<p>This problem is quite specific to economics. The correct statement is:</p>

<p><strong>Proposition</strong> 
If <span class="math-container">$u(\cdot)$</span> is quasiconcave, strictly increasing, and continuous, then <span class="math-container">$\forall x$</span>, there exists <span class="math-container">$p \gg 0$</span> and <span class="math-container">$w \geq 0$</span> such that <span class="math-container">$x \in x^*(p, w)$</span>, where <span class="math-container">$x^*(p, w)$</span> is the Marshallian demand correspondence.</p>

<p><strong>Proof</strong></p>

<p>Quasiconcavity of <span class="math-container">$u$</span> means the upper-contour set <span class="math-container">$\succeq\!\!(x) = \{x': u(x')\geq u(x)\}$</span> of bundles weakly preferred over <span class="math-container">$x$</span> is convex.</p>

<p>By continuity of <span class="math-container">$u$</span>, <span class="math-container">$\succeq\!\!(x)$</span> is closed.</p>

<p>By strict monotonicity of <span class="math-container">$u$</span>, <span class="math-container">$x$</span> lies on the boundary of <span class="math-container">$\succeq\!\!(x)$</span>.
Consider the supporting hyperplane <span class="math-container">$p\cdot (x' - x) = 0$</span> of the closed convex set <span class="math-container">$\succeq\!\!(x)$</span> at boundary point <span class="math-container">$x$</span> with normal vector <span class="math-container">$p$</span>.
Since <span class="math-container">$u$</span> is increasing, then we can take <span class="math-container">$p \gg 0$</span>. </p>

<p>Since <span class="math-container">$\succ\!\!(x) \subset \{ p\cdot (x' - x) &gt; 0 \}$</span> (general relationship between the interior of a convex set and a half space given by a supporting hyperplane),
<span class="math-container">$x \in x^*(p, \,p\cdot x)$</span>. <span class="math-container">$\Box$</span></p>

<p>Differentiability is not needed. If <span class="math-container">$u$</span> is strictly quasiconcave, the Marshallian demand correspondence is a function, and <span class="math-container">$x = x^*(p, w)$</span>.
The argument above also tells us that <span class="math-container">$x \in x^*( p, \, e(p, u(x)) )$</span>, where <span class="math-container">$e(p, u)$</span> is the minimum expenditure required to achieve utility level <span class="math-container">$u$</span> given price <span class="math-container">$p$</span>.</p>

<p><strong>Questions</strong></p>

<p><strong>Q</strong> <em>"...<span class="math-container">$x$</span> lies on the boundary of <span class="math-container">$\succeq\!\!(x)$</span>..."</em></p>

<p><strong>A</strong>
By definition, <span class="math-container">$x$</span> lies on the boundary of <span class="math-container">$\succeq\!\!(x)$</span> if every open neighborhood of <span class="math-container">$x$</span> contains some <span class="math-container">$x' \in \succeq\!\!(x)$</span> and some 
<span class="math-container">$x'' \in \prec\!\!(x)$</span>. If <span class="math-container">$u$</span> is strictly increasing, this is clear. Every open neighborhood of <span class="math-container">$x$</span> contains some <span class="math-container">$x' \gg x$</span> and some <span class="math-container">$x'' \ll x$</span>.</p>

<p><strong>Q</strong> <em>"Since <span class="math-container">$u$</span> is increasing, then we can take <span class="math-container">$p \gg 0$</span>."</em> </p>

<p><strong>A</strong>
Suppose <span class="math-container">$p \geq 0$</span> but has a zero entry, say the first one. Let <span class="math-container">$e_1$</span> be the standard basis vector with 1 in the first entry and zero elsewhere. Then <span class="math-container">$x + \alpha e_1$</span> lies in <span class="math-container">$\succeq\!\!(x)$</span> for any <span class="math-container">$\alpha \geq 0$</span>, since <span class="math-container">$u$</span> is increasing. Strict quasiconcavity of <span class="math-container">$u$</span> then implies that <span class="math-container">$x + \alpha' e_1 \succ\!\!(x) $</span> for some <span class="math-container">$\alpha' &gt; 0$</span> (take a convex combination on this half line). This means the plane with normal vector <span class="math-container">$p$</span> cannot be tangent to <span class="math-container">$\succeq\!\!(x)$</span>---this plane intersects the interior <span class="math-container">$\succ\!\!(x)$</span>.</p>

CC BY-SA 4.0
2019-11-18T22:16:53.050Z
thanks. I have a few question: 1. Could you explain the connection between the continuity of $u(.)$ and why $x$ has to be on the boundary? 2. Could you explain why we can choose $p>>0$ if $u(.)$ is strictly increasing?
Rainroad
CC BY-SA 4.0
2019-11-22T19:57:33.513Z
Response to edited answer: Thanks a lot for the clarifcation. I still don't quite get your answer to question #2. Why does the supporting hyperplane have to be tangent to the upper contour set? Is this where differentiability come in? Technically the supporting hyperplane just has to contain $x$ and the whole upper contour set right?
Rainroad
CC BY-SA 4.0
2019-11-22T21:31:06.573Z
There's nothing being claimed about the supporting hyperplane being "tangent" to anything. All that is being claimed is that its normal vector must have all non-zero entries.
Michael
CC BY-SA 4.0
2019-11-22T21:34:52.030Z
If you assume, on top of strict quasiconcavity, that $u$ is $C^1$, then the Lagrangian necessary FOC gives you tangency. But that is ex post---after you've already shown that $x$ is the Marshallian demand for some $p \gg 0$.
Michael
CC BY-SA 4.0
2019-11-16T06:17:06.980Z
2019-11-22T21:36:12.453Z
2019-11-22T21:36:12.453Z
Michael
<p>My notes says that if <span class="math-container">$u(.)$</span> is strictly quasiconcave and differentiable, by the supporting hyperplane theorem, there exists <span class="math-container">$p &gt;&gt;0$</span> and <span class="math-container">$w \geq 0$</span> such that <span class="math-container">$ x = x(p,w)$</span> <span class="math-container">$\forall x$</span>. I am having a little trouble understanding this. Here's how I think:</p>

<p><span class="math-container">$u(.)$</span> quasiconcave,so the set <span class="math-container">$X = \{x \in R^L: x  \succeq x_0\}$</span> is convex. We also know that <span class="math-container">$u(.)$</span> is differentiable, and thus continuous, so <span class="math-container">$X$</span> is closed. Thus, <span class="math-container">$x_0$</span> is on the boundary of <span class="math-container">$X$</span>. Then according to the theorem, there is some <span class="math-container">$p$</span> such that if <span class="math-container">$x \succeq x_0$</span> then <span class="math-container">$p.x &gt; p.x_0$</span> This means that with this <span class="math-container">$p$</span> and <span class="math-container">$w = p.x_0$</span>, <span class="math-container">$x_0 = x(p,w)$</span>. </p>

<p>Is my understanding correct? In addition, if the set <span class="math-container">$X$</span> is strictly convex, does that mean that theorem is now "if <span class="math-container">$X$</span> is a convex set and <span class="math-container">$x_0$</span> is a point in the boundary of <span class="math-container">$X$</span> there must exist a <span class="math-container">$p$</span> such that <span class="math-container">$p.x &gt; p.x_0 \forall x \in X$</span> and not <span class="math-container">$\geq$</span>?</p>

CC BY-SA 4.0
2019-09-26T04:18:37.450Z
I think your understanding is correct, because a strictly convex better set $A(q)$ such that the indifference curve is the lower bound would generate a unique tangency and hence $p.x$ would be strictly larger than $p.x_{0}$ if my understanding is also correct
Brennan
CC BY-SA 4.0
2019-09-26T21:31:52.127Z
Does $X$ have to be closed for $x_0$ to be on the boundary or is it always on the boundary?
Rainroad
CC BY-SA 4.0
2019-09-26T23:22:04.040Z
Okay as long as preferences are monotonic,  $x_0$ should always be on the boundary. So what's the point of u(.) being differentiable here? Just so that there exists x(.)?
Rainroad
CC BY-SA 4.0
2019-11-21T17:44:16.950Z
Not sure what the site rules around here are, but I imagine this would be a better fit for mathematics.stackexchange.com – this may come up in economics, but it doesn’t seem to me that the problem is particularly specific to economics.
Cubic
CC BY-SA 4.0
2019-11-22T19:00:49.143Z
Fair enough. I was debating whether to post this on Mathematics or Economics forum, but eventually settled for Economics, since there are concepts that are only used in Economics such as utility function and Walrasian demand function . It would take me quite a long time to explain the relations and concepts so I posted this on Economics forum instead.
Rainroad
<microeconomics><mathematical-economics><real-analysis>
Supporting Hyperplane Theorem and quasiconcave utility function
Rainroad
# Supporting Hyperplane Theorem and quasiconcave utility function

(Asked by: username_0 on 2019-09-26)

My notes says that if $u(.)$ is strictly quasiconcave and differentiable, by the supporting hyperplane theorem, there exists $p >>0$ and $w \geq 0$ such that $ x = x(p,w)$ $\forall x$. I am having a little trouble understanding this. Here's how I think:

$u(.)$ quasiconcave,so the set $X = \{x \in R^L: x \succeq x\_0\}$ is convex. We also know that $u(.)$ is differentiable, and thus continuous, so $X$ is closed. Thus, $x\_0$ is on the boundary of $X$. Then according to the theorem, there is some $p$ such that if $x \succeq x\_0$ then $p.x > p.x\_0$ This means that with this $p$ and $w = p.x\_0$, $x\_0 = x(p,w)$.

Is my understanding correct? In addition, if the set $X$ is strictly convex, does that mean that theorem is now "if $X$ is a convex set and $x\_0$ is a point in the boundary of $X$ there must exist a $p$ such that $p.x > p.x\_0 \forall x \in X$ and not $\geq$?

**Tags:** microeconomics, mathematical-economics, real-analysis
> **Comments:**
> **username_1**: I think your understanding is correct, because a strictly convex better set $A(q)$ such that the indifference curve is the lower bound would generate a unique tangency and hence $p.x$ would be strictly larger than $p.x_{0}$ if my understanding is also correct  
> **username_0**: Does $X$ have to be closed for $x_0$ to be on the boundary or is it always on the boundary?  
> **username_0**: Okay as long as preferences are monotonic,  $x_0$ should always be on the boundary. So what's the point of u(.) being differentiable here? Just so that there exists x(.)?  
> **username_2**: Not sure what the site rules around here are, but I imagine this would be a better fit for mathematics.stackexchange.com – this may come up in economics, but it doesn’t seem to me that the problem is particularly specific to economics.  
> **username_0**: Fair enough. I was debating whether to post this on Mathematics or Economics forum, but eventually settled for Economics, since there are concepts that are only used in Economics such as utility function and Walrasian demand function . It would take me quite a long time to explain the relations and concepts so I posted this on Economics forum instead.  

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## 1 answers

### Answer by username_3 on 2019-11-16. Score: 1 | Accepted answer

This problem is quite specific to economics. The correct statement is:

**Proposition**
If $u(\cdot)$ is quasiconcave, strictly increasing, and continuous, then $\forall x$, there exists $p \gg 0$ and $w \geq 0$ such that $x \in x^\*(p, w)$, where $x^\*(p, w)$ is the Marshallian demand correspondence.

**Proof**

Quasiconcavity of $u$ means the upper-contour set $\succeq\!\!(x) = \{x': u(x')\geq u(x)\}$ of bundles weakly preferred over $x$ is convex.

By continuity of $u$, $\succeq\!\!(x)$ is closed.

By strict monotonicity of $u$, $x$ lies on the boundary of $\succeq\!\!(x)$.
Consider the supporting hyperplane $p\cdot (x' - x) = 0$ of the closed convex set $\succeq\!\!(x)$ at boundary point $x$ with normal vector $p$.
Since $u$ is increasing, then we can take $p \gg 0$.

Since $\succ\!\!(x) \subset \{ p\cdot (x' - x) > 0 \}$ (general relationship between the interior of a convex set and a half space given by a supporting hyperplane),
$x \in x^\*(p, \,p\cdot x)$. $\Box$

Differentiability is not needed. If $u$ is strictly quasiconcave, the Marshallian demand correspondence is a function, and $x = x^\*(p, w)$.
The argument above also tells us that $x \in x^\*( p, \, e(p, u(x)) )$, where $e(p, u)$ is the minimum expenditure required to achieve utility level $u$ given price $p$.

**Questions**

**Q** *"...$x$ lies on the boundary of $\succeq\!\!(x)$..."*

**A**
By definition, $x$ lies on the boundary of $\succeq\!\!(x)$ if every open neighborhood of $x$ contains some $x' \in \succeq\!\!(x)$ and some
$x'' \in \prec\!\!(x)$. If $u$ is strictly increasing, this is clear. Every open neighborhood of $x$ contains some $x' \gg x$ and some $x'' \ll x$.

**Q** *"Since $u$ is increasing, then we can take $p \gg 0$."*

**A**
Suppose $p \geq 0$ but has a zero entry, say the first one. Let $e\_1$ be the standard basis vector with 1 in the first entry and zero elsewhere. Then $x + \alpha e\_1$ lies in $\succeq\!\!(x)$ for any $\alpha \geq 0$, since $u$ is increasing. Strict quasiconcavity of $u$ then implies that $x + \alpha' e\_1 \succ\!\!(x) $ for some $\alpha' > 0$ (take a convex combination on this half line). This means the plane with normal vector $p$ cannot be tangent to $\succeq\!\!(x)$---this plane intersects the interior $\succ\!\!(x)$.

> **Comments:**
> **username_0**: thanks. I have a few question: 1. Could you explain the connection between the continuity of $u(.)$ and why $x$ has to be on the boundary? 2. Could you explain why we can choose $p>>0$ if $u(.)$ is strictly increasing?  
> **username_0**: Response to edited answer: Thanks a lot for the clarifcation. I still don't quite get your answer to question #2. Why does the supporting hyperplane have to be tangent to the upper contour set? Is this where differentiability come in? Technically the supporting hyperplane just has to contain $x$ and the whole upper contour set right?  
> **username_3**: There's nothing being claimed about the supporting hyperplane being "tangent" to anything. All that is being claimed is that its normal vector must have all non-zero entries.  
> **username_3**: If you assume, on top of strict quasiconcavity, that $u$ is $C^1$, then the Lagrangian necessary FOC gives you tangency. But that is ex post---after you've already shown that $x$ is the Marshallian demand for some $p \gg 0$.  

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