Prove the second convex conjugate of a function is itself. First, we define the conjugate $^*$ as $f^*(y)=\sup_{x\in\mathbb{R}^n} \{y^Tx−f(x)\}$, where $f$ is a convex function.
The second convex conjugate is defined as $f^{**}:=(f^*)^*$.
I know if $f$ is closed, then $f^{**}=f$.
But I saw a statement saying that if lim$_{\vert x\vert\to\infty}\frac{f(x)}{\vert x\vert}=\infty$, then $f^{**}=f$. I have no idea how to prove it. Any help will be appreciated.
 A: $\newcommand{\R}{\mathbb{R}}$This statement is not true because $f^{**}=f$ if and only if $f$ is lsc, but the condition you mentioned only describes the behaviour of the function at infinity. 
This condition is typically employed for constrained optimization problems and problems where the cost function is the sum of two (extended-real-valued) lsc functions.
The property you are referencing, $\lim_{\|x\|\to\infty}\frac{f(x)}{\|x\|}$, is called supercoercivity. Supercoercivity is a condition under which the sum of two lower semicontinuous function has an exact infimal convolution, that is, if $f, g:\R^n\to\R$ are two lsc functions and $f$ is supercoercive then
$$
f {\,}\square{\,} g = f \boxdot g,
$$
where $(f {\,}\square{\,} g)(x) = \inf_{z\in\R^n} f(z) + g(x-z)$ and $(f \boxdot g)(x) = \min_{z\in\R^n} f(z) + g(x-z)$, implying that the minimum is attained at some point $z$, i.e., there is a $\bar{z}(x)$ such that 
$$(f {\,}\square{\,} g)(x) = f(\bar{z}(x)) + g(x-\bar{z}(x)).$$
So, if $f$ and $g$ are lsc and $f$ is supercoercive, we have
\begin{align}
\inf_{x} f(x) + g(x) {}={}& - (f+g)^*(0)
\\ {}={}& -(f^* \, \square\, g^*)(0) 
\\ {}={}& -(f^* \, \boxdot g^*)(0) 
\\ {}={}& -\min_{z\in\R^n} f^*(z) + g^*(-z),
\end{align}
which is the Fenchel dual of the original problem.
If $C$ is a nonemtpy convex closed set, then function $g := \delta_C:\R^n\to\overline{\R}$ is lsc, so if $f$ is lsc and supercoercive
$$
\inf_{x\in C} f(x) = -\min_{z\in\R^n} f^*(x) + \sigma_C(-x),
$$
where $\sigma_C$ is the support function of $C$ (which is easy to compute for some simple sets).
Note. For problems of the form $\min_x f(x)$ a similar condition, namely coercivity (that is, $\lim_{\|x\|\to\infty} f(x) = \infty$), is used to establish existence of minimizers. But again, $f$ must be lsc so that $f^{**} = f$. The fact that $f^{**}=f$ does not imply that $f$ has a minimizer (counterexample: $f(x) = e^x$).
References: 
[1] H.H. Bauschke and P.L. Combettes, Convex analysis and monotone operator theory in Hilbert spaces, Springer, 2011.
