2
$\begingroup$

Let $K \subset \mathbb{R}^n$ be a closed convex set. Given a point $u \in K$, the tangent cone of $K$ at $u$ is defined as (or characterized by) $$ T_K(u) := \mathrm{cl}(\left\{ t (v - u) \mid v \in K, t \geq 0 \right\}). $$

My question is about the cases where $K$ is a closed convex cone. The equation right after Eq. (2.1) of this paper says that, if $K$ is a closed convex cone, we have $$ T_K(u) = \left\{ v - tu \mid v \in K, t \geq 0 \right\}. $$ I've tried to prove this identity, but I'm having trouble with the last step of the proof. So can anyone help me figure out the right way?

What I've tried so far:

The inclusion $\left\{ v - tu \mid v \in K, t \geq 0 \right\} \subset \left\{ t (v - u) \mid v \in K, t \geq 0 \right\} \subset T_K(u)$ is easily proven with some algebra, so I will prove the opposite inclusion. It is straightforward to see that $$ \left\{ t (v - u) \mid v \in K, t \geq 0 \right\} \subset \left\{ v - tu \mid v \in K, t \geq 0 \right\} $$ so taking the closures of both sides, we obtain $$ T_K(u) \subset \mathrm{cl}(\left\{ v - tu \mid v \in K, t \geq 0 \right\}). $$ So it suffices to show that the set $\left\{ v - tu \mid v \in K, t \geq 0 \right\}$ is closed, but I can't figure out how I can do it. Given an arbitrary sequence $\{v_n - t_n u\}_{n=1}^\infty$ ($v_n \in K, t_n \geq 0$) that converges to some point, my goal is to prove the limit can be written in the form of $v - tu$ with $v \in K, t \geq 0$. If both $\{v_n\}$ and $\{t_n\}$ converge, this is easy because $K$ is closed. But we can easily construct examples in which neither $\{v_n\}$ nor $\{t_n\}$ doesn't converge, e.g., $v_n = nu$ and $t_n = n$, so we cannot assume the convergence of $\{v_n\}$ or $\{t_n\}$.

$\endgroup$

1 Answer 1

3
$\begingroup$

The equality $$T_K(u)=\{v-tu\colon v\in K,\, t\ge0\}$$ is false in general, because $T_K(u)$ is closed by definition, whereas $$S_K(u):=K-\mathbb R_+ u=\{v-tu\colon v\in K,\, t\ge0\}$$ can be not closed.

A counterexample is provided by this answer. Indeed, let $$K=\big\{(x,y,z)\in\mathbb R^3\colon z\ge\sqrt{x^2+y^2}\big\}$$ and $u=(-1,0,1)$. Then $K$ is a closed convex cone in $\mathbb R^3$ and $u\in K$.

Also, $w:=(0,1,0)\notin S_K(u)$ -- otherwise, we would have $w+tu=(-t,1,t)\in K$ for some real $t$. However, letting $$w_t:=v_t-tu$$ for real $t>0$ and $v_t:=(-t,1+1/t,\sqrt{t^2+(1+1/t)^2})$, we have $v_t\in K$ and hence $w_t\in K-\mathbb R_+ u=S_K(u)$, whereas $w_t\to w$ as $t\to\infty$. $\quad\Box$

$\endgroup$
1
  • $\begingroup$ Thank you very much for your nice & concise answer! Now I've got it. $\endgroup$
    – aest
    Commented Apr 30, 2023 at 4:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.