defined on the variable x, the Legendre transform of f, denoted by f ^∗, is the function on the variable p defined by

  $$\displaystyle{f^{{\ast}}(p) =\sup _{ x\in \mathbb{R}}\{p\,x -\, f(x)\}.}$$

Theorem 10 (Fenchel-Rockafellar)

Suppose f(x) is smooth strictly convex, $f: \mathbb{R} \rightarrow \mathbb{R}$ , and, g(x) is smooth strictly concave, $g: \mathbb{R} \rightarrow \mathbb{R}$ . Denote by f ^∗ and g ^∗ the corresponding Legendre transforms on the variable p. Then,

$$\displaystyle{\inf _{x\in \mathbb{R}}\,\{\,f(x)\, -\, g(x)\,\}\, =\,\sup _{p\in \mathbb{R}}\,\{\,g^{{\ast}}(p)\, -\, f^{{\ast}}(p)\,\}}$$

Proof

By convexity and concavity properties we have that there exists x ₀ such that

$$\displaystyle{\inf _{x\in \mathbb{R}}\,\{\,f(x)\, -\, g(x)\,\} = f(x_{0}) - g(x_{0}).}$$

It is also true that f′(x ₀) − g′(x ₀) = 0. Denote by $\overline{p}$ that value $\overline{p} = f'(x_{0})$. We illustrate the proof via two pictures in a certain particular case. Figure 11 shows a geometric picture of the position and values of f(x ₀) − g(x ₀), $g^{{\ast}}(\overline{p})$ and $f^{{\ast}}(\overline{p}).$ Note that in this picture we have that f(x ₀) − g(x ₀) > 0. This picture also shows the graph of $\overline{p}\,x$ as a function of x. We observe that the Legendre transform is not linear on the function. Let’s consider different values of p and estimate f ^∗(p) and g ^∗(p). Suppose first $p\, >\, \overline{p}$. In Fig. 12 we show the graph of p x, and the values of f ^∗(p) and g ^∗(p). We denote by x ₂ the value such that

$$\displaystyle{f^{{\ast}}(p) =\sup _{ x\in \mathbb{R}}\{p\,x -\, f(x)\}\, =\, p\,x_{2} - f(x_{2}).}$$

Note that x ₂  > x ₀. We denote by x ₁ the value such that

$$\displaystyle{0\, <\, g^{{\ast}}(p) =\sup _{ x\in \mathbb{R}}\{p\,x -\, g(x)\}\, =\, p\,x_{1} - g(x_{1}).}$$

Note that x ₁  < x ₀.

Fig. 11

The infimum

Fig. 12

The supremum

Note also that f ^∗(p) and g ^∗(p) have different signs. From this picture one can see that g ^∗(p) − f ^∗(p) < f(x ₀) − g(x ₀). In the case $p\, <\, \overline{p}$ a similar reasoning can be done.