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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 $\geq$?

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  • $\begingroup$ 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 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 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 at 23:22

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