Why is this the optimal value of the dual problem? 
Let the primal problem be to determine the feasibility of the system $f_0 =0, f_i(x) \le 0$ for $i=1, 2, \dots, m$ and
   $h_i(x)=0$ for $i = 1, 2, \dots, m$.
The dual problem associated with this is to maximize $g(\lambda, v) =
 \inf_{x \in D}(\lambda^Tf(x) + v^Th(x))$.
The optimal value of the dual problem is given by 
$d^* =
\begin{cases}
\infty,  & \text{ if } \space\lambda \ge 0, g(\lambda, v)\gt 0 \text{ is feasible } \\[2ex]
0, & \text{ if } \space\lambda \ge 0, g(\lambda, v)\gt 0 \text{ is infeasible }
\end{cases}$

Why is the optimal value of the dual problem given as $d^*$?  Why does $g(\lambda, v) \gt 0$ have to be feasible or infeasible?  Why can't we have $g(\lambda, v) \le 0$ in any situation?  If $g \lt 0$ then how can $d^*=0$?
Below is the full text from where this is taken.  From the book Convex Optimization.


 A: Let $f(x) = (f_1(x), ..., f_m(x))$ and $h(x)=(h_1(x), ..., h_p(x))$. Let $D$ be a set.  The primal problem is to find $x \in D$ that satisfies $f(x)\leq 0, h(x)=0$. We say the primal problem is feasible if there is such an $x \in D$. 
Define 
$$ M = \{(\lambda, v) : \lambda \in \mathbb{R}^m, v \in \mathbb{R}^p, \lambda \geq 0\}$$
The dual function $g(\lambda, v)$ should be maximized subject to the constraint $(\lambda, v) \in M$. Define $d^* = \sup_{(\lambda, v) \in M} g(\lambda, v)$. It is not difficult to show that 
$$d^* = \left\{ \begin{array}{ll}
\infty &\mbox{ if the primal is not feasible } \\
0  & \mbox{ if the primal is feasible} 
\end{array}
\right.$$
So there are only two possibilities for $d^*$ (either $d^*=0$ or $d^*=\infty)$. 
It follows that if there is a $(\lambda, v) \in M$ such that $g(\lambda, v)>0$, then certainly the supremum of $g(\lambda, v)$ cannot be zero, so the supremum must be $\infty$. Conversely, if there is no $(\lambda, v) \in M$  such that $g(\lambda, v)>0$, then the supremum certainly cannot be $\infty$ and so the supremum must be $0$. Thus
$$ d^* = \left\{ \begin{array}{ll}
\infty &\mbox{ if there is a $(\lambda, v) \in M$ such that $g(\lambda,v)>0$. } \\
0  & \mbox{ else} 
\end{array}
\right.$$
This is likely what the equation in your question means. 

I agree with Tony S.F. comments that the way of writing $d^*$ in your question (which I think is equivalent to my "second way" of writing $d^*$ above) is a bit unusual.  However, in your cut-and-paste it seems the authors use the second way to prove the first way (not the other way round as I suggest above). 
Actually it is not a bad way to prove it:  The authors observe: 


*

*$(\lambda, v) \in M \implies (\alpha \lambda, \alpha v) \in M$ for any real number $\alpha \geq 0$. 

*$g(\alpha \lambda, \alpha v) = \alpha g(\lambda, v)$ for any real number $\alpha \geq 0$. 

*$(0,0) \in M$.
So if there is a $(\lambda, v) \in M$ such that $g(\lambda, v)>0$, we observe $\lim_{\alpha\rightarrow\infty}g(\alpha\lambda, \alpha v)\rightarrow \infty$ and so $\sup_{(\lambda, v) \in M} g(\lambda, v) = \infty$. On the other hand if there is no $(\lambda, v) \in M$ for which $g(\lambda, v)>0$ then the supremum must be less than or equal to 0; The supremum cannot be negative because $g(0,0)=0$, and so the supremum must be 0.
