How to solve an optimization problem whose optimization variable is a function?

Question

I would like to find an optimal probability density function (PDF) $f$. Given $b$,

$$ \begin{array}{ll} \underset {f} {\text{minimize}} & C \\ \text{subject to} & 1 + \frac{b}{x} \displaystyle\int_{0}^{x}f(t)dt \leq C, \text{ for any } x \in [0,\infty) \\\\ & \frac{1}{b} \left(\displaystyle\int_{0}^{x}(t+b)\cdot f(t)dt+x\cdot\int_{x}^{\infty}f(t)dt \right) \leq C, \text{ for any }x\in(b,\infty) \\\\& \displaystyle\int_{0}^{\infty}f(t)dt=1, \\ & f(t)\geq 0, \ \text{for any $t\geq 0$.}\end{array} $$

I do not know how to solve it. Are there any systematic ways or general ideas of dealing with such a functional optimization problem?

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My idea

Here I can introduce another optimization variable $B\geq b$ such that the probability density is restricted within the range of $[0,B]$, i.e. $\int_{0}^{B}f(t)dt=1$. Then the original optimization problem can be rewritten as

$\min\limits_{B,f(t)} C$

s.t. $1+\frac{b\cdot\int_{0}^{x}f(t)dt}{x}\leq C, \text{ for any }x\in[0,B]$

$\quad\frac{\int_{0}^{x}(t+b)\cdot f(t)dt+x\cdot\int_{x}^{B}f(t)dt}{b}\leq C, \text{ for any }x\in(b,B]$

$\quad\int_{0}^{B}f(t)dt=1$,

$\quad B\geq b$

$\quad f(t)\geq 0, \ \text{for any $t\geq 0$.}$

We take a look at the second constraint $\frac{\int_{0}^{x}(t+b)\cdot f(t)dt+x\cdot\int_{x}^{B}f(t)dt}{b}\leq C$. Taking the derivative of the left hand side with respect to x, we find that its derivative is $f(x)\geq 0$, which implies that the left hand side is non-decreasing with respect to $x\in (b,B]$. Hence the second constraint can be reduced to $1+\frac{\int_{0}^{B}t\cdot f(t)dt}{b}\leq C$ (by taking $x=B$). Hence the optimization problem can be rewritten as

$\min\limits_{B,f(t)} C$

s.t. $1+\frac{b\cdot\int_{0}^{x}f(t)dt}{x}\leq C, \text{ for any }x\in[0,B]$

$\quad 1+\frac{\int_{0}^{B}t\cdot f(t)dt}{b}\leq C$

$\quad\int_{0}^{B}f(t)dt=1$,

$\quad B\geq b$

$\quad f(t)\geq 0, \ \text{for any $t\geq 0$.}$

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The reason why I have this optimization problem

The reason why I have this optimization problem is that I am trying to solve a cost optimization problem for a distributed storage system, constant $b$ here is the cost of transferring the data object among different servers. $f(t)$ here is the pdf of the storage duration of the data copy in a server (and storage cost is proportional to the storage duration, here we assume that the storage cost rate is 1 per unit of time), and after time $t$ the data will be dropped in the server so that there will be no storage cost anymore for our server. So our algorithm is something like a randomized online algorithm, whose target is to find the optimal pdf $f(t)$ to minimized the overall cost of storing and transferring data in the system. Through some simplifications of my modelling, I finally derive this functional optimization problem, but do not know what to do next.

This problem has infinitely many constraints (i.e., for each continuous value $x\in [0,\infty)$, we have constraints like the first and the second constraint), and this is the mainly obstacle of solving this optimization problem.

Answer 1

Differentiating the left-hand side of the second condition I get $b^{-1}(f(x) + \int_x^\infty f(t) dt)$, not just $b^{-1} f(x)$, but it is still $\ge 0$. So the second condition can be replaced by its version at $x=\infty$, $X:=\int_0^\infty t f(t) dt \le b(C-1)$. Here we have named $X$ the yet unknown first moment of $f(t)dt$. Since $X$ gives a lower bound on $C$, you'd want $X$ to be as small as possible, in other words concentrating the weight of $f(t)dt$ near $t=0$.

Next, the first condition can be rewritten as $\int_0^x f(t) dt \le b^{-1} (C-1) x$. It counterbalances the idea of concentrating the weight of $f(t)dt$ near the origin by the bound $f(t)\le b^{-1}(C-1)$ (more precisely, if $f(t)\ge F>0$ on some interval $[0,T]$, then it must be that $F\le b^{-1}(C-1)$).

At this point, one reasonable assumption would be that the first condition does not increase the lower bound on $C$ obtained from the second condition, which translates to the bound $f(t) \le b^{-2} X$. Saturating this bound gives a step function distribution $f(t) = b^{-2} X \Theta(B-t)$ for some $B>0$. Using this $f(t)$ to check its normalization and compute the moment $X$ results in the relations $b^{-2} BX = 1$, $X = \frac{1}{2} (B/b)^2 X$, or $B=\sqrt{2} b$, $X=b/\sqrt{2}$. Then, solving $X=b(C-1)$ gives $C=1+1/\sqrt{2} \approx 1.7$, corresponding to $f(t) = \frac{b^{-1}}{\sqrt{2}} \Theta(\sqrt{2}b - t)$.

The above argument used a heuristic assumption, so it is not airtight, but it does give you a feasible pair of $C$ and $f(t)$. Perhaps showing that the assumption is necessary for optimality would give you a complete argument.