Now that I understand the context better I am going to promote my own comments to an answer.
There is no general principle for artificially restricting the domain of a function when computing its convex conjugate. In fact, you should not do so unless you have a compelling and mathematically sound reason in your specific application. The domain of the conjugate function---that is, the set of $y$ for which the supremum is finite---is an important part of the conjugate itself, and should not be discarded. For example, when computing the dual of a convex optimization problem with an objective $f(x)$, the domain information imposes implicit constraints on the dual variables. If you remove that information, the dual problem is incorrect; it no longer provides bounds for the primal problem.
So what's going on in Chapter 5 of Ms. Fazel's thesis, or in the case of the so-called $\ell_0$ norm? Note that the interest is not in the conjugate of these functions, but rather in their convex envelopes. It just so happens that the convex envelope of a function can computed using the conjugate of the conjugate. Unfortunately, the convex envelopes of $f(x)=\mathop{\textbf{card}}(x)$ or $g(X)=\mathop{\textbf{rank}}(X)$ are not very interesting---in fact, I think they are identically zero. On other hand, the modified, extended-valued functions
$$\bar{f}(x) = \begin{cases} \mathop{\textbf{card}}(x) & \|x\|_\infty \leq 1 \\ +\infty &\|x\|_\infty > 1 \end{cases}, \qquad
\bar{g}(X) = \begin{cases} \mathop{\textbf{rank}}(x) & \|X\|_2 \leq 1 \\ +\infty &\|X\|_2 > 1 \end{cases}$$
have non-trivial convex envelopes.
As for why one would want these convex envelopes, it is because they provide some theoretical justification for why $\bar{f}^{**}(x)=\|x\|_1$ and $\bar{g}^{**}(X)=\|X\|_*$ are effective convex proxies for their non-convex counterparts. As you know, Ms. Fazel's thesis is entitled Matrix Rank Minimization with Applications, and it makes heavy use of trace minimization and nuclear norm minimization to find low-rank matrices that satisfy the modeling conditions. Section 5.1 is devoted to providing justifications for the convex heuristics that she uses throughout the work.
Added to clarify: you have expressed an interest in a comment above in finding the "tightest" convex envelope of a function. It's very important to note that if you truly want the tightest convex envelope for the entire function, you cannot restrict the domain in any way before taking the double conjugate. When you impose a domain restriction, you are computing the convex envelope of a different function. It will no longer serve as a lower bound for the original function. It will, however, be a tighter envelope over the domain you have selected.
You can see this for yourself: look at $f(x)=\mathop{\textbf{card}}(x)$ and $g(x)=\|x\|_1$. Clearly, there are values of $x$ for which $f(x)<g(x)$. So $g$ cannot be the convex envelope for $f$. It is, however, the convex envelope if you restrict the domain of $f$ to $\{x\,|\,\|x\|_\infty \leq 1\}$ (as we did with $\bar{f}$ above).
In general, $g_i(x)$ is not convex. We can construct a counterexample as follows.
Consider the case of three categories, where $\boldsymbol{x} = (x_1, x_2, x_3)$. If $g_1(\boldsymbol{x})$ is convex, then $h(t) \triangleq g_1((0, t, -t))$ should also be convex.
Since
$$
\begin{aligned}
g_1(\boldsymbol{x}) &= \log \frac{\exp(x_1) + \exp(x_2) + \exp(x_3)}{\exp(x_2) + \exp(x_3)}\\
&= \log \left(1 + \frac{1}{\exp(x_2 - x_1) + \exp(x_3 - x_1)}\right)
\end{aligned}
$$
we know that
$$
h(t) = \log\left(1 + \frac{1}{e^{t} + e^{-t}}\right),
$$
and it is not convex.
Indeed, if $h(t)$ is convex, note that
$$
\lim_{t \to \infty} h(t) = 0,
$$
then the convexity will imply $h(t) \equiv 0$, which is impossible.
Best Answer
The Conjugate Function is a Concave Function in its domain.
You can easily see that the derivative implies:
$$ {y}_{i} = \frac{ {e}^{ {x}_{i} } }{ \sum_{j = 1}^{n} {e}^{ {x}_{j} } } $$
This requires $ {y}_{i} > 0 $ (Which can be later relaxes by the definition of $ 0 \ln \left( 0 \right) $).