Dual problem with integrals I am reading a paper where the author derives the following Lagrangian dual problem :
$\min_v \int_R \frac{1}{4} \frac{\beta^2}{v-2\|x\|}dx+v\;\;\;\text{s.t.}\;\;\;v\geq 2\|x\|\;\;\;\forall x \in R$
from the primal problem :
$\max_{f(.)} \int_R (2\|x\| f(x) + \beta \sqrt{f(x)})dx\;\;\; 
\text{s.t.}\;\;\;\int_R f(x) dx= 1 \;\;\;\text{and}\;\;\;f(x) \geq 0\;\;\;\forall x \in R$
where $f(.)$ belongs to the Banach space $L^2$ over a compact set $R$ (a distribution function).
Do you know how to construct a dual problem in case the objective and constraints of the primal include integrals. It was said that standard techniques of vector space optimization could be used to approach the function $f(.)$, but this is maybe not obvious. I could not pinpoint the starting point.
 A: $\newcommand\R{\mathbb R}$
A convenient way to derive the dual problem from a primal one is by using the minimax duality for the Lagrangian, which is given here by the formula  $$L(f,v):=\int_R\big[2|x|f(x)+b\sqrt{f(x)}\big]\,dx-v\Big(\int_Rf(x)\,dx-1\Big),$$
where $|x|:=\|x\|$ and $b:=\beta$. Clearly,
$$\sup_{f\ge0}\inf_{v\in\R} L(f,v)
=\sup\Big\{\int_R\big[2|x|f(x)+b\sqrt{f(x)}\big]\,dx\colon f\ge0,\int_R f(x)\,dx=1\Big\},$$
which is value of the primal problem.
The value of the dual problem is
$$\inf_{v\in\R}\sup_{f\ge0} L(f,v)
=\inf_{v\in\R}\Big(v+\sup\Big\{\int_R\big[2|x|f(x)+b\sqrt{f(x)}-vf(x)\big]\,dx\colon f\ge0\Big\}\Big)
=\inf_{v\in\R}\Big(v+\int_R s(|x|,v)\,dx\Big),
$$
where
$$s(a,v):=\sup\{2at+b\sqrt t-vt\colon t\ge0\}.$$
For any real $a>0$, it is easy to see that
$$s(a,v)=\frac{b^2}{4 (v-2 a)}$$
if $b>0$ and $v>2a$, $s(a,v)=0$ if $b\le0$ and $v\ge2a$, and $s(a,v)=\infty$ otherwise; in particular, $s(a,v)=\infty$ if $v<2a$. Thus, with $|R|:=\max\{|x|\colon x\in R\}$, the value of the dual problem is
$$\inf_{v>2|R|}\Big(v+\int_R \frac{\max(0,b)^2}{4 (v-2|x|)}\,dx\Big).
$$
