This isn't what was asked, but there's nice intuition for the fact that if $f$ is a closed convex function, then $f^{**} = f$. (Here $f^*$ is the convex conjugate of $f$.)
I think the source of all duality results in convex analysis is the fact that a closed convex set $C$ is the intersection of all closed half spaces that contain $C$.
If we apply this idea to the epigraph of a closed convex function $f$, we see that $f$ is the supremum of all affine functions that are majorized by $f$.
For any given slope $m$, there may be many different constants $b$ such that the affine function $\langle m, x \rangle - b$ is majorized by $f$. We only need the best such constant.
That's what the convex conjugate $f^*$ does for us. Given a slope $m$, $f^*$ returns the best constant $b$ such that $\langle m, x \rangle - b$ is majorized by $f$. And (clearly) the best choice of $b$ is:
\begin{equation}
f^*(m) = \sup_x \quad \langle m, x \rangle - f(x).
\end{equation}
($\langle m, x \rangle$ exceeds $f(x)$ by at most $f^*(m)$, so
$\langle m, x \rangle - f^*(m)$ exceeds $f(x)$ by at most $0$.)
From this perspective it's immediate that for any $x$
\begin{equation}
f(x) = \sup_m \quad \langle m, x \rangle - f^*(m).
\end{equation}
And this says that $f(x) = f^{**}(x)$.
Here there was a natural way to associate a closed convex set with the function $f$, and it led to a duality result. Is there a similar way to associate a closed convex set with a convex optimization problem, and obtain a duality result for convex optimization problems?
If the optimization problem is:
\begin{align}
\operatorname*{minimize}_x &\quad f(x) \\
\text{subject to} & \quad Ax = b \\
& \quad h_i(x) \leq 0 \quad \text{for } i = 1,\ldots, m
\end{align}
where $f$ and $h_i,i=1,\ldots, m$ are closed convex functions, then we can associate an "epigraph" to this convex optimization problem as follows:
\begin{equation}
C = \{ (u,v,t) \mid \exists \, x \,\text{such that } f(x) \leq t, Au = b, h_i(x) \leq v_i \text{ for } i = 1,\ldots,m \}.
\end{equation}
By thinking of $C$ in terms of closed half-spaces that contain $C$, or in terms of hyperplanes that support $C$, we can get a geometric interpretation of the dual problem, and a nice proof of a strong duality theorem.
