About supporting hyperplanes of convex sets Let $K \subset R^n$ a convex set, and $x \in \partial K$ such that that there exists  a closed ball $B(x_0,R) \subset K$ of positive radius  with $x \in B(x_0,R) $. 
My intuition tells me that there exist exactly one supporting hyperplane for $K$ at $x$. How can I prove this? 
I know that exist a hyperplane of $K$ at  $x$ by the convexity of $K$ . I don't know how prove that it's unique.
Thanks in advance!
 A: If $H$ is a supporting hyperplane to $K$ at $x$, that means $x\in H$, and $K$ lies entirely on one side of $H$ (including $H$ itself). Describing the hyperplane by a linear equation, $H = \{ y\in \mathbb{R}^n : \lambda(y) = 1\}$ (every Hyperplane has such a representation), we have either $\lambda(k) \leqslant 1$ for all $k\in K$ or $\lambda(k) \geqslant 1$ for all $k\in K$. Let us suppose we have $\lambda(k) \leqslant 1$ for all $k\in K$.
Since $B(x_0,R) \subset K$, we have $\lambda(y) \leqslant 1$ [in fact $\lambda(y) < 1$, since the ball is open] for all $y\in B(x_0,R)$. That means the ball $B(x_0,R)$ also lies entirely on one side of $H$, and since $x\in \partial B(x_0,R)$, the hyperplane $H$ is a supporting hyperplane for $B(x_0,R)$ at $x$.
But we know (hopefully, otherwise prove it, it's not difficult using the normal vector to the hyperplane) that supporting hyperplanes to balls are unique. Thus there is also only one supporting hyperplane to $K$ at $x$.
A: HINT: 
A supporting hyperplane of $K$ at $x$ is also a supporting hyperplane of $B$ at $x$. 
