0

In page 3 of https://link.springer.com/content/pdf/10.1007%2Fs10107-005-0690-4.pdf . It is stated that "The dual of any set is a closed convex cone". I want to know how to prove this. We formulate the problem as follows:

Given a set $K\in\mathbb{R}^n$, the dual of $K$ is defined as $$K^*=\{w\in\mathbb{R}^n:x^Tw\ge0,\forall x \in K\}$$ The problem is how to prove that $K^*$ is a closed convex cone?

My thinkings

We should prove close and convex respectively. For closeness, we can follow this: dual cone is closed , which states that each $x^Tw\ge0$is a half space, therefore, $K^*$ is a intersection of halfspaces. But how to prove convex?

maple
  • 2,883
  • 2
  • 29
  • 37

1 Answers1

1

Let $v,w \in K^*$, $x \in K$, and $\lambda \in [0,1]$.

Since $x^Tw \ge 0$ and $x^Tv \ge 0$, we have that

\begin{align*} x^T(\lambda v + (1- \lambda)w)&= \lambda x^Tv + (1-\lambda)x^Tw \\ &\ge 0. \end{align*}

$x$ was arbitrary, so it must be that $\lambda v + (1- \lambda)w \in K^*$.

Theoretical Economist
  • 6,472