How to prove the conjugate of the conjugate function is itself? Suppose $f$ is closed and convex. $f^*$ is the conjugate of $f$. 
$$f^*(y)=\sup_{x\in\mathbb{R}^n} \{y^Tx−f(x)\}$$
How to prove that $f=f^{**}$?
My thought:
I can write the conjugate by substitute terms into the definition as
$$\sup_a\{z^T a-\sup_x\{a^Tx-f(x)\}\}$$
But it seems cannot get any more simplification.
 A: This problem will be made simpler by translating to the language of convex sets.
Consider $\operatorname{epi} f$ and $\operatorname{epi} f^{**}$, the epigraphs of $f$ and $f^{**}$. 
To start, we have that both epigraphs are convex because $f$ and $f^{**}$ are closed and convex. To show that $f^{**}$ is closed and convex, consider that its epigaph is the intersection of (closed, convex) halfspaces of the form $\{z^{T} y - f^{*}(y): z \in \mathbb{R}^n\}$, because supremum of functions results in the intersections of epigraphs.
We have that $f^{**} \leq f$. From its definition,
$$f^{**}(x) = \sup_{y} x^{T} y - f^{*}(y)$$
$$= \sup_{y} \{ x^{T} y - \sup_{z} \{ y^{T} z - f(z) \} \}$$
$$= \sup_{y} \, \inf_{z} \,y^{T}(z-x) + f(z) \leq \inf_{z} \, \sup_{y} \,y^{T}(z-x) + f(z) = f(x)$$
where the inequality comes from exchanging the infimum and supremum (you may also recall this maneuver from the proof of weak duality).
Now assume for contradiction that $f \neq f^{**}$. From our just-derived inequality, this means that $\exists x$ with $f^{**}(x) < f(x)$. By the closed/compact version of the hyperplane separation theorem, there must be hyperplane in $\mathbb{R}^{n+1}$ that strictly separates $\operatorname{epi} f$ from $(x, f^{**}(x))$. 
This hyperplane cannot be vertical and strictly separate $\operatorname{epi} f$ from $(x, f^{**}(x))$, so we can normalize the normal vector of the hyperplane to be $1$ in the vertical component. This strict separation gives, for some $\epsilon > 0$ and non-vertical component $y \in \mathbb{R}^n$ of our hyperplane,
$$f(z) - \epsilon \geq y^T(z-x) + f^{**}(x) \quad \forall z \in \mathbb{R}^{n}.$$
Some manipulations give
$$y^{T}x - f^{**}(x) - \epsilon \geq y^{T} z - f(z) \quad \forall z$$
and taking the supremum in $z$ yields
$$y^{T} x - f^{**}(x) - \epsilon \geq f^{*}(y).$$
Another manipulation gives
$$y^{T} x - f^{*}(y) - \epsilon \geq f^{**}(x).$$
Expanding the definition of $f^{**}$, we have just shown that
$$y^{T} x - f^{*}(y) - \epsilon \geq \sup_{v}\, v^{T} x - f^{*}(v).$$
Obviously the choice of $y$ on the LHS cannot exceed the supremum on the right, so we have our contradiction.
