How do linear constraints affect the convexity of my OLS-like optimisation problem?

Question

I would like to augment a linear regression (so a convex OLS problem) with some additional constraints on the coefficients to match the subject I'm working on.

Having $x\in \mathbb{R}^n$, the solution of my linear regression, and my constraints restricting $x$ with are

$$\text{low}_\text{bound} \le A.x \le \text{up}_\text{bound}.$$

$A\in \mathbb{R}^{i\times n}$, $i$ being the number of constraints I'm defining, and $\text{low}_\text{bound}, \text{up}_\text{bound} \in \mathbb{R}^i$ the bounds of my problem, respecting $\text{up}_\text{bound}-\text{low}_\text{bound}\in \mathbb{R}^{+i}$.

Are there any such constraints that would break the convexity of my problem? Or can I just say that since each constraint can be expressed as a reduction of the optimisation space to the space between two hyperplanes, I guarantee the convexity of the problem if the problem was convex at the beginning?

Answer 1

Your intuition is correct.

By definition, an optimization problem (that is, maximizing or minimizing the value of a function $f:\mathcal{X}\to \mathbb R$) is convex when both $f$ is a convex (or, as appropriate, a concave) function and $\mathcal X$ is a convex set in a Euclidean space $E^n.$

One definition of a convex set is that it is the intersection of a (possibly empty) collection of half-planes in $E^n.$ Each inequality in your constraints specifies a half-plane (again, by definition). Ergo, they define a convex subset of $\mathcal X,$ which is automatically a convex set in $E^n.$

At the same time, convexity of $f$ is a global property. Thus, the restriction of $f$ to any subset of its domain is automatically convex--there's nothing to check.

This shows that the constrained version of the original optimization problem is also convex.