How equality in Fenchel-Young inequality characterizes subdifferential? I am not able to see why equality in the Fenchel-Young inequality
characterizes subgradients.
As per Fenchel-Young inequality:
\begin{equation}
f(x)+f^*(u) \geq \langle x,u \rangle
\end{equation}
while the definition of subdifferential set says:
\begin{equation}
\partial f(x) = \{u: f(z) \geq f(x) + \langle u, z-x\rangle \}
\end{equation}
Now it holds that $u \in \partial f(x)$ at equality of Fenchel-Young inequality.
Assume whatever is necessary for defining functions involved above.
Thanks.
 A: We will show that $$f(x)+f^*(u) = \langle x,u \rangle \Longleftrightarrow u \in \partial f(x).$$
Indeed, we have
\begin{align*}
u \in \partial f(x) &\Longleftrightarrow f(z) \geq f(x) + \langle u, z-x\rangle \quad\forall z \\
&\Longleftrightarrow \langle u, x\rangle - f(x) \ge \langle u, z\rangle - f(z) \quad\forall z \\
&\Longleftrightarrow \langle u, x\rangle - f(x) = \sup_z\left\{ \langle u, z\rangle - f(z)\right\} \\
&\Longleftrightarrow \langle u, x\rangle - f(x) = f^*(u) \\
&\Longleftrightarrow f(x)+f^*(u) = \langle x,u \rangle, \text{QED}.
\end{align*}
A: Let $f:X\rightarrow \mathbb R$ and $X^*$ is the dual of $X$.
First recall the definition of $f^*: X^*\rightarrow \mathbb R$
$f^*(u^*)=\sup\limits_{x\in X}{\{\langle u^*,x\rangle - f(x)\}}$, where $u^*\in X^*$ and $\langle .,.\rangle$ is the duality pairing in $X^*\times X$.
Then the definition for subdifferential set is:
$\partial f (x)=\{u^*\in X^*: f(z)\ge f(x)+\langle u^*,z-x\rangle\}$.
Proposition:
$f(x)+f^*(u^*)-\langle u^*,x\rangle =0\Leftrightarrow u^*\in\partial f(x)$.
Proof:
1) Let $f(x)+f^*(u^*)-\langle u^*,x\rangle =0$. From the definition $f^*(u^*)=\sup\limits_{x\in X}{\{\langle u^*,x\rangle - f(x)\}}\Rightarrow$
$0=f(x)+f^*(u^*)-\langle u^*,x\rangle \ge f(x)+\langle u^*,z\rangle -f(z)-\langle u^*,x\rangle\quad \forall z\in X$ which is $f(z)\ge f(x)+\langle u^*,z-x\rangle \quad \forall z\in X$
2) Let $u^*\in \partial f(x)$. By definition $f(z)\ge f(x)+\langle u^*,z-x\rangle\quad \forall z\in X$. Consequently $\langle u^*,x\rangle-f(x)\ge \langle u^*,z\rangle - f(z)\quad\forall z\in X$
$\Rightarrow \langle u^*,x\rangle-f(x)\ge \sup\limits_{z\in X}{\{\langle u^*,z\rangle - f(z)\}}=f^*(u^*)\quad\quad   (1)$. 
But from the definition of $f^*(u^*)\Rightarrow f^*(u^*)\ge \langle u^*,x\rangle-f(x)\quad\quad (2)$. 
From $(1)$ and $(2)$ follows the equality.
