Epigraphical Cones, Fenchel Conjugates, and Duality I'm trying to derive a result relating cones conceived as epigraphs of convex functions, duality, and Fenchel conjungates. Let me state exactly what I'm looking for:
Let $\mathbb{E}$ be an Euclidean space and let $f\colon\mathbb{E}\to\mathbb{R}$ be a function such that 
$$K:=\text{epi}(f)=\{(x,t)\in\mathbb{E}\times\mathbb{R}\,\colon f(x)\leq t\}$$
is a proper (convex,closed,pointed, and with nonempty interior) cone. What is the relation between the Fenchel conjugate $f^\ast$ of $f$ and the dual cone $K^\ast$?
This was my (trivial) attempt until I got stuck:
Let $(y,w)\in\mathbb{E}\times\mathbb{R}$. Then $(y,w)\in K^\ast$ if and only if:
\begin{align*}
&\langle (x,t),(y,w)\rangle\geq 0 \text{ for each } (x,t)\in K\\\iff& \langle x,y\rangle+tw\geq 0\text{ for each } (x,t)\in K\\\iff& \langle x,y\rangle + f(x)w\geq 0\text{ for each }(x,t)\in K\\\iff& \inf_{(x,t)\in K}\{\langle x,y\rangle +f(x)w\}\geq 0 
\end{align*}
First, I'm not sure if I can substitute $t$ for $f(x)$ in the third line above and why this is correct, but I've seen several people taking this approach. For instance, Glineur uses the exact same technique in page 123 of this text. Moreover, I know that I'm quite close of finding something but I must be missing some silly detail on how to deal with $w$.
Does anyone have any source with a proof of this result or can help me finish my proof? 
Thanks
 A: $\newcommand{\R}{\mathbb{R}}\newcommand{\<}{\left\langle}\newcommand{\>}{\right\rangle}\newcommand{\epi}{\operatorname{epi}}$
I do not think there is an elegant relationship. As an example, take $f(x) = |x|$, with $x\in \R$. Then $f^*(y) = \delta_{[-1,1]}(y)$, whose epigraph is not a cone.
There is a correspondence between the epigraph of $f$ and that of $f^*$, that is $(y, b) \in \epi f^*$ if and only if 
$$
\<y, x\> - a \leq b, \forall (x, a) \in \epi f.
$$
If $f$ is lsc, then $f=f^{**}$ and we can apply this to $f^*$. Then $(x, a) \in \epi f$ if and only if 
$$
\<y, x\> - b \leq a, \forall (y, b) \in \epi f^*.
$$
We can try to use this fact to determine $K^*$, but I think it becomes clear that there is no direct relationship between $K$ and $\epi f^*$:
\begin{align}
K^* 
{}={}&
\{(y,w) \in \R^{n+1} {}:{} \<(y, w), (x, t)\>\geq 0, \forall (x,t) \in \epi f\}
\\
{}={}&
\{(y,w) \in \R^{n+1} {}:{} \<y,x\> + tw\geq 0, \forall (x,t) \in \epi f\}
\\
{}={}&
\{(y,w) \in \R^{n+1} {}:{} \<y,x\> + tw\geq 0, \forall (x,t) \in \epi f\}
\\
{}={}&
\{(y,w) \in \R^{n+1} {}:{} \<y,-x\> - tw\leq 0, \forall (x,t) \in \epi f\}
\\
{}={}&
\{(y,w) \in \R^{n+1} {}:{} \<y,z\> - tw\leq 0, \forall (z,t) \in \epi \breve{f}\}
\\
{}={}&\{(y,w) \in \R^{n+1} {}:{} \<y,z\> - tw\leq 0,  \forall (z,t) : \breve{f}(z) \leq t\},
\end{align}
where $\breve{f}$ is the reversal of $f$ defined as $\breve{f}(x) = f(-x)$. The issue here is that we do not know the sign of $w$ in advance. For example, if we knew that $w > 0$, we would have
$$
K^* = \{(y, w): (y,0) \in \epi(w\breve{f})^*\}.
$$
Your question reminds me of the following property: $(u, -1) \in N_{\epi f}(x, f(x))$ if and only if $u \in \partial f(x)$. In this case, we would be interested in $N_{\epi f}(0,0)$. However, this result allows us to tell whether $(u,-1)$ is in $N_{\epi f}(x, f(x))$, so it does not offer a full characterisation of the normal cone. Note that $K^* = -N_{\epi f}(0,0)$.
In case you are interested, we can show that $K^*$ can be written as 
$$
-K^* =
\{\lambda (v, -1) {}:{} v \in \partial f^*(0)\}
\cup 
\{(v, 0) {}:{} 0 \in \partial^\infty f(v)\},
$$
where $\partial^\infty f$ is the horizon subgradient of $f$, provided that $f$ is lsc. This is Theorem 8.9 in the book of Rockafellar and Wets.
By the way, since $\epi f$ is a cone, we know that it is positively homogeneous.
