Slater's condition for closedness of the linear image of a closed convex cone

I would like to understand what is known as Slater's condition for the closedness of the linear image of a closed, convex cone.

Let $V=\mathbb{R}^n$ and $W=\mathbb{R}^m$ (i.e. finite dimensional real inner product spaces), let $A:V\rightarrow W$ be a linear map, and let $K\subseteq V$ be a closed convex cone.

First some notation:

• The dual cone of $K$ is the set $$K^* = \{v\in V \,:\, \langle v,x\rangle \geq 0\}.$$
• The adjoint map of $A$ is the unique linear map $A^*:W\rightarrow V$ such that $\langle A x,y\rangle = \langle x, A^* y\rangle$ for all $x,y$.
• The interior of $K^*$ is denoted $\mathrm{int}(K^*)$ and can be defined as follows: For $v\in\mathrm{int}(K^*)$ and $x\in K$, if $\langle v,x\rangle = 0$ then $x=0$.
• The image of $K$ under $A$ will be denoted $$A(K) = \{Ax \,:\, x\in K\}.$$

What I would like to prove is the following:

Theorem. Suppose there exists a $y\in W$ such that $A^*y\in\mathrm{int}(K^*)$. Then $A(K)$ is closed.

Here is my attempt to prove:

Let $\{y_i\}$ be a Cauchy sequence in $A(K)$, and let $\overline{y}\in W$ such that $y_i\rightarrow \overline{y}$. (We want to show that $\overline{y}\in A(K)$ to show that $A(K)$ is closed.) Since $y_i\in A(K)$, we can find $x_i\in K$ such that $Ax_i=y_i$. Since $\{y_i\}$ is Cauchy, it is bounded so there exists a $c\geq 0$ such that $$\langle A^*y,x_i\rangle = \langle y,y_i\rangle \leq c$$ for all $i$. We can assume without loss of generality that each $y_i\neq 0$, then each $x_i$ also satisfies $$0<\langle A^*y,x_i\rangle$$ since $x_i\in K$ and we assume $A^*y\in\mathrm{int}(K^*)$. Thus, each $x_i$ satisfies $$0<\langle A^*y,x_i\rangle \leq c.$$ Somehow this implies that the sequence $\{x_i\}$ must be bounded? This is what I do not understand and I might be missing something.

Finally, since $\{x_i\}$ is a bounded sequence in a closed $K$, it has an accumulation point $x$ which satisfies $Ax=\overline{y}$.

All that I do not understand is how to show that $\{x_i\}$ is bounded. Any help is appreciated.

(Note: The existence of $y$ such that $A^*y\in\mathrm{int}(K^*)$ is known as Slater's condition).

• Essentially, what I want is to prove Proposition 1.3 on page 7 of these notes (which is given there as an exercise). Jun 13, 2017 at 16:48
• Do you know how this connects to what is normally called Slater's​ condition in convex optimization? Jun 13, 2017 at 19:40
• Can this result be used to prove that $\{Ax \mid x\geq 0\}$ is closed? (Some proofs of this fact are given here:math.stackexchange.com/questions/1831401/… ) Jun 13, 2017 at 19:43
• Something you call here relative interior is not actually relative interior of $K^*$. They don't coincide! See en.wikipedia.org/wiki/Relative_interior for definition of relint Jun 13, 2017 at 20:58
• @Ashkan. You're right. I suppose I really just mean the true interior, not the relative interior. The interior of the dual cone does in fact coincide with the set I describe. Jun 14, 2017 at 15:02

I've figured it out. It appears to be a fairly standard argument in duality theorems for conic programming.

Suppose instead that $$\{x_i\}$$ is unbounded. We will derive a contradiction.

Define the sequence $$\tilde x_i = \frac{x_i}{\lVert x_i\rVert}$$, which is in $$K$$. It is clearly bounded so it converges (or at least has a convergent subsequence) $$\tilde x_i \rightarrow\tilde x$$ to some $$\tilde x\in K$$ and $$\tilde x \neq 0$$. Hence $$0<\langle A^*y, \tilde x_i \rangle\leq \frac{1}{\lVert x_i\rVert}c$$
for all $$\tilde x_i$$. Since $$\{x_i\}$$ is unbounded, it follows that $$\langle A^*y, \tilde x_i \rangle\rightarrow 0$$. Hence $$\langle A^*y, \tilde x \rangle = \lim_{i\rightarrow\infty}\langle A^*y, \tilde x_i \rangle =0.$$ But $$A^*y\in\mathrm{int}(K^*)$$ implies that $$\tilde x=0$$, a contradiction.

• I can see only how to prove what you said when we are actually looking for $\text{int}(K^{*})$, but not $\text{relint} (K*)$. See PROPOSITION 1.1.4. of the book global.oup.com/academic/product/… Sep 14, 2023 at 10:49
• His answer does not hold when $\text{int} (K^{*})$ is substituted by $\text{relint} (K^*)$. For the relint version, I recommend following the article "On the closedness of the linear image of a closed convex cone." Directly finding yourself proof is not something one should pursue due to its difficulty. Instead, the article I recommended would enlighten your ideas even more than finding a proof by yourself. Sometimes, we need to stand on the shoulders of giants Sep 19, 2023 at 9:42