A Theory for Estimating Consumer’s Preference from Demand

Abstract

This study shows that if the estimate error of a demand function satisfying the weak axiom of revealed preference is sufficiently small with respect to local C ¹ topology, then the estimate error of the corresponding preference relation (which is possibly nontransitive, but uniquely determined from demand function, and transitive under the strong axiom) is also sufficiently small. Furthermore, we show a similar relation for the estimate error of the inverse demand function with respect to the local uniform topology. These results hold when the consumption space is the positive orthant, but are not valid in the nonnegative orthant.

JEL Classification: D11.

Mathematics Subject Classification (2010): 91B16, 62P20.

Appendices

Appendix 1: Proof of Theorem 1

We at first introduce a lemma.

Lemma 1.

Let $(\succsim _{k})_{k}$ be a sequence on $\mathcal{P}$ and $\succsim \in \mathcal{P}$. Suppose

$$ \displaystyle\begin{array}{rcl} & x \succsim _{k}v \Leftrightarrow u_{k}(x,v) \geq 1& {}\\ & x \succsim v \Leftrightarrow u(x,v) \geq 1 & {}\\ \end{array} $$

for some real-valued function u_k,u on $\Omega ^{2}$. Assume u,u_k are continuous and0Note that under these conditions, $\succsim,\succsim _{k}$ must satisfy completeness and continuity.

$$ \displaystyle\begin{array}{rcl} & u_{k}(x,v) > 1 \Leftrightarrow u_{k}(v,x) < 1,u(x,v) > 1 \Leftrightarrow u(v,x) < 1,& {}\\ & u_{k}(x,av) = \frac{1} {a}u_{k}(x,v),u(x,av) = \frac{1} {a}u(x,v), & {}\\ & u_{k}(x,v) \geq 1\text{ and }u(x,v) \geq 1\text{ if }x \geq v. & {}\\ \end{array} $$

Then, $\text{Lim}_{k} \succsim _{k} =\succsim$ with respect to the closed convergence topology if $(u_{k})_{k}$ converges to u uniformly on any compact set.

Proof.

Suppose $(u_{k})_{k}$ converges to u uniformly on any compact set, and choose any $(x,v) \in \text{Limsup}_{k} \succsim _{k}$. Then, for any neighborhood U of (x, v) and any $k_{0} \in \mathbb{N}$, there exists $k \geq k_{0}$ such that $\succsim ^{k} \cap U\neq \varnothing$. Therefore, there exists an increasing sequence (k(m)) such that $u_{k(m)}(x_{m},v_{m}) \geq 1$ and $\|(x_{m},v_{m}) - (x,v)\| \leq \frac{1} {m}$. By the assumption of uniform convergence, we have $\vert u_{k(m)}(x_{m},v_{m}) - u(x_{m},v_{m})\vert \rightarrow 0$. Because u is continuous, $u(x_{m},v_{m}) \rightarrow u(x,v)$, and thus $u_{k(m)}(x_{m},v_{m}) \rightarrow u(x,v)$. As $u_{k(m)}(x_{m},v_{m}) \geq 1$, we have u(x, v) ≥ 1, and thus $\text{Limsup}_{k} \succsim _{k} \subset \succsim$.

Next, under the same assumption, choose any $(x,v) \in \succsim$. Then, we have u(x, v) ≥ 1. This implies $u(x,(1 - \frac{1} {m})v) > 1$ for any m. Now, take any neighborhood U of (x, v). Then, there exists m > 0 such that $(x,(1 - \frac{1} {m})v) \in U$. Because $u_{k}(x,(1 - \frac{1} {m})v) \rightarrow u(x,(1 - \frac{1} {m})v) > 1$, there exists $k_{0} \in \mathbb{N}$ such that $(x,(1 - \frac{1} {m})v) \in \succsim _{k}$ for any k ≥ k ₀. Hence, we have $\succsim \subset \text{Liminf}_{k} \succsim _{k}$. Therefore, we have $\text{Lim}_{k} \succsim _{k} =\succsim$. ■ 

Hence, it suffices to show that $(u^{g_{k}})_{ k}$ converges to u ^g uniformly on any compact set.

The next lemma was proved in Hosoya [6]. Thus, we omit the proof of this result.

Lemma 2.

Let $g: \Omega \rightarrow \mathbb{R}_{++}^{n}$ be C ¹ -class, and y(t;x,v) be the nonextendable solution of the following problem:

$$\displaystyle\begin{array}{rcl} \dot{y}& =& (g(y) \cdot x)v - (g(y) \cdot v)x, {}\\ y(0)& =& x, {}\\ \end{array}$$

where $x,v \in \Omega$ . Define $w(x,v) = (v \cdot x)v - (v \cdot v)x$ . Then, for any $z \in \text{span}\{x,v\}$ , z ⋅ w(x,v) = 0 if and only if z is proportional to v. Let ]a,b[ be the domain of $y(\cdot;x,v)$ . Then, $t(x,v) =\min \{ t \geq 0\vert y(t;x,v) \cdot w(x,v) \geq 0\}$ is well-defined. Moreover, there exists a continuous function $y_{1}(x,v),y_{2}(x,v)$ such that $y_{1}(x,v) = y_{2}(x,v) = x$ if x is proportional to v, and $y(t(x,v);x,v) \in [y_{1}(x,v),y_{2}(x,v)]$ for any $(x,v) \in \Omega ^{2}$ . Let

$$\displaystyle{A(x,v) =\| x\|\|v -\frac{v \cdot x} {\|x\|^{2}} x\|.}$$

Then, w(x,v) is orthogonal to v, w(x,v) ≠ 0 if and only if x is not proportional to v, and if w(x,v) ≠ 0,

$$\displaystyle{\dot{y} \cdot w(x,v) = A(x,v)^{2}(g(y) \cdot v) > 0.}$$

Hence, y(t;x,v) ⋅ w(x,v) is increasing in t, and thus t(x,v) is a unique t such that y(t;x,v) ⋅ w(x,v) = 0.

Lemma 3.

Suppose $(g_{k})_{k}$ converges to g uniformly on any compact set. Then, $(u^{g_{k}})_{ k}$ converges to u ^g uniformly on any compact set.

Proof.

Let y(t; x, v) be the nonextendable solution of the following equation:

$$\displaystyle\begin{array}{rcl} \dot{y}& =& (g(y) \cdot x)v - (g(y) \cdot v)x, {}\\ y(0)& =& x, {}\\ \end{array}$$

and y _k (t; x, v) be the nonextendable solution of the following equation:

$$\displaystyle\begin{array}{rcl} \dot{y}& =& (g_{k}(y) \cdot x)v - (g_{k}(y) \cdot v)x, {}\\ y(0)& =& x. {}\\ \end{array}$$

Let y(⋅ ; x, v) be defined on ]a(x, v), b(x, v)[. We separate this proof into four steps.

Step 1::

fix any $(x,v) \in \Omega ^{2}$ and c ∈ ]0, b(x, v)[. Then, there exists a compact neighborhood V of (x, v) such that y(⋅ ; y, z) is defined on [0, c] for any (y, z) ∈ V. The set $\{y(t;y,z)\vert t \in [0,c],(y,z) \in V \}$ is a compact set of $\Omega$, and thus there exists a compact set $C \subset \Omega$ such that the above set is included in the interior of C. We shall prove that, for sufficiently large k, $y_{k}(t;y,z)$ is defined for any $(t,y,z) \in [0,c] \times V$, $y_{k}(t;y,z) \in C$, and

$$\displaystyle{ \|y_{k}(t;y,z) - y(t;y,z)\| \leq \frac{\|g_{k} - g\|} {M} (e^{2M\|y\|\|z\|t} - 1), }$$

(1)

where M > 0 is a Lipschitz constant of g on C and $\|g_{k} - g\| =\sup _{w\in C}\|g_{k}(w) - g(w)\|$.

First,

$$\displaystyle\begin{array}{rcl} y_{k}(t;y,z) - y(t;y,z)& =& (y_{k}(t;y,z) - y) - (y(t;y,z) - y) {}\\ & =& \int _{0}^{t}[\dot{y}_{ k}(s;y,z) -\dot{ y}(s;y,z)]ds. {}\\ \end{array}$$

Hence, if t ∈ [0, c] is in the domain of $y_{k}(\cdot;y,z)$ and $y_{k}(s;y,z) \in C$ for any s ∈ [0, t], then we have

$$\displaystyle\begin{array}{rcl} & & \|y_{k}(t;y,z) - y(t;y,z)\| \\ & & \qquad \leq \int _{0}^{t}\|g_{ k}(y_{k}(s;y,z)) - g(y(s;y,z))\| \cdot 2\|y\|\|z\|ds \\ & & \qquad \leq \int _{0}^{t}\|g(y_{ k}(s;y,z)) - g(y(s;y,z))\| \cdot 2\|y\|\|z\|ds \\ & & \qquad +\int _{ 0}^{t}\|g_{ k}(y_{k}(s;y,z)) - g(y_{k}(s;y,z))\| \cdot 2\|y\|\|z\|ds \\ & & \qquad \leq 2\|y\|\|z\|\int _{0}^{t}[\|g(y_{ k}(s;y,z)) - g(y(s;y,z))\| +\| g_{k} - g\|]ds \\ & & \qquad \leq 2\|y\|\|z\|\int _{0}^{t}(M\|y_{ k}(s;y,z) - y(s;y,z)\| +\| g_{k} - g\|)ds, {}\end{array}$$

(2)

which implies equation (1).0See Sect. 2.3. Hence, if k is sufficiently large (and thus, $\|g_{k} - g\|$ is sufficiently small), then $\|y_{k}(t;y,z) - y(t;y,z)\| \leq \delta$, where δ > 0 is sufficiently small so that the following inequality holds:

$$\displaystyle{\inf _{w\notin C,t\in [0,c],(y,z)\in V }\|w - y(t;y,z)\| \geq 2\delta.}$$

Now, define $t^{{\ast}} =\sup \{ s \in [0,c]\vert \forall \tau \in [0,s],y_{k}(\tau;y,z) \in C\}$. Then y _k (t; y, z) ∈ C for any $t \in [0,t^{{\ast}}[$. If t ^∗ is not included in the domain of $y_{k}(\cdot;y,z)$, then there exists $\hat{t} < t^{{\ast}}$ such that $y_{k}(t;y,z)\notin C$ for any $t \in [\hat{t},t^{{\ast}}[$, a contradiction. Hence, t ^∗ is in the domain of $y_{k}(\cdot;y,z)$ and $y_{k}(t^{{\ast}};y,z) \in C$.

It suffices to show that t ^∗ = c. Suppose t ^∗ < c. Then, the above arguments implies $\|y_{k}(t^{{\ast}};y,z) - y(t^{{\ast}};y,z)\| \leq \delta$, and thus $y_{k}(t^{{\ast}};y,z)$ is included in the interior of C. Hence, y _k (⋅ ; y, z) is defined on $[t^{{\ast}},t^{{\ast}}+\varepsilon ]$ and $y_{k}([0,t^{{\ast}}+\varepsilon ];y,z) \subset C$ for sufficiently small $\varepsilon > 0$, a contradiction. This completes the proof of our claim.

Step 2::

define w(x, v) as in Lemma 2, and t(x, v) (resp. t _k (x, v)) as the minimum constant t ≥ 0 such that y(t; x, v) (resp. y _k (t; x, v)) is proportional to v. By Lemma 2, if x is not proportional to v, then t = t(x, v) if and only if y(t; x, v) ⋅ w(x, v) = 0 and t = t _k (x, v) if and only if $y_{k}(t;x,v) \cdot w(x,v) = 0$. Also, again by Lemma 2, t > t(x, v) if and only if y(t; x, v) ⋅ w(x, v) > 0 and t > t _k (x, v) if and only if $y_{k}(t;x,v) \cdot w(x,v) > 0$. Note that w(x, v) = 0 and $t(x,v) = t_{k}(x,v) = 0$ if and only if x is proportional to v.

Step 3::

choose any compact subset V of $\Omega ^{2}$. Fix any $\varepsilon > 0$. Let $W =\{ (x,v) \in V \vert \|y_{1}(x,v) - y_{2}(x,v)\| <\| v\|\varepsilon \}$, where $y_{1},y_{2}$ are as in Lemma 2. Because $y_{1},y_{2}$ is continuous, W is open in V, and thus $V ^{{\prime}} = V \setminus W$ is compact. By Lemma 2, if (x, v) ∈ W, then both $y_{k}(t_{k}(x,v);x,v)$ and y(t(x, v); x, v) are included in $[y_{1}(x,v),y_{2}(x,v)]$, and thus,

$$\displaystyle{\|y_{k}(t_{k}(x,v);x,v) - y(t(x,v);x,v)\| \leq \| v\|\varepsilon.}$$

Note that

$$\displaystyle{ u^{g}(x,v) = \frac{1} {\|v\|^{2}}y(t(x,v);x,v) \cdot v,\ u^{g_{k} }(x,v) = \frac{1} {\|v\|^{2}}y_{k}(t_{k}(x,v);x,v) \cdot v. }$$

(3)

Hence, if (x, v) ∈ W,

$$\displaystyle\begin{array}{rcl} \vert u^{g}(x,v) - u^{g_{k} }(x,v)\vert & =& \frac{1} {\|v\|^{2}}\vert (y_{k}(t_{k}(x,v);x,v) - y(t(x,v);x,v)) \cdot v\vert {}\\ &\leq & \frac{\|y_{k}(t_{k}(x,v);x,v) - y(t(x,v);x,v)\|} {\|v\|} {}\\ & \leq &\varepsilon. {}\\ \end{array}$$

Meanwhile, suppose $(x,v) \in V ^{{\prime}}$. If x is proportional to v, then $y_{1}(x,v) = y_{2}(x,v) = x$, and thus (x, v) ∈ W, a contradiction. Hence, x is not proportional to v. Because t(x, v) is the unique solution of the equation y(t; x, v) ⋅ w(x, v) = 0, by the implicit function theorem, the function t(x, v) is continuous on V ^′. Choose any $(x,v) \in V ^{{\prime}}$ and any $c(x,v) \in ]t(x,v),b(x,v)[$, and let V _(x, v) be a compact neighborhood of (x, v) in V ^′ such that $y(\cdot;y,z)$ is defined on [0, c(x, v)] and t(y, z) < c(x, v) for any $(y,z) \in V _{(x,v)}$. By step 1, we can take a compact set $C_{(x,v)} \subset \mathbb{R}_{++}^{n}$ such that

• (i) y(t; y, z) ∈ C _(x, v) for any $(t,y,z) \in [0,c(x,v)] \times V _{(x,v)}$,

• (ii) There exists k ₀ such that for any k ≥ k ₀ and $(y,z) \in V _{(x,v)}$, y _k (⋅ ; y, z) is also defined on [0, c(x, v)] and $y_{k}(t;y,z) \in C_{(x,v)}$ for any t ∈ [0, c(x, v)], and

• (iii) For any k ≥ k ₀, Eq. (1) holds for any t ∈ [0, c(x, v)].

Because V ^′ is compact, there is a finite collection $\{(x_{1},v_{1}),\ldots,(x_{m},v_{m})\}$ of V ^′ such that $\cup _{i}V _{(x_{i},v_{i})} = V ^{{\prime}}$. Hence, it suffices to show that for any $i \in \{ 1,\ldots,m\}$, there exists k _i such that for any k ≥ k _i , $\sup _{(x,v)\in V _{(x_{ i},v_{i})}}\vert u^{g_{k}}(x,v) - u^{g}(x,v)\vert \leq \varepsilon$. For notational convention, we abbreviate $V _{(x_{i},v_{i})}$ as V _i , $C_{(x_{i},v_{i})}$ as C _i , and $c(x_{i},v_{i})$ as c _i .

Step 4::

by definition of V _i , we have t(x, v) is in the domain of $y_{k}$ if k is sufficiently large. Hence,

$$\displaystyle\begin{array}{rcl} \vert u^{g_{k} }(x,v) - u^{g}(x,v)\vert & =& \frac{1} {\|v\|^{2}}\vert (y_{k}(t_{k}(x,v);x,v) - y(t(x,v);x,v)) \cdot v\vert {}\\ &\leq & \frac{1} {\|v\|^{2}}\vert (y_{k}(t_{k}(x,v);x,v) - y_{k}(t(x,v);x,v)) \cdot v\vert {}\\ & & + \frac{1} {\|v\|^{2}}\vert (y_{k}(t(x,v);x,v) - y(t(x,v);x,v)) \cdot v\vert. {}\\ & & {}\\ \end{array}$$

By Eq. (1),

$$\displaystyle\begin{array}{rcl} & & \frac{1} {\|v\|^{2}}\vert y_{k}(t(x,v);x,v) - y(t(x,v);x,v) \cdot v\vert {}\\ &\leq & \frac{1} {\|v\|}\|y_{k}(t(x,v);x,v) - y(t(x,v);x,v)\| {}\\ & \leq & \frac{\|g_{k} - g\|} {M\|v\|} (e^{2M\|x\|\|v\|c_{i} } - 1), {}\\ \end{array}$$

where the right-hand side is less than $\frac{\varepsilon }{2}$ uniformly on V _i if k is sufficiently large. Hence, it suffices to show that, for any sufficiently large k,

$$\displaystyle{\max _{(x,v)\in V _{i}} \frac{1} {\|v\|^{2}}\vert (y_{k}(t_{k}(x,v);x,v) - y_{k}(t(x,v);x,v)) \cdot v\vert \leq \frac{\varepsilon } {2}.}$$

By the Cauchy-Schwarz inequality, it suffices to show that, for any sufficiently large k,

$$\displaystyle{\max _{(x,v)\in V _{i}}\|y_{k}(t_{k}(x,v);x,v) - y_{k}(t(x,v);x,v)\| \leq \frac{\|v\|\varepsilon } {2}.}$$

Recall that $y(t;x,v) \cdot w(x,v) > 0$ (resp. $y_{k}(t;x,v) \cdot w(x,v) > 0$) if and only if t > t(x, v). (resp. $t > t_{k}(x,v)$.) As c _i  > t(x, v), we have there exists a positive constant C > 0 such that $y(c_{i};x,v) \cdot w(x,v) \geq C$ for any (x, v) ∈ V _i . Hence, for sufficiently large k, $y_{k}(c_{i};x,v) \cdot w(x,v) > 0$ for any (x, v) ∈ V _i . Therefore, $c_{i} > t_{k}(x,v)$, and thus t _k (x, v) is in the domain of y(⋅ ; x, v).

Now, define

$$\displaystyle\begin{array}{rcl} & M_{1} =\inf _{k\in \mathbb{N},(x,v)\in V _{i},y\in C_{i}}\{((g_{k}(y) \cdot x)v - (g_{k}(y) \cdot v)x) \cdot w(x,v)\},& {}\\ & M_{2} =\min _{(x,v)\in V _{i},y\in C_{i}}\{((g(y) \cdot x)v - (g(y) \cdot v)x) \cdot w(x,v)\}. & {}\\ \end{array}$$

By Lemma 2, when x is not proportional to v,

$$\displaystyle\begin{array}{rcl} & ((g_{k}(y) \cdot x)v - (g_{k}(y) \cdot v)x) \cdot w(x,v) = A(x,v)^{2}(g_{k}(y) \cdot v) > 0,& {}\\ & ((g(y) \cdot x)v - (g(y) \cdot v)x) \cdot w(x,v) = A(x,v)^{2}(g(y) \cdot v) > 0. & {}\\ \end{array}$$

Because $g_{k} \rightarrow g$ uniformly on C _i , we can show that $M_{1},M_{2} > 0$. As both y(t(x, v); x, v) and $y_{k}(t_{k}(x,v);x,v)$ are proportional to v, we have $y(t(x,v);x,v) \cdot w(x,v) = y_{k}(t_{k}(x,v);x,v) \cdot w(x,v) = 0$. If $t(x,v) \geq t_{k}(x,v)$, then

$$\displaystyle\begin{array}{rcl} y_{k}(t(x,v);x,v) \cdot w(x,v)& =& \int _{t_{k}(x,v)}^{t(x,v)}\dot{y}_{ k}(t;x,v) \cdot w(x,v)dt {}\\ & \geq &\int _{t_{k}(x,v)}^{t(x,v)}M_{ 1}dt {}\\ & =& M_{1}(t(x,v) - t_{k}(x,v)). {}\\ \end{array}$$

Hence, we have

$$\displaystyle{t(x,v) \leq t_{k}(x,v) + \frac{1} {M_{1}}y_{k}(t(x,v);x,v) \cdot w(x,v).}$$

Similarly, if $t(x,v) \leq t_{k}(x,v)$, then

$$\displaystyle\begin{array}{rcl} y(t_{k}(x,v);x,v) \cdot w(x,v)& =& \int _{t(x,v)}^{t_{k}(x,v)}\dot{y}(t;x,v) \cdot w(x,v)dt {}\\ & \geq &\int _{t(x,v)}^{t_{k}(x,v)}M_{ 2}dt {}\\ & =& M_{2}(t_{k}(x,v) - t(x,v)). {}\\ \end{array}$$

Hence, we have

$$\displaystyle{t(x,v) \geq t_{k}(x,v) - \frac{1} {M_{2}}y(t_{k}(x,v);x,v) \cdot w(x,v).}$$

Now, y(t(x, v); x, v) is proportional to v, and thus $y(t(x,v);x,v) \cdot w(y,z) = 0$. Hence, by Eq. (1), for any c > 0,

$$\displaystyle\begin{array}{rcl} & & \max _{(x,v)\in V _{i}}\vert y_{k}(t(x,v);x,v) \cdot w(x,v)\vert {}\\ &&\qquad =\max _{(x,v)\in V _{i}}\vert (y_{k}(t(x,v);x,v) - y(t(x,v);x,v)) \cdot w(x,v)\vert {}\\ &&\qquad \leq \max _{(x,v)\in V _{i}}\|y_{k}(t(x,v);x,v) - y(t(x,v);x,v)\|\|w(x,v)\| {}\\ & & \qquad \leq c, {}\\ \end{array}$$

for sufficiently large k. Similarly,

$$\displaystyle\begin{array}{rcl} & & \max _{(x,v)\in V _{i}}\vert y(t_{k}(x,v);x,v) \cdot w(x,v)\vert {}\\ &&\qquad =\max _{(x,v)\in V _{i}}\vert (y_{k}(t_{k}(x,v);x,v) - y(t_{k}(x,v);x,v)) \cdot w(x,v)\vert {}\\ &&\qquad \leq \max _{(x,v)\in V _{i}}\|y_{k}(t_{k}(x,v);x,v) - y(t_{k}(x,v);x,v)\|\|w(x,v)\| {}\\ & & \qquad \leq c, {}\\ \end{array}$$

for sufficiently large k. For such k, we have $\vert t(x,v) - t_{k}(x,v)\vert \leq \frac{c} {M_{3}}$, where $M_{3} =\min \{ M_{1},M_{2}\}$. Then, if c > 0 is sufficiently small,

$$\displaystyle\begin{array}{rcl} & & \|y_{k}(t_{k}(x,v);x,v) - y_{k}(t(x,v);x,v)\| {}\\ & & \qquad = \left \|\int _{t(x,v)}^{t_{k}(x,v)}\dot{y}_{ k}(t;x,v)dt\right \| {}\\ & & \qquad \leq \left \vert \int _{t(x,v)}^{t_{k}(x,v)}[\max _{ y\in C_{i}}\|(g(y) \cdot x)v - (g(y) \cdot v)x\|]dt\right \vert {}\\ &&\qquad \leq \max _{y\in C_{i}}\|(g(y) \cdot x)v - (g(y) \cdot v)x\| \frac{c} {M_{3}} {}\\ & & \qquad \leq \frac{\|v\|\varepsilon } {2}, {}\\ \end{array}$$

which completes the proof. ■ 

Lemmas 1 and 3 ensure that the claim of Theorem 1 holds. This completes the proof of Theorem 1. ■ 

Appendix 2: Proof of Theorem 2

Define $h(q) = f(q^{1},\ldots,q^{n-1},1,q^{n})$ and $h_{k}(q) = f_{k}(q^{1},\ldots,q^{n-1},1,q^{n})$. Then all $h,h_{k}: \mathbb{R}_{++}^{n} \rightarrow \mathbb{R}_{++}^{n}$ are invertible0That is, all h, h _k are bijective. Injectivity follows from the uniqueness of the inverse demand function. Surjectivity arises from the surjectivity and homogeneity of f and f _k . and, for any $q \in \mathbb{R}_{++}^{n}$, there exists a neighborhood U of q such that

$$\displaystyle\begin{array}{rcl} \lim _{k\rightarrow \infty }\sup _{r\in U}\|h_{k}(r) - h(r)\|& =& 0,{}\end{array}$$

(4)

$$\displaystyle\begin{array}{rcl} \lim _{k\rightarrow \infty }\sup _{r\in U}\|Dh_{k}(r) - Dh(r)\|& =& 0.{}\end{array}$$

(5)

Moreover, by the rank condition, we have all Dh(q), Dh _k (q) are regular.0See the mathematical appendix of Samuelson [14], or the proof of Proposition 1 in Hosoya [6]. Clearly,

$$\displaystyle{\lim _{k\rightarrow \infty }\|Dh_{k}(q)^{-1} - Dh(q)^{-1}\| = 0}$$

for any $q \in \mathbb{R}_{++}^{n}$.

Next, choose any $x \in \Omega$ and corresponding $q \in \mathbb{R}_{++}^{n}$ such that h(q) = x. Then, there exists $k_{0} \in \mathbb{N}$ such that some neighborhood of q is included in the domain of h _k for any k ≥ k ₀. We now introduce a lemma.

Lemma 4.

For any sufficiently small δ > 0, there exists M > 0 such that $\|h(r) - h(q)\| \leq \delta$ or $\|h_{k}(r) - h_{k}(q)\| \leq \delta$ for some k ≥ k ₀ implies $\|r - q\| \leq M\delta$ . Moreover, M can be chosen independent to δ.

Proof.

Let

$$\displaystyle\begin{array}{rcl} & \xi (r) = Dh(q)(r - q) - (h(r) - h(q)), & {}\\ & \xi _{k}(r) = Dh_{k}(q)(r - q) - (h_{k}(r) - h_{k}(q)).& {}\\ \end{array}$$

Then, by the assumption of convergence of h _k , $M = 2\sup \{\|Dh(q)^{-1}\|,$ $\|Dh_{k_{0}}(q)^{-1}\|,\ldots \}< +\infty$. Since $D\xi (r) = Dh(q) - Dh(r)$ and $D\xi _{k}(r) = Dh_{k}(q) - Dh_{k}(r)$, for any sufficiently small δ > 0, $\|r - q\| \leq M\delta$ implies that r is included in the domain of h _k for any k ≥ k ₀ and0See Eq. (5).

$$\displaystyle{\sup \{\|Dh(q)^{-1}\|\|D\xi (r)\|,\|Dh_{ k_{0}}(q)^{-1}\|\|D\xi _{ k_{0}}(r)\|,\ldots \}\leq \frac{1} {2}.}$$

Now, let $S =\{ s\vert \|s\| \leq M\delta \}$ and choose any $r \in \frac{1} {M}S$. Define,

$$\displaystyle{\phi _{r}(s) = Dh(q)^{-1}(r +\xi (s + q)),\phi _{ r}^{k}(s) = Dh_{ k}(q)^{-1}(r +\xi _{ k}(s + q)).}$$

Then, if s ∈ S,

$$\displaystyle\begin{array}{rcl} \|\phi _{r}(s)\|& \leq & \|Dh(q)^{-1}\|\|r\| +\| Dh(q)^{-1}\|\|\xi (s + q)\| {}\\ & \leq & \frac{M} {2} \delta +\| Dh(q)^{-1}\|\|D\xi (ts + q)\|\|s\| {}\\ & \leq & \frac{M} {2} \delta + \frac{1} {2}M\delta \leq M\delta, {}\\ \end{array}$$

where t ∈ [0, 1].0Use the mean value theorem for k(t) = ξ(ts + q). Hence, ϕ _r (S) ⊂ S. Moreover,

$$\displaystyle\begin{array}{rcl} \|\phi _{r}(s_{1}) -\phi _{r}(s_{2})\|& \leq & \|Dh(q)^{-1}\|\|\xi (s_{ 1} + q) -\xi (s_{2} + q)\| {}\\ & \leq & \|Dh(q)^{-1}\|\|D\xi (s_{ 3} + q)\|\|s_{1} - s_{2}\| \leq \frac{1} {2}\|s_{1} - s_{2}\|, {}\\ \end{array}$$

where $s_{3} \in [s_{1},s_{2}]$.0Again, use the mean value theorem for $k(t) =\xi ((1 - t)s_{1} + ts_{2} + q)$. Hence, ϕ _r is a contraction. By contraction mapping fixed point theorem, there uniquely exists s ∈ S such that ϕ _r (s) = s, that is,

$$\displaystyle{s = Dh(q)^{-1}r + s - Dh(q)^{-1}(h(s + q) - h(q)),}$$

and thus h(s + q) = h(q) + r. Because h is an injection, we have $\|h(s + q) - h(q)\| \leq \delta$ implies $\|s\| \leq M\delta$. Similarly, we have $\|h_{k}(s + q) - h_{k}(q)\| \leq \delta$ implies $\|s\| \leq M\delta$. This completes the proof of Lemma 4. ■ 

Note that y = h(r) if and only if $r = (g^{1}(y),\ldots,g^{n-1}(y),g(y) \cdot y)$ and y = h _k (r) if and only if $r = (g_{k}^{1}(y),\ldots,g_{k}^{n-1}(y),g_{k}(y) \cdot y)$. For any $\varepsilon > 0$ there exists δ > 0 such that $\|h(r) - h(q)\| \leq \delta$ or $\|h_{k}(r) - h_{k}(q)\| \leq \delta$ for some k ≥ k ₀ implies $\|r - q\| \leq M\delta$ and $M\delta \leq \varepsilon$. Let $U =\{ y \in \Omega \vert \|x - y\| \leq \delta \}$. Then for any y ∈ U,

$$\displaystyle{\|g_{k}(y) - g_{k}(x)\| \leq \| h_{k}^{-1}(y) - h_{ k}^{-1}(x)\| \leq \varepsilon.}$$

Hence we conclude that $(g_{k})_{k}$ is equicontinuous at x. As x is an arbitrary point of $\Omega$, we conclude that $(g_{k})_{k}$ is equicontinuous.

Next, we will show that $g_{k}(x) \rightarrow g(x)$ pointwise. Let $h(q) = h_{k}(q_{k}) = x$. It suffices to show that $q_{k} \rightarrow q$. By assumption, there exists a neighborhood V of q and $k_{0} \in \mathbb{N}$ such that V is included in the domain of h _k for any $k \geq k_{0}$. Without loss of generality, we can assume that V is compact. By assumption, $(h_{k})_{k}$ converges to h uniformly on V and thus, for any δ > 0, there exists $k_{1} \in \mathbb{N}$ such that $\|h_{k}(q) - h(q)\| \leq \delta$ for any k ≥ k ₁. However, as $h(q) = x = h_{k}(q_{k})$, we have $\|h_{k}(q) - h_{k}(q_{k})\| \leq \delta$. Hence, by Lemma 4, if δ > 0 is sufficiently small, then $\|q - q_{k}\| \leq M\delta$. Thus, $q_{k} \rightarrow q$.

Therefore, $(g_{k})_{k}$ is equicontinuous and $g_{k}(x) \rightarrow g(x)$ for any $x \in \Omega$. Fix any $x \in \Omega$ and let h(q) = x. Choose any sufficiently small δ > 0 such that $\|h(r) - h(q)\| \leq \delta$ or $\|h_{k}(r) - h_{k}(q)\| \leq \delta$ for some k ≥ k ₀ implies $\|r - q\| \leq M\delta$. Then, if $\|y - x\| \leq \delta$,

$$\displaystyle\begin{array}{rcl} \|g_{k}(y)\|& \leq & \|h_{k}^{-1}(y)\| + 1 {}\\ & \leq & \|h_{k}^{-1}(y) - h_{ k}^{-1}(x)\| +\| h_{ k}^{-1}(x)\| + 1 {}\\ & \leq & M\delta +\sup _{\ell\geq k_{0}}\|h_{\ell}^{-1}(x)\| + 1 < +\infty, {}\\ \end{array}$$

and thus $(g_{k})_{k}$ is uniformly bounded on $B =\{ y \in \Omega \vert \|y - x\| \leq \delta \}$. If δ > 0 is sufficiently small, then B is compact. Hence, by Ascoli-Arzela theorem, any subsequence of $(g_{k})_{k}$ has a convergent subsubsequence.

Suppose that $(g_{k})_{k}$ does not converge to g uniformly on B. Then, there exists $\varepsilon > 0$ and a subsequence $(g_{m(k)})_{k}$ such that

$$\displaystyle{\sup _{y\in B}\|g_{m(k)}(y) - g(y)\| \geq \varepsilon.}$$

By the above result, there exists a uniformly convergent subsubsequence $(g_{\ell(k)})_{k}$. Because g _k converges to g pointwise, we conclude that

$$\displaystyle{\sup _{y\in B}\|g_{\ell(k)}(y) - g(y)\| <\varepsilon }$$

for some k, a contradiction. This completes the proof. ■