Let $X,Y$ two Banach spaces, $G:Y \to R\cup\{+\infty\}$ a convex l.s.c function and $A:dom(A)\subset X \to Y$ a linear closed (graph(A) is closed in $X\times Y$) linear operator. Define \begin{eqnarray} G_A(x,y) &=& G(y), \;\;if\;\;y=Ax \\ &=& +\infty,\;\; otherwise \end{eqnarray} and
\begin{equation} \overline{G}(x,y)=G(y). \end{equation}
So the fenchel transform is \begin{eqnarray} \overline{G}^*(x^*,y^*)&=& G^*(y^*) + \delta_0(x^*)\\ \delta_{graph(A)}^* &=& \delta_{graph(-A^*)}, \end{eqnarray} where $\delta$ is the indicator function.
I am having a hard time to calculate the fenchel transform. I tried first the indicator function and I got this: \begin{eqnarray} \delta_{graph(A)}^*(x^*,y^*) &=& \sup_{(x,y) \in X\times Y}\{\langle (x,y), (x^*,y^*)\rangle - \delta_{graph(A)}(x,y)\} \\ &=& \sup_{(x,y) \in graph(A)}\{\langle (x,y), (x^*,y^*)\rangle\} \\ &=& \sup_{x \in dom(A)}\{\langle x, x^*+A^*y^*\rangle\}, \end{eqnarray} so, if $(x^*,y^*) \in graph(-A^*)$ then $\langle x, x^*+A^*y^*\rangle = 0$ and we have $\delta_{graph(A)}^*(x^*,y^*) = 0$, but if $(x^*,y^*) \notin graph(-A^*)$ why we are going to have $\delta_{graph(A)}^*(x^*,y^*) = \infty?$
And the first fenchel transform I did not understand the $\delta_0(x^*)$.