Short-Term Land use Planning and Optimal Subsidies

Abstract

Urban planning is a complex problem which includes choosing a social objective for a city, finding the associated optimal allocation of agents and identifying instruments like subsidies to decentralize this allocation as a market equilibrium. We split the problem in two independent steps. First, we find the short-term optimal allocation for a social objective and, second, we derive subsidies that reproduce this optimal allocation as a market equilibrium. This splitting is supported by a fundamental result asserting that the optimal allocation of any social objective can be decentralized by applying feasible subsidies, which can be computed even in the case with location externalities and transportation costs. In the first step, we compute the optimal allocation using an algorithm to solve a convex urban planning problem, which is applicable to a wide class of objective functions. In the second step, we compute optimal subsidies in several political situations for the planner, like budget constraints and limited impact on specific agents, zones, rents and/or utilities. As an example, we simulate a prototype city which aims at improving social inclusion.

Appendix

1.1 Notation and preliminaries

Let $({{\mathcal H}},\|\cdot \|)$ be a finite dimensional Euclidean space and denote by ${\Gamma }_{0}({{\mathcal H}})$ the family of lower semicontinuous convex functions $\varphi \colon {{\mathcal H}}\to {\left ]-\infty ,+\infty \right ]}$ such that ${\operatorname {dom}}\varphi =\left \{{x\in {{\mathcal H}}} \big | {\varphi (x)<{{+\infty }}}\right \}\neq {{\varnothing }}$. A function $\varphi \colon {{\mathcal H}}\to {\left ]-\infty ,+\infty \right ]}$ is coercive if $\lim _{\|x\|\to {{+\infty }}}\varphi (x)={{+\infty }}$. Now let $\varphi \in {\Gamma }_{0}({{\mathcal H}})$. The conjugate of φ is the function $\varphi ^{*}\in {\Gamma }_{0}({{\mathcal H}})$ defined by $\varphi ^{*}\colon u\mapsto \sup _{x\in {{\mathcal H}}}({\left \langle {{x}\mid {u}}\right \rangle }-\varphi (x))$. Moreover, for every $x\in {{\mathcal H}}$, φ + ∥x −⋅∥²/2 possesses a unique minimizer, which is denoted by prox_φx. Alternatively,

$$ {\operatorname{prox}}_{\varphi}=({\operatorname{Id}} +\partial\varphi)^{-1}, $$

(45)

where

$$ \partial\varphi\colon{{\mathcal H}}\to2^{{{\mathcal H}}}\colon x\mapsto\{u\in{{\mathcal H}}| {(\forall y\in{{\mathcal H}})\quad{\left\langle{{y-x}\mid{u}}\right\rangle}+\varphi(x)\leq\varphi(y)}\} $$

(46)

is the subdifferential of φ. In the particular case when φ is differentiable in some subset C of ${{\mathcal H}}$, we have, for every x ∈ C, ∂φ(x) = {∇φ(x)}. For every convex subset C of ${{\mathcal H}}$, the indicator function of C, denoted by ι_C, is the function which is 0 in C and +∞ in ${{\mathcal H}}\setminus C$.

The following result will be useful in the following sections and some parts of it can be derived from the proof given in Rockafellar (1970, Corollary 26.3.1).

Lemma 1

Let$\psi \colon {\operatorname {dom}}\psi \subset {\mathbb {R}}\to {\left ]-\infty ,+\infty \right ]}$bestrictly convex, differentiable in intdomψ, and such that${\operatorname {ran}}(\psi ^{\prime })={\mathbb {R}}$. Then,ψ^∗is strictly convex, differentiable in${\operatorname {dom}}\psi ^{*}={\mathbb {R}}$, and ranψ^∗⊂domψ. Moreover,

$$ \psi^{*}\colon \eta\mapsto (\psi^{\prime})^{-1}(\eta)\eta-\psi\left( (\psi^{\prime})^{-1}(\eta)\right) \quad\text{and}\quad (\psi^{*})^{\prime}=(\psi^{\prime})^{-1}. $$

(47)

1.2 Dual problem computation

For computing the dual formulation of Problem 1, we need the following definitions and preliminaries. Define

$$ \left\{\begin{array}{ll} \boldsymbol{\Psi}\colon{\mathbb{R}}^{|C|\times|N|}\to{\left]-\infty,+\infty\right]}\\ \hspace{1.34cm}{\boldsymbol{x}}\quad\mapsto\quad \left\{\begin{array}{ll} \displaystyle{\sum\limits_{h\in C}\sum\limits_{i\in N}\psi_{hi}(z_{hi},x_{hi})}, &\text{if} {\boldsymbol{x}}\in\underset{h\in C}{{\raisebox{-0.5mm} {\text{\LARGE{\(\times\)}}}}\!}\underset{i\in N}{{\raisebox{-0.5mm} {\text{\LARGE{\(\times\)}}}}\!}{\operatorname{dom}}\psi_{hi}(z_{hi},\cdot)\\ {{+\infty}},&\text{otherwise} \end{array}\right.\\ \boldsymbol{{\Lambda}} \colon{\mathbb{R}}^{|C|\times|N|}\to{\mathbb{R}}^{|C|+|N|}\\ \hspace{1.3cm}{\boldsymbol{x}}\quad\mapsto\quad\displaystyle{\left( \left( \sum\limits_{i\in N}x_{hi}\right)_{h\in C},\left( \sum\limits_{h\in C} x_{hi}\right)_{i\in N}\right)}, \end{array}\right. $$

(48)

where x = (x_hi)_{h ∈ C, i ∈ N} is a generic element of ${\mathbb {R}}^{|C|\times |N|}$.

Proposition 6 (Properties of Ψ and Λ)

LetΨandΛbedefined as in Eq. 48. Then, the following statements hold.

• 1. Ψis strictly convex, coercive, and

  $$ (\forall \gamma\in{\left]0,+\infty\right[})\quad {\operatorname{prox}}_{\gamma\boldsymbol{\Psi}} =\left( {\operatorname{prox}}_{\gamma\psi_{hi}(z_{hi},\cdot)}\right)_{h\in C, i\in N}. $$

  (49)

• 2. We have

  $$ \boldsymbol{\Psi}^{*}\colon{\mathbb{R}}^{|C|\times|N|}\to{\left]-\infty,+\infty\right]} \colon\boldsymbol{u}\mapsto \sum\limits_{h\in C}\sum\limits_{i\in N}\varphi_{hi}(z_{hi},u_{hi}), $$

  (50)

  where, for everyh ∈ Candi ∈ N,

  $$ \varphi_{hi}\colon (z_{hi},u)\mapsto\psi_{hi}(z_{hi},\cdot)^{*}(u) =\sup_{x\in\left[0,a_{hi}\right[}({\left\langle{{x}\mid{u}}\right\rangle} -\psi_{hi}(z_{hi},x)) $$

  (51)

  is differentiable. Moreover, Ψ^∗is differentiable and ∇Ψ^∗ = (φ_hi(z_hi, ⋅)^′)_{h ∈ C, i ∈ N}. In addition, suppose that, for everyh ∈ Candi ∈ N, ψ_hi(z_hi, ⋅) is differentiable in ]0,a_hi[ and${\operatorname {ran}}(\psi _{hi}(z_{hi},\cdot )')={\mathbb {R}}$. Then, for everyh ∈ Candi ∈ N,

  $$ (\forall u\in{\mathbb{R}})\quad \varphi_{hi}(z_{hi},u)=(\psi_{hi}(z_{hi},\cdot)')^{-1} (u)u-\psi_{hi}\left( z_{hi},(\psi_{hi}(z_{hi},\cdot)')^{-1}(u)\right), $$

  (52)

  φ_hi(z_hi, ⋅)^′ = (ψ_hi(z_hi, ⋅)^′)^− 1, andΨ^∗is strictly convex.

• 3. Λis linear, bounded, Λ^∗: (b, r)↦(b_h + r_i)_{h ∈ C, i ∈ N}, where (b, r) = ((b_h)_{h ∈ C}, (r_i)_{i ∈ N}) is a generic element of${\mathbb {R}}^{|C|+|N|}$, and$\|\boldsymbol {{\Lambda }} \|=\sqrt {|C|+|N|}$.

Proof

1: Is a consequence of Eq. 48, the properties of the class $\mathscr{C}$ in Eq. 3, and Combettes and Wajs (2005, Lemma 2.9). 2: It follows from Bauschke and Combettes (2011, Proposition 13.27) that $\boldsymbol {\Psi }^{*}={\sum }_{h\in C}{\sum }_{i\in N} \psi _{hi}(z_{hi},\cdot )^{*}={\sum }_{h\in C}{\sum }_{i\in N}\varphi _{hi}(z_{hi},\cdot )$. The differentiability follows from Hiriart-Urruty and Lemaréchal (1993, Proposition 6.2.1) and the last result follows from Lemma 1. 3: It is clear that Λ is linear and bounded. For every ${\boldsymbol {x}}\in {\mathbb {R}}^{|C|\times |N|}$ and $({\boldsymbol {b}},{\boldsymbol {r}})\in {\mathbb {R}}^{|C|+|N|}$, we have

$$ \begin{array}{@{}rcl@{}} {\left\langle{{({\boldsymbol{b}},{\boldsymbol{r}})}\mid{\boldsymbol{{\Lambda}} {\boldsymbol{x}}}}\right\rangle}&=&\sum\limits_{h\in C}b_{h}\left( \sum\limits_{i\in N}x_{hi}\right)+ \sum\limits_{i\in N}r_{i}\left( \sum\limits_{i\in N}x_{hi}\right)\\ &=&\sum\limits_{h\in C}\sum\limits_{i\in N}x_{hi}(b_{h}+r_{i})\\ &=&{\left\langle{{\boldsymbol{{\Lambda}}^{*}({\boldsymbol{b}},{\boldsymbol{r}})}\mid{{\boldsymbol{x}}}}\right\rangle}. \end{array} $$

(53)

On the other hand, using the inequality 2xy ≤ x² + y² we obtain, for every ${\boldsymbol {x}}\in {\mathbb {R}}^{|C|\times |N|}$,

$$ \begin{array}{@{}rcl@{}} \|\boldsymbol{{\Lambda}} {\boldsymbol{x}}\|^{2} &=&\sum\limits_{h\in C}\left( \sum\limits_{i\in N}x_{hi}\right)^{2}+\sum\limits_{i\in N}\left( \sum\limits_{h\in C}x_{hi}\right)^{2}\\ &=&\sum\limits_{h\in C}\left( \sum\limits_{i\in N}x_{hi}^{2}+\sum\limits_{j\neq i}2x_{hi}x_{hj}\right) +\sum\limits_{i\in N}\left( \sum\limits_{h\in C}x_{hi}^{2}+\sum\limits_{g\neq h}2x_{hi}x_{gi}\right)\\ &\leq&\sum\limits_{h\in C}\left( \sum\limits_{i\in N}x_{hi}^{2}+(|N|-1)\sum\limits_{i\in N}x_{hi}^{2}\right) +\sum\limits_{i\in N}\left( \sum\limits_{h\in C}x_{hi}^{2}+(|C|-1)\sum\limits_{h\in C}x_{hi}^{2}\right)\\ &=&(|C|+|N|)\|{\boldsymbol{x}}\|^{2}, \end{array} $$

(54)

which yields $\|\boldsymbol {{\Lambda }} \|\leq \sqrt {|C|+|N|}$. The equality follows by taking, in particular, for every (h, i) ∈ C × N, x_hi = 1, which yields

$$ \|\boldsymbol{{\Lambda}} {\boldsymbol{x}}\|^{2}=\sum\limits_{h\in C}|N|^{2}+\sum\limits_{i\in N}|C|^{2} =(|C|+|N|)|C||N|=(|C|+|N|)\|{\boldsymbol{x}}\|^{2}. $$

(55)

Hence $\|\boldsymbol {{\Lambda }} \|=\sqrt {|C|+|N|}$. □

Proof of Proposition 1

Indeed, Problem 1 can be written equivalently as

$$ \underset{\underset{\boldsymbol{{\Lambda}} {\boldsymbol{x}} =(\boldsymbol{H},\boldsymbol{S})}{\boldsymbol{x} \in{\mathbb{R}}^{|C|\times|N|}}}{\text{minimize}} \boldsymbol{\Psi}({\boldsymbol{x}}), $$

(56)

where Ψ and Λ are defined in Eq. 6. Therefore, from Bauschke and Combettes (2011, Proposition 19.19) we have that the dual problem is

$$ {\underset{({\boldsymbol{b}},{\boldsymbol{r}})\in{\mathbb{R}}^{|C|+|N|}} {\text{minimize}} \boldsymbol{\Psi}^{*}(-\boldsymbol{{\Lambda}}^{*}({\boldsymbol{b}},{\boldsymbol{r}})) +{\left\langle{{({\boldsymbol{b}},{\boldsymbol{r}})}\mid{(\boldsymbol{H},\boldsymbol{S})}}\right\rangle}} , $$

(57)

or equivalently, from Proposition 62-3,

$$ {\underset{({\boldsymbol{b}},{\boldsymbol{r}})\in{\mathbb{R}}^{|C|+|N|}} {\text{minimize}} \sum\limits_{h\in C}\sum\limits_{i\in N} \varphi_{hi}(z_{hi},-b_{h}-r_{i})+\sum\limits_{h\in C}H_{h}b_{h}+\sum\limits_{i\in N}S_{i}r_{i}} , $$

(58)

and the proof is finished. □

Proof of Proposition 2

Since $\boldsymbol {\Psi }\in {\Gamma }_{0}({\mathbb {R}}^{|C|\times |N|})$ is coercive, Ξ is closed and convex, and Eq. 5 yields ${\Xi }\cap {\operatorname {dom}}\boldsymbol {\Psi }\neq {{\varnothing }}$, Bauschke and Combettes (2011, Proposition 11.4(i)) asserts that the primal problem has solutions. It follows from Eq. 56 that Problem 1 can be written equivalently as

$$ {\underset{{\boldsymbol{x}}\in{\mathbb{R}}^{|C|\times|N|}} {\text{minimize}} \boldsymbol{\Psi}({\boldsymbol{x}}) +\iota_{\{(\boldsymbol{H},\boldsymbol{S})\}}(\boldsymbol{{\Lambda}} {\boldsymbol{x}})} . $$

(59)

Note that $\iota _{\{(\boldsymbol {H},\boldsymbol {S})\}}\in {\Gamma }_{0}({\mathbb {R}}^{|C|+|N|})$ is polyhedral. Now, since (H, S) ∈ ]0, +∞[^|C|+|N|, it follows from Eqs. 5 and 48 that

$$ (\boldsymbol{H},\boldsymbol{S})\in{\operatorname{int}}\left( \boldsymbol{{\Lambda}} \left( \underset{h\in C}{{\raisebox{-0.5mm} {\text{\LARGE{\(\times\)}}}}\!}\underset{i\in N} {{\raisebox{-0.5mm} {\text{\LARGE{\(\times\)}}}}\!}\left[0,a_{hi}\right[\right)\right)=\underset{h\in C}{{\raisebox{-0.5mm} {\text{\LARGE{\(\times\)}}}}\!}\left]0,a_{h}\right[\times\underset{i\in N}{{\raisebox{-0.5mm} {\text{\LARGE{\(\times\)}}}}\!}\left]0,a_{i}\right[, $$

(60)

where, for every h ∈ C, $a_{h}={\sum }_{i\in I}a_{hi}$ and, for every i ∈ N, $a_{i}={\sum }_{h\in C}a_{hi}$. Hence, from Bauschke and Combettes (2011, Fact 15.25) we have $\inf (\boldsymbol {\Psi }+ \iota _{\{(\boldsymbol {H},\boldsymbol {S})\}}\circ \boldsymbol {{\Lambda }} )=-\min (\boldsymbol {\Psi }^{*} \circ -\boldsymbol {{\Lambda }}^{*}+\iota ^{*}_{\{(\boldsymbol {H},\boldsymbol {S})\}})$. Therefore, we have existence of solutions to the dual problem. Moreover, it follows from Proposition 62 that Ψ^∗ is differentiable in $\mathbb {R}^{|C|+|N|}$. Altogether, Bauschke and Combettes (2011, Proposition 19.3) asserts that Problem 1 has a unique solution

$$ \overline{\boldsymbol{x}}=\nabla\boldsymbol{\Psi}^{*}(\boldsymbol{{\Lambda}}^{*}(\overline{\boldsymbol{b}}, \overline{\boldsymbol{r}})), $$

(61)

where $(\overline {\boldsymbol {b}},\overline {\boldsymbol {r}})$ is a solution to the dual problem (7). Moreover, it follows from Lemma 1 and Proposition 63 that Eq. 61 is equivalent to

$$ (\forall h\in C)(\forall i\in N)\quad \overline{x}_{hi} =(\psi_{hi}(z_{hi},\cdot)^{*})'(\overline{b}_{h}+\overline{r}_{i}) =\left( \varphi_{hi}(z_{hi},\cdot)\right)'(\overline{b}_{h}+\overline{r}_{i}). $$

(62)

Finally let us prove that, under one of the constraints 1–4, Φ is strictly convex and, hence, the dual problem (7) has a unique solution. Indeed, it follows from Lemma 1 that, for every h ∈ C and i ∈ N, φ_hi is strictly convex. Let (b¹, r¹)≠(b², r²) be vectors in ${\mathbb {R}}^{|C|+|N|}$ and let α ∈ ]0, 1[. We have

$$ \begin{array}{@{}rcl@{}} \boldsymbol{\Phi}(\alpha({\boldsymbol{b}}^{1}\!,{\boldsymbol{r}}^{1}\!) + (1 - \alpha)({\boldsymbol{b}}^{2},{\boldsymbol{r}}^{2})) &=&\sum\limits_{h\in C}H_{h}(\alpha {b_{h}^{1}}+(1-\alpha){b_{h}^{2}})+ \sum\limits_{i\in N}S_{i}(\alpha {r_{i}^{1}}+(1-\alpha){r_{i}^{2}})\\ &&\hskip.4cm+\!\sum\limits_{h\in C}\sum\limits_{i\in N}\varphi_{hi}\left( z_{hi}, -(\alpha {b_{h}^{1}} + (1-\alpha){b_{h}^{2}})-(\alpha {r_{i}^{1}} + (1-\alpha){r_{i}^{2}})\right)\\ &=&\alpha\left( \!\sum\limits_{h\in C}H_{h}{b_{h}^{1}}+\sum\limits_{i\in N}S_{i}{r_{i}^{1}}\right) + (1-\alpha)\left( \!\sum\limits_{h\in C}H_{h}{b_{h}^{2}}+\sum\limits_{i\in N}S_{i}{r_{i}^{2}}\right)\\ &&\hskip.4cm+\!\sum\limits_{h\in C}\sum\limits_{i\in N}\varphi_{hi}\left( z_{hi}, \alpha(-{b_{h}^{1}}-{r_{i}^{1}})+(1-\alpha)(-{b_{h}^{2}}-{r_{i}^{2}})\right). \end{array} $$

(63)

Since, for every h ∈ C and i ∈ N, φ_hi(z_hi, ⋅) is strictly convex, it is enough to prove that, under one of the constraints 1–4, there exist h₀ ∈ C and i₀ ∈ N such that $-b_{h_{0}}^{1}-r_{i_{0}}^{1}\neq -b_{h_{0}}^{2}-r_{i_{0}}^{2}$, in which case from Eq. 63 we obtain that

$$ \boldsymbol{\Phi}(\alpha({\boldsymbol{b}}^{1},{\boldsymbol{r}}^{1})+(1-\alpha)({\boldsymbol{b}}^{2}, {\boldsymbol{r}}^{2}))<\alpha\boldsymbol{\Phi}({\boldsymbol{b}}^{1},{\boldsymbol{r}}^{1}) +(1-\alpha)\boldsymbol{\Phi}({\boldsymbol{b}}^{2},{\boldsymbol{r}}^{2}), $$

(64)

and the result follows. Let us proceed by contradiction. Suppose that

$$ (\forall h\in C)(\forall i\in N)\quad -{b_{h}^{1}}-{r_{i}^{1}}=-{b_{h}^{2}}-{r_{i}^{2}}. $$

(65)

If 1 holds, we have ${b_{1}^{1}}={b_{1}^{2}}=\eta$ and we deduce from Eq. 65 in the particular case h = 1 that, for every i ∈ N, ${r_{i}^{1}}={r_{i}^{2}}$. Hence, it follows again from Eq. 65 that, for every h ∈ C ∖{1}, ${b_{h}^{1}}={b_{h}^{2}}$, which contradicts (b¹, r¹)≠(b², r²). Now suppose that 2 holds. Then we have ${\sum }_{h\in C}{b_{h}^{1}}={\sum }_{h\in C}{b_{h}^{2}}=\eta$ and, by summing in h in Eq. 65, we deduce, for every i ∈ N, ${r_{i}^{1}}={r_{i}^{2}}$. The contradiction is obtained in the same way as before. The cases 3 and 4 are analogous. □

1.3 Planning algorithms

Proposition 7 (Planning problem algorithm)

For everyh ∈ Candi ∈ N, let$(e_{hi,n})_{n\in {\mathbb N}}$bean absolutely summable sequence in${\mathbb {R}}$, let$x_{hi,0}\in {\mathbb {R}}$, let$(b_{h,0},r_{i,0})\in {\mathbb {R}}^{2}$, let$\varepsilon \in \left ]0,1/(\sqrt {|C|+|N|}+ 1)\right [$, let$(\gamma _{n})_{n\in {\mathbb N}}$bea sequence in$[\varepsilon ,(1-\varepsilon )/\sqrt {|C|+|N|} ]$, and set

$$ (\forall n\in{\mathbb N})\quad \begin{array}{l} \left\lfloor \begin{array}{l} \textup{For every } h\in C \text{and} i\in N\\ \begin{array}{l} \left\lfloor \begin{array}{l} y_{1hi,n}=x_{hi,n}-\gamma_{n}(b_{h,n}+r_{i,n})\\ p_{1hi,n}={\operatorname{prox}}_{\gamma_{n}\psi_{hi}(z_{hi},\cdot)}y_{1hi,n}+e_{hi,n} \end{array} \right. \end{array}\\ \textup{For every } h\in C\\ \begin{array}{l} \left\lfloor \begin{array}{l} p_{2h,n}=b_{h,n}+\gamma_{n}\left( {\sum}_{i\in N}x_{hi,n}-H_{h}\right)\\ b_{h,n + 1}=b_{h,n}+\gamma_{n}\left( {\sum}_{i\in N}p_{1hi,n}-H_{h}\right) \end{array} \right. \end{array}\\ \textup{For every } i\in N\\ \begin{array}{l} \left\lfloor \begin{array}{l} p_{2i,n}=r_{i,n}+\gamma_{n}\left( {\sum}_{h\in C}x_{hi,n}-S_{i}\right)\\ r_{i,n + 1}=r_{i,n}+\gamma_{n}\left( {\sum}_{h\in C}p_{1hi,n}-S_{i}\right) \end{array} \right. \end{array}\\ \textup{For every } h\in C \text{and} i\in N\\ \begin{array}{l} \left\lfloor \begin{array}{l} q_{hi,n}=p_{1hi,n}-\gamma_{n}(p_{2h,n}+p_{2i,n})\\ x_{hi,n + 1}=x_{hi,n}-y_{1hi,n}+q_{hi,n} \end{array} \right. \end{array} \end{array} \right. \end{array} $$

(66)

Then the following statements hold for the solution$((\overline {x}_{hi})_{h\in C})_{i\in N}$to Problem 1 and some solution$\left ((\overline {b}_{h})_{h\in C},(\overline {r}_{i})_{i\in N}\right )$to its dual in Eq. 7.

• 1. For everyh ∈ Candi ∈ N, x_{hi, n} − p_{1hi, n} → 0, b_{h, n} − p_{2h, n} → 0, andr_{i, n} − p_{2i, n} → 0.

• 2. For everyh ∈ Candi ∈ N, $x_{hi,n}\to \overline {x}_{hi}$, $p_{1hi,n}\to \overline {x}_{hi}$, $b_{h,n}\to \overline {b}_{h}$, $p_{2h,n}\to \overline {b}_{h}$, $r_{i,n}\to \overline {r}_{i}$, and$p_{2i,n}\to \overline {r}_{i}$.

Proof

See (Briceño-Arias and Combettes 2011). □

The difficulty of the algorithm proposed in Proposition 7 lies in the computation, for every h ∈ C, i ∈ N, and $n\in {\mathbb N}$, of ${\operatorname {prox}}_{\gamma _{n}\psi _{hi}(z_{hi},\cdot )}$. Several examples in which the proximity operator can be computed explicitly can be found in Combettes and Wajs (2005). The following result shows some interesting cases in which an explicit computation of the proximity operator can be obtained.

Lemma 2

Let$z\in {\mathbb {R}}$,a ∈ ]0, +∞[,b ∈ ]0, +∞[,γ ∈ ]0, +∞[, andμ ∈ ]0, +∞[.

• 1. Let$\psi \colon (z,x)\mapsto -zx+\frac {1}{\mu } x(\ln x-1)$.Then$\psi \in \mathscr{C}$and${\operatorname {prox}}_{\gamma \psi (z,\cdot )}\colon x\mapsto \frac {\gamma }{\mu } W(\frac {\mu }{\gamma } e^{\mu (x/\gamma +z)})$,where W is the product log function.

• 2. Letψ: (z, x)↦ι_[0,+∞[− zx + a(x − b)².Then$\psi \in \mathscr{C}$andprox_γψ(z,⋅): x↦ max{(x + γz + 2γab)/(1 + 2γa), 0}.

Proof

Let $(x,p)\in {\mathbb {R}}^{2}$. It is clear from Eq. 3 that both functions are in $\mathscr{C}$. 1: We have $p={\operatorname {prox}}_{\gamma \psi (z,\cdot )}x \Leftrightarrow x-p=\gamma \psi (z,\cdot )'(p) \Leftrightarrow x+\gamma z=p+\frac {\gamma }{\mu }\ln p \Leftrightarrow \frac {\mu }{\gamma }x+ \mu z=\frac {\mu }{\gamma }p+\ln p \Leftrightarrow e^{\mu (x/\gamma + z)}=pe^{\frac {\mu }{\gamma }p} \Leftrightarrow \frac {\mu }{\gamma }p =W(\frac {\mu }{\gamma }e^{\mu (x/\gamma + z)})$, and the result follows. 2: We have p = prox_γψ(z,⋅)x ⇔ x−p ∈ γ∂ψ(z,⋅)(p) ⇔ x−p ∈ N_[0,+∞[(p)−γz+ 2γa(p−b) ⇔ x+γz+ 2γab ∈ N_[0,+∞[(p)+p(1 + 2γa) ⇔ (x+γz+ 2γab)/(1 + 2γa) ∈ N_[0,+∞[(p)+p ⇔ p = P_[0,+∞[((x+γz+ 2γab)/(1 + 2γa)), which yields the result. □

Example 5

(Social inclusion algorithm) As an example of Proposition 7, we consider the social inclusion problem (38) by using the algorithm proposed in Eq. 66, which, by applying Lemma 22, becomes (we set e_{hi, n} ≡ 0)

$$ (\forall n\in{\mathbb N})\quad \begin{array}{l} \left\lfloor \begin{array}{l} \text{For every } h\in C \text{and} i\in N\\ \begin{array}{l} \left\lfloor \begin{array}{l} y_{1hi,n}=x_{hi,n}-\gamma_{n}(b_{h,n}+r_{i,n})\\ p_{1hi,n}=\max\left\{\frac{y_{1hi,n}+\gamma_{n} z_{hi}+\frac{2\gamma_{n}I_{h}H_{h}}{\alpha S_{i}T}}{1+\frac{2\gamma_{n}I_{h}}{\alpha {S_{i}^{2}}}},0\right\} \end{array} \right. \end{array}\\ \text{For every } h\in C\\ \begin{array}{l} \left\lfloor \begin{array}{l} p_{2h,n}=b_{h,n}+\gamma_{n}\left( {\sum}_{i\in N}x_{hi,n}-H_{h}\right)\\ b_{h,n + 1}=b_{h,n}+\gamma_{n}\left( {\sum}_{i\in N}p_{1hi,n}-H_{h}\right) \end{array} \right. \end{array}\\ \text{For every } i\in N\\ \begin{array}{l} \left\lfloor \begin{array}{l} p_{2i,n}=r_{i,n}+\gamma_{n}\left( {\sum}_{h\in C}x_{hi,n}-S_{i}\right)\\ r_{i,n + 1}=r_{i,n}+\gamma_{n}\left( {\sum}_{h\in C}p_{1hi,n}-S_{i}\right) \end{array} \right. \end{array}\\ \text{For every } h\in C \text{and} i\in N\\ \begin{array}{l} \left\lfloor \begin{array}{l} q_{1hi,n}=p_{1hi,n}-\gamma_{n}(p_{2h,n}+p_{2i,n})\\ x_{hi,n + 1}=x_{hi,n}-y_{1hi,n}+q_{1hi,n}. \end{array} \right. \end{array} \end{array} \right. \end{array} $$

(67)

If the sequence $(\gamma _{n})_{n\in {\mathbb N}}$ is in ]0, (|C| + |N|)^− 1/2[, Proposition 7 asserts that, for every h ∈ C and i ∈ N, the sequence $(x_{hi,n})_{n\in {\mathbb N}}$ converges to some $x_{hi}^{*}$ and $\boldsymbol {x}^{*}=(x_{hi}^{*})_{h\in C, i\in N}$ is the solution to Eq. 38 and, additionally, the sequence $(b_{h,n},r_{i,n})_{n\in {\mathbb N}}$ converges to some $(b_{h}^{*},r_{i}^{*})$ and $((b_{h}^{*})_{h\in C},(r_{i}^{*})_{i\in N})$ is a solution to the associated dual problem.