Abstract
Quasilinear utility functions are often met in various economic applications. We provide an axiomatization of preferences that are representable by a continuous quasilinear utility function. Following the topological approach and building on existing results (Wakker in J Math Psychol 32:421–435, 1988), we obtain representation theorems on connected topological spaces. A direct proof for general topological spaces is achieved.
Similar content being viewed by others
Notes
Let \(x \in \mathcal{C}\). Assume U(x, .) is continuous on \({\mathbb {R}}^+\) and continuously differentiable on \((0, + \infty )\). Then, according to the finite increments theorem, for all \(z > 0\), there exists some \(z_x \in (0,z)\) such that \(U(x,z)  U(x, 0) = \frac{\partial U}{ \partial z} (x,z_x) \times (z 0) = a z\). Defining \(v(x) = U(x, 0)\), we obtain, \(U(x,z) = v(x) + a z\).
Cauchy’s functional equation is usually given on \({\mathbb {R}}\). We have \(f(0)= 0\). Define the oddextension \({\tilde{f}}\) by \({\tilde{f}} (x) = f (x)\) for \(x \ge 0\) and \({\tilde{f}} (x) =  f (x)\) for \(x < 0\). \({\tilde{f}}\) is continuous and increasing. If \(x,y \ge 0\) or \(x,y \le 0\), then \({\tilde{f}} (x + y) = {\tilde{f}} (x) + {\tilde{f}} (y) \) holds. If \(x \ge 0, y < 0\) and \(x+y \ge 0\), then \({\tilde{f}} (x) = f(x) = f(x+y) + f(y) = {\tilde{f}} (x + y)  {\tilde{f}} (y)\). If \(x \ge 0, y < 0\) and \(x+y < 0\), then \({\tilde{f}} (y) =  f ( y) =  f ( x  y)  f(x) = {\tilde{f}} (x + y)  {\tilde{f}} (x)\). The other cases are similar. Hence, \({\tilde{f}}\) satisfies Cauchy’s functional equation on \({\mathbb {R}}\).
References
Aczél, J.: Lectures on Functional Equations and Their Applications. Academic Press, NewYork (1966)
Bosi, G., Mehta, G.B.: Existence of a semicontinuous or continuous utility function: a unified approach and an elementary proof. J. Math. Econ. 38, 311–328 (2002)
Brown, D.J., Calsamiglia, C.: Alfred Marshall’s cardinal theory of value: the strong law of demand. Econ. Theory Bull. 2, 65–76 (2014)
Cauchy, A.L. : Cours d’analyse de l’École Royale Polytechnique. In: Gabay, J. (ed.) 1ère partie: analyse algébrique. De l’Imprimerie royale, Paris (1821)
Debreu, G.: Representation of a preference ordering by a numerical function. In: Thrall, R., Coombs, C., Davies, R. (eds.) Decision Processes, pp. 159–166. Wiley, New York (1954)
Debreu, G.: Topological methods in cardinal utility theory. In: Arrow, K.J., Karlin, S., Suppes, P. (eds.) Mathematical methods in the social sciences 1959, pp. 16–26. Stanford University Press, Stanford, CA (1960)
Debreu, G.: Continuity properties of a Paretian utility. Int. Econ. Rev. 5, 285–293 (1964)
Eilenberg, S.: Ordered topological spaces. Am. J. Math. 63, 39–45 (1941)
Fishburn, P.: Utility Theory for Decision Making. Wiley, New York (1970)
Fishburn, P., Rubinstein, A.: Time preference. Int. Econ. Rev. 23, 677–694 (1982)
Gonzales, C.: Two factor additive conjoint measurement with one solvable component. J. Math. Psychol. 44, 285–309 (2000)
Krantz, D.H., Luce, R.D., Suppes, P., Tversky, A.: Foundations of Measurement Vol. I : Additive and Polynomial Representations. Academic Press, New York (1971). (2nd edn. Dover Publications, New York, 2007)
Mehta, G.B.: Preference and utility. In: Barbera, S., et al. (eds.) Handbook of Utility Theory, vol. 1, pp. 1–47. Kluwer Academic Publishers, Dordrecht (1998)
Rader, T.: The existence of a utility function to represent preferences. Rev. Econ. Stud. 30, 229–232 (1963)
Rubinstein, A.: Lecture Notes in Microeconomic Theory: The Economic Agent, 2nd edn. Princeton University Press, Princeton (2012)
Varian, H.: Microeconomic Analysis, 3rd edn. Norton and Company Inc., New York (1992)
Wakker, P.: The algebraic versus the topological approach to additive representations. J. Math. Psychol. 32, 421–435 (1988)
Wakker, P.: Additive Representations of Preferences: A New Foundation of Decision Analysis. Kluwer Academic Publishers, Dordrecht (1989)
Author information
Authors and Affiliations
Corresponding author
Appendices
Appendix
A A version of Cauchy’s functional equation
Let us recall Cauchy’s functional equation (CFE) and its solutions. We state it on \({\mathbb {R}}^+\).
Theorem
(Cauchy 1821) Let \(f : {\mathbb {R}}^+ \longrightarrow {\mathbb {R}}\) be a continuous function. Then,
if and only if there exists \( a > 0\) such that \(f = a \ Id\).
Necessarily \(f(0) = 0\) and a is unique. We may state a generalized Cauchy’s functional equation (GCFE).
Theorem A.1
Let \(\mathcal{C}\) be a connected topological space, \(0_{\mathcal{C}}\in \mathcal{C}\). Let \(v_1 : \mathcal{C}\longrightarrow {\mathbb {R}}\) be a continuous function with \(v_1 (0_{\mathcal{C}}) = 0, \ \sup _{\mathcal{C}} v_1 > 0\) and \(v_2 : {\mathbb {R}}^+ \longrightarrow {\mathbb {R}}\) with \(v_2 (0) = 0\). Then, the following conditions hold,

(i)
\(v_2\) is increasing,

(ii)
\(range(v_1) \subset range(v_2)\),

(iii)
(GCFE) : \(\forall y,z \in {\mathbb {R}}^+\), \(v_2 (z) \in range(v_1) \Rightarrow v_2 (z+ y) = v_2 (z) + v_2 (y) \),
if and only if there exists \(a > 0\) such that \(v_2 = a \ Id\).
Proof
(If). It is immediate to check.
(Only if). Since \(\sup _{\mathcal{C}} v_1 > 0\) there is some \(\overline{x}\) such that \(v_1 (\overline{x}) > 0 \). By (ii) \(range(v_1) \subset range(v_2)\), there exists \(\overline{z} \in {\mathbb {R}}^+\) such that \(v_2 (\overline{z} ) = v_1 (\overline{x})\) and by (i) \(v_2\) is increasing and \(v_2 (\overline{z} ) > 0 = v_2 (0)\), so we obtain \(\overline{z} > 0\).
Now, \(v_1 (\overline{x}) = v_2 (\overline{z} )\), thus by (iii) (GCFE,) but for \(\overline{z} \) fixed, we have for all \(y \in {\mathbb {R}}^+\),
Let us prove that
It is true for \(n =0\). Assume it is true for some \(n \in {\mathbb {N}}\). For \(y = n \overline{z} \), it comes \(v_2 ((n+1) \ \overline{z} ) = v_2 (\overline{z} ) + v_2 (n \ \overline{z} ) = v_2 (\overline{z} ) + n v_2 (\overline{z} ) = (n +1) \ v_2 (\overline{z} ) \).
This establishes the property for \(n + 1\).
We have \(v_1 (\overline{x}) > 0 = v_1 (0_{\mathcal{C}})\). Let \(p \in {\mathbb {N}}^*\). Since \(v_1\) is continuous and \(\mathcal{C}\) is connected, by the intermediate value theorem there exists some \(\overline{x}_p \in \mathcal{C}\) such that \(v_1 (\overline{x}_p) = v_1 (\overline{x}) / p \in [0, v_1 (\overline{x})]\).
By (ii) \(range(v_1) \subset range(v_2)\), then there exists \(\overline{z} _p \in {\mathbb {R}}^+\) such that \(v_2 (\overline{z} _p) = v_1 (\overline{x}_p) \).
Applying (iii) (GCFE), we may obtain \((*)\) with \(\overline{z} _p\) instead, that is,
Again, we may obtain \((**)\) with \(\overline{z} _p\) instead, so,
In particular, for \(n =p\), it comes
Since \(v_2\) is increasing and thus injective, we have \(p \ \overline{z} _p = \overline{z} \), that is, \(\overline{z} _p = \overline{z} / p\). Then,
and \((**p)\) becomes \(v_2 (n \ \overline{z} /p) = n \ v_2 (\overline{z} /p)\) for all \(n \in {\mathbb {N}}\); thus,
Let \(\alpha >0\). By denseness of \({\mathbb {Q}}^+\) in \({\mathbb {R}}^+\), there are sequences \((r_n)_n \subset {\mathbb {Q}}^+\) and \((s_n)_n \subset {\mathbb {Q}}^+\) such that \(r_n \uparrow \alpha \) and \( s_n \downarrow \alpha \), so \(r_n \overline{z} \uparrow \alpha \overline{z} \) and \( s_n \overline{z} \downarrow \alpha \overline{z} \). By increasingness of \(v_2\), it comes,
and
So, \(v_2 (\alpha \overline{z} ) = \alpha v_2( \overline{z} )\) holds for all \(\alpha > 0\).
Equivalently, we have, \(v_2 (y) = (y/\overline{z} ) \ v_2 (\overline{z} ) = (v_2 (\overline{z} )/\overline{z} ) \ y , \forall y > 0\). Now, \(v_2 (0) = 0\). Hence, \(v_2 = a \ Id \) with \(a = v_2 (\overline{z} )/\overline{z} > 0\). \(\square \)
Remark A.1
Continuity of \(v_2\) is not needed, though it is a byproduct.
Remark A.2
One can consider more generally some set \(\mathcal{C}\) and some \(v_1\) on \(\mathcal{C}\) satisfying an intermediate value property, i.e., \([0, v_1 (\overline{x})] \subset range (v_1)\).
Example A.1
(Cauchy’s functional equation) Let \(f : {\mathbb {R}}^+ \longrightarrow {\mathbb {R}}\) be an increasing function with \(f(0) = 0\).
Then, there exists \(a > 0\) such that \(f = a \ Id\) if either (i) or (ii) holds:

(i)
there exists some \(\overline{z} > 0\) such that f is continuous on \([ 0, \overline{z} ]\) and
$$\begin{aligned} \forall y \in {\mathbb {R}}^+ , \forall z \in [ 0, \overline{z} ], f(z+ y) = f(z) + f(y) \, \end{aligned}$$ 
(ii)
there exists some \(\overline{b}> 0\) such that \([ 0, \overline{b}] \subset range(f) \) and
$$\begin{aligned} \forall y, z \in {\mathbb {R}}^+ , f(z) \in [ 0, \overline{b}] \Rightarrow f(z+ y) = f(z) + f(y). \end{aligned}$$
Proof
It suffices to apply Theorem A.1.

(i)
Take \(\mathcal{C}= [ 0, \overline{z} ], 0_{\mathcal{C}}= 0\), \(v_1 = f\) on \([ 0, \overline{z} ]\) and \(v_2 = f\) on \({\mathbb {R}}^+\).

(ii)
Take \(\mathcal{C}= [ 0, \overline{b}], 0_{\mathcal{C}}= 0\), \(v_1 = Id_{[ 0, \overline{b}]}\) and \(v_2 = f\) on \({\mathbb {R}}^+\).
\(\square \)
In (ii), f is not supposed to be continuous but satisfies a range condition instead. The introduction of \(\mathcal{C}\) and \(v_1\) appears as an “auxiliary space.” Since \(v_1\) is continuous and nonconstant, \(range (v_1)\) is a nondegenerate interval containing 0. Hence, if \(v_2\) is increasing and continuous, \(v_2^{1} (range (v_1)) \) is a nondegenerate interval containing 0 too. So, what is precisely required is that \(v_2 (z+ y) = v_2 (z) + v_2 (y)\), for all \(y\in {\mathbb {R}}^+\) and \(z \in I\) with I a nondegenerate interval and \(0 \in I\).
B Proofs
Proof
Theorem 1
(If). It is immediate to check.
(Only if). Take \(\overline{x}\) given by (SENS), so \(v_1 (\overline{x}) > 0 \) and hence \(\sup _{\mathcal{C}} v_1 > 0\).
Let us check that the hypotheses (i)–(ii)–(iii) of Theorem A.1 are fulfilled.

(i)
It is straightforward to check that (i) \(v_2\) is increasing by (MD) since U is additive.

(ii)
Let \(x \in \mathcal{C}\). By (MEQ), there exists \(z_x \in {\mathbb {R}}^+\) such that \((x , 0) \sim (0_{\mathcal{C}}, z_x)\) and thus \(v_1 (x) = v_2 (z_x)\). So (ii) \(range(v_1) \subset range(v_2)\) is satisfied.

(iii)
Let \(z \in {\mathbb {R}}^+\) and \(v_2 (z) \in range(v_1)\). There exists \(x_z \in \mathcal{C}\) such that \(v_1 (x_z) = v_2 (z)\). So, \(U (x_z, 0) = v_1 (x_z) = v_2 (z) = U (0_{\mathcal{C}}, z) \). Or equivalently, \((x_z, 0) \sim (0_{\mathcal{C}}, z)\), and thus by (LM) we have for all \(y \in {\mathbb {R}}^+\), \( (x_z, y) \sim (0_{\mathcal{C}}, z + y)\). Hence,
$$\begin{aligned} v_1 (x_z) + v_2 (y) = U (x_z, y) = U (0_{\mathcal{C}}, z + y) = v_1 (0_{\mathcal{C}}) + v_2 (z + y) = v_2 (z + y), \end{aligned}$$that is, \(v_2 (z) + v_2 (y) = v_2 (z + y), \forall y \in {\mathbb {R}}^+ \). This establishes (iii) (GCFE) for \(v_2\).
Thus, by Theorem A.1, there exists some \(a > 0\) such that \(v_2 = a \ Id\). Hence, U is quasilinear. \(\square \)
Proof
Proposition 1

(i)
(If). Let \(x \in \mathcal{C}, z \in {\mathbb {R}}^+\) with \((x, 0) \succeq (0_{\mathcal{C}}, z)\). Then, \(v(x) \ge a z\), so \((v'(x)/a') a \ge a z\), that is, \(v'(x) \ge a' z\). Hence, \((x, 0) \succeq ' (0_{\mathcal{C}}, z)\). (Only if). Let \(x \in \mathcal{C}, z \in {\mathbb {R}}^+\). We have \((x, 0) \succeq (0_{\mathcal{C}}, z)\) if and only if \(v(x) \ge a z\). Take \(z = v(x)/a\). Then, \((x, 0) \succeq (0_{\mathcal{C}}, v(x)/a)\) and thus \((x, 0) \succeq ' (0_{\mathcal{C}}, v(x)/a)\). Hence, \(U' ((x, 0)) = v' (x) \ge U' ( (0_{\mathcal{C}}, v(x)/a) ) = a' v(x)/a\).

(ii)
(If). Let \(x,x' \in \mathcal{C}\) with \((x, 0) \succeq (x', 0)\). Then, \(v(x)/a \ge v(x')/a\), so \(v'(x)/a' = \phi (v(x)/a) \ge \phi (v(x')/a) = v'(x')/a' \). Thus, \((x, 0) \succeq ' (x', 0)\). (Only if). For \(x \in \mathcal{C}\), put \(\phi ((v (x) /a) = v' (x) /a' \). Let \(x,x' \in \mathcal{C}\). If \(v(x) = v(x')\), then \((x, 0) \sim (x', 0)\) so \((x, 0) \sim ' (x', 0)\), that is, \(v'(x)/a' = v'(x')/a'\). So \(\phi \) is unambiguously defined. Let us show that \(\phi \) is nondecreasing. Take, \(\beta \ge \beta '\) in \((v/a) (\mathcal{C})\). There are \(x,x' \in \mathcal{C}\) such that \(\beta = v(x)/a\) and \(\beta ' = v(x')/a\). So, \((x, 0) \succeq (x', 0)\), and then \((x, 0) \succeq ' (x', 0)\). That is, \(v'(x)/a' \ge v'(x')/a'\). Hence, \(\phi ( \beta ) = \phi ((v (x) /a) \ge \phi ((v (x') /a) = \phi ( \beta ')\).

(iii)
Combining (i) and (ii), we have that there exists some nondecreasing function such that \(v' /a' = \phi \circ v /a\) and then \(\phi \ge 1\) on \((v/a) (\mathcal{C})\) is equivalent to \((v'/a') = \phi \circ (v/a) \ge v/a\).

(iv)
(If). Let \(v'/a' = \varphi \ v/a\) with \(\varphi >0\). Let \( x, x' \in \mathcal{C}, z,z' \in {\mathbb {R}}^+, \alpha > 0, \) such that \( (x, 0) \sim (0_{\mathcal{C}}, z), (x, 0) \sim ' (0_{\mathcal{C}}, \alpha \ z), (x', 0) \sim (0_{\mathcal{C}}, z') \). Then, \(v(x) = a z\), \(v' (x) = a' \alpha z \), \(v(x') = a z'\). Thus, \(v' (x)/a' = a' \alpha z / a' = \alpha z\) and also \(v' (x)/a' = \varphi \ v (x)/a = \varphi \ (a z)/a = \varphi \ z\). So, with \(z > 0\), it comes \(\alpha = \varphi \). Finally, \(v' (x') = a' \varphi v(x')/a = a' \varphi (az')/a = a' \varphi z' = a' \alpha z'\), that is, \((x', 0) \sim ' (0_{\mathcal{C}}, \alpha \ z') \). (Only if). Let \( \overline{x}\in \mathcal{C}\) such that \(v(\overline{x}) > 0\). Then, \( (\overline{x}, 0) \sim (0_{\mathcal{C}}, v(\overline{x})/a)\) and \( (\overline{x}, 0) \sim ' (0_{\mathcal{C}}, v'(\overline{x})/a')\). So, \( (\overline{x}, 0) \sim ' (0_{\mathcal{C}}, \varphi \ v(\overline{x})/a)\) with \(\varphi = (v'(\overline{x})/a') / (v(\overline{x})/a)\) and \(\varphi \ge 0\) since \(v' \ge 0\). So for all, \(x' \in \mathcal{C}, z' \in {\mathbb {R}}^+\) such that \((x', 0) \sim (0_{\mathcal{C}}, z') \) we have by (A3), \((x', 0) \sim ' (0_{\mathcal{C}}, \varphi \ z') \). Since \(v(x') = a z'\) and \(v' (x') = a' (\varphi \ z') \), we obtain \(v' (x')/a' = \varphi \ z' = \varphi \ v(x')/a \). Since \(x'\) is arbitrary chosen, we have established that \(v'/a' = \varphi \times v/a \) with \(\varphi \ge 0\).

(v)
By conjunction of (i) and (iv).
\(\square \)
Proof
Lemma 1
It is clear that (MEQ) implies (BD).
Let \(x \in \mathcal{C}\). Put \({{\mathcal {U}}}(x) = \{ z \in {\mathbb {R}}^+ : (0_{\mathcal{C}}, z) \succeq (x , 0) \}\) and \(\mathcal{D}(x) = \{ z \in {\mathbb {R}}^+ : (0_{\mathcal{C}}, z) \preceq (x , 0) \}\). Let us prove that \({{\mathcal {U}}}(x)\) and \(\mathcal{D}(x)\) are closed.Thus, by Lemma 2 that we will prove hereafter we may conclude that (MEQ) is satisfied.
Let \(z_0 \notin {{\mathcal {U}}}(x)\). Then, \((x,0) \succ (0_{\mathcal{C}}, z_0)\) by (WO). By (CONT), \(\{ (s, z) : (x , 0) \succ (s, z) \}\) is open in the product topology. So for some open sets \((O_1 , O_2)\) with \( 0_{\mathcal{C}}\in O_1 \subset \mathcal{C}\) and \(z_0 \in O_2 \subset {\mathbb {R}}^+\) we have \((x , 0) \succ (s , z)\) for \((s , z) \in O_1 \times O_2\). In particular, \((x , 0) \succ (0_{\mathcal{C}}, z)\) for \(z \in O_2\). Hence, \(\{ z \in {\mathbb {R}}^+ : (0_{\mathcal{C}}, z) \prec (x , 0) \}\) is open, so \({{\mathcal {U}}}(x)\) is closed.
The case of \(\mathcal{D}(x)\) is treated in a similar way. \(\square \)
This proves that (CONT) \(\Rightarrow \) (CL) on topological spaces.
Proof
Theorem 2
(If). It is immediate to check.
(Only if). By (SENS), the first factor goods are essential. By (MD), the second factor money is essential. (INDG) and (INDM) are assumed, that is, coordinate independence in Wakker (1988). Since (WO), (CONT), (THC) are assumed, we may apply P. Wakker’s theorem. So, by Theorem 4.4 p. 427 in Wakker (1988), there exists an additive representation \(f = f_1 + f_2\) for \(\succeq \) with \(f_1, f_2\) continuous.
We may consider \(U = v_1 + v_2\) the additive utility function where \(v_1 = f_1  f_1 (0_{\mathcal{C}})\) and \(v_2 = f_2  f_2 (0)\). By Lemma 1, (MEQ) follows from (CONT) and (BD). Hence, \(0_{\mathcal{C}}\) is an origin. So, according to Theorem 1, U is quasilinear with \(v_1\) nonconstant, \(v_1 (0_{\mathcal{C}}) = 0\) and \(v_2 = a \ Id\) with \(a > 0\). Take \(v = a^{1} v_1\), then \(\succeq \) is represented by \(v + Id\).
(Moreover). Let us prove that v is unique. Let \(v'\) be another representation of \(\succeq \) with \(v' (0_{\mathcal{C}}) = 0\). Then, for all \(x \in \mathcal{C}\), \((x , 0) \sim (0_{\mathcal{C}}, v(x))\) and \((x , 0) \sim (0_{\mathcal{C}}, v'(x))\), thus \(v(x) = v' (x)\) by (WO) and (MD). \(\square \)
Proof
Corollary 1
(If). It is immediate to check.
(Only if). Let \(C , C' \in [1, + \infty )\). We have \((C,0) \sim (1 , v(C))\), thus by (LM) \((C, v(C')) \sim (1 , v(C) + v(C'))\). And also, \((C',0) \sim (1 , v(C'))\), thus by (PROP) \((C \ C',0) \sim (C , v(C'))\). So, by (WO), \((1 , v(C) + v(C')) \sim (C \ C',0)\). But, \((C \ C',0) \sim (1 , v(C \ C'))\). And this establishes,
In particular, \(v(1) = v(1) + v(1)\), so \(v(1) = 0\). We may extend this equation to \((0, + \infty )\). For this extend v to \({\tilde{v}}\) on \((0, + \infty )\). For \(c \in (0, 1), {\tilde{v}} (c) =  v(1/c)\). \({\tilde{v}}\) is continuous on \((0, + \infty )\). It is now standard to check that
This is the logarithmic form of Cauchy’s equation: \(f (yz) = f(y) + f(z)\) for all \(y, z > 0\).
v is increasing, for \(C' > C \ge 1\), with \(\lambda = C' /C > 1\) we have \(v(C') > v(C)\) since \((C', 0) \succ (C,0)\). Since \({\tilde{v}} (.) =  v (1 / (.))\) on (0, 1], we have \({\tilde{v}}\) increasing on (0, 1]. So \({\tilde{v}}\) increasing on \((0, + \infty )\). Since \({\tilde{v}}\) is continuous and increasing there exists some \(b >1\) such that \({\tilde{v}} = \log _b (.) = \ln (.) / \ln (b)\). So, we may take \(\rho = \ln (b) > 0\). \(\square \)
Proof
Lemma 2
It is clear that (MEQ) implies (BD).
Let \(x \in \mathcal{C}\). By (CL) the sets \({{\mathcal {U}}}(x) = \{ z \in {\mathbb {R}}^+ : (0_{\mathcal{C}}, z) \succeq (x , 0) \}\) and \(\mathcal{D}(x) = \{ z \in {\mathbb {R}}^+ : (0_{\mathcal{C}}, z) \preceq (x , 0) \}\) are closed in \({\mathbb {R}}^+\). By (BD) both are nonempty. By (WO), for any \(z \in {\mathbb {R}}^+\), either \((0_{\mathcal{C}}, z) \succeq (x , 0)\) or \((0_{\mathcal{C}}, z) \preceq (x , 0)\) and thus \({{\mathcal {U}}}(x) \cup \mathcal{D}(x) = {\mathbb {R}}^+\). So, by connectedness of \({\mathbb {R}}^+\) it comes \({{\mathcal {U}}}(x) \cap \mathcal{D}(x) \ne \emptyset \). Take \(v(x) \in {{\mathcal {U}}}(x) \cap \mathcal{D}(x)\), we have then \((0_{\mathcal{C}}, v(x)) \sim (x , 0)\) and (MEQ) is established. \(\square \)
Proof
Theorem 3
(If). We only check for (CL) the rest is immediate.
Let \(x \in \mathcal{C}\). Then, \(\{ z : z \in {\mathbb {R}}^+, (0_{\mathcal{C}}, z) \succeq (x, 0)\} = \{ z : z \in {\mathbb {R}}^+, z \ge v(x) \} = [ v(x) , + \infty ) \) and \(\{ z : z \in {\mathbb {R}}^+, (0_{\mathcal{C}}, z) \preceq (x, 0)\} = \{ z : z \in {\mathbb {R}}^+, z \le v(x) \} = [0 , v(x) ] \) which are closed in \({\mathbb {R}}^+\).
(Only if). Let \(0_{\mathcal{C}}\in \mathcal{C}\) be an “origin.” By Lemma 2, (MEQ) holds; thus, \(0_{\mathcal{C}}\) is an origin indeed. So for any \(x \in \mathcal{C}\) there exists some \(v(x) \in {\mathbb {R}}^+\) such that \((x,0) \sim (0_{\mathcal{C}}, v(x))\). Clearly, \(v(0_{\mathcal{C}}) = 0\). By (MD), such \(v(x) \in {\mathbb {R}}^+\) is unique. Let \((x,y) \in \mathcal{C}\times {\mathbb {R}}^+\). We have \((x,0) \sim (0_{\mathcal{C}}, v(x))\), so by (LM), \((x,y) \sim (0_{\mathcal{C}}, v(x) + y) \).
Then, for all \(x,x' \in \mathcal{C}\) and for all \(y,y' \in {\mathbb {R}}^+\), by (WO) and (MD),
So, \(U = v + Id\) represents the preferences. By (SENS), v is nonconstant.
(Moreover). Same proof as proof of Theorem 2. \(\square \)
Proof
Theorem 4
(If). It is immediate to check.
(Only if). According to Theorem 3, there exists a quasilinear utility function \(U = v + Id \) representing \(\succeq \), with v nonconstant and \(v(0_{\mathcal{C}}) = 0\).
Let us prove that v is continuous. Let \(x_0 \in \mathcal{C}\) and \(\epsilon >0\).
Then, \((x_0, \epsilon ) \succ (x_0, 0)\) since U is quasilinear. By (CONT), \(\{ (t , k) : (t , k) \succ (x_0, 0) \}\) is open, so there exists an open set \(O_0\) with \(x_0 \in O_0\) and \(\eta _0 > 0\) with \(\eta _0 < \epsilon \) such that \((x , z) \succ (x_0, 0)\) for all \((x,z) \in O_0 \times (\epsilon  \eta _0, \epsilon + \eta _0)\) and hence \(v(x) + z > v(x_0)\). So, \(v(x)> v(x_0)  z> v(x_0)  (\epsilon + \eta _0) > v(x_0)  2 \epsilon \). This proves that v is lower semicontinuous at \(x_0\).
Let us proceed similarly for upper semicontinuity. Let \(x_0 \in \mathcal{C}\) and \(\epsilon >0\) with \(\epsilon < 1/3\). We have \((x_0, 1  \epsilon ) \prec (x_0, 1)\) since U is quasilinear. By (CONT), \(\{ (t , k) : (t , k) \prec (x_0, 1) \}\) is open, so there exist an open set \(O'_0\) with \(x_0 \in O'_0\) and \(\eta '_0 > 0\) with \(\eta '_0 < \epsilon \) such that \((x , z) \prec (x_0, 1)\) for all \((x,z) \in O'_0 \times (1  \epsilon  \eta '_0, 1  \epsilon + \eta '_0)\) and hence \(v(x) + z < v(x_0) + 1\). So, \(v(x)< v(x_0) + 1  z< v(x_0) + 1  (1  \epsilon  \eta '_0) < v(x_0) + 2 \epsilon \). This proves that v is upper semicontinuous at \(x_0\).
(Moreover). Same proof as proof of Theorem 2. \(\square \)
Rights and permissions
About this article
Cite this article
Rébillé, Y. An axiomatization of continuous quasilinear utility. Decisions Econ Finan 40, 301–315 (2017). https://doi.org/10.1007/s102030170202z
Received:
Accepted:
Published:
Issue Date:
DOI: https://doi.org/10.1007/s102030170202z