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Suppose we have the following variational problem. Given $\mu>1$ and the function $g \in C[0,1]$,

$$ \begin{array}{ll} \underset{f \in C[0,1]}{\text{maximize}} & \overbrace{\displaystyle\int_0^1 x f(x) \, {\rm d} x}^{=: I(f)} \\ \text{subject to} & \displaystyle\int_0^1f(x) \, {\rm d} x = 1 \\ & 0 \leq f(x)\leq \mu g(x), \quad \forall x\in[0,1] \end{array} $$

If we write the integrals as Riemann sums, discretised with equal steps $1/n$ the interval $[0,1]$, we can see that the problem can be seen as maximising over all $f\in C[0,1]$ the limiting problem, as $n$ goes to infinity, $\sum_{i=1}^n\frac{i}{n^2}f(i/n)$ constrained on $\sum_{i=1}^n\frac{f(i/n)}{n}=1$ and $0\leq f(x)\leq \mu g(x)$.

Let us now swap the limit with the maximisation, that is, let us consider a linear program defined as follows: let $y_i$ replace $f(i/n)$, and maximise, over all coordinates $(y_1,\ldots,y_n)$, $\sum_{i=1}^n\frac{i}{n^2}y_i$ constrained on $\sum_{i=1}^n\frac{y_i}{n}=1$ and $0\leq y_i\leq \mu g(i/n)$ for all $1\leq i\leq n$. Assume that the solution exists and denote it as $y^n$.

I am curious about the following:

  1. Intuitively, since we discretised the variational problem, as $n$ grows large we would expect that the linear program yields an approximation of the maximal value $I^*$ of the original objective. Is it true, under mild assumptions, that as $n$ tends to infinity, the value yielded by the solution of the linear program, say $I^n$, tends to $I^*$? If yes, how does one prove so?

  2. Say that $I^*$ is yielded by some functions. Is it true that the step function corresponding to the values of the coordinates of the vector $y^n$, which maximises the linear program, tends to one of these functions as $n$ tends to infinity?

The problem above is just a toy example. Essentially, the question is if discretising a simple variational problem yields solutions of the original problem, under mild assumptions. It is essentially a question about exchanging the order of the maximisation over functions and the limit of the discretisation.

Rodrigo de Azevedo
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xyz
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  • Why discretize? – Rodrigo de Azevedo Jul 23 '23 at 18:55
  • Well, for example to elaborate methods that solve the variational problem numerically, within arbitrary precision dependent on the discretisation parameter. Linear programming algorithms offer very reliable numerics. But it is also interesting in itself: if it is true, technically you would not need much of the sophistication used in variational calculus, as long as the linear program obtained has feature that make it easily solvable. – xyz Jul 23 '23 at 21:18
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    You might want to take a look at Matthew West's thesis. It's not exactly what you need, but the references may be relevant – Rodrigo de Azevedo Jul 23 '23 at 21:57
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    One way of ensuring the non-negativity is to work with $\psi^2 := f$ instead. You could then get rid of the other inequality constraint using a slack variable $s$, i.e., $\psi^2 + s^2 = \mu g$. You are left with only equality constraints – Rodrigo de Azevedo Jul 23 '23 at 22:00
  • sure, that is not a problem, it is a standard step of the LP solution. The question is more about the approximation aspect of the problem. – xyz Jul 24 '23 at 23:24
  • Which is why I linked to West's thesis, where the discretisation happens sooner than usual – Rodrigo de Azevedo Jul 25 '23 at 05:07
  • I had a look, unfortunately it does not treat linear programming methods for solutions of the discretised problem. – xyz Jul 25 '23 at 11:04
  • Why do you need to use LP? – Rodrigo de Azevedo Jul 25 '23 at 11:49
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    The point of the post is that, upon discretisation of the problem and the constraints in this fashion, we end up with an LP. This is easy to solve with known methods (i.e. no need for a discrete variational theory, which is the one developed in the thesis). The question is: does the linear program approximates the original problem in this specific case, as it seems natural to assume, given mild regularity assumptions if needed? (Note that lagrangian has a much more general structure, so it is likely that that discrete theory is overkill, for such a simple structure as in the problem I stated). – xyz Jul 25 '23 at 18:35
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    You could use your naive discretisation and then the sophisticated "Lagrangian discretisation" and then compare them – Rodrigo de Azevedo Jul 25 '23 at 18:54

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