If there are strict inequalities can we always replace them with non-strict ones? I'm inclined to say yes, but I'm struggling to think of an example where we could do this.

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    $\begingroup$ One has to guess a little about your context. When the domain of a function to optimize is discrete, then the difference between a strict equality and a non-strict one can be crucial. For instance, the minima of the function $f(x)=x^2$ on the sets $\{n\in\mathbb{Z}\mid 0 \lt n\}$ and $\{n\in\mathbb{Z}\mid 0 \le n\}$ differ substantially. $\endgroup$ – whuber Jun 14 '18 at 13:27
  • $\begingroup$ How about in the case where the domain of the function to optimize is assumed to be continuous? $\endgroup$ – Fruh Jun 14 '18 at 20:48
  • $\begingroup$ Then it depends on whether $f$ is continuous. If it's continuous, then there's only an infinitesimal difference between its value on the boundary and its values near the boundary. $\endgroup$ – whuber Jun 14 '18 at 21:03

I am going to assume you are interested in a constrained optimisation problem with a continuous objective function. In its general form, this problem can be written as:

$$\text{Maximise }f(\mathbf{x}) \quad \text{subject to} \quad \mathbf{x} \in \mathcal{G},$$

where $\mathcal{G}$ is the constraint set in the problem. The constraint set can consist of equality constraints or strict/non-strict inequality constraints.

One of the difficulties with these problems is that there might not be any maximum value of the objective function over the constraint set. This can occur when the function has no upper bound over the constraint set, or when the function approaches a supremum at a boundary point of the constraint set, but that boundary point that is not actually in the constraint set. So the first issue you have in these problems is that there might not be a solution at all.

There is a famous mathematical result called the Weierstrauss extreme-value theorem, which gives sufficient conditions for the existence of a maximising value. If $f$ is continuous and $\mathcal{G}$ is a compact set (bounded and closed) then there will exist one or more maximising values in the problem, and so the supremum can be obtained exactly as a maximum.

The Weierstrauss theorem is applicable when your constraint set is bounded by a bunch of equality constraints and/or non-strict inequality constraints, so that the set of values obeying the constraints is closed and bounded. Strict inequality constraints give you open points on the boundary of the constraint set. So in the latter case, you will sometimes get situations where there is no maximising value, and you just get closer and closer to the supremum without getting there.


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