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Let $C \subset R^n$ be a closed, convex cone. Let $T: R^n \to R^m$ be a linear transformation.

Is $T(C) \subseteq R^m$ closed?

I'm very positive that the answer is NO. But I couldn't come up to a counter example so far. I also realized that any counter example (if exists) has to be in Dimension $n \geq 3$, otherwise $C$ becomes a polyhedral and so is $T(C)$.

Red shoes
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  • What is your thinking behind your belief that the answer is "no?" – Cheerful Parsnip Jun 24 '19 at 21:41
  • Observe that it is sufficient to consider the special case $m < n$ and $T$ = projection of $\mathbb R^n$ to the first $m$ coordinates. – Paul Frost Jun 24 '19 at 22:01
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    I think the answer is indeed no. Look at the standard cone in $\Bbb R^3$, e.g., the set of points where $x^2+y^2 \leq 2 z^2$. Then take $T:R^3 \to R^2$ to project orthogonally to the direction $(1,1,1)$ which is on the boundary of the cone (technically $T$ doesn't map to $R^2$ but I'm identifying $R^2$ with some 2d subspace of $R^3$ here). – shalin Jun 24 '19 at 23:07
  • The answer is no. Find a counterexample in the following link https://math.stackexchange.com/questions/1497944/a-linear-transform-of-a-closed-set-is-closed – Arindam Jun 24 '19 at 23:17
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    @Arindam Note that $C$ here is convex cone, not a typical closed set. – Red shoes Jun 24 '19 at 23:19
  • Can you please explain what do you mean by project orthogonally to the direction $(1,1,1)$ ? @Shalop – Red shoes Jun 24 '19 at 23:38
  • In any Hilbert space if you have a closed subspace $M$ then there's a unique linear map which projects orthogonally onto it. I'm saying to take $M$ to be the orthogonal complement of the vector $(1,1,1)$. – shalin Jun 24 '19 at 23:41
  • @Shalop I drew it on the computer and I'm convinced, the projection seems to be half of the plane with the dividing line never quite being reached. Do you know how to show this more rigorously, it seem like a lot of nasty algebra? – J_P Jun 24 '19 at 23:43
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    Without any algebra here is the main idea. In Euclidean geometry there are conic sections which comprise of four shapes: circle, ellipse, parabola, and hyperbola. In this case we consider conic sections which are parallel to the boundary of the cone itself, so they will be parabolas. Now, if you look at the projected cone in my example, it is the union of the convex hulls of all of its conic sections, in other words a union of convex hulls of parabolas which are getting wider and wider. So the union will never reach the x-axis except at the origin, hence is not closed. – shalin Jun 24 '19 at 23:50
  • I don't think in your specific example you have parabolas but hyperbolas because the cone's angle isn't $90$ degrees. Perhaps the example $x^2+y^2=z^2$ and the line $(1,0,1)$ are simpler? That does give parabolas. Or even just $y^2=2zx$ which is a cone slanted at $45$ degrees. – J_P Jun 24 '19 at 23:56
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    Indeed, with this last example even the algebra is simple. If a point gets projected onto the $y$-axis, $y^2=0$ and so the only solutions are on the vertical ray $(0,0,z);z\geq 0$. Otherwise, if $x>0$, $z=\frac{y^2}{2x}$ is on the cone and gets projected onto $(x,y)$. So indeed the projection of this slanted cone is ${(x,y):x>0}\cup{(0,0,0)}$ which is not closed. – J_P Jun 25 '19 at 00:04
  • Yup thanks for the corrections and simplifications @J_P – shalin Jun 25 '19 at 00:08
  • You should post this as an answer so that the OP can accept it if they wish. – J_P Jun 25 '19 at 00:09
  • You or the OP may do so, I am too tired. Also you thought of a nice way to write it up. – shalin Jun 25 '19 at 00:12
  • Thanks both of you. – Red shoes Jun 25 '19 at 00:14

2 Answers2

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Big thanks to @Shalop who had the idea of using plain old cones as a counterexample, I just cleaned up the algebra.

So, by rotating the cone defined by $x^2+y^2=z^2;z\geq0$ by $45$ degrees toward the positive $x$-axis, we get the cone $y^2=2xz$ (this is just an exercise in $2D$ rotations). The projection of this cone to the $xy$ plane is the set $$S=\{(x,y):x>0\}\cup\{(0,0)\}$$ This is simple to see. If $x=0$, then $y^2=0$ and so only the vertical ray $(0,0,z);z\geq0$ get projected onto the $y$-axis onto the point $(0,0)$. If $x>0$, we get that the point $(x,y,y^2/2x)$ is on the cone and gets projected into $(x,y)$. This covers all possibilities, so the projection of the cone is $S$ but $S$ is not closed.

J_P
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The question was already answered by @J_P. Here is another way of thinking: There are two closed convex cone, $C_1$ and $C_2$ in $R^3$ whose their sum i.e., $C_1 + C_2$ is not closed see "https://math.stackexchange.com/a/3273305/219176"

Clearly $C_1 \times C_2$ is closed convex cone. Now take the linear transformation $T(x,y)= x+y$ then we have $T(C_1 \times C_2) = C_1 + C_2$.

Red shoes
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