Show that polar of the unit ball is the unit ball of the dual space I know the definition of the polar set is:
$$C^0 = \left\{ y \in \mathbb{R}^n \mid y^Tx \leq 1 , \forall x \in C\right\},$$
and the dual norm is defined as:
$$||z||_* = \text{sup} \left\{ z^Tx \mid ||x|| \leq 1\right\}.$$
But I cant see any connection between the definition. How can I show that the polar of the unit ball for a general norm $||\cdot||$ is the unit ball of the dual space? Where do I start from? Thanks. 
 A: Let $B:=\{x\in\mathbf{R}^n:\|x\|\leq 1\}$.
We want to show that for the polar set defined as $B^\circ:=\{y\in\mathbf{R}^n:y^\top x\leq1,\,x\in B\}$, it holds that $B^\circ=\{y\in\mathbf{R}^n:\|y\|_*\leq 1\}$.
We will do so by showing that if $y\in B^\circ$, then $y\in \{y\in\mathbf{R}^n:\|y\|_*\leq 1\}$ and also the reverse.
($\subseteq$):
By definition, we have that
\begin{align}
B^\circ&:=\{y\in\mathbf{R}^n:y^\top x\leq1,\,x\in B\} = \{y\in\mathbf{R}^n:y^\top x\leq1,\,\|x\|\leq1\}.
\end{align}
Inspecting the last set, we have that for all $ y\in B^\circ$, it holds that
\begin{align}
y^\top x\leq1,\; \forall x:\|x\|\leq1.
\end{align}
Since this holds for all $y\in B^\circ$, it must hold for the most extreme such $y$; that is, we can convert the "for all $y\in B^\circ$" condition into an optimization problem:
\begin{align}
\sup_y\{y^\top x :\|x\|\leq1\}\leq 1
\end{align}
but the LHS is $\|y\|_*$, so we see that $\|y\|_*\leq1$ for the most extreme $y\in B^\circ$, meaning that $\|y\|_*\leq1,\;\forall y\in B^\circ$.
Thus we have that if $y\in B^\circ$, then $y\in \{y':\|y'\|_*\leq1\}$.
($\supseteq$):
On the other hand, if $y\in \{y':\|y'\|_*\leq1\}$, then
\begin{align}
\|y\|_*\leq1\iff\sup_x\{y^\top x :\|x\|\leq1\}\leq 1 \implies y\in B^\circ
\end{align}
by using the above argument but replacing the optimization problem in $x$ with the "for all $x$" quantifier.
