# Supporting Hyperplane Theorem and quasiconcave utility function

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 $$\geq$$?

• 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 Sep 26 at 4:18
• Does $X$ have to be closed for $x_0$ to be on the boundary or is it always on the boundary? – Rainroad Sep 26 at 21:31
• 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 Sep 26 at 23:22