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 $\geq$?

  • $\begingroup$ 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 $\endgroup$
    – Brennan
    Sep 26 '19 at 4:18
  • $\begingroup$ Does $X$ have to be closed for $x_0$ to be on the boundary or is it always on the boundary? $\endgroup$
    – Rainroad
    Sep 26 '19 at 21:31
  • $\begingroup$ 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(.)? $\endgroup$
    – Rainroad
    Sep 26 '19 at 23:22
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    $\begingroup$ 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. $\endgroup$
    – Cubic
    Nov 21 '19 at 17:44
  • $\begingroup$ 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. $\endgroup$
    – Rainroad
    Nov 22 '19 at 19:00

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.


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 $p$.


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 $\succ\!\!(x)$.

  • $\begingroup$ 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? $\endgroup$
    – Rainroad
    Nov 18 '19 at 22:16
  • $\begingroup$ 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? $\endgroup$
    – Rainroad
    Nov 22 '19 at 19:57
  • $\begingroup$ 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. $\endgroup$
    – Michael
    Nov 22 '19 at 21:31
  • $\begingroup$ 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 $p \gg 0$. $\endgroup$
    – Michael
    Nov 22 '19 at 21:34

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