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A positive-semidefinite Hessian need not give strict convexity
Statement refuted
If a function on an open convex set has a positive-semidefinite Hessian everywhere, then it is strictly convex.
The counterexample below establishes: The Hessian of is positive semidefinite everywhere, but is not strictly convex.
Facts & Assumptions
Given: The positive-semidefinite quadratic-form convention Positive definite, negative definite, semidefinite, and indefinite quadratic forms and strict convexity Convex and strictly convex functions on Euclidean convex sets.
A function on an open convex set is convex if and only if its Hessian is positive semidefinite everywhere (A function is convex exactly when its Hessian is positive semidefinite).
The derivative of is for every integer (For a natural the function is differentiable everywhere with derivative ; for it is the constant , with derivative ; for a natural the function is differentiable at every with derivative ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term).
Scalar multiples of differentiable real functions are differentiable with the expected derivatives (Sums, scalar multiples, products and quotients: , , , and when ).
Counterexample
By [L2] and [L3], the Hessian of is , which is positive semidefinite. By [L1], is convex.
The distinct points and , and every point of the segment joining them, all have value zero. Thus the strict convexity inequality is an equality on that segment. The Hessian of is positive semidefinite everywhere, but is not strictly convex.
Depends on
- A $C^2$ function is convex exactly when its Hessian is positive semidefinite
- Positive definite, negative definite, semidefinite, and indefinite quadratic forms
- Convex and strictly convex functions on Euclidean convex sets
- For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$; for $n = 0$ it is the constant $1$, with derivative $0$; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term
- Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$, $(\alpha f)'(c) = \alpha f'(c)$, $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$, and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$
Used by
- FALSE: a positive-semidefinite Hessian gives strict convexity False statement
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Sources
- S. Boyd and L. Vandenberghe, Convex Optimization, §3.1.4 (standard reference, not scraped)