Let's say we have smooth function $\phi:[a,b]\to \Bbb R$ and we would like to find a measurable function $f$ such that $c\leq f(x) \leq d$ for all $x\in[a,b]$, $\int_a^b f(x)\phi(x)\mathrm dx = e$ and it maximizes $\int_a^b xf(x)\phi(x)\mathrm dx$. Is there an explicit solution in this case? It reminds of linear programming, but in somewhat infinite-dimensional setting.
Simple linear functional optimization
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functional-analysis
optimization
linear-programming
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0I wonder if a similar approach can be used to linear programming (i.e. Lagrangian multipliers), but perhaps using techniques from variadic calculus. For instance, maybe using the functional derivative instead of the typical derivative? – 2017-02-22
1 Answers
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Due to the inequality constraints, the problem is solvable. From Lagrange multiplier theory, we get the following necessary and sufficient optimality conditions: $f\in L^1$ is optimal iff there is $\lambda\in\mathbb R$ such that $$ \int_a^b (x-\lambda)\phi(x)( g(x)-f(x))dx \ge0 $$ for all $L^1$-functions $g$ with $c\le g\le d$. If $\lambda$ is known and $\phi(x)\ne0$ almost everywhere, this inequality determines $f=f_\lambda$ as a function of $\lambda$.
Together with the condition $$ \int_a^b f(x)\phi(x)dx=e $$ you end up with a non-smooth system in the parameter $\lambda$: Find $\lambda$ such that $$ \int_a^b f_\lambda(x)\phi(x)dx=e. $$ It might be possible to calculate the solution explicitly for special choices of the parameter.
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0It seems to me, just getting inspiration from discrete version of the problem, that in case $c = 0$ I need to have $f = d$ for all $x\geq \lambda$ and $0$ otherwise, perhaps that applies to general $c$ as well. – 2017-02-23
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0Yes, that is true: however, $\lambda$ still needs to be determined from the integral equation constraint. – 2017-02-23