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 $p &gt;&gt;0$ 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 $p.x &gt; p.x_0$ 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 $p.x &gt; p.x_0 \forall x \in X$ and not $\geq$?
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 $\succ\!\!(x) \subset \{ p\cdot (x' - x) &gt; 0 \}$ (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 $\alpha' &gt; 0$ (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 $\succ\!\!(x)$.