Let $C$ be a closed, convex cone in $R^n$. Define $$ A := \{ y(.) \in L_1 [0,T]\; |\; y(t) \in C ~ \mbox{for almost all} ~t\in [0 ,T] \} .$$
My question is that can we derive an explicit formulation for the Dual of $A$ defined by $$ A^* = \{ z(.) \in L_{\infty} [0,T] \; | \; \langle z , y \rangle = \int z(t) y(t) \;dt \leq 0 \; \text{for all}\; y(.) \in A \} $$
Note that, $A$ is closed convex cone.
Clearly $$ \{x\in L^\infty: \ z(t)\in C^* \text{ for almost all } t\} \subset A^*. $$ Let me show equality. Take $z\in A^*$, $t\in (0,T)$. Let $\chi_{B_r(t)}$ be the characteristic function of the ball $B_r(t)$ for $r>0$ small.
Let $\tilde C\subset C$ be a countable and dense subset. Take $c_i\in \tilde C$. Define $y(s):= \chi_{B_r(t)}(s)c_i$. Then $$ 0\ge \int y(s)z(s)ds = \int_{B_r(t)} z(s)c_i ds . $$ By the Lebesgue differentiation theorem, we have the convergence $$ \frac1{|B_r(t)|}\int_{B_r(t)} z(s)c_i ds \to z(t)c_i $$ for all $t\in [0,T]\setminus N_i$ with $N_i$ being of zero measure. This implies that for all $i$ $$ z(s)c_i \le 0 \quad \forall s\not\in N_i. $$ Now, $N:=\cup_iN_i$ has zero measure as well, so $$ z(s)c_i \le 0 \quad \forall i \text{ and } s\not\in N. $$ The set $\tilde C$ is dense in $C$, hence $$ z(s)c \le 0 \quad \forall c\in C \text{ and } s\not\in N. $$ This implies $$ z\in \{x\in L^\infty: \ z(t)\in C^* \text{ for almost all } t\} . $$