# Proving a transformation preserves a property

Let $$f(x_1,x_2,\ldots,x_n)=(x_1/(\sum_{i=1}^n x_i),x_2/(\sum_{i=1}^n x_i),\ldots,x_n/(\sum_{i=1}^n x_i))$$ be in $$\mathbb{R}^n_{++}$$, and $$X$$ be a set of points in $$\mathbb{R}^n_{++}$$.

I was trying to prove if the following holds:

$$hull(f(X))=f(hull(X))$$ where $$hull(X)$$ is the convex hull of $$X$$. $$f$$ is a projection onto probability simplex in $$n$$ dimensions, but I was not able to prove the equality.

Another question that I tried to prove is, does this transformation preserve convexity? Thanks a lot.

The answer to the second question is yes, the image under $$f$$ of a convex set is convex. In general, the image of a segment $$AB$$ under $$f$$ is the segment $$f(A)f(B)$$ (provided $$f$$ is well-defined throughout the segment $$AB$$, which is certainly the case if $$A, B \in (0,+\infty)^n$$).

Let $$H$$ be the hyperplane $$x_1 + \dots + x_n = 1$$. Then $$f$$ is the central projection onto $$H$$ with respect to the origin $$O$$. The domain of $$f$$ is the complement of the hyperplane $$H_0 \colon x_1 + \dots + x_n = 0$$, that is, the union of the two open half-spaces determined by $$H_0$$.

Now assume the segment $$AB$$ is contained in the domain of $$f$$, i.e., $$A$$ and $$B$$ are on the same side of $$H_0$$. If $$O, A, B$$ lie on the same line, there is nothing to prove, as $$f$$ is constant on the segment $$AB$$. Otherwise, the whole problem is situated within the plane $$\alpha$$ through $$O, A, B$$.

Let $$l, l_0$$ be the lines of intersection of $$\alpha$$ with $$H, H_0$$. Then $$A, B$$ are points in $$\alpha$$ on the same side of $$l_0$$, which we can assume without loss of generality to be the same side where $$l$$ is. The problem is to see that if $$C$$ is a point between $$A$$ and $$B$$, then $$f(C)$$ is between $$f(A)$$ and $$f(B)$$ on line $$l$$ (and all points of the segment $$f(A)f(B)$$ are of this form).

Now this is reduced to a well-known fact of plane geometry: given three rays $$r_1$$, $$r_2$$ and $$r_3$$ all in the same plane, if at least one segment drawn from $$r_1$$ to $$r_2$$ intersects $$r_3$$, then all segments drawn from $$r_1$$ to $$r_2$$ intersect $$r_3$$.

It is enough to apply this fact to the rays $$r_1 = OA$$, $$r_2 = OB$$, $$r_3 = OC$$ and the segments $$AB$$, $$f(A)f(B)$$.

To answer the first question, it is enough to prove that if $$C$$ is a convex subset of $$H$$, then the portion of $$f^{-1}(C)$$ lying in a chosen half-space with respect to $$H_0$$ is also convex. This can be given a similar proof by restricting to a plane as above. Then it is enough to see that the region lying between two rays is convex.

• Thanks a lot for your response. I could not follow the following: Now assume the segment $AB$ is contained in the domain of $f$, i.e., $A$ and $B$ are on the same side of $H_0$.", specifically $A$ and $B$ being on the same side of $H_0$, and is the plane $\alpha$ defined through the three points $0, A, B?$? Thank you. Commented Jul 9, 2020 at 13:39
• Saying that the segment $AB$ is contained in the domain of $f$ means that it doesn't intersect $H_0$. That's the same as saying that $A$ and $B$ are in the same half-space determined by $H_0$. Yes, $\alpha$ is the plane determined by $O$, $A$ and $B$. Commented Jul 9, 2020 at 17:01