An axiomatization of continuous quasilinear utility

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.

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,

$$\begin{aligned} f \text { is increasing and satisfies} \ (CFE) \ \forall y,z \in {\mathbb {R}}^+ , f (y+z) = f(y) + f(z) \end{aligned}$$

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

$$\begin{aligned} v_2 ( \ \overline{z} _p) = p \ v_2 (\overline{z} _p) = p \ v_1 (\overline{x}_p) = p \ v_1 (\overline{x}) / p = v_1 (\overline{x}) = v_2 (\overline{z} ). \end{aligned}$$

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,

$$\begin{aligned} v_2 ((n/p) \ \overline{z} ) = n \ v_2 ( \overline{z} /p ) = n \ v_2 (\overline{z} ) /p = (n/p) \ v_2 (\overline{z} ), \forall (n,p) \in {\mathbb {N}}\times {\mathbb {N}}^* \ . \end{aligned}$$

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,

$$\begin{aligned} v_2 (\alpha \overline{z} ) \ge \lim _n v_2(r_n \overline{z} ) = \lim _n r_n v_2( \overline{z} ) = \alpha v_2( \overline{z} ) \end{aligned}$$

and

$$\begin{aligned} v_2 (\alpha \overline{z} ) \le \lim _n v_2(s_n \overline{z} ) = \lim _n s_n v_2( \overline{z} ) = \alpha v_2( \overline{z} ). \end{aligned}$$

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 by-product.

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 non-constant, $range (v_1)$ is a non-degenerate interval containing 0. Hence, if $v_2$ is increasing and continuous, $v_2^{-1} (range (v_1))$ is a non-degenerate 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 non-degenerate 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 non-decreasing. 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 non-decreasing 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 $v_1$ non-constant, $v_1 (0_{\mathcal{C}}) = 0$ 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,

$$\begin{aligned} v(C \ C') = v(C) + v(C') , \text { for all } C , C' \in [1, + \infty ). \end{aligned}$$

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

$$\begin{aligned} {\tilde{v}} (C \ C') = {\tilde{v}} (C) + {\tilde{v}} (C') , \text { for all } C , C' \in (0, + \infty ). \end{aligned}$$

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 non-empty. 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),

$$\begin{aligned} (x,y) \succeq (x' , y')\iff & {} (0_{\mathcal{C}}, v(x) + y) \succeq (0_{\mathcal{C}}, v(x') + y')\\ {}\iff & {} v(x) + y \ge v(x') + y'. \end{aligned}$$

So, $U = v + Id$ represents the preferences. By (SENS), v is non-constant.

(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 non-constant 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 semi-continuous at $x_0$.

Let us proceed similarly for upper semi-continuity. 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 semi-continuous at $x_0$.

(Moreover). Same proof as proof of Theorem 2. $\square$