If $K \subset R^n$ and $0 \in K$. Define $K'=\{u|\langle u,x\rangle \leq1, \forall x \in K\}$. Note: $\langle u,x\rangle = u^Tx$
Prove:
If $K$ is a cone, then $K' = -K^*$, where $K^*$ is the dual cone of $K$.
I use contradiction:
If $K' \neq -K^*$, i.e. there exist $y \in K^*$, and $y \notin K'$. That is $x^T(-y) > 1 \to x^Ty < 1$. $\forall x \in K$.
I still cannot find any way to use the cone property to say $K$ is not a cone. Obviously, I get an intersection of halfspace, which is stll a cone.
Any better way to prove this?
Suppose $u \in K'$. If $x \in K$, then $cx \in K$ for all $c > 0$, and so \begin{align} & \langle u, cx \rangle \leq 1 \quad \forall c > 0 \\ \implies & \langle u, x \rangle \leq \frac{1}{c} \quad \forall c > 0 \\ \implies & \langle u, x \rangle \leq 0. \end{align} Hence, $u \in -K^*$. This shows that \begin{equation} K' \subset -K^*. \end{equation}
Now suppose that $u \in -K^*$, so $-u \in K^*$. If $x \in K$, then \begin{align} & \langle -u, x \rangle \geq 0 \\ \implies & \langle u, x \rangle \leq 0 \\ \implies & \langle u, x \rangle \leq 1 \end{align} which shows that $u \in K'$. Thus, \begin{equation} -K^* \subset K'. \end{equation}
Putting these pieces together, we see that \begin{equation} K' = -K^*. \end{equation}