Indeed, all except perhaps property (2) can be proved without functional analysis. It leads to a completely deterministic algorithm to compute a basis for the dual cone (not a very good algorithm though as its complexity
is exponential in the initial basis size).
In the sequel RCP is shorthand for “rational convex polyhedral
cone”.
Nearly everything reduces to the iteration of the following lemma:
Fundamental lemma. Let $\sigma={\sf Cone}(u_1,\ldots,u_k)$ be a RPC cone. Let
$w\in{{\mathbb Z}^d}$, and $\sigma_w=\lbrace x\in\sigma \,|\, \langle x,w\rangle\ \geq 0\rbrace$. Then
$\sigma_w$ is again a RPC cone. In fact, a completely explicit generating set can be
found, as follows. Let
$$
\begin{array}{lcl}
I_{-} &=& \big\lbrace i \ \big|\ \langle u_i,w\rangle \ < 0 \big\rbrace \\
I_{0} &=& \big\lbrace i \ \big|\ \langle u_i,w\rangle \ = 0 \big\rbrace \\
I_{+} &=& \big\lbrace i \ \big|\ \langle u_i,w\rangle \ < 0 \big\rbrace \\
\end{array}
$$
Then $\sigma_w={\sf Cone}({\cal F})$ where
$$
{\cal F}=\big\lbrace u_i\ \big|\ i \in I_0 \cup I_{+}\big\rbrace \cup
\big\lbrace \langle u_j,w\rangle u_i-\langle u_i,w\rangle u_j\ \big| \ i \in I_{-}, j\in I_{+}\big\rbrace
$$
To show this fundamental lemma we need another :
Lemma 1. Let $n,m \geq 1$ be two integers. Let $x_1,x_2,\ldots,x_n$
and $y_1,y_2,y_3, \ldots ,y_m$ be nonnegative numbers. Then the
following are equivalent :
(i) $x_1+x_2+ \ldots +x_n \geq y_1+\ldots +y_m$
(ii) There exists nonnegative numbers $\xi_k(1\leq k\leq n)$ and
$t_{ij}(1\leq i \leq n,1\leq j \leq n)$ such that
$x_i=\xi_i+\sum_{j}t_{ij}$ for every $i$ and
$y_j=\sum_{i}t_{ij}$ for every $j$.
Proof of lemma 1. The direction (ii)$\Rightarrow$(i) is easy because
under the hypothesis of (ii) we have
$$
\big(\sum_{i} x_i\big)-\big(\sum_{j} y_j\big)=
\big(\sum_{i} \xi_i\big) \geq 0
$$
Conversely, we are going to show (i)$\Rightarrow$(ii) by induction on
$N=n+m$.
The base case $N=2$ is easy : we have $x_1\geq y_1$ and need just take
$\xi_1=0,t_{11}=y_1$.
Now suppose the property is true at level $N$ ; we are going to show it still
holds at level $N+1$. So suppose
$x_1+x_2+ \ldots +x_n \geq y_1+\ldots +y_m$ for some nonnegative
numbers $x_1,x_2,\ldots,x_{n},y_1,y_2,\ldots,y_m$, with $n+m=N+1$. There are two cases.
Case 1. $x_n \geq y_m$. Then we can write $x_n=y_m+X_n$ where
$X_n$ is nonnegative. We then have an inequality with one
variable less : $x_1+x_2+\ldots+x_{n-1}+X_n \geq
y_1+y_2+\ldots +y_{m-1}$ and we can apply the induction hypothesis.
Case 2. $y_m \geq x_n$. Then we can write $y_m=x_n+Y_m$ where
$Y_m$ is nonnegative. We then have an inequality with one
variable less : $x_1+x_2+\ldots+x_{n-1} \geq
y_1+y_2+\ldots +y_{m-1}+Y_m$ and we can apply the induction hypothesis.
This concludes the proof of lemma 1.
Proof of fundamental lemma from lemma 1. Let $\tau={\sf Cone}({\cal F})$.
As all vectors in $\cal F$ are nonnegative linear combinations
of the $u_i$, whose scalar product with
$w$ is nonnegative, we have ${\cal F} \subseteq \sigma_w$ and hence
$\tau \subseteq \sigma_w$. Conversely, let $v\in\sigma_w$. Let us put
$n=|I_+|,m=|I_{-}|,p=I_{0}$. Reordering the $u_i$ if necessary,
we may assume without loss that
$I_+=[1,n],I_-=[n+1,n+m],I_0=[n+m,d]$. Then, there are nonnegative coefficients
$a_1,a_2,\ldots, a_n,b_1,b_2,\ldots, b_m,c_1,c_2, \ldots, c_p$ such that
$$
v=\sum_{i=1}^{n} a_iu_i-
\sum_{j=1}^{m} b_ju_{n+j}+
\sum_{k=1}^{p} c_ku_{n+m+k} \tag{1}
$$
Then, the condition $\langle v,w\rangle \geq 0$ can be expressed as
$$
\sum_{i=1}^{n} a_i\langle u_i,w\rangle-
\sum_{j=1}^{m} b_j\langle u_{n+j},w\rangle \geq 0 \tag{2}
$$
Or, if we put $x_i=a_i\langle u_i,w\rangle$ and $y_j=b_j\langle u_{n+j},w\rangle $,
$$
\sum_{i=1}^{n} x_i-
\sum_{j=1}^{m} y_j \geq 0 \tag{3}
$$
This is the hypothesis of (i) in lemma 1, so let
$\xi_k(1\leq k\leq n)$ and $t_{ij}(1\leq i \leq n,1\leq j \leq n)$
as in (ii) of this lemma. We then have
$$
\begin{array}{lcl}
a_i&=&\frac{\xi_i+\sum_{j}t_{ij}}{\langle u_i,w\rangle } (1\leq i\leq n),\\
b_j&=&\frac{\xi_i+\sum_{i}t_{ij}}{\langle u_{n+j},w\rangle } (1\leq j\leq m)
\end{array} \tag{4}
$$
If we put $z_i=\frac{\xi_i}{\langle u_i,w\rangle }$ and
$s_{ij}=\frac{t_{ij}}{\langle u_i,w\rangle\,|\langle u_{n+j},w\rangle|}$, we then have
$$
v=\sum_{i=1}^n z_i u_i
+\sum_{i=1}^n \sum_{j=1}^m
s_{ij}(\langle u_{n+j},w\rangle u_i-\langle u_i,w\rangle u_{n+j})+
\sum_{k=1}^{p} c_ku_k \tag{5}
$$
which is plainly an element of $\tau$. This concludes the proof
of the fundamental lemma.
Proof of property (1) from fundamental lemma Notice that
${\cal C}_0={\mathbb R}^d$ is itself a RCP cone (it can be written as ${\sf Cone}(e_1,-e_1,e_2,-e_2,
\ldots ,e_d,-e_d)$ where $(e_1,e_2,\ldots ,e_d)$) is the canonical
basis of ${\mathbb R}^d$). Then, let
$$
{\cal C}_i=\lbrace v\in {\mathbb R}^d \ | \
\langle v,u_j\rangle \ \geq 0 \ (1\leq j \leq i)\rbrace
$$
Iterating the fundamental lemma, we see successively that ${\cal C}_0,
{\cal C}_1,\ldots,$ are all RCP cones. In the end ${\sigma}^\vee={\cal C}_k$
is a RCP cone.
Best Answer
It has been three and a half years since this question was asked. I hope my answer still helps somehow.
By definition, the dual cone of a cone $K$ is:
$$K^* = \{y | x^Ty \ge 0, \forall x \in K\}$$ Denote $Ax \in K$, and directly using the definition, we have:
$$K^* = \{y|(Ax)^Ty, x\succeq 0\} = \{y|(A^Ty)^Tx\ge0, x\succeq0\}$$ Now lets have a close look at the conditions of the set.
For concise, first denote $A^Ty$ as $u$.
For $\forall x\succeq0$, to make $u^Tx \ge 0$, all the components of $u$ must be greater than $0$. If $u_i<0$, you can choose a $x$ looks like $(x_0=0, ..., x_i=1, x_{i+1}=0, ...)$, which makes $u^Tx \lt 0$.
Thus the final form of the dual cone is:
$$K^*=\{y|A^Ty\succeq0\}$$