How is the subdifferential of the $l_2$ norm at $x=0$ the polar of the unit ball? I was reviewing subdifferentials, and my professor writes on the board that the subdifferential of the $l_2$ norm at $x=0$ is the polar of the unit ball?
I tried to work through this myself:
Let $\Phi(x):\mathbb{R}^n\rightarrow\mathbb{R}$ and $C$ is a closed, non-empty convex set.
$$\Phi(x)=||x||$$
I know the following:
$$\Phi^*(z)=\sup_{x\in C}\{\langle z,x\rangle-\Phi(x)\}$$
and I know that the support function of the set $C$ is the dual of the indicator function:
$$\sigma_C(x)=\delta^*_C(x):=   \left\{
\begin{array}{ll}
      0 & x\in C \\
     \infty &\text{else} \\
\end{array} 
\right. $$
and I also know that it is equal to the sup:
$$\sigma_C(x) = \sup_z\{\langle z,x\rangle\}$$
And I know the definition of the subdifferential at the point $x$ for the set $C$ is this:
$$\partial\Phi(x)=\{z\hspace{0.2cm}|\hspace{0.2cm} \Phi(y)\ge \Phi(x)+ \langle z,y-x\rangle \hspace{0.2cm}\forall y \}$$
But I do not understand how to put these definitions together to get this:
$$\partial\Phi(0)=\mathbb{B}^o$$  which is the polar of the unit ball.  How are the steps used to get this?
edit :: I am also interested in how these things relate to each other:  the $\sigma_C(x)$, and $\partial\Phi(x)$, (if at all) ?
 A: Following the derivation in Example 3.3 of the book by Beck (First-Order Methods in Optimization), I will provide the derivation for a general norm and then answer specifically for $\ell_2$ norm.
We have
$$
\Phi (x) = \| x \|
$$
and we are interested in its subdifferential at $x=0$ i.e. $\partial \, \Phi (0)$.
The correct answer is
$$
\partial \, \Phi (0) = \{ g \in \mathbb{R}^n : \| g \|_* \leq 1 \}.
$$
This is the unit ball of the dual norm $\| \cdot \|_*$. It is not polar of the unit ball.
Note that $g \in \partial \, \Phi (0)$ if and only if
$$
\Phi (y) \geq \Phi (0) + \langle g , y - 0 \rangle  \, \forall y \in \mathbb{R}^n
$$
i.e.
$$
\| y \| \geq \langle g , y \rangle \; \forall y \in \mathbb{R}^n
$$
In other words
$$
g \in \partial \, \Phi (0) \iff \langle g , y \rangle \leq \| y \| \; \forall y \in \mathbb{R}^n.
$$
We first show that if $g$ belongs to the unit ball of the dual norm, then it is a subgradient.
If $\| g \|_* \leq 1$ then, by generalized Cauchy-Schwarz inequality
$$
\langle g , y \rangle \leq \| g \|_* \| y \| \leq \| y \| \; \forall y \in \mathbb{R}^n.
$$
Thus, if $\| g \|_* \leq 1$ then, $g$ is subgradient of $\Phi$ at $x=0$.
In the reverse direction, we will show that if $g$ is a subgradient, then
$\| g \|_* \leq 1$ must hold.
Assuming $g \in \mathbb{R}^n$ to be a subgradient, we have:
$$
  \langle g , y \rangle \leq \| y \| \; \forall y \in \mathbb{R}^n
$$
Then it will also hold for all $ \| y \| \leq 1$. Maximizing the L.H.S. over the unit ball, we get
$$
\| g \|_* = \underset{y : \| y \| \leq 1}{\sup} \langle g, y \rangle \leq \| y \| 
$$
holds for some $y=y_0$ s.t. $ \| y_0 \| \leq 1$. Thus, $\| g \|_* \leq 1$
must hold.
Combining, we get the result that:
$$
\partial \, \Phi (0) = \{ g \in \mathbb{R}^n : \| g \|_* \leq 1 \}.
$$
Finally, for the special case of $\ell_2$ norm $\Phi (x) = \| x \|_2$, the norm is self dual hence the answer in this case is the unit ball of the $\ell_2$ norm  $\{ g \in \mathbb{R}^n : \| g \|_2 \leq 1 \}$.
A: By definition of the subdifferential we get for $\Phi(0)$ the following
$$\partial\Phi(0)=\{z\in\mathbb{R}^n|\Phi(y)\geqslant \Phi(0)+\langle z,y-0\rangle,\forall y\in \mathbb{R}^n\}=\{z\in\mathbb{R}^n|\hspace{0.2cm}||y||\geqslant \langle z,y\rangle,\forall y\in \mathbb{R}^n\}\\=\{z\in\mathbb{R}^n|\hspace{0.2cm}||y||\geqslant \langle z,y\rangle,\forall y\in \mathbb{R}^n, ||y||<1\}\cup \{z\in\mathbb{R}^n|\hspace{0.2cm}||y||\geqslant \langle z,y\rangle,\forall y\in \mathbb{R}^n, ||y||\geqslant 1\}
\\=\{z\in\mathbb{R}^n|\hspace{0.2cm}||y||\geqslant \langle z,y\rangle,\forall y\in \mathbb{R}^n, ||y||<1\}\cup\{z\in\mathbb{R}^n|\hspace{0.2cm}1\geqslant \langle z,\frac{y}{||y||}\rangle,\forall y\in \mathbb{R}^n\}=\{z\in\mathbb{R}^n|\hspace{0.2cm}1\geqslant \langle z,y\rangle,\forall y\in \mathbb{B}\}:=\mathbb{B}^o$$
Note that $\forall y\in \mathbb{R}^n\setminus\{0\}$ we have $y/||y||\in \mathbb{B}$ where $\mathbb{B}$ is the unit ball. 
