Entropy of Non-equilibrium Stationary Measures of Boundary Driven TASEP

Abstract

We examine the entropy of non-equilibrium stationary states of boundary driven totally asymmetric simple exclusion processes. As a consequence, we obtain that the Gibbs–Shannon entropy of the non equilibrium stationary state converges to the Gibbs–Shannon entropy of the local equilibrium state. Moreover, we prove that its fluctuations are Gaussian, except when the mean displacement of particles produced by the bulk dynamics agrees with the particle flux induced by the density reservoirs in the maximal phase regime.

Appendices

Appendix A: Proof of Theorem 2.2

Recall the definition of D from (4.2). We can rewrite the set D in a more convenient form by introducing

$$ a_m = \frac{E(m)-1}{1-\xi_0} \quad\text{and}\quad b_m= \frac{m-E(m)}{\xi_0}. $$

It is clear that if E(m)≥sup{ξ ₀,m−ξ ₀} then a _m ≥−1 and b _m ≤1. Also, if E(m)≤m+(1−m)ξ ₀ then a _m ≤b _m . We have a _m =−1 (resp. b _m =1) if and only if E(m)=ξ ₀ (resp. m−E(m)=ξ ₀).

We have

$$ (y,m) \in D \quad\Leftrightarrow\quad \begin{cases} 0\leq m\leq2, \quad\sup\{\xi_0, m-\xi_0\} \le E(m) \le m- (m-1)\xi_0,\\ a_m \le y \le b_m. \end{cases} $$

It is easy to check that D≠∅ is equivalent to {m∈[0,2];sup{ξ ₀,m−ξ ₀}≤E(m)≤m−(m−1)ξ ₀}≠∅ which is equivalent to $E\in[E_{+\infty}^{-}, E_{+\infty}^{+}]$.

Assume from now on that $E\in[E_{+\infty}^{-}, E_{+\infty}^{+}]$.

We denote by α:=α(E) (resp. β:=β(E), resp. γ:=γ(E)) the solution to the linear equation E(m)=ξ ₀ (resp. m−E(m)=ξ ₀, resp. E(m)−m+(m−1)ξ ₀=0). We have that {m∈[0,2];sup{ξ ₀,m−ξ ₀}≤E(m)≤m−(m−1)ξ ₀} is given by m∈[0,2] such that

$$ \begin{cases} E(\alpha) \le E(m)\\ m-E(m) \le\beta-E(\beta)\\ E(m)-m+(m-1)\xi_0\le E(\gamma)-\gamma+(\gamma-1)\xi_0. \end{cases} $$

Since E(m) is a linear function of m, {m∈[0,2];sup{ξ ₀,m−ξ ₀}≤E(m)≤m−(m−1)ξ ₀} is a closed interval [m ₋,m ₊] (with m ₋:=m ₋(E) and m ₊:=m ₊(E)) of [0,2] and it is easy to show by inspection of the different cases that we have Table 1.

Table 1 Values of m ₋ and m ₊ in terms of α,β and γ

This shows that γ is always equal to m ₋ or m ₊, that α or β is the other boundary of the interval [m ₋,m ₊] and that the remaining point among {α,β,γ} does not belong to (m ₋,m ₊).

Let $f: [m_{-}, m_{+}]\to\mathbb{R}$ be defined by

$$ f(m)=\sup_{y \in[a_m, b_m]} F(y,m) $$

(A.1)

so that

$$ \begin{aligned} {S}^+ (E) &=\frac{1}{2} \sup _{m \in[m_-, m_+] } f(m). \end{aligned} $$

(A.2)

Observe that α≠β≠γ apart from a finite (at most three) number of explicit values of E. If E is different from these values we say that E is a regular value of the energy. For simplicity we restrict the study to the case where E is a regular value but the same analysis could be performed for the non-regular values of E. Since α=β is equivalent to α=β=2ξ ₀ we will always assume that it is not the case.

Lemma A.1

Let $E \in[E_{+\infty}^{-}; E_{+\infty}^{+}]$ be a regular value of the energy.

For any m∈(m ₋,m ₊), we have that f(m)=F(y(m),m) for a unique y(m)∈(a _m ,b _m ) which is solution to the equation ∂ _y F(y(m),m)=0.

If α (resp. β) belongs to {m ₋,m ₊} then f(α)=−2s(α/2) (resp. f(β)=−2s(β/2)) and the supremum appearing in the definition of f is uniquely realized for y=−1 (resp. y=1).

We have f(γ)=0 and the supremum in the definition of f is uniquely realized for y=γ−1.

Proof

We notice that if m=γ then a _m =b _m =m−1 and f(γ)=0. This shows the last sentence of the lemma.

Now, let m∈(m ₋,m ₊). Then a _m <b _m . For any y∈(a _m ,b _m ), we have

$$\begin{aligned} \partial_{y} F (y,m) =& -\frac{\xi_0 -E(m)}{y+1} s^{\prime} \biggl( \frac{y \xi_0 + E(m)}{y +1} \biggr) - s \biggl( \frac{y \xi _0 + E(m)}{y +1} \biggr) \\ &{}- \frac{m-E(m)-\xi_0}{1-y}s^{\prime} \biggl( \frac{m- (y \xi _0 + E(m))}{1-y} \biggr) + s \biggl( \frac{m- (y \xi_0 + E(m))}{1-y} \biggr), \\ \partial_{y}^2 F (y,m) =&- \frac{(\xi_0 -E(m))^2}{(y+1)^3} s^{\prime\prime} \biggl( \frac{y \xi_0 + E(m)}{y +1} \biggr) \\ &{}- \frac{(m-E(m)-\xi_0)^2}{(1-y)^3}s^{\prime\prime} \biggl( \frac {m- (y \xi_0 + E(m))}{1-y} \biggr). \end{aligned}$$

Since s is a strictly convex function, $\partial_{y}^{2} F (y,m)<0$, i.e. F(⋅,m) is strictly concave on (a _m ,b _m ) so that $\sup_{[a_{m},b_{m}]} F (\cdot,m)$ is attained for a unique point y(m) of [a _m ,b _m ]. If m≠α then a _m ≠−1 and if m≠β then b _m ≠1. Noticing that (yξ ₀+E(m))/(y+1) (resp. (m−(yξ ₀+E(m)))/(1−y)) goes to 1 (resp. 0) as y goes to a _m (resp. b _m ), we conclude that ∂ _y F(y,m) goes to +∞ (resp. −∞). This implies that y(m)∈(a _m ,b _m ) and (∂ _y F)(y(m),m)=0. If m=α∈[m ₋,m ₊], then

$$F(y,\alpha)= -(y+1)s (\xi_0) -(1-y) s \biggl( \frac{\alpha- \xi_0 (y + 1)}{1-y} \biggr) $$

for y∈[−1,b _α ]⊂[−1,1), F(⋅,α) is strictly concave on[−1,b _α ] and

$$ \begin{aligned} \lim_{ y \to-1} \partial_y F(y,\alpha) = (-s) (\xi_0) - (-s) (\alpha/2) - (\xi_0 - \alpha/2) (-s)' (\alpha/2) < 0 \end{aligned} $$

because s is strictly convex and α≠2ξ ₀ (otherwise E is not regular). The maximum of F(⋅,α) is uniquely attained for y(α):=−1 and F(y(α),α)=−2s(α/2).

Similarly, if m=β∈[m ₋,m ₊], the supremum is uniquely attained for y(β)=1 and F(y(β),β)=−2s(β/2). □

Lemma A.2

Let $E \in[E_{+\infty}^{-}; E_{+\infty}^{+}]$ be a regular value of the energy. The function f:[m ₋,m ₊]→[0,2log(2)] is a continuously differentiable function on (m ₋,m ₊) and, when α,β∈[m ₋,m ₊],

$$ f'(\alpha)= - s'( \alpha/2)\quad \textit{and} \quad f'(\beta)= - s'( \beta/2). $$

Proof

It is clear that f is smooth on (m ₋,m ₊) and that the implicit function theorem applies. Thus the continuity and differentiability problems are only around the points m ₋ and m ₊.

Let us prove that f(m) goes to f(α) and that f′(m) has a limit as m goes to α (assuming that α∈{m ₋,m ₊}) equal to −s′(α/2). The other case can be treated similarly. We will show that

$$ \begin{aligned} &\lim_{m \to\alpha} y(m)=-1, \\ &\lim_{m \to\alpha} \frac {y(m) \xi_0 + E(m)}{y(m) +1}=\lim_{m \to+\infty} \frac{m- (y(m) \xi_0 + E(m))}{1-y(m)} = \alpha/2. \end{aligned} $$

(A.3)

By the implicit function theorem, for any m≠α,β we have

$$ \begin{aligned} f' (m) &= \partial_m F \bigl(y(m),m\bigr) \\ &= - \frac{\varphi_+}{\varphi_+ - \varphi_-} s' \biggl( \frac {y(m) \xi_0 + E(m)}{y(m) +1} \biggr) + \frac{\varphi_-}{\varphi _+ - \varphi_-} s' \biggl(\frac{m- (y(m) \xi_0 + E(m))}{1-y(m)} \biggr). \end{aligned} $$

From (A.3) we deduce that lim _m→α f(m)=f(α) and lim _m→α f′(m)=−s′(α/2).

Let (m _n ) _n≥0 be a sequence in (m ₋,m ₊) (m _n ≠α,β,γ for any n) converging to α. Since for any m,

$$y(m) \in[-1,1], \qquad\frac{y(m) \xi_0 + E(m)}{y(m) +1} \in[0,1] ,\qquad\frac{m- (y(m) \xi_0 + E(m))}{1-y(m)} \in[0,1] , $$

up to a subsequence we can assume that y _n :=y(m _n ) converges to some a∈[−1,1] and that $\frac{y_{n} \xi_{0} + E(m_{n})}{y_{n} +1}$ converges to u∈[0,1] and $\frac{m_{n}- (y_{n} \xi_{0} + E(m_{n}))}{1-y_{n}}$ converges to v∈[0,1].

If a∈(−1,1), then u=ξ ₀∈(0,1), $v= \xi_{0} + \frac {\alpha-2 \xi_{0}}{1-a}$ and v≠u since α≠2ξ ₀. By continuity of the functions involved and taking into account that ∂ _y F(y _n ,m _n )=0, we get that

$$ -s (u) -\frac{\alpha-2\xi_0}{1-a} s' (v) +s(v) =0. $$

The term on the left hand side of the previous equality can be written as s(v)−s(u)−(v−u)s′(v) which is negative, by the convexity of the function s (if v=0 or v=1 then s′(v)=−∞ or s′(v)=+∞ and the inequality is still valid). Therefore a∈{−1,1}.

If a=1, then since $\frac{m_{n}- (y_{n} \xi_{0} + E(m_{n}))}{1-y_{n}}$ converges to v∈[0,1] and y _n →a=1, m _n →α it implies that α=2ξ ₀ that is in contradiction with our assumptions.

It follows that a=−1, so that v=α/2. Observe that $\frac {E(m_{n}) -\xi_{0}}{y_{n} +1}$ converges to u−ξ ₀. Using the fact that ∂ _y F(y _n ,m _n )=0, we get

$$ \begin{aligned} 0&=(u-\xi_0) s' (u) -s(u) - \frac{\alpha-2\xi_0}{2} s' (\alpha/2) + s(\alpha/2). \end{aligned} $$

Since s is convex, the function z→(z−ξ ₀)s′(z)−s(z) is monotone, so that the equality is uniquely satisfied for u=α/2. This proves (A.3). □

Lemma A.3

Let $E \in[E_{+\infty}^{-}; E_{+\infty}^{+}]$ be a regular value of the energy. The function f is strictly concave on (m ₋,m ₊).

Proof

On $\mathring{D}$ we have

$$\begin{aligned} \partial^{2}_m F (y,m) =& - \biggl( \frac{\varphi_+}{\varphi_+ -\varphi_-} \biggr)^2 \frac{1}{y+1} s^{\prime\prime} \biggl( \frac{\xi_0 -E(m)}{y+1} \biggr) \\ &{} - \biggl( \frac{\varphi_-}{\varphi_+ -\varphi_-} \biggr)^2 \frac{1}{1-y} s^{\prime\prime} \biggl( \frac{m- (y \xi_0 + E(m))}{1-y} \biggr) \\ \partial^2_{y,m} F (y,m) =& - \biggl( \frac{\varphi_+}{\varphi_+ -\varphi_-} \biggr) \frac{\xi_0 -E(m)}{(y+1)^2} s^{\prime\prime} \biggl( \frac{\xi_0 -E(m)}{y+1} \biggr) \\ &{}+ \biggl( \frac{\varphi_-}{\varphi_+ -\varphi_-} \biggr) \frac {(m-E(m)-\xi_0)}{(1-y)^2} s^{\prime\prime} \biggl( \frac{m- (y \xi_0 + E(m))}{1-y} \biggr). \end{aligned}$$

One easily checks that

$$ \begin{aligned} &\bigl(\partial_{y}^2 F\bigr) \bigl(\partial_m^2 F\bigr) - \bigl(\partial^2_{y,m} F\bigr)^2 \\ &\quad{}= \frac{s^{\prime\prime} (a) s^{\prime\prime} (b)}{(\varphi_+ -\varphi_-)^2(y^2-1)^2} \biggl[ \varphi_+ \sqrt{\frac{1+y}{1-y}} \bigl(m-E(m)-\xi_0\bigr) + \varphi_- \sqrt{\frac{1-y}{1+y}} \bigl(\xi_0 -E(m)\bigr) \biggr]^2, \end{aligned} $$

where

$$a= \frac{y\xi_0 -E(m)}{y+1} \quad\textrm{and} \quad b= \frac{m- (y \xi_0 + E(m))}{1-y}. $$

By convexity of the function s it follows that $(\partial_{y}^{2} F)(\partial_{m}^{2} F) - (\partial^{2}_{y,m} F)^{2} >0$ if m≠α,β.

By the implicit function theorem, the function m→F(y(m),m) is smooth on (m ₋,m ₊) and

$$ \begin{aligned} &f'' (m)= \partial^2_m F\bigl(y(m),m\bigr) + \partial^2_{y,m} F \bigl( y(m),m\bigr) y' (m) \\ &\frac{d}{dm} \bigl[ \partial_y F \bigl(y(m),m\bigr) \bigr]=0=\bigl(\partial _{m,y}^2 F\bigr) \bigl(y(m),m\bigr)+ \bigl(\partial_y^2 F\bigr) \bigl(y(m),m\bigr) y' (m), \end{aligned} $$

so that

$$ f'' (m) = \frac{ [ (\partial_{y}^2 F)(\partial_m^2 F) - (\partial^2_{y,m} F)^2 ](y(m),m) }{\partial_y^2 F(y(m),m)} . $$

Recalling from the proof of Lemma A.1 that $\partial_{y}^{2} F <0$, we get that f″(m)<0. □

We have to compute sup _m∈[0,2] f(m). Since f is strictly concave there exists a unique m ₀∈[m ₋,m ₊] for which the supremum of f is attained.

The point m ₀ belongs to (m ₋,m ₊) if and only if there exists m∈(m ₋,m ₊) such that f′(m)=0. This is equivalent to the existence of m∈(m ₋,m ₊) and y∈(a _m ,b _m ) such that

$$ \partial_m F(y,m) =0, \qquad\partial_{y} F (y,m)=0. $$

(A.4)

To simplify notations we introduce

$$ \tilde{a}= s^{\prime} \biggl( \frac{y\xi_0 +E(m)}{y+1} \biggr)\quad\textrm{and} \quad\tilde{b}= s^{\prime} \biggl( \frac{m- (y \xi_0 + E(m))}{1-y} \biggr), $$

so that $\tilde{a}:=s'(a)$ and $\tilde{b}:=s'(b)$, where a and b were introduced above. Then (A.4) is equivalent to

$$ \tilde{b}=\frac{\varphi_+}{\varphi_-} \tilde{a}\quad\textrm {and} \quad\xi_0 \biggl( 1-\frac{\varphi_-}{\varphi_+} \biggr) \tilde{b} + \log \biggl( \frac{1+e^{\frac{\varphi_-}{\varphi _+}\tilde{b} }}{1+e^{\tilde{b}}} \biggr) =0. $$

There are two solutions to the second equation, $\tilde{b}=0$ and $\tilde{b}=\varphi_{+}$.

If $\tilde{b}=0$ then from a=b=1/2 we get that m=1. As a consequence we obtain that y=(1/2−E(1))/(ξ ₀−1/2). The condition y∈(a ₁,b ₁) implies that E(1)<sup{1−ξ ₀,ξ ₀}, which is in contradiction with the fact that m=1 shall satisfy sup{ξ ₀,m−ξ ₀}≤E(m)≤m−(m−1)ξ ₀.

If $\tilde{b}=\varphi_{+}$ then

$$ a=\frac{y\xi_0 +E(m)}{y+1}=\rho_- \quad\textrm{and} \quad b=\frac{m-(y \xi_0 +E(m))}{1-y} = \rho_+. $$

As a consequence we obtain that $y= \frac{m-\rho_{+} - \rho_{-} }{\rho _{-} -\rho_{+}}$. Then we get that

$$ E(m) - \xi_0 = (\rho_- - \xi_0) (y+1), \qquad\bigl(m- E(m) -\xi_0\bigr) =(\rho_+ -\xi_0) (1-y). $$

Thus the conditions sup{ξ ₀,m−ξ ₀}≤E(m) and y∈(−1,1) imply ρ ₋≥ξ ₀≥ρ ₊. But we assumed ρ ₋<ρ ₊ and we have a contradiction.

Therefore m ₀∈{m ₋,m ₊}. Consequently, for any regular value $E \in[E_{+\infty}^{-}; E_{+\infty}^{+}]$,

$$ S^+ (E) = \frac{1}{2} \sup \bigl\{ f(m_-) , f(m_+) \bigr\} . $$

Recall that the set {m ₋,m ₊} is equal to {α,γ} or to {β,γ} and that f(γ)=0, f(α)=−2s(α/2) and f(β)=−2s(β/2). Thus, by using the results in Table 1, we have

$$ S^+ (E)= \begin{cases} - s (\sup\{\alpha,\beta\}/2), \quad0 <\varphi_- < \varphi_+,\\ - s (\inf\{\alpha,\beta\}/2), \quad\varphi_- < \varphi_+<0,\\ - s (\alpha/2), \quad0 < -\varphi_- \le\varphi_+,\\ - s(\beta/2), \quad0 < \varphi_+ \le-\varphi_-. \end{cases} $$

By definition of α and β we have that

$$ \frac{\alpha}{2}= \frac{1}{\varphi_+} \bigl\{ \bigl(E+{\bar{V}}^+\bigr) + \log \bigl(1+e^{\varphi_+}\bigr) \bigr\} \quad\textrm{and} \quad\frac{\beta }{2} = \frac{1}{\varphi_-} \bigl\{ \bigl(E+{\bar{V}}^+\bigr) + \log \bigl(1+e^{\varphi_-} \bigr) \bigr\} . $$

Recall from (2.5) the definition of W(φ ₋,φ ₊) and let φ=log(ρ/(1−ρ)). Observe that the function x→S _ρ,ρ (−x) is a concave function equal to −∞ outside [i ₁,i ₂]:=[−log(1+e ^−|φ|);−log(1+e ^|φ|)], positive inside, vanishing at the boundaries of the interval, attaining its maximum equal to log(2) for x ₀:=(i ₁+i ₂)/2=φ/2−log(1+e ^φ). It is increasing on [i ₁,x ₀] and decreasing on [x ₀,i ₂].

1.1 A.1 The Case $\frac{1}{2}\le1-\rho_{-} \le\rho_{+}$

Recall that the energy band is given by

$$-{\bar{V}}^+ + \bigl[ -\log\bigl(1+e^{\varphi_+}\bigr) ; W(\varphi_-, \varphi_+)\bigr]. $$

The entropy function S ⁺, in the energy band, is given by ${S}^{+} (E) =S_{\rho_{+}, \rho_{+}} (-(E+{\bar{V}}^{+}))$. Remark that we have $W(\varphi_{-}, \varphi_{+}) + \log(1+e^{\varphi _{-}}) = \varphi_{-} \xi_{0} <0$ since ξ ₀∈(0,1). Thus, $W(\varphi _{-},\varphi_{+}) < -\log(1+e^{-\varphi_{+}})$ and the function S ⁺ is concave, smooth in the interior of the energy band, but does not vanish at the top of the energy band.

1.2 A.2 The Case $\frac{1}{2}\le \rho_{+} \le1-\rho_{-}$

Recall that the energy band is given by

$$-{\bar{V}}^+ + \bigl[ -\log\bigl(1+e^{-\varphi_-}\bigr) ; W(\varphi_-, \varphi_+)\bigr]. $$

The entropy function S ⁺, in the energy band, is given by ${S}^{+} (E) =S_{\rho_{-}, \rho_{-}} (-(E+{\bar{V}}^{+}))$. The function S ⁺ is concave, smooth in the interior of the energy band, but does not vanish at the top of the energy band.

It remains now to prove the last statement of Theorem 2.2. Let us assume that ρ is a maximizer of S ⁺(E), E belonging to the energy band. We use the notations of the proof of Proposition 4.1. In the proof of this proposition, we have seen that ρ being a maximizer of S ⁺(E) is equivalent to the fact that (y _ρ ,H _ρ (1))∈D(E) being a maximizer of the function F over D(E) and ρ is such that H _ρ is linear on [−1,y _ρ ] and on [y _ρ ,1] with H _ρ (y _ρ )=y _ρ ξ ₀+E(m _ρ ). Moreover, we have seen above that such a maximizer (y,m) satisfies y=±1 and

• (i) ,

• (ii) ,

• (iii) m=α 1/2≤1−ρ ₋≤ρ ₊,

• (iv) m=β 1/2≤ρ ₊≤1−ρ ₋.

This implies in particular that if ρ is a maximizer of S ⁺(E) then H _ρ is linear with a slope equal to m _ρ /2, i.e. ρ is constant equal to 1−m _ρ /2. Since, by definition, we have

$$ \alpha= 2 \frac{\log(1+e^{\varphi_+}) + (E+{\bar{V}}^+)}{\varphi _+}\quad\textrm{and} \quad\beta= 2 \frac{\log(1+e^{\varphi _-}) + (E+{\bar{V}}^+)}{\varphi_-}, $$

we get the result.

1.3 A.3 The Case $\rho_{-} < \rho_{+} \le\frac{1}{2}$

Recall that the energy band is given by

$$-{\bar{V}}^+ + \bigl[-\log\bigl(1+e^{-\varphi_-}\bigr) ; -\log \bigl(1+e^{\varphi_+}\bigr)\bigr]. $$

The condition α/2≤β/2 is equivalent to

$$ E +{\bar{V}}^+ \ge W(\varphi_-, \varphi_+). $$

Observe that $W(\varphi_{-}, \varphi_{+}) + \log(1+e^{\varphi_{+}}) = \varphi_{+} \xi_{0}$ and ξ ₀∈(0,1), so that

$$ -\log\bigl(1+e^{-\varphi_-}\bigr) < -\log\bigl(1+e^{-\varphi_+}\bigr) \le W( \varphi_-, \varphi_+) \le-\log\bigl(1+e^{\varphi_+}\bigr) < -\log\bigl(1+ e^{\varphi_-}\bigr). $$

Moreover, we have ξ ₀<1/2 because there exists ${\tilde{\varphi}} \in[\varphi_{-}, \varphi_{+}]$ such that $\xi_{0}= \frac{e^{\tilde{\varphi}}}{1+e^{\tilde{\varphi}}}$. This implies that $W(\varphi_{-}, \varphi_{+}) \ge \frac{\varphi_{+}}{2} - \log(1+e^{\varphi_{+}})$. We recall that $\frac{\varphi_{+}}{2} - \log(1+e^{\varphi_{+}})$ is the first coordinate of the point for which the concave function $x \to S_{\rho_{+},\rho_{+}} (-x)$ attains its maximum given by log(2). It follows that S ⁺ is a concave function. On the energy band it is given by

$$\begin{aligned} {S}^+ (E) =& S_{\rho_-, \rho_-} \bigl(-\bigl(E+{\bar{V}}^+\bigr)\bigr) {\bf1} \bigl\{ \bigl(E+{\bar{V}}^+\bigr) \le W(\varphi_-,\varphi_+)\bigr\} \\ &{}+ S_{\rho_+, \rho_+} \bigl(-\bigl(E+{\bar{V}}^+\bigr)\bigr) {\bf1}\bigl\{ \bigl(E+{\bar{V}}^+\bigr) > W(\varphi_-,\varphi _+)\bigr\} . \end{aligned}$$

The function S ⁺ is not differentiable at the point $W(\varphi _{-},\varphi_{+}) - {\bar{V}}^{+}$.

1.4 A.4 The Case $\frac{1}{2}\le\rho_{-} < \rho_{+}$

Recall that the energy band is given by

$$-{\bar{V}}^+ + \bigl[-\log\bigl(1+e^{\varphi_+}\bigr) ; -\log \bigl(1+e^{-\varphi _-}\bigr) \bigr]. $$

The condition α/2≤β/2 is equivalent to $E +{\bar{V}}^{+} \ge W(\varphi_{-}, \varphi_{+})$. Observe that $W(\varphi_{-}, \varphi_{+}) + \log(1+e^{\varphi_{-}}) = \varphi_{-} \xi_{0}$ and ξ ₀∈(0,1), so that

$$ -\log\bigl(1+e^{\varphi_+}\bigr) \le-\log\bigl(1+e^{\varphi_-}\bigr) \le W(\varphi_-, \varphi_+) \le-\log\bigl(1+e^{-\varphi_-}\bigr) \le-\log \bigl(1+e^{-\varphi_+}\bigr). $$

Moreover, we have ξ ₀>1/2 because there exists ${\tilde{\varphi}} \in[\varphi_{-}, \varphi_{+}]$ such that $\xi_{0}= \frac{e^{\tilde{\varphi}}}{1+e^{\tilde{\varphi}}}$. This implies that $W(\varphi_{-}, \varphi_{+}) \ge\frac{\varphi_{-}}{2} - \log(1+e^{\varphi_{-}})$. We recall that $\frac{\varphi_{-}}{2} - \log(1+e^{\varphi_{-}})$ is the first coordinate of the point for which the concave function $x \to S_{\rho_{-},\rho_{-}} (-x)$ attains its maximum given by log(2). It follows that S ⁺ is a concave function. On the energy band it is given by

$$\begin{aligned} {S}^+ (E) =& S_{\rho_-, \rho_-} \bigl(-\bigl(E+{\bar{V}}^+\bigr)\bigr) {\bf1} \bigl\{ \bigl(E+{\bar{V}}^+\bigr) \ge W(\varphi_-,\varphi_+)\bigr\} \\ &{}+ S_{\rho_+, \rho_+} \bigl(-\bigl(E+{\bar{V}}^+\bigr)\bigr) {\bf1}\bigl\{ \bigl(E+{\bar{V}}^+\bigr) < W(\varphi_-,\varphi_+)\bigr\} . \end{aligned}$$

The function S ⁺ is not differentiable at the point $W(\varphi _{-},\varphi_{+}) - {\bar{V}}^{+}$.

Appendix B: Proof of Theorem 2.3

In order to prove the theorem, we have simply to compute the Legendre transform of S ⁺ whose explicit form is given in Theorem 2.2. Recall also that the Legendre transform of the function S _ρ,ρ defined by (1.4) is given by the function P(φ,⋅) defined by (1.6).

2.1 B.1 The Case $\frac{1}{2}\leq1-\rho_{-}< \rho_{+}$

For any $E \in(E_{+\infty}^{-} ; E_{+\infty}^{+})$ we have that $\frac{d S^{+}}{dE}$ is a decreasing function and

$$\lim_{E \to E_{+\infty}^-} \frac{d S^+}{dE} =+\infty, \qquad \lim _{E \to{E}_{+\infty}^+} \frac{d S^+}{dE}=\theta_0^-. $$

We get that

$$ P^+ (\theta) = \begin{cases} P(\varphi_-, -\theta) -\theta{\bar{V}}^+, \quad\theta\ge\theta_0^-,\\ P^+ (\theta_0) + E_{+\infty}^+ (\theta- \theta_0^-), \quad\theta<\theta_0^-. \end{cases} $$

2.2 B.2 The Case $\frac{1}{2}\leq\rho_{+} < 1-\rho_{-}$

For any $E \in(E_{+\infty}^{-} ; E_{+\infty}^{+})$ we have that $\frac{d S^{+}}{dE}$ is a decreasing function and

$$\lim_{E \to E_{+\infty}^-} \frac{d S^+}{dE} =+\infty, \qquad \lim _{E \to{E}_{+\infty}^+} \frac{d S^+}{dE}=\theta_0^+. $$

We get that

$$ P^+ (\theta) = \begin{cases} P(\varphi_+, -\theta) -\theta{\bar{V}}^+, \quad\theta\ge\theta_0^+,\\ P^+ (\theta_0) + E_{+\infty}^+ (\theta- \theta_0), \quad\theta<\theta_0^+. \end{cases} $$

2.3 B.3 The Case $\rho_{-} < \rho_{+}\leq\frac{1}{2}$

In this case we have that $\theta_{0}^{+} \le \theta_{0}^{-}$ because ξ ₀∈(0,1/2). We get similarly that

$$ P^+ (\theta) = \begin{cases} P(\varphi_-, -\theta) -\theta{\bar{V}}^+, \quad\theta\ge\theta_0^-,\\ P(\varphi_+, -\theta) -\theta{\bar{V}}^+, \quad\theta\le\theta_0^+,\\ P^+ (\theta_0^+) + \frac{P^+(\theta_0^-) - P^+ (\theta_0^+)}{\theta_0^- - \theta_0^+}(\theta-\theta_0^+), \quad\theta\in(\theta_0^+, \theta_0^-). \end{cases} $$

2.4 B.4 The Case $\frac{1}{2} \le\rho_{-} < \rho_{+}$

The function S ⁺ is differentiable everywhere in the interior of the energy band apart from the point $W(\varphi_{-}, \varphi_{+}) -{\bar{V}}^{+}$. For $E=W(\varphi_{-}, \varphi_{+}) -{\bar{V}}^{+}$, S ⁺ has a left-tangent and a right-tangent. Moreover, $\frac{d S^{+}}{dE}$ is decreasing on $(E_{+\infty}^{-} ; W(\varphi_{-},\varphi_{+}) -{\bar{V}}^{+})$ and increasing on $(W(\varphi_{-},\varphi_{+}) -{\bar{V}}^{+} ; E_{+\infty}^{+})$. We have

$$\lim_{E \to E_{+\infty}^\pm} \frac{d S^+}{dE} =\mp\infty\quad \textrm{and} \quad \lim_{E \to[W(\varphi_-, \varphi_+) -{\bar{V}}^+]^\pm} \frac{d S^+}{dE}=\theta_0^\pm. $$

Observe that $\theta_{0}^{-} \le \theta_{0}^{+}$ because ξ ₀∈(1/2,1). We get easily that

$$ P^+ (\theta) = \begin{cases} P(\varphi_+, -\theta) -\theta{\bar{V}}^+, \quad\theta\ge\theta_0^+,\\ P(\varphi_-, -\theta) -\theta{\bar{V}}^+, \quad\theta\le\theta_0^-,\\ P^+ (\theta_0^-) + \frac{P^+(\theta_0^+) - P^+ (\theta_0^-)}{\theta_0^+ - \theta_0^-}(\theta-\theta_0^-), \quad\theta \in(\theta_0^-, \theta_0^+). \end{cases} $$

Appendix C: Proof of Theorem 3.2

In this section we determine the extremal points of the domain Δ according to the position of E along the energy band, we find the supremum of F(⋅,⋅,y ₋,y ₊) among those points and then we maximize over y ₋ and y ₊.

3.1 C.1 The Case $\rho_{-} \leq\frac{1}{2}\leq\rho_{+}$

This case corresponds to φ ₋≤0≤φ ₊, therefore m=(−s)(ρ ₀) and $M=(-s) (1/2)=\log(2)=-E_{-\infty}^{-}-\bar{V}^{-}$. Since $\gamma_{+}'(y_{+})=\varphi_{+}>0$, then the function γ ₊ is increasing. On the other hand, since $\gamma_{-}'(y_{-})=\varphi_{-}<0$, the function γ ₋ is decreasing. Since $\gamma_{+}(\rho_{+}):=\varphi_{+}/(1+e^{-\varphi_{+}})-\log(1+e^{\varphi _{+}})$ and the function t→t/(1+e ^−t)−log(1+e ^t) is increasing in (0,+∞), we obtain that −M≤γ ₊(ρ ₊)<γ ₊(1). On the other hand, since $\gamma_{-}(\rho_{-})=\varphi_{-}/(1+e^{-}\varphi_{-})-\log(1+e^{\varphi _{-}})$ and the function t→t/(1+e ^−t)−log(1+e ^t) is decreasing in (−∞,0) we obtain that −M≤γ ₋(ρ ₋)≤γ ₋(0).

We only consider the case $\frac{1}{2} \leq1-\rho_{-}< \rho_{+}$ (which corresponds to φ ₀=φ ₊), the case $\frac{1}{2} \leq \rho_{+}< 1-\rho_{-}$ (which corresponds to φ ₀=−φ ₋) being similar.

Since, −m=γ ₊(ρ ₊) we have that $-M<-m<\gamma _{+}(1)=E_{-\infty}^{+}+\bar{V}^{-}$. As a consequence, γ ₊(y ₊)≥−m for all y ₊∈[ρ ₊,1]. Notice that the function t→t/(1+e ^−t)−log(1+e ^t) is even and increasing in (0,+∞). Therefore, γ ₋(ρ ₋)≤γ ₊(ρ ₊). On the other hand, γ ₋(0)≤γ ₊(1), which implies that −M≤γ ₋(ρ ₋)≤−m=γ ₊(ρ ₊)≤γ ₊(1). Now, two things can happen, either γ ₋(0)>−m or γ ₋(0)<−m. We start by the former.

(a) γ ₋(0)>−m : Since we do not know the sign of γ ₋−γ ₊ we split again into two cases: γ ₋<γ ₊ and γ ₋>γ ₊. We start by the former.

(a.1) Case γ ₋>γ ₊ : Recall the intersection points of the lines D _m and D _M from Sect. 3.2. In this case X ₀ is in the square [−1,1]² if and only if $\gamma_{+} \le E+{\bar{V}}^{-} \le \gamma_{-}$. We first restrict to the case X ₀∈[−1,1]², i.e. $\gamma_{+} \le E+{\bar{V}}^{-} \le\gamma_{-}$. Now, we check wether X _m , X _M , Y _m and Y _M are in the domain D. We start with X _m , and the same computations holds for X _M . For that purpose, it is enough to notice that X _m satisfies the third equation in (4.10), that is

$$ \frac{(E+{\bar{V}}^-) +m }{\gamma_+ +m} \le\frac{(E +{\bar{V}}^-) +M}{\gamma_+ +M}. $$

Since the function $t\to(E+\bar{V}^{-}+t)/(\gamma_{+}+t)$ is decreasing and m<M we conclude that X _m is not in the domain D. Analogously one shows that X _M is not in D. By replacing γ ₊ by γ ₋ in the computations above, one shows that Y _m and Y _M are in the domain D.

Now, if X ₀∉[−1,1]², i.e. $E+{\bar{V}}^{-}>\gamma_{-}$ then the same computation as done above shows that Y _m and Y _M are not in D and as a consequence D is empty; and if $E+{\bar{V}}^{-}<\gamma_{+}$ then X _m and X _M are not in D and as a consequence D is empty.

So, we are restricted to the case $\gamma_{+} \le E+{\bar{V}}^{-} \le \gamma_{-}$. It remains to compute

$$ \sup_{(y_-,y_+) \in{\varGamma}} \bigl\{ F(X_0,y_-,y_+), F(Y_m,y_-,y_+), F(Y_M,y_-,y_+) \bigr\} , $$

where $\varGamma:=\{(y_{-},y_{+}) : \gamma_{+}\leq{E+\bar{V}^{-}}\leq {\gamma_{-}}\}$. Observe now that whatever the value of y ₋ is, we have that

$$ \begin{aligned} &F(Y_m,y_-,y_+)= \frac{(E+{\bar{V}}^-) +m }{\gamma_- +m} \bigl( (-s) (y_-) + \gamma_- \bigr), \\ &F (Y_M,y_-,y_+)= \frac{(E+{\bar{V}}^-) +M }{\gamma_-- +M} \bigl( (-s) (y_-) + \gamma_- \bigr). \end{aligned} $$

Now we notice that:

$$ \begin{aligned} \bigl(-s(y_-)+\gamma_-\bigr)&=- \biggl(s \biggl( \frac{\gamma_-+\log(1+e^{\varphi _-})}{\varphi_-} \biggr)-\gamma_-\biggr) \\ &=- \biggl(s \biggl(\frac{-\gamma_--\log(1+e^{-\varphi_-})}{\varphi _-} \biggr)-\gamma_- \biggr) \\ &=- \bigl(-S_{1-\rho_-,1-\rho_-}(-\gamma_-)-\gamma_- \bigr) \\ &=-J_{\rho_-,\rho_-}(-\gamma_-). \end{aligned} $$

In the second equality above, we wrote y ₋ in terms of γ ₋, in the third equality we used the fact that s(θ)=s(1−θ) for all θ∈(0,1) and in fourth equality we used that for any ρ∈[0,1],

$$ S_{\rho,\rho} =S_{1-\rho,1-\rho} $$

(C.1)

together with the definition of J _ρ,ρ given above. The same argument also shows that $(-s(y_{+})+\gamma_{+})=-J_{\rho_{+},\rho _{+}}(-\gamma_{+})$. Then we conclude that

$$ \begin{aligned} &F(Y_m,y_-,y_+)= - \frac{(E+{\bar{V}}^-) +m }{\gamma_- +m} J_{\rho _-,\rho_-} (-\gamma_-) \\ &F (Y_M,y_-,y_+)= - \frac{(E+{\bar{V}}^-) +M }{\gamma_- +M} J_{\rho _-,\rho_-} (- \gamma_-). \end{aligned} $$

Observe that since the function $t\to- ((E+{\bar{V}}^{-}) +t ) / (\gamma _{-} +t)$ is decreasing, m≤M and $J_{\rho_{-},\rho_{-}} (-\gamma _{-}) \ge0$, we get that F(Y _m ,y ₋,y ₊)≥F(Y _M ,y ₋,y ₊).

Now we recall some properties of the function J _ρ,ρ , for ρ∈[0,1]. At first we notice that by (C.1), we have that J _ρ,ρ =J _{1−ρ,1−ρ} . The function J _ρ,ρ is convex and positive apart from the point −s(ρ) where it vanishes. As a consequence the function $J_{\rho_{-},\rho _{-}}$ is convex, non-negative and finite on $[\log(1+e^{\varphi_{-}}) ;\log(1+e^{-\varphi_{-}})]$ and has a minimum equal to 0 at the point −s(ρ ₋)=γ ₋(ρ ₋). Analogously, the function $J_{\rho_{+},\rho_{+}}$ is convex, non-negative and finite on $[\log (1+e^{-\varphi_{+}});\log(1+e^{\varphi_{+}})]$ and has a minimum equal to 0 at the point −s(ρ ₊)=γ ₊(ρ ₊). Since −s(ρ ₋)≤m=−s(ρ ₊) then for all $y\in{I}:=[\log (1+e^{-\varphi_{+}});m]$ we have that $J_{\rho_{-},\rho_{-}}(y) \geq {J_{\rho_{+},\rho_{+}}(y)}$. In particular, since γ ₋∈I we obtain that $J_{\rho_{-},\rho_{-}}(-\gamma_{-}) \geq{J_{\rho_{+},\rho_{+}}(-\gamma_{-})}$. Putting together the previous observations, the fact that $E+{\bar{V}}^{-}\geq{\gamma_{+}}\geq{m}$ and γ ₋>−m, we conclude that

$$ F(Y_m,y_-,y_+)= - \frac{(E+{\bar{V}}^-) +m }{\gamma_- +m} J_{\rho _-,\rho_-} (-\gamma_-) \leq{- \frac{(E+{\bar{V}}^-) +m }{\gamma_- +m} J_{\rho_+,\rho_+} (-\gamma_-)}. $$

On the other hand, by computing the derivative of the function

$$G(\gamma_-)=- \frac{(E+{\bar{V}}^-) +m }{\gamma_- +m} J_{\rho _+,\rho_+} (-\gamma_-) $$

with respect to γ ₋ and using the fact that the function $J_{\rho_{+},\rho_{+}} (-\gamma_{-})$ is convex at the point γ ₋, we conclude that G(⋅) is decreasing. Then,

$$ \begin{aligned} \sup_{(y_-,y_+) \in{\varGamma}}F(Y_m,y_-,y_+) & \leq\sup_{ \gamma _-\geq E+\bar{V}^-}G(\gamma_-) \\ &=G\bigl(E+{\bar{V}}^-\bigr) \\ &=-J_{\rho_+,\rho_+} \bigl(-\bigl(E+{\bar{V}}^-\bigr) \bigr). \end{aligned} $$

Now we rewrite F(X ₀,y ₋,y ₊) as

$$ F (X_0, y_-, y_+ )=-\frac{(E+{\bar{V}}^-) -\gamma_+ }{\gamma_--\gamma_+} J_{\rho_-,\rho_-} (- \gamma_-)-\frac{\gamma _- -(E+\bar{V}^-)}{\gamma_--\gamma_+} J_{\rho_+,\rho_+} (-\gamma_+). $$

By computing the derivative with respect to γ ₊ of F(X ₀,y ₋,y ₊), and noticing that both $J_{\rho_{+},\rho_{+}}$ and $J_{\rho_{-},\rho_{-}}$ are convex, we obtain that F(X ₀,y ₋,y ₊) is increasing as a function of γ ₊. Then

$$\begin{aligned} &\sup \bigl\{ F(X_0,y_-,y_+): (y_-,y_+) \in{\varGamma} \bigr\} \\ &\quad{}\leq \sup \bigl\{ F(X_0,y_-,y_+): \gamma_-\geq E+\bar{V}^- \bigr\} \\ &\quad{}=-J_{\rho_+,\rho_+} \bigl(-\bigl(E+{\bar{V}}^-\bigr) \bigr). \end{aligned}$$

Putting together the previous computations we obtain that

$$\begin{aligned} &\sup_{(y_-,y_+) \in{\varGamma}} \bigl\{ F(X_0,y_-,y_+), F(Y_m,y_-,y_+), F(Y_M,y_-,y_+) \bigr\} \\ &\quad{}=\sup_{ \gamma_+ \le E+{\bar{V}}^- \le \gamma_-}F(X_0,y_-,y_+) \\ &\quad{}=\sup_{\gamma_+=E+{\bar{V}}^-} F(X_0,y_-,y_+) \\ &\quad{}=-J_{\rho_+,\rho_+} \bigl(-\bigl(E+{\bar{V}}^-\bigr) \bigr). \end{aligned}$$

(a.2) Case γ ₋<γ ₊ : A simple computation shows that X ₀ belongs to the square [−1,1]² if and only if $\gamma_{-} \le E+{\bar{V}}^{-} \le \gamma_{+}$. As above, we first restrict to $\gamma_{-} \le E+{\bar{V}}^{-} \le \gamma _{+}$. A simple computation as performed above, shows that X _m and X _M are in D and Y _m and Y _M are not in D.

Now, if X ₀∉[−1,1]², i.e. $E+{\bar{V}}^{-}>\gamma_{+}$ then the same computation as done above shows that X _m and X _M are not in D and as a consequence D is empty; and if $E+{\bar{V}}^{-}<\gamma_{-}$ then Y _m and Y _M are not in D and as a consequence D is empty.

So, we are restricted to the case $\gamma_{-} \le E+{\bar{V}}^{-} \le \gamma_{+}$. Then, it remains to compute

$$\sup_{(y_-,y_+) \in{\varGamma}} \bigl\{ F(X_0,y_-,y_+), F(X_m,y_-,y_+), F(X_M,y_-,y_+) \bigr\} , $$

where $\varGamma:=\{(y_{-},y_{+}): \gamma_{-}\leq{E+\bar{V}^{-}}\leq {\gamma_{+}}\}$. By inverting the role of ρ ₋ with ρ ₊ and of γ ₋ with γ ₊ in the proof of the previous case, we obtain here that the previous supremum equals to $-J_{\rho_{+},\rho_{+}} (-(E+{\bar{V}}^{-}) )$.

(b) γ ₋(0)<−m : In this case we have that γ ₋<−m<γ ₊. As above, we have to check whether the points X ₀,X _m ,X _M ,Y _m ,Y _M are in the domain D or not.

The point X ₀ belongs to [−1,1]² if and only if $\gamma_{-} \le E+{\bar{V}}^{-} \le\gamma_{+}$. If X ₀∉[−1,1]², i.e. $E+{\bar{V}}^{-}>\gamma_{+}$, then X _m and X _M are not in the domain D and as a consequence D is empty.

Then we restrict to $\gamma_{-} \le E+{\bar{V}}^{-} \le\gamma_{+}$. A simple computation shows that X _m and X _M belong to the domain D, but Y _M ,Y _m are not in D. So we have to compute

$$\sup_{(y_-,y_+) \in{\varGamma}} \bigl\{ F(X_0,y_-,y_+), F(X_m,y_-,y_+), F(X_M,y_-,y_+) \bigr\} , $$

where $\varGamma:=\{(y_{-},y_{+}): \gamma_{-}\leq{E+\bar{V}^{-}}\leq {\gamma_{+}}\}$.

As above easily we can show that F(X _m ,y ₋,y ₊)≥F(X _M ,y ₋,y ₊). Now we have to compare F(X _m ,y ₋,y ₊) with F(X ₀,y ₋,y ₊). A simple computation shows that F(X ₀,y ₋,y ₊) can be written as

$$ F (X_0, y_-, y_+ )=-\frac{\gamma_+-(E+{\bar{V}}^-) }{\gamma _+-\gamma_-} J_{\rho_-,\rho_-} (- \gamma_-) -\frac{\gamma_- -(E+\bar{V}^-)}{\gamma_--\gamma_+} J_{\rho _+,\rho_+} (-\gamma_+). $$

Since $J_{\rho_{-},\rho_{-}}$ is a positive function, $E+{\bar{V}}^{-} \le \gamma_{-}$ and γ ₊>γ ₋ we have that

$$ F (X_0, y_-, y_+ )\leq-\frac{\gamma_- -(E+\bar {V}^-)}{\gamma_--\gamma_+} J_{\rho_+,\rho_+} (- \gamma_+). $$

Now, since the function $t\to-(E+{\bar{V}}^{-}+t)/(\gamma_{+}+t)$ is decreasing and m>−γ ₋ we obtain that F(X ₀,y ₋,y ₊)≤F(X _m ,y ₋,y ₊).

It follows that

$$\begin{aligned} &\sup_{(y_-,y_+) \in{\varGamma}} \bigl\{ F(X_0,y_-,y_+), F(X_m,y_-,y_+), F(X_M,y_-,y_+) \bigr\} \\ &\quad{}=\sup_{(y_-,y_+) \in{\varGamma}} F(X_m,y_-,y_+) \\ &\quad{}= \sup_{\gamma _+ \geq E+ {\bar{V}}^-}F(X_m,y_-,y_+) \\ &\quad{}= -J_{\rho_+,\rho_+} \bigl(-\bigl(E+{\bar{V}}^-\bigr) \bigr). \end{aligned}$$

By the conclusions above together with (4.9), we obtain that the restriction of the entropy function S ⁻ to $[E_{-\infty}^{-} ; E_{-\infty}^{+}]$ is given by

$$ \begin{aligned} &S^-(E) \\ &\quad{}= \begin{cases} -(E+\bar{V}^-),\qquad -\log(2)\leq{E+\bar{V}^-}\leq{\frac{\varphi _0}{1+e^{-\varphi_0}}-\log(1+e^{\varphi_0})},\\ S_{\rho_0,\rho_0} (-(E+\bar{V}^-) ),\qquad \frac{\varphi _0}{1+e^{-\varphi_0}}-\log(1+e^{\varphi_0})<E+\bar{V}^-\leq{- \log (1+e^{-\varphi_0})}. \end{cases} \end{aligned} $$

Above we used the equality $S_{\rho_{+},\rho_{+}}(E)=E-J_{\rho_{+},\rho _{+}}(E)$. To conclude, we notice that for any ρ=e ^φ/(1+e ^φ)∈[0,1],

$$ s(\rho)=\frac{\varphi}{1+e^{-\varphi}}-\log\bigl(1+e^{\varphi}\bigr)= \frac {-\varphi}{1+e^{\varphi}}-\log\bigl(1+e^{-\varphi}\bigr). $$

Now, we prove the last assertion of the theorem. As above, we consider the case φ ₀=φ ₊ the other case being similar. We have to split now into two cases, whether $E+\bar{V}^{-}> s(\rho_{+})$ or $E+\bar{V}^{-}\leq s(\rho_{+})$. We start by the later.

Assume $E+\bar{V}^{-}\leq s(\rho_{+})$ and let ρ be a profile such that ${{\mathbb{S}}} (\rho) = S^{-} (E)$ and ${{\mathbb{S}}} (\rho) +V^{-} (\rho) =E+{\bar{V}}^{-}$. With the notations of Proposition 4.2, we have ${S}^{-} (E)= {{\mathbb{S}}} (\rho) = {{\mathbb{S}}} (H^{\prime}_{\rho}) \le{{\mathbb{S}}} (G^{\prime}_{\rho})$. Moreover, we have seen in the proof of Proposition 3.1 that $({{\mathbb{S}}} +V^{-}) (G^{\prime}_{\rho}) = ({{\mathbb{S}}} +V^{-}) (\rho)$. We claim now that $\rho={G}^{\prime}_{\rho}$. Indeed, let (a,b)∈[−1,1] be a maximal interval where G _ρ <H _ρ (which implies H _ρ (a)=G _ρ (a) and H _ρ (b)=G _ρ (b)). Since G _ρ is the convex envelope of H _ρ , it implies that G _ρ is linear on [a,b]. By Jensen’s inequality one has that

$$ \begin{aligned} \int_a^b (-s) \bigl(H^{\prime}_{\rho} (x)\bigr) dx &> (b-a) (-s) \biggl( \frac {1}{b-a} \int_a^b H_{\rho}^{\prime} (x) dx \biggr) \\ &= (b-a) (-s) \biggl( \frac{H_{\rho} (b) -H_{\rho} (a) }{b-a} \biggr) \\ &= (b-a) (-s) \biggl( \frac{G_{\rho} (b) -G_{\rho} (a) }{b-a} \biggr) \\ &= \int_a^b (-s) \bigl(G^{\prime}_{\rho} (x)\bigr) dx. \end{aligned} $$

Thus, if H _ρ ≠G _ρ we can find a profile ${\tilde{\rho}}$ (i.e. $1-G^{\prime}_{\rho}$) which satisfies the constraint ${{\mathbb{S}}} ({\tilde{\rho}}) +V^{-} ({\tilde{\rho}}) =E+{\bar{V}}^{-}$ and such that ${{\mathbb{S}}} ({\tilde{\rho}}) > {{\mathbb{S}}} (\rho )=S^{-} (E)$. Since this is not possible we get that H _ρ =G _ρ , i.e. ρ is a non-increasing profile. In particular it implies that 1−ρ=g where g is a maximizer of (4.6). In the proof of Proposition 4.2 we have seen that $S^{-} (E)= -(E+{\bar{V}}^{-})$ corresponds to the case where the supremum sup _k∈K F(k)=0, which is equivalent to (x _±∓1)(y _±−ρ _±)=0. For such a 4-tuple (x ₋,x ₊,y ₋,y ₊)∈K, a maximizer g of (4.6) is then given by any non-decreasing function on [x ₋,x ₊] taking values in [ρ ₋,ρ ₊], constant equal to ρ ₋ on [−1,x ₋] and to ρ ₊ on [x ₊,1] and such that (4.7) is satisfied. Thus the set of maximizers ρ of S ⁻(E), when $E +{\bar{V}}^{-} \in[s (\rho_{+}) , s (\rho_{-})]$ is given by the set of non-increasing profiles ρ such that 1−ρ ₊≤ρ≤1−ρ ₋ and satisfying ${{\mathbb{S}}} (\rho) =-(E+{\bar{V}}^{-})$.

Now assume that $E+\bar{V}^{-}> s(\rho_{+})$. In all the cases (a ₁), (a ₂) and (b) above, the supremum of F is attained at X ₀,X ₀ and X _m , respectively, and $\gamma_{+}=E+\bar{V}^{-}$, which implies that x ₋=x ₊=−1. Then, the function g realizing the supremum $\sup_{\rho}\int_{-1}^{1}(-s)(g(x))dx$ with the constraint $\int_{-1}^{1}g(x)dx=(1-x_{+})y_{+}$ is constant and equals to y ₊. Therefore, the profile $u_{\rho_{+}}$ is given by 1−y ₊. Using the definition of y ₊ and the fact that $\gamma_{+}=E+\bar{V}^{-}$, it follows that

$$u_{\rho_+}\equiv\frac{\log(\rho_+)-E+\bar{V}^-}{\log(\rho_+)-\log (1-\rho_+)}. $$

Finally, putting together (1.7), (1.8), (1.4) and the expression for $u_{\rho_{+}}$ we recover the expression for S ⁻(E) that is $S^{-} (E)= S_{\rho_{+},\rho _{+}}(-(E+\bar{V}^{-}))$.

3.2 C.2 The Case $\rho_{-} < \rho_{+} \leq\frac{1}{2}$

This case corresponds to φ ₋<φ ₊≤0, therefore m=(−s)(ρ ₋) and M=(−s)(ρ ₊). Since γ′(y _±)=φ _±<0, then both functions γ _± are decreasing. Notice that $\gamma_{+}(1)=E_{-\infty}^{-}+\bar{V}^{-}$ and $\gamma _{-}(0)=E_{-\infty}^{+}+\bar{V}^{-}$. Then $-M=\gamma_{+}(\rho_{+})\geq{\gamma_{+}(1)}=-\log(1+e^{-\varphi _{+}})$. Analogously, $-m=\gamma_{-}(\rho_{-})\leq{\gamma_{-}(0)}=-\log(1+e^{\varphi_{-}})$. As a consequence we have the following inequalities:

$$ -\log\bigl(1+e^{-\varphi_+}\bigr)=\gamma_+(1)\leq \gamma_+(\rho_+)= -M < -m \le\gamma_-(\rho_-)\leq{\gamma_-(0)} =-\log\bigl(1 +e^{\varphi_-}\bigr). $$

These inequalities imply that γ ₊<γ ₋. As in the previous case we have to check whether the intersection points are in the domain D. At first we notice that X ₀ belongs to [−1,1]² if and only if $\gamma_{+} \le E+{\bar{V}}^{-} \le\gamma_{-}$. Since $-(E+{\bar{V}}^{-}) \notin[m,M]$ we have only to distinguish two cases$: \gamma_{+} \le E+{\bar{V}}^{-} < -M$ and $-m\le E+{\bar{V}}^{-} < \gamma_{-}$.

If X ₀ is not in [−1,1]², i.e$. E+{\bar{V}}^{-}> \gamma_{-}$ then X _m and X _M are not in D and as a consequence D is empty; and if $E+{\bar{V}}^{-}< \gamma_{+}$ then Y _m and Y _M are not in D and as a consequence D is empty. So we restrict to $\gamma_{+} \le E+{\bar{V}}^{-} < -M$ and $-m\le E+{\bar{V}}^{-} < \gamma_{-}$. We start by the former.

(a) $\gamma_{+} \le E+{\bar{V}}^{-} < -M$ : In this case, a simple computation shows that X _m and X _M are in D, but Y _m and Y _M are not in D. Then we have to compute

$$\sup_{ (y_-,y_+) \in{\varGamma}} \bigl\{ F(X_0,y_-,y_+), F(X_m,y_-,y_+), F(X_M,y_-,y_+) \bigr\} , $$

where $\varGamma:=\{(y_{-},y_{+}): \gamma_{+} \le E+{\bar{V}}^{-} < -M\}$.

As above, by noticing that the function $t\to- ((E+{\bar{V}}^{-}) +t)/ (\gamma_{+} +t)$ is increasing, m≤M and $J_{\rho_{+},\rho_{+}} (-\gamma_{+}) \ge0$, we get that F(X _m ,y ₋,y ₊)≤F(X _M ,y ₋,y ₊). On the other hand

$$ F (X_0, y_-, y_+ )= -\frac{(E+{\bar{V}}^-) -\gamma_+ }{\gamma_--\gamma_+} J_{\rho_-,\rho_-} (- \gamma_-)-\frac{\gamma _- -(E+\bar{V}^-)}{\gamma_--\gamma_+} J_{\rho_+,\rho_+} (-\gamma_+). $$

Since the function $J_{\rho_{-},\rho_{-}}$ is positive, γ ₋>γ ₊ and $E+\bar{V}^{-}>\gamma_{+}$, we have that

$$ F (X_0, y_-, y_+ )\leq-\frac{\gamma_- -(E+\bar {V}^-)}{\gamma_--\gamma_+} J_{\rho_+,\rho_+} (- \gamma_+)= -\frac{(E+\bar{V}^-)-\gamma_- }{\gamma_+-\gamma_-} J_{\rho _+,\rho_+} (-\gamma_+). $$

Finally, since the function $t\to - ((E+{\bar{V}}^{-}) +t) / (\gamma_{+} +t)$ is increasing, −γ ₋<M and $J_{\rho_{+},\rho_{+}} (-\gamma _{+})\geq{0}$ we obtain that F(X ₀,y ₋,y ₊)≤F(X _M ,y ₋,y ₊). Since the function $J_{\rho_{+},\rho_{+}}$ is convex, with a unique minimum at M equal to 0, it follows that

$$\begin{aligned} &\sup_{(y_-,y_+) \in{\varGamma}} \bigl\{ F(X_0,y_-,y_+), F(X_m,y_-,y_+), F(X_M,y_-,y_+) \bigr\} \\ &\quad{}=\sup_{(y_-,y_+) \in{\varGamma }} F(X_M,y_-,y_+) \\ &\quad{}=\sup_{\gamma_+ \le(E+ {\bar{V}}^-)}F(X_M,y_-,y_+) \\ &\quad{}=-J_{\rho_+,\rho_+} \bigl(-\bigl(E+{\bar{V}}^-\bigr) \bigr). \end{aligned}$$

(b) $-m < E+{\bar{V}}^{-} \le\gamma_{-}$ : In this case, a simple computation shows that Y _m and Y _M are in D, but X _m and X _M are not in D. Then we have to compute

$$\sup_{(y_-,y_+) \in{\varGamma}} \bigl\{ F(X_0,y_-,y_+), F(Y_m,y_-,y_+), F(Y_M,y_-,y_+) \bigr\} , $$

where $\varGamma:=\{(y_{-},y_{+}): -m < E+{\bar{V}}^{-} \le\gamma_{-}\}$.

As above, by noticing that the function $t\to-((E+{\bar{V}}^{-}) +t) / (\gamma_{-} +t)$ is increasing, m≤M and $J_{\rho_{-},\rho_{-}} (-\gamma_{-}) \ge0$, we get F(Y _m ,y ₋,y ₊)≤F(Y _M ,y ₋,y ₊).

As above we can show that F(X ₀,y ₋,y ₊)≤F(Y _M ,y ₋,y ₊) and as a consequence

$$\begin{aligned} &\sup_{(y_-,y_+) \in{\varGamma}} \bigl\{ F(X_0,y_-,y_+), F(Y_m,y_-,y_+), F(Y_M,y_-,y_+) \bigr\} \\ &\quad{}=\sup_{(y_-,y_+) \in{\varGamma }} F(Y_M,y_-,y_+) \\ &\quad{}=\sup_{\gamma_- \le(E+ {\bar{V}}^-)}F(Y_M,y_-,y_+) \\ &\quad{}=-J_{\rho_-,\rho_-} \bigl(-\bigl(E+{\bar{V}}^-\bigr) \bigr). \end{aligned}$$

Collecting the previous facts and by (4.9), we have that the restriction of the entropy function S ⁻ to $[E_{-\infty}^{-} ; E_{-\infty}^{+}]$ is given by

$$ \begin{aligned} &S^-(E) \\ &\quad{}= \begin{cases} S_{\rho_+,\rho_+} (-(E+{\bar{V}}^-) ),\qquad -\log(1+e^{-\varphi _+})\leq{E+\bar{V}^-} <\frac{-\varphi_+}{1+e^{\varphi_+}}-\log(1+e^{-\varphi_+}),\\ -(E+\bar{V}^-),\qquad \frac{-\varphi_+}{1+e^{\varphi_+}}-\log (1+e^{-\varphi_+})\leq{E+\bar{V}^-}\leq{\frac{-\varphi _-}{1+e^{\varphi_-}}-\log(1+e^{-\varphi_-})},\\ S_{\rho_-,\rho_-} (-(E+{\bar{V}}^-) ),\qquad \frac{-\varphi _-}{1+e^{\varphi_-}}-\log(1+e^{-\varphi_-})<E+\bar{V}^-\leq{-\log (1+e^{\varphi_-})} \end{cases} \end{aligned} $$

Now, we prove the last assertion of the theorem. As above, we have to split into several cases, whether $E+\bar{V}^{-}< s(\rho_{+})$, $s(\rho _{+})\leq E+\bar{V}^{-}\leq s(\rho_{-})$ or $E+\bar{V}^{-}>s(\rho_{-})$. We start by the first case, the third being completely similar. Analogously to what we have done for ρ ₋≤1/2≤ρ ₊, it is enough to notice that in the cases (a) and (b) above, the supremum is attained at X _M and Y _M , respectively, with $\gamma_{+}=E+\bar{V}^{-}$, which implies that x ₋=x ₊=−1. The rest of the argument follows as above. The second case, follows by reasoning as in the case $\rho_{-} \leq \frac{1}{2}\leq\rho_{+}$.

3.3 C.3 The Case $\frac{1}{2}\leq\rho_{-} < \rho_{+}$

Repeating the same computations as performed in the previous situation, we can show that the restriction of the entropy function S ⁻ to $[E_{-\infty}^{-} ; E_{-\infty}^{+}]$ is given by

$$ \begin{aligned} &S^-(E) \\ &\quad{}= \begin{cases} S_{\rho_-,\rho_-} (-(E+{\bar{V}}^-) ),\qquad -\log (1+e^{\varphi_-})\leq{E+\bar{V}^-}<\frac{-\varphi_-}{1+e^{\varphi _-}}-\log(1+e^{-\varphi_-}),\\ -(E+\bar{V}^-),\qquad \frac{-\varphi_-}{1+e^{\varphi_-}}-\log (1+e^{-\varphi_-})\leq{E+\bar{V}^-}\leq{\frac{-\varphi _+}{1+e^{\varphi_+}}-\log(1+e^{-\varphi_+})}\\ S_{\rho_+,\rho_+} (-(E+{\bar{V}}^-) ),\qquad \frac{-\varphi _+}{1+e^{\varphi_+}}-\log(1+e^{-\varphi_+}) <E+\bar{V}^-\leq{-\log(1+e^{-\varphi_+})}. \end{cases} \end{aligned} $$

We notice that to prove the last assertion of the theorem is enough to invert the role of ρ ₋ with ρ ₊ in the proof of the previous case.

Appendix D: Proof of Theorem 3.3

To prove this theorem we compute explicitly the Legendre transform of P ⁻, namely, ${\tilde{S}}^{-} (E)=\inf_{\theta\in\mathbb{R}} \{ \theta E -P^{-} (\theta) \}$ and we show that it coincides with the expression for S ⁻(E) obtained in the previous section. Since the Legendre transform is a one to one correspondence between concave functions, this is sufficient to conclude. We denote by P ₀ the function defined by P ₀(θ)=−θlog(m(θ)).

4.1 D.1 The Case $\rho_{-} \leq\frac{1}{2} \leq \rho_{+}$

This case corresponds to φ ₋≤0≤φ ₊. Recall that φ ₀=sup(|φ ₋|,|φ ₊|). Since

$$ m(\theta)=\min\bigl(f_{\theta} (\varphi_{-}), f_{\theta} ({\varphi_+})\bigr)= \begin{cases} f_{\theta} (\varphi_0), \quad\theta<-1,\\ 1, \quad\theta=-1,\\ f_{\theta} (0), \quad\theta>-1, \end{cases} $$

we have that

$$ P _0 (\theta) = \begin{cases} -\theta\log(1+e^{-{\varphi}_0}) - \log(1+e^{\theta{\varphi}_0}), \quad\theta<-1,\\ 0, \quad\theta=-1,\\ - (\theta+1)\log2 , \quad\theta>-1. \end{cases} $$

As a consequence

$$ P' _0 (\theta) = \begin{cases} -\log(1+e^{-{\varphi}_0}) - \frac{\varphi_0}{1+e^{-\theta\varphi _0}}, \quad\theta<-1,\\ -\log(2) , \quad\theta>-1. \end{cases} $$

The function P ₀ is differentiable everywhere except for θ=−1. A simple computation show that $P_{0}'$ is decreasing in (−∞,−1). We have that

$$ \begin{aligned} &P_0'\bigl(-1^-\bigr)=\lim _{\substack{\theta\to-1\\ \theta<-1}}{P}_0^{\prime } (\theta)=-\log \bigl(1+e^{-{\varphi}_0}\bigr)-\frac{\varphi _0}{1+e^{\varphi_0}}, \\ &P_0'(-\infty)=\lim_{\theta\to-\infty} { P}_0^{\prime} (\theta )=- \log\bigl(1+e^{-\varphi_0}\bigr). \end{aligned} $$

Observe that $P_{0}'(-\infty)>P_{0}'(-1^{-})>- \log(2)$, which is a consequence of the function t→−log(1+e ^−t)−t/(1+e ^t) being decreasing on (−∞,0]. This implies that the function P ₀ is concave. Now we compute ${\tilde{S}}^{-}(E)$. By the previous observations we have that

$$ {\tilde{S}}^-(E)=\inf \Bigl\{ \inf_{\theta>-1}\bigl\{ \theta\bigl(E+ \bar {V}^-\bigr)-P_0(\theta)\bigr\} ,-\bigl(E+\bar{V}^-\bigr),\inf _{\theta<-1}\bigl\{ \theta \bigl(E+\bar{V}^-\bigr)-P_0( \theta)\bigr\} \Bigr\} . $$

Now, for θ>−1 we have that

$$ \inf_{\theta>-1} \bigl\{ \theta\bigl(E+\bar{V}^- \bigr)-P_0(\theta)\bigr\} =\log (2)+\inf_{\theta>-1}\bigl\{ \theta\bigl(\bigl(E+\bar{V}^-\bigr)+\log(2)\bigr) \bigr\} . $$

This equals to $-(E+\bar{V}^{-})$ if $(E+\bar{V}^{-})>-\log(2)$ and equals −∞ if $(E+\bar{V}^{-})<-\log(2)$. On the other hand, for θ<−1 we have that

$$ \inf_{\theta<-1} \bigl\{ \theta\bigl(E+\bar{V}^- \bigr)-P_0(\theta)\bigr\} =\inf_{\theta<-1}\bigl\{ \theta \bigl(E+\bar{V}^-\bigr)-P(-\varphi_0,-\theta) \bigr\} , $$

where P(⋅,⋅) is defined in (1.6). Let $I_{1}:= [\frac{-\varphi_{0}}{1+e^{\varphi_{0}}}-\log(1+e^{-\varphi_{0}}) ; - \log(1+e^{-\varphi_{0}}) ] \subset(-\infty,0)$. We have that

$$\begin{aligned} &\inf_{\theta\in\mathbb{R}} \bigl\{ \theta\bigl(E+\bar{V}^-\bigr) -P(-\varphi _0, -\theta) \bigr\} \\ &\quad{}=\frac{1}{2}\inf_{\theta\in\mathbb{R}} \bigl\{ \theta2\bigl(E+ \bar{V}^-\bigr) -2P(-\varphi_0, -\theta) \bigr\} \\ &\quad{}=\frac{1}{2}\inf_{\theta\in\mathbb{R}} \bigl\{ \theta\bigl(-2 \bigl(E+\bar {V}^-\bigr)\bigr) -2P(-\varphi_0, \theta) \bigr\} \\ &\quad{}=-s \biggl(\frac{-(E+\bar {V}^-) -\log(1+e^{-\varphi_0})}{\varphi_0} \biggr) \\ &\quad{}=S_{\rho_0,\rho_0} \bigl(-\bigl(E+\bar{V}^-\bigr) \bigr), \end{aligned}$$

for any $E+\bar{V}^{-}\in I_{1}$. Then, we conclude that

$$ \inf_{\theta>-1} \bigl\{ \theta\bigl(E+\bar{V}^- \bigr)-P_0(\theta) \bigr\} =S_{\rho_0,\rho_0} \bigl(-\bigl(E+ \bar{V}^-\bigr) \bigr), $$

for $E+\bar{V}^{-}\in I_{1}$. Now we look for ${\tilde{S}}^{-}(E)$ for $E+\bar{V}^{-}$ outside I ₁. Let $I_{1}^{-}:= (-\infty ;\frac{-\varphi_{0}}{1+e^{\varphi_{0}}}-\log (1+e^{-\varphi_{0}}) )$ and $I_{1}^{+}:= (- \log(1+e^{-\varphi _{0}}) ;+\infty )$. Then, if $E+\bar{V}^{-}\in{I_{1}^{-}}$, we have that $\operatorname{sign}(E+\bar{V}^{-}-P_{0}'(\theta))=\operatorname{sign}((E+\bar{V}^{-})-P_{0}'(-\infty ))=\operatorname{sign}((E+\bar{V}^{-})+\log(1+e^{-\varphi_{0}}))<0$, and as a consequence the infimum is attained at θ=−1 and in this case ${\tilde{S}}^{-}(E)= -(E+\bar{V}^{-})$. On the other hand if $E+\bar {V}^{-}\in{I_{1}^{+}}$, we have that $\operatorname{sign}(E+\bar{V}^{-}-P_{0}'(\theta ))=\operatorname{sign}((E+\bar{V}^{-})-P_{0}'(-\infty))>0$, and as a consequence the infimum is attained at θ=−∞ and in this case ${\tilde{S}}^{-}(E)=-\infty$.

Finally we conclude that ${\tilde{S}}^{-}$ when restricted to the energy band $[E_{-\infty}^{-} ; E_{-\infty}^{+}]$ is given by

$$ \begin{aligned} &{\tilde{S}}^-(E) \\ &\quad{}= \begin{cases} -(E+\bar{V}^-), \qquad-\log(2)\leq{E+\bar{V}^-}\leq{\frac {-\varphi_0}{1+e^{\varphi_0}}-\log(1+e^{-\varphi_0})},\\ S_{\rho_0,\rho_0} (-(E+\bar{V}^-) ), \qquad\frac{-\varphi _0}{1+e^{\varphi_0}}- \log(1+e^{-\varphi_0})<E+\bar{V}^-\leq{- \log(1+e^{-\varphi_0})} \end{cases} \end{aligned} $$

and equal to −∞ outside the energy band. Thus ${\tilde{S}}^{-}$ coincides with S ⁻.

4.2 D.2 The Case $\rho_{-}<\rho_{+}\leq\frac{1}{2}$

In this case we have that

$$ m(\theta)=\min\bigl(f_{\theta} (\varphi_{-}), f_{\theta} ({\varphi_+})\bigr)= \begin{cases} f_{\theta} (\varphi_-), \quad\theta<-1,\\ 1, \quad\theta=-1,\\ f_{\theta} (\varphi_+), \quad\theta>-1, \end{cases} $$

and as a consequence

$$ { P}_0'(\theta)= \begin{cases} -\frac{\varphi_-}{1+e^{-\theta\varphi_-}} -\log(1+e^{-\varphi _-}),\quad\theta<-1,\\ -\frac{\varphi_+}{1+e^{-\theta\varphi_+}}-\log(1+e^{-\varphi _+}), \quad\theta>-1. \end{cases} $$

The function P ₀ is differentiable everywhere except for θ=−1. We have that ${P}_{0}^{\prime}$ is decreasing on (−∞,−1) and

$$ P_0'(-\infty)>P_0' \bigl(-1^-\bigr)=\lim_{\substack{\theta\to-1\\ \theta <-1}} P_0^{\prime} ( \theta) > \lim_{\substack{\theta\to-1\\ \theta>-1}} P_0^{\prime}, ( \theta)=P_0'\bigl(-1^+\bigr)>P_0'(+ \infty) $$

which is a consequence of the function t→log(1+e ^−t)+t/(1+e ^t) being increasing on (−∞,0). This implies that the function P ₀ is concave. Now we compute ${{\tilde{S}}}^{-}(E)$. From the previous observations it follows that

$$ {\tilde{S}}^-(E)=\inf \Bigl\{ \inf_{\theta>-1}\bigl\{ \theta\bigl(E+ \bar {V}^-\bigr)-P_0(\theta)\bigr\} ,-\bigl(E+\bar{V}^-\bigr),\inf _{\theta<-1}\bigl\{ \theta \bigl(E+\bar{V}^-\bigr)-P_0( \theta)\bigr\} \Bigr\} . $$

We start by the case θ>−1. Since P ₀ is concave, then $P_{0}'$ is decreasing. A simple computation shows that $P'_{0}(\theta)=E+\bar {V}^{-}$ for

$$\theta(E):=-\frac{1}{\varphi_+}\log \biggl(\frac{-\varphi_+-(E+\bar {V}^-)-\log(1+e^{-\varphi_+})}{E+\bar{V}^-+\log(1+e^{-\varphi _+})} \biggr). $$

Since $P_{0}'$ is decreasing, its image is given by $I_{1}:=[P_{0}'(+\infty ) ; P_{0}'(-1)]$, that is $I_{1}=[-\log(1+e^{-\varphi_{+}}) ; \frac {-\varphi_{+}}{1+e^{\varphi_{+}}}-\log(1+e^{-\varphi_{+}})]$. Then, for $E+\bar{V}^{-}\in{I_{1}}$ we have that

$$ {\tilde{S}}^-(E)=\theta(E) \bigl(E+\bar{V}^-\bigr)-P_0\bigl( \theta(E)\bigr)=S_{\rho _+,\rho_+} \bigl(-\bigl(E+\bar{V}^-\bigr) \bigr). $$

Now we look for ${\tilde{S}}^{-}(E)$ for $E+\bar{V}^{-}$ outside I ₁. Let $I_{1}^{-}:=(-\infty ; -\log(1+e^{-\varphi_{+}}))$ and $I_{1}^{+}:=( \frac{-\varphi_{+}}{1+e^{\varphi_{+}}}-\log(1+e^{-\varphi _{+}}) ; +\infty)$. Then, if $E+\bar{V}^{-}\in{I_{1}^{-}}$, we have that $\operatorname{sign}(E+\bar {V}^{-}-P_{0}'(\theta))=\operatorname{sign}(E+\bar{V}^{-}-P_{0}'(+\infty))=\operatorname{sign}(E+\bar {V}^{-}+\log(1+e^{-\varphi_{+}}))<0$, and as a consequence the infimum is attained at θ=+∞ and in this case ${\tilde{S}}^{-}(E)=-\infty$. On the other hand if $E+\bar{V}^{-}\in{I_{1}^{+}}$, we have that $\operatorname{sign}(E+\bar{V}^{-}-P_{0}'(\theta))=\operatorname{sign}(E+\bar {V}^{-}-P_{0}'(+\infty))>0$, and as a consequence the infimum is attained at θ=−1 and in this case ${\tilde{S}}^{-}(E)=-\infty$.

Now we look to the case θ<−1. Since P ₀ is concave, then $P_{0}'$ is decreasing. A simple computation shows that $P'_{0}(\theta )=E+\bar{V}^{-}$ for

$$\theta(E):=-\frac{1}{\varphi_-}\log \biggl(\frac{-\varphi_--(E+\bar {V}^-)-\log(1+e^{-\varphi_-})}{E+\bar{V}^-+\log(1+e^{-\varphi _-})} \biggr). $$

Since $P_{0}'$ is decreasing, its image is given by $I_{2}:=[P_{0}'(-1) ; P_{0}'(-\infty)]$, that is $I_{2}=[\frac{-\varphi_{-}}{1+e^{\varphi _{-}}}-\log(1+e^{-\varphi_{-}}) ; -\log(1+e^{\varphi_{-}})]$. Then, for $E+\bar{V}^{-}\in{I_{2}}$ we have that ${S}^{-}(E):= \theta (E)(E+\bar{V}^{-})-P_{0}(\theta(E))$ which can be written as $S_{\rho _{-},\rho_{-}} (-(E+\bar{V}^{-}) )$. Now we look for ${\tilde{S}}^{-}(E)$ for $E+\bar{V}^{-}$ outside I ₂. Let $I_{2}^{-}:=(-\infty ; \frac{-\varphi_{-}}{1+e^{\varphi_{-}}}-\log (1+e^{-\varphi_{-}}))$ and $I_{2}^{+}:=(-\log(1+e^{\varphi_{-}}) ; +\infty)$. Then, if $E+\bar{V}^{-}\in{I_{2}^{-}}$, we have that $\operatorname{sign}(E+\bar {V}^{-}-P_{0}'(\theta))=\operatorname{sign}(E+\bar{V}^{-}-P_{0}'(-\infty))=\operatorname{sign}(E+\bar {V}^{-}+\log(1+e^{\varphi_{-}}))<0$, and as a consequence the infimum is attained at θ=−1 and in this case ${\tilde{S}}^{-}(E)=-(E+\bar {V}^{-})$. On the other hand if $E+\bar{V}^{-}\in{I_{2}^{+}}$, we have that $\operatorname{sign}(E+\bar{V}^{-}-P_{0}'(\theta))=\operatorname{sign}(E+\bar{V}^{-}-P_{0}'(\theta))>0$, and as a consequence the infimum is attained at θ=−∞ and in this case ${\tilde{S}}^{-}(E)=-\infty$.

Now, a simple computation shows that the function t→−log(1+e ^−t)−t/(1+e ^t) is decreasing on (−∞,0] so that the sets I ₁ and I ₂ do not intersect. The restriction of the function ${\tilde{S}}^{-}$ to $[E_{-\infty}^{-} ; E_{-\infty}^{+}]$ has the expression:

$$ \begin{aligned} &{\tilde{S}}^-(E) \\ &\quad{}= \begin{cases} S_{\rho_+,\rho_+} (-(E+\bar{V}^-) ), \qquad-\log (1+e^{-\varphi_+})\leq{E+\bar{V}^-}<\frac{-\varphi_+}{1+e^{\varphi _+}}-\log(1+e^{-\varphi_+}),\\ -(E+\bar{V}^-), \qquad\frac{-\varphi_+}{1+e^{\varphi_+}}-\log (1+e^{-\varphi_+})\leq{E+\bar{V}^-}\leq{\frac{-\varphi _-}{1+e^{\varphi_-}}-\log(1+e^{-\varphi_-}),}\\ S_{\rho_-,\rho_-} (-(E+\bar{V}^-) ), \qquad\frac{-\varphi _-}{1+e^{\varphi_-}}-\log(1+e^{-\varphi_-})<E+\bar{V}^-\leq{-\log (1+e^{\varphi_-})} \end{cases} \end{aligned} $$

and is equal to −∞ outside the energy band. Thus ${\tilde{S}}^{-}$ coincides with S ⁻.

4.3 D.3 The Case $\frac{1}{2}\leq\rho_{-}<\rho_{+}$

In this case, by inverting the role of ρ ₋ with ρ ₊ and of φ ₋ with φ ₊ in the previous case, we obtain that ${\tilde{S}}^{-}$ restricted to $[E_{-\infty}^{-} ; E_{-\infty}^{+}]$ is given by

$$ \begin{aligned} &{\tilde{S}}^-(E) \\ &\quad{}= \begin{cases} S_{\rho_-,\rho_-} (-(E+\bar{V}^-) ), \qquad-\log (1+e^{\varphi_-})\leq{E+\bar{V}^-}<\frac{-\varphi_-}{1+e^{\varphi _-}}-\log(1+e^{-\varphi_-}),\\ -(E+\bar{V}^-), \qquad\frac{-\varphi_-}{1+e^{\varphi_-}}-\log (1+e^{-\varphi_-})\leq{E+\bar{V}^-}\leq{\frac{-\varphi _+}{1+e^{\varphi_+}}-\log(1+e^{-\varphi_+})},\\ S_{\rho_+,\rho_+} (-(E+\bar{V}^-) ), \qquad\frac{-\varphi _+}{1+e^{\varphi_+}}-\log(1+e^{-\varphi_+})<E+\bar{V}^-\leq{-\log (1+e^{-\varphi_+})} \end{cases} \end{aligned} $$

and is equal to −∞ outside the energy band. Thus ${\tilde{S}}^{-}$ coincides with S ⁻.