Mathematics · Advanced university
Convexity and constrained optimization
Values on a line segment lie below the corresponding chord of function values.
Prerequisites
Gradients, inequalities and matrices.
Study notes
Convex function
Values on a line segment lie below the corresponding chord of function values.
First-order criterion
For differentiable convex f, f(y)≥f(x)+∇f(x)·(y−x).
Global minima
A stationary point of a differentiable convex function is a global minimizer; strict convexity gives at most one.
Constraints
Lagrange multipliers and KKT conditions require their own regularity and feasibility checks.
Worked examples
Find the x-coordinate minimizing f(x,y)=x²+2y²−6x.
The x derivative is 2x−6; setting it to zero gives 3, and the Hessian is positive definite.
3
For a differentiable convex function, a zero gradient implies what?
The supporting-hyperplane inequality gives f(y)≥f(x).
A global minimum
Check your understanding
Prove that a strictly convex function has at most one minimizer on a convex domain.
Minimize x²+y² subject to x+y=6. Find the minimum value.
Is the feasible set x+y≤1 in R² convex?
- Only its boundary is convex
- Only for positive coordinates
- Yes
- No, because it is unbounded
Why must a proposed constrained optimum be checked for feasibility?
Answers and marking guidance
Answers and marking notes
- If two distinct minimizers existed, strict convexity at their midpoint would give a value strictly below their common minimum, a contradiction.
Credit the midpoint lying in the domain and strict inequality.
- 18
The minimum occurs at x=y=3 by completing a square or Lagrange multipliers; value 9+9=18.
- Yes
A linear inequality is preserved under convex combinations.
- Stationarity equations alone can produce points outside the allowed set; an infeasible point cannot solve the constrained problem.
Credit distinction between algebraic candidates and admissible solutions.
Further reading and review references
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