Dual cone of $K = \{(x,t) \mid \| \boldsymbol{x} \|_1 \le t \}$

Question

This slide shows the dual cone of $K = \{(x,t) \mid \| \boldsymbol{x} \|_1 \le t \}$ is $K^{*} = \{(x,t) \mid \| \boldsymbol{x} \|_{\infty} \le t\}$. Is it right? How is it proved?

Answer 1

Suppose that $(y,t)$ is such that $\|y\|_\infty \leq t$. Select any $(x,t') \in K$. We apply Hölder's inequality to find $$ \langle(x,t'),(y,t)\rangle =\\ \langle x,y \rangle + t't \geq\\ - \|x\|_1 \|y\|_{\infty} + t't \geq\\ -\|x\|_1 \|y\|_{\infty} + \|x\|_1 \|y\|_{\infty} = 0 $$

Now, suppose that $(y,t)$ is such that $\langle(x,t'),(y,t)\rangle \geq 0$ for all $(x,t') \in K$.

Select $(x,t') = (\pm e_i,1)$, where $e_i$ is the $i$th standard basis vector in $\Bbb R^n$. By the above, we have $$ \langle(x,t'),(y,t)\rangle = \langle(\pm e_i,1),(y,t)\rangle = \pm y_i + t \geq 0 \implies\\ t \geq \mp y_i $$ Thus, we conclude $|y_i| \leq t$ for all $i$, so that $\|y\|_\infty \leq t$, as desired.

The conclusion follows.