Outer normal of a convex Let $K$ be a non emty convex in $\mathbb{R}^{2}$
I would like to know an explicit expression of the outer normal of  the boundary of $K$?
In other words how can I define a convex in a way to have the explicit expression of the outer normal. I think the classiccal method like the dual function or the gauge function ... of a convex dont give a positive answer to my question  
Thanks in advance. 
 A: Any explicit expression for the outer normal depends on a certain explicit expression for
the boundary of your convex (body), and then in most cases it will follow classical differential geometry. This means that you have to assume some differentiability of the boundary in order to have general results. In other words, you need to have a parametrization of the boundary of your convex body in order to get an explicit expression for the outer normal.

Parametrization via the gauge function, or equivalently, by the radial function

Lets assume that $K$ contains the origin. Define the radial function of $K$ as follows. Given a direction vector $u(\theta)=(\cos\theta,\sin\theta)$, define $r(\theta)$ to be the length of the vector in the direction $u(\theta)$ that connects the origin to the boundary of $K$. In other words,
$$r(\theta)=\sup\{t>0: tu(\theta)\in K\}$$
Using the radial function, we have a parametrization of the boundary of $K$:
$$(x(\theta),y(\theta))=r(\theta)u(\theta)=r(\theta)(\cos\theta,\sin\theta)$$
Now, assuming appropriate smoothness, we can use elementary differential geometry to compute the normal vector. First,
the tangent vector is simply
$$T(\theta)=(x'(\theta),y'(\theta))=r'(\theta)u(\theta)+r(\theta)u'(\theta)$$
and it is not hard
to see that $r'(\theta)u'(\theta)-r(\theta)u(\theta)$ is orthogonal to $T(\theta)$. Since $\{u(\theta),u'(\theta)\}$ is an orthonormal basis of $\mathbb{R}^2$ for every $\theta$, the length of the vector $r(\theta)u(\theta)-r'(\theta)u'(\theta)$ is $\sqrt{r^2(\theta)+(r'(\theta))^2}$, so we have an explicit formula for the normal vector
$$N(\theta)=\frac{r(\theta)u(\theta)-r'(\theta)u'(\theta)}{\sqrt{r^2(\theta)+(r'(\theta))^2}}$$
Note that the radial function is linked to the gauge function (i.e., the norm of the convex body $K$) by the formula $r(\theta)=\frac{1}{\|u(\theta)\|}$, so in principle, the formula above gives you a way to compute the normal vector from the norm function.
There are of course other parametrizations that may suggest themselves to the problem. For example, the boundary may be given as a zero-set of some function $f(x,y)=0$, and in that case the tools of Differential Geometry are available once again to compute the moving frame of tangent and normal vectors at every point.
