Suppose one has a convex, bounded polytope P $\subset R^n$ and a strictly convex function $f$ defined everywhere on $R^n$. $f$ has a unique minimum; and suppose this minimum occurs somewhere strictly outside P, at a point x $\in R^n$.
Let's define two points on the boundary of P, which are in general distinct.
The first point, N, is the projection of x onto P, by which I mean that N is a point on P's boundary which minimizes the distance (in any norm, but say $l_2$) to x.
The second point, M, is the point on P's boundary which minimizes the function $f$ when restricted to P.
I think that N and M are always on the same facet of P. (By "on a facet" I include the (lower dimensional) "edges" of the facet.) My question is: is this true in general?
Here is an illustration of the following concrete example. Suppose we are in $R^2$, with $f(a,b) = a^2 + 4(b-4)^2$. Suppose that P is given by the four half-planes $a+b \le 7$, $2b-a \le 4$, $a \gt 0$, $b \gt 0$. This looks like:
Here's a picture of this situation:
The minimum of $f$ occurs at x = (0,4). In this case, we see that N and M are indeed on the same facet of P. Is this relationship true in general?
It's false. To see that, introduce the new constraint $10b - 7a \leq 17$ to your example to create a new polytope $P'$. This constraint cuts off the point $N$ but keeps $M$ in $P'$. Thus $M$ minimizes $f$ over $P'$. The closest point to $x$ on $M$'s facet is now $(1.5, 2.75)$, which is the intersection of the lines $10b-7a=17$ and $2b-a=4$. However, the point $(1,2.4)$, which is on the new facet defined by $10b-7a=17$, is closer to $x = (0,4)$ than is $(1.5, 2.75)$ (a distance of $\approx 1.89$ vs. $\approx 1.95$).
For example, the image below shows the new polytope $P'$. The minimizer of $f$ over $P$ is $M$, which is in blue. The point $(1,2.4)$, in black, is closer to $(0,4)$, in red, than is any point on $M$'s facet.