<p>It follows indeed from the first three conditions, though I did not find a simple proof. Here is a messy one:</p>
<p>Observe first that, <span class="math-container">$v(tx)&gt;p\cdot tx$</span> with <span class="math-container">$t&gt;0$</span> is equivalent to <span class="math-container">$v(tx)/t&gt;px$</span>. By replacing <span class="math-container">$p$</span> by some multiple, we can see that the right-side can take any value without changing the left value. The function <span class="math-container">$t\mapsto v(tx)$</span> is strictly concave, strictly increasing, and has value <span class="math-container">$0$</span> at <span class="math-container">$0$</span>. The problem is then equivalent to showing that its right derivative is infinite for all <span class="math-container">$x\gg 0$</span>. That the right derivative exists <a href="https://en.wikipedia.org/wiki/Convex_function#Functions_of_one_variable" rel="nofollow noreferrer">follows from the concavity</a> of <span class="math-container">$v$</span>.</p>
<p>As it turns out, it is enough to find a single <span class="math-container">$x^*\gg0$</span> such that <span class="math-container">$t\mapsto v(tx^*)$</span> has an infinite right derivative at <span class="math-container">$0$</span>.</p>
<p>To see this, let <span class="math-container">$x\gg 0$</span>. There is some <span class="math-container">$\delta&gt;0$</span> such that <span class="math-container">$x\gg \delta x^*$</span>. Take some <span class="math-container">$K&gt;0$</span>. For <span class="math-container">$t&gt;0$</span> small enough, we have <span class="math-container">$v\big(\delta t~x^*\big)/(\delta t)&gt;K/\delta $</span> or <span class="math-container">$v(\delta t x^*)/t&gt;K.$</span>
Since <span class="math-container">$v$</span> is strictly increasing on the interior,
<span class="math-container">$v(tx)/t&gt;v(t\delta x^*)/t&gt;K$</span> for <span class="math-container">$t&gt;0$</span> small enough. Therefore, the function <span class="math-container">$t\mapsto v(tx)$</span> must have infinite right derivative at <span class="math-container">$0$</span>.</p>
<p>It remains to find the damn <span class="math-container">$x^*$</span>. Let <span class="math-container">$\Delta$</span> be the <span class="math-container">$K-1$</span>-dimensional unit simplex. For each <span class="math-container">$N$</span>, there must exist some <span class="math-container">$x\in\Delta$</span> such that the function <span class="math-container">$t\mapsto v(tx)$</span> has a right derivative of at least <span class="math-container">$N$</span> at <span class="math-container">$0$</span>. Otherwise, for <span class="math-container">$p=(N,N,\ldots,N)$</span> we would have <span class="math-container">$A(p)-\{0\}=\emptyset.$</span> Also, <span class="math-container">$A(p)-\{0\}\subseteq\mathbb{R}^K_{++}$</span> implies that such an <span class="math-container">$x$</span> must be positive in each coordinate. So let <span class="math-container">$x_n\in\Delta$</span> be such that <span class="math-container">$t\mapsto v(tx_n)$</span> has a right derivative of at least <span class="math-container">$n$</span> at <span class="math-container">$0$</span>. Now let
<span class="math-container">$$x^*=\sum_{i=1}^\infty\frac{1}{2^i}x_{2^i}.$$</span>
That the relevant series converges follows from any norm being bounded on the compact set <span class="math-container">$\Delta$</span> and <a href="https://en.wikipedia.org/wiki/Absolute_convergence#Proof_that_any_absolutely_convergent_series_in_a_Banach_space_is_convergent" rel="nofollow noreferrer">absolute convergence implying convergence</a>. Clearly, <span class="math-container">$x^*\gg 0$</span> since <span class="math-container">$x_{2^i}\gg 0$</span> for all <span class="math-container">$i$</span>. It remains to show that <span class="math-container">$t\mapsto v(tx^*)$</span> has an infinite right derivative at <span class="math-container">$0$</span>. Since
<span class="math-container">$$t^{-1}~v\bigg(t\sum_{i=1}^\infty\frac{1}{2^i}x_{2^i}\bigg)&gt;t^{-1}~v\bigg(t\sum_{i=1}^n\frac{1}{2^i}x_{2^i}\bigg)$$</span>
for all <span class="math-container">$n$</span> by <span class="math-container">$v$</span> being strictly increasing on the interior, it suffices to show that for each <span class="math-container">$K&gt;0$</span>, there exists some <span class="math-container">$n$</span> such that
<span class="math-container">$$t^{-1}~v\bigg(t\sum_{i=1}^n\frac{1}{2^i}x_{2^i}\bigg)&gt;K$$</span> for <span class="math-container">$t&gt;0$</span> small enough. Let <span class="math-container">$S_n=\big(\sum_{i=1}^n 1/2^i\big)^{-1}$</span> and note that
<span class="math-container">$$t^{-1}v\bigg(tS_n\sum_{i=1}^n\frac{1}{2^i}x_{2^i}\bigg)=t^{-1}v\bigg(\sum_{i=1}^n S_n\frac{1}{2^i}t x_{2^i}\bigg)\geq\sum_{i=1}^n S_n\frac{1}{2^i}t^{-1}v(tx_{2^i})$$</span>
by the concavity of <span class="math-container">$v$</span>. Since <span class="math-container">$\lim_{t\downarrow 0}t^{-1}v(t x_{2^i})\geq 2^i$</span>, this sum is almost <span class="math-container">$S_n n$</span> for <span class="math-container">$t&gt;0$</span> small enough. Now <span class="math-container">$S_n$</span> converges to <span class="math-container">$1$</span>, so <span class="math-container">$S_n n$</span> is larger than <span class="math-container">$K$</span> for <span class="math-container">$n$</span> large enough and <span class="math-container">$t&gt;0$</span> small enough.</p>

CC BY-SA 4.0
2021-03-29T16:41:44.190Z
+1 for finding "the damn $x^*$."
Herr K.
CC BY-SA 4.0
2021-03-30T06:18:16.637Z
The construction of $x^*$ is very clever. The statement of existing $x$ that make the map $t \to v(t x)$ has arbitrarily large right derivative at 0 is simple. But the statement of existing $x$ that make the map $t \to v(t x)$ has infinite right derivative at 0 is not simple. Thanks for a very nice proof.
Khánh Toàn
CC BY-SA 4.0
2021-03-30T07:15:22.233Z
@KhánhToàn if the answer answered your question do not forget to accept it (you can do that by clicking on the tick under votes)
1muflon1
CC BY-SA 4.0
2021-03-30T07:53:48.653Z
@1muflon1 Oops I never know about that. Thanks for telling me.
Khánh Toàn
CC BY-SA 4.0
2021-03-29T15:08:56.467Z
2021-03-29T17:43:18.913Z
2021-03-29T17:43:18.913Z
Michael Greinecker
<p>A function <span class="math-container">$v: \mathbb{R}^K_+ \xrightarrow{} \mathbb{R}_+$</span> is is said to be a valuation function if</p>
<ol>
<li>The value of function <span class="math-container">$v$</span> at <span class="math-container">$x = \textbf{0}$</span> is <span class="math-container">$0$</span>: <span class="math-container">$v(\textbf{0}) = 0$</span></li>
<li><span class="math-container">$v$</span> is continuous on the domain <span class="math-container">$\mathbb{R}^K_+$</span>, strictly increasing and strictly concave on <span class="math-container">$\mathbb{R}^K_{++}$</span></li>
<li>For any <span class="math-container">$p \in \mathbb{R}^K_{++}$</span>, the set <span class="math-container">$A(p) = \{x \in \mathbb{R}^K_+| v(x) \geq p \cdot x\}$</span> is compact, the set <span class="math-container">$A^*(p) = A (p)-\{\bf{0}\}$</span> is nonempty and lies inside <span class="math-container">$\mathbb{R}^K_{++}$</span>.</li>
</ol>
<p>Show that a if <span class="math-container">$v$</span> is a valuation function, for every <span class="math-container">$p \in \mathbb{R}^K_{++}$</span>, for every <span class="math-container">$x \in \mathbb{R}^K_{++}$</span>, there exist <span class="math-container">$t&gt;0$</span> small enough that
<span class="math-container">$$
v(tx) &gt; p\cdot tx
$$</span></p>
<p>In other word, for every <span class="math-container">$p&gt;\bf{0}$</span>, if you zoom the set <span class="math-container">$A(p)$</span> sufficiently close at the origin, it contains all the points near the origin in first quadrant.</p>
<p><strong>Remark 1</strong>: This is the problem I reduce from a microeconomics problem I am working on.  I want to extend the Inada condition to multivariate case in a topological way. My prototype function is <span class="math-container">$f(x,y)=x^{\alpha}y^{\beta}$</span> where <span class="math-container">$\alpha+\beta &lt;1, \alpha&gt;0,\beta&gt;0$</span>.
This family of functions satisfy (1),(2),(3) and the property I want to prove, say (4). I am trying to show either (1),(2),(3) =&gt; (4) or should I include (4) as an additional axiom?</p>
<p><strong>Remark 2</strong>: A few examples of <span class="math-container">$A(p)$</span></p>
<ol>
<li>An example of A(p) when p is small</li>
</ol>
<p><a href="https://i.sstatic.net/GkNop.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/GkNop.jpg" alt="An example of A(p) when p is small " /></a></p>
<ol start="2">
<li>An example of A(p) when p is large</li>
</ol>
<p><a href="https://i.sstatic.net/CHZeB.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/CHZeB.jpg" alt="An example of A(p) when p is large" /></a></p>
<p>(4) is demonstrated by the geometric fact that the contours approach vertical and horizontal axis near <span class="math-container">$0$</span>.</p>

<microeconomics>
Seemingly simple consumer theory problem
Khánh Toàn
# Seemingly simple consumer theory problem

(Asked by: username_0 on 2021-03-28)

A function $v: \mathbb{R}^K\_+ \xrightarrow{} \mathbb{R}\_+$ is is said to be a valuation function if

1. The value of function $v$ at $x = \textbf{0}$ is $0$: $v(\textbf{0}) = 0$
2. $v$ is continuous on the domain $\mathbb{R}^K\_+$, strictly increasing and strictly concave on $\mathbb{R}^K\_{++}$
3. For any $p \in \mathbb{R}^K\_{++}$, the set $A(p) = \{x \in \mathbb{R}^K\_+| v(x) \geq p \cdot x\}$ is compact, the set $A^\*(p) = A (p)-\{\bf{0}\}$ is nonempty and lies inside $\mathbb{R}^K\_{++}$.

Show that a if $v$ is a valuation function, for every $p \in \mathbb{R}^K\_{++}$, for every $x \in \mathbb{R}^K\_{++}$, there exist $t>0$ small enough that
$$
v(tx) > p\cdot tx
$$

In other word, for every $p>\bf{0}$, if you zoom the set $A(p)$ sufficiently close at the origin, it contains all the points near the origin in first quadrant.

**Remark 1**: This is the problem I reduce from a microeconomics problem I am working on. I want to extend the Inada condition to multivariate case in a topological way. My prototype function is $f(x,y)=x^{\alpha}y^{\beta}$ where $\alpha+\beta <1, \alpha>0,\beta>0$.
This family of functions satisfy (1),(2),(3) and the property I want to prove, say (4). I am trying to show either (1),(2),(3) => (4) or should I include (4) as an additional axiom?

**Remark 2**: A few examples of $A(p)$

1. An example of A(p) when p is small

[![An example of A(p) when p is small ](https://i.sstatic.net/GkNop.jpg)](https://i.sstatic.net/GkNop.jpg)

2. An example of A(p) when p is large

[![An example of A(p) when p is large](https://i.sstatic.net/CHZeB.jpg)](https://i.sstatic.net/CHZeB.jpg)

(4) is demonstrated by the geometric fact that the contours approach vertical and horizontal axis near $0$.

**Tags:** microeconomics

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## 1 answers

### Answer by username_1 on 2021-03-29. Score: 8 | Accepted answer

It follows indeed from the first three conditions, though I did not find a simple proof. Here is a messy one:

Observe first that, $v(tx)>p\cdot tx$ with $t>0$ is equivalent to $v(tx)/t>px$. By replacing $p$ by some multiple, we can see that the right-side can take any value without changing the left value. The function $t\mapsto v(tx)$ is strictly concave, strictly increasing, and has value $0$ at $0$. The problem is then equivalent to showing that its right derivative is infinite for all $x\gg 0$. That the right derivative exists [follows from the concavity](https://en.wikipedia.org/wiki/Convex_function#Functions_of_one_variable) of $v$.

As it turns out, it is enough to find a single $x^\*\gg0$ such that $t\mapsto v(tx^\*)$ has an infinite right derivative at $0$.

To see this, let $x\gg 0$. There is some $\delta>0$ such that $x\gg \delta x^\*$. Take some $K>0$. For $t>0$ small enough, we have $v\big(\delta t~x^\*\big)/(\delta t)>K/\delta $ or $v(\delta t x^\*)/t>K.$
Since $v$ is strictly increasing on the interior,
$v(tx)/t>v(t\delta x^\*)/t>K$ for $t>0$ small enough. Therefore, the function $t\mapsto v(tx)$ must have infinite right derivative at $0$.

It remains to find the damn $x^\*$. Let $\Delta$ be the $K-1$-dimensional unit simplex. For each $N$, there must exist some $x\in\Delta$ such that the function $t\mapsto v(tx)$ has a right derivative of at least $N$ at $0$. Otherwise, for $p=(N,N,\ldots,N)$ we would have $A(p)-\{0\}=\emptyset.$ Also, $A(p)-\{0\}\subseteq\mathbb{R}^K\_{++}$ implies that such an $x$ must be positive in each coordinate. So let $x\_n\in\Delta$ be such that $t\mapsto v(tx\_n)$ has a right derivative of at least $n$ at $0$. Now let
$$x^\*=\sum\_{i=1}^\infty\frac{1}{2^i}x\_{2^i}.$$
That the relevant series converges follows from any norm being bounded on the compact set $\Delta$ and [absolute convergence implying convergence](https://en.wikipedia.org/wiki/Absolute_convergence#Proof_that_any_absolutely_convergent_series_in_a_Banach_space_is_convergent). Clearly, $x^\*\gg 0$ since $x\_{2^i}\gg 0$ for all $i$. It remains to show that $t\mapsto v(tx^\*)$ has an infinite right derivative at $0$. Since
$$t^{-1}~v\bigg(t\sum\_{i=1}^\infty\frac{1}{2^i}x\_{2^i}\bigg)>t^{-1}~v\bigg(t\sum\_{i=1}^n\frac{1}{2^i}x\_{2^i}\bigg)$$
for all $n$ by $v$ being strictly increasing on the interior, it suffices to show that for each $K>0$, there exists some $n$ such that
$$t^{-1}~v\bigg(t\sum\_{i=1}^n\frac{1}{2^i}x\_{2^i}\bigg)>K$$ for $t>0$ small enough. Let $S\_n=\big(\sum\_{i=1}^n 1/2^i\big)^{-1}$ and note that
$$t^{-1}v\bigg(tS\_n\sum\_{i=1}^n\frac{1}{2^i}x\_{2^i}\bigg)=t^{-1}v\bigg(\sum\_{i=1}^n S\_n\frac{1}{2^i}t x\_{2^i}\bigg)\geq\sum\_{i=1}^n S\_n\frac{1}{2^i}t^{-1}v(tx\_{2^i})$$
by the concavity of $v$. Since $\lim\_{t\downarrow 0}t^{-1}v(t x\_{2^i})\geq 2^i$, this sum is almost $S\_n n$ for $t>0$ small enough. Now $S\_n$ converges to $1$, so $S\_n n$ is larger than $K$ for $n$ large enough and $t>0$ small enough.

> **Comments:**
> **username_2**: +1 for finding "the damn $x^*$."  
> **username_0**: The construction of $x^*$ is very clever. The statement of existing $x$ that make the map $t \to v(t x)$ has arbitrarily large right derivative at 0 is simple. But the statement of existing $x$ that make the map $t \to v(t x)$ has infinite right derivative at 0 is not simple. Thanks for a very nice proof.  
> **username_3**: @KhánhToàn if the answer answered your question do not forget to accept it (you can do that by clicking on the tick under votes)  
> **username_0**: @1muflon1 Oops I never know about that. Thanks for telling me.  

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