and C be the space of continuous and bounded real-valued functions endowed with the sup-norm 1.

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1 1 Proof T : C C Let T be the following mapping: Tϕ = max {u (x, a)+βeϕ [f (x, a, ε)]} (1) a Γ(x) and C be the space of continuous and bounded real-valued functions endowed with the sup-norm 1. Proposition 1 T maps continuous and bounded real-valued functions into the space of continuous and bounded real-valued functions i.e. T : C C Proof. Assumptions: (iv) Γ (x) is a non-empty, continuous (u.h.c and l.h.c.), and compact-valued correspondence. 2. in addition to ϕ C. Since u (x, a) is bounded and ϕ ( ) is bounded, the sum of two bounded functions is also bounded. The maximum of a bounded function is also bounded. Therefore, Tϕis bounded. To deal with continuity, let us consider the Theorem of the Maximum. First of all, we need to show that u (x, a)+βeϕ [f (x, a, ε)] (2) is continuous. As f ( ) and ϕ are continuous functions, the composite function (ϕ f) is also continuous and, moreover, as the expectation is a linear operator, E [ϕ f] is a continuous function. Therefore, (2) is continuous and, given the assumptions on the constraint correspondence Γ (x), we can invoke the Theorem of the Maximum 3 to ensure that Tϕis continuous. 1 C is a complete metric space. 2 f ( ) being continuous implies that the stochastic structure satisfies the Feller property. The Feller property states that Z E [ϕ (x t+1 ) x t = x, a t = a] = ϕ [f (x, a, ε)] df (ε x, a) 3 Theorem of the Maximum. Let X R l,y R m, h (x) = max y Γ(x) f (x, y) and G (x) ={y Γ (x) :f (x, y) =h (x)} such that f : X Y R is a continuous function and let Γ : X Y be a compactvalued and continuous correspondence. Then the function h : X R is continuous, and the correspondence G : X Y is non-empty, compact-valued, and u.h.c. 1

2 ALTERNATIVE PROOF: Since the space of continuous and bounded real-valued functions endowed with the sup-norm is a complete metric space and T is a contraction, we can use the Contraction Mapping Theorem to ensure that there exists a unique fixed point i.e.! ϕ C s.t. ϕ = Tϕ Therefore, by the previous proof we have that Tϕ C, then, as there exists a unique fixed point, ϕ must also belong to C. 2 Proving T maps INCREASING functions into increasing functions In particular, we want to show T : D D where D is the space of increasing, continuous, and bounded real-valued functions (endowed with the sup-norm i.e. D is a complete metric space. Thus. we could use the alternative way of proving the statement). Proof. Standard assumptions: In order to show that T maps increasing functions into increasing functions, we need the following additional assumptions: (vi) u ( ) is increasing (vii) Γ (x) is increasing in the following sense x, x 0 X s.t. x 0 x = Γ (x 0 ) Γ (x) (viii) f (x, a, ε) is increasing in x. Let ϕ be a continuous, bounded, and increasing real-valued function. Given the standard assumptions, we can invoke either the Theorem of the Maximum or the Extreme Value Theorem, to argue the existence of an optimal solution to the optimization problem stated in (1) for any x X. Let x, x 0 X s.t. x 0 x, and a, a 0 be the corresponding optimal solution when the state is given by x and x 0. By assumption (vii), we have a Γ (x) Γ (x 0 )= a Γ (x 0 ) By assumptions (vi) and (viii), u ( ) and f ( ) are increasing functions, therefore u (x, a)+βeϕ[f (x, a, ε)] u (x 0,a)+βEϕ[f (x 0,a,ε)] (3) 2

3 where u (x 0,a)+βEϕ[f (x 0,a,ε)] u (x 0,a 0 )+βeϕ[f (x 0,a 0,ε)] (4) Hence, by (3) and (4), we can conclude that where u (x, a)+βeϕ[f (x, a, ε)] u (x 0,a 0 )+βeϕ[f (x 0,a 0,ε)] u (x, a)+βeϕ[f (x, a, ε)] = (T ϕ)(x) u (x 0,a 0 )+βeϕ[f (x 0,a 0,ε)] = (T ϕ)(x 0 ) So, x x 0 = Tϕ(x) Tϕ(x 0 ). 3 Proving T maps STRICLY INCREASING functions into strictly increasing functions Note that the space of continuous, bounded, and strictly increasing real-valued functions endowed with the sup-norm is not a complete metric space. Here, theproofhastobedonebyusingatwostepsprocedure. Proof. We need an extra assumption: (ix) u ( ) is strictly increasing First step: show T maps increasing functions into increasing functions. Second step: As u ( ) is strictly increasing and βeϕ[f (x, a, ε)] is increasing, u (x, a) +βeϕ[f (x, a, ε)] is stricly increasing since the sum of an increasing and a strictly increasing function is a strictly increasing function (CHECK!!!!!). The max operator preserves such a property, therefore Tϕis strictly increasing. Since we already have ϕ D (D is the space of continuous, increasing, anb bounded real-valued functions) s.t. ϕ = Tϕ, and that Tϕis strictly increasing, it directly follows that ϕ is strictly increasing. 4 Proving T maps CONCAVE functions into concave functions Proof. Let ϕ be a concave, continuous, and bounded real-valued function. Let x 0 >xand a 0,abe the corresponding optimal solutions. Standard assumptions: Additional assumptions: 3

4 *forconcavity: (x) u ( ) is concave in (x, a) (xi) Γ (x) is convex in the following sense θ [0, 1], a Γ (x), a 0 Γ (x 0 ) θa+(1 θ ) a 0 Γ (θ x+(1 θ ) x 0 ) (xii) f (a, x, ε) is concave in (x, a) (xiii) ϕ ( ) is increasing (in addition to concave, continuous, and bounded) We want to show the following Tϕ[θ x+(1 θ ) x 0 ] θtϕ( x)+(1 θ ) Tϕ(x 0 ) By assumption (xi), [θa+(1 θ ) a 0 ] is feasible but not necessarily optimal Tϕ[θ x+(1 θ ) x 0 ] u [θ x+(1 θ ) x 0,θa+(1 θ) a 0 ]+ +βeϕ[f (θ x+(1 θ ) x 0,θa+(1 θ) a 0,ε)] By concavity of u ( ) Tϕ[θ x+(1 θ ) x 0 ] θu (x, a)+(1 θ) u (x 0,a 0 )+ (5) +βeϕ[f (θ x+(1 θ ) x 0,θa+(1 θ) a 0,ε)] By concavity of f ( ) and increasigness of ϕ ( ) we have βeϕ[f (θ x+(1 θ ) x 0,θa+(1 θ) a 0,ε)] βeϕ[θf(x, a, ε)+(1 θ) f (x 0,a 0,ε)] β [θeϕ [f (x, a, ε)] + (1 θ) Eϕ[f (x 0,a 0,ε)]] Therefore, (5) will be as follows Proof. Tϕ[θ x+(1 θ ) x 0 ] θ [u (x, a)+βeϕf(x, a, ε)]] + +(1 θ) [u (x 0,a 0 )+βeϕf(x 0,a 0,ε)]] = Tϕ[θ x+(1 θ ) x 0 ] θtϕ ( x)+(1 θ ) Tϕ ( x 0 ) 5 Proving T maps STRICTLY CONCAVE functions into strictly concave functions Proof. We have to use a two-step proof. First of all, let ϕ be a strictly concave, continuous, and bounded real-valued function. Secondly, consider the following assumptions: Standard assumptions: 4

5 Additional assumptions: *forconcavity: (xi) Γ (x) is convex in the following sense θ [0, 1], a Γ (x), a 0 Γ (x 0 ) θa+(1 θ ) a 0 Γ (θ x+(1 θ ) x 0 ) (xii) f (a, x, ε) is concave in (x, a) (xiii) ϕ ( ) is increasing (in addition to strictly concave, continuous, and bounded) (xiv) u ( ) is strictly concave in (x, a) FIRST STEP: Given the above assumptions we can show that Tϕ is a continuous, concave, and bounded real-valued functions. Now, since the space of continuous, concave, and bounded real-valued functions (D) endowed with the sup-norm is a complete metric space, and T is a contraction by assumption, we can use the Contraction Mapping Theorem to ensure that there exists a unique fixed point ϕ D (i.e.! ϕ D s.t. ϕ = Tϕ). SECOND STEP: Since the space of continuous, strictly concave, and bounded real-valued functions is not complete, we cannot invoke the Contraction Mapping Theorem. However, we will use the following reasoning. By assumption (xiv) we have that u ( ) is strictly concave. Since the sum of a concave function (βeϕ[f (x, a, ε)]) and a strictly concave function (u (x, a)), is a strictly concave function. Then, u (x, a) +β E ϕ[f (x, a, ε)] is strictly concave. The max operator preserves such a curvature property. So, Tϕis strictly concave. Therefore, since we already have ϕ = T ϕ, and T ϕ is stricly concave, it follows that ϕ is strictly concave. 5