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We assume a unit hypercube. There is a convex set in this hypercube and we want to find the minimum Euclidian distance between the average of k points (centroid) and the boundary of this convex set.

I would like to ask if the minimum Euclidean Distance between the average of k points and the boundary of the convex set mentioned above can be deduced to finding the minimum Euclidean Distance between each of these k points and the boundary of the convex set, add these minimum Euclidean Distances and divide the sum with k which is the number of points in the convex set?

(I ask this because in my problem the average of k points cannot be easily defined due to the nature of the problem, but the minimum Euclidean Distance between each point and the boundary of a convex set in a unit hypercube can be defined and computed more easily).

Except my alternative method is any other method to deduce the minimum Euclidean distance of the average to the individual Euclidean Distances?

Thanks in advance!

Novemberland
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    distance of the average of k points Please clarify (i) whether the average of k points means the centroid (barycenter), and (ii) distance to what - the origin, an arbitrary point etc. – dxiv Jan 02 '18 at 19:47
  • The edited question now asks to find the *minimum* Euclidian distance of the average of k points, but that still doesn't parse. Please define what is the distance of the average of k points. P.S. Does your question reduce to this one in 2D, by any chance? Then the right wording could be average shortest distance. – dxiv Jan 03 '18 at 02:20
  • @dxiv i)To find the average of k points the paper to which I refer adds the projections on each dimension for the k points and divides with k, ii) I corrected the use of Euclidean distance, it is the minimum Euclidean distance of the average on the boundary of a convex curve in the unit hypercube. If I am correct the minimum Euclidean Distance of that point on the boundary of a convex curve is the projection of that point on the convex curve. – Novemberland Jan 03 '18 at 02:22
  • OK, then i) means the average point is the centroid. But if ii) means the (shortest) distance to the boundary of the convex region then each of the points which are on the boundary will have distance $0$, while the centroid most likely not - so that part is still unclear. – dxiv Jan 03 '18 at 02:27
  • @dxiv No each point does not lie on the boundary of the convex curve. If it confusing I will correct it. On the contrary I can compute the projection of each one of k points on the curve, but I cannot compute the centroid due to the nature of the problem. Can I deduce the projection of the centroid to individual projections? Thanks in advance. – Novemberland Jan 03 '18 at 02:32
  • If I understand it correctly (after the latest edit), the answer is negative. Take the simple case of a 2D unit square, with the convex curve being the square itself. Take $k=2$ and let the two points $A,B$ initially coincide, at a distance $\epsilon$ from the nearest side of the square. Keep $A$ fixed, and start moving $B$ around the square while keeping it at the same constant distance $\epsilon$ from the nearest side. Then the two distances from $A,B$ to the square curve are constant, but the distance of the centroid varies from $\epsilon$ (initially) to $2(1-\epsilon)$ (when opposite). – dxiv Jan 03 '18 at 02:50

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