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A Hessian lower bound gives strong convexity and gradient-gap bounds
Statement
Let be open and convex, let , and fix . Suppose for every and . Then for every ,
and
If is a global minimizer and , it is unique and every satisfies
Here is the matrix of second partial derivatives and is the vector of first partial derivatives, as in The Hessian matrix and critical points of a scalar field and The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case.
Facts & Assumptions
Given: An open convex , , , and for every and .
Multivariable Taylor formula with a Lagrange remainder along a line segment: for a function on an open convex set and a segment in that set, the first-order Taylor formula has a second-order Hessian remainder at a point on the segment.
A function is convex exactly when its Hessian is positive semidefinite: a function on an open convex set is convex if and only if its Hessian is positive semidefinite everywhere.
The Hessian matrix and critical points of a scalar field and The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case: is the matrix of second partial derivatives; has the first partial derivatives as its standard coordinates.
Directional derivatives and partial derivatives of a map : the derivative of at is the directional derivative ; for a standard basis vector it is the corresponding partial derivative.
Proof
For any , convexity keeps the segment from to in . Applying [F1] with Taylor order one gives for some . The Hessian hypothesis makes the last term at least , proving the first inequality. Since , [F2] also makes convex on .
Let be an interior minimizer. For each standard basis vector , openness gives an interval around on which . The one-variable function has a minimum at , so its derivative there is ; [F4] identifies this derivative with . Thus every coordinate of is zero.
Apply the first inequality from step 1.1 once to and once to , and add. The function-value terms cancel, leaving .
Apply step 1.1 with and . Using gives . This also proves uniqueness, because equality of the minimum values forces .
Apply step 1.1 with the ordered pair and write . Rearrangement gives . By [F5] and completing the square, the right-hand side is at most , since . This is the gradient-gap inequality.
Depends on
- The Hessian matrix and critical points of a scalar field
- The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case
- Directional derivatives and partial derivatives of a map $U\subseteq\mathbb{R}^m\to\mathbb{R}^n$
- A $C^2$ function is convex exactly when its Hessian is positive semidefinite
- Multivariable Taylor formula with a Lagrange remainder along a line segment
- Cauchy-Schwarz $\lvert\langle x,y\rangle\rvert \le \lVert x\rVert_2\lVert y\rVert_2$ with its equality case, the triangle inequality for $\lVert\cdot\rVert_2$, the parallelogram law and polarisation
Used by
- An inexact gradient residual certifies distance under strong convexity Theorem
- Armijo gradient descent has an explicit global linear rate Theorem
- Damped Newton reaches its local quadratic phase in finitely many continued updates Theorem
- Exact gradient line search has a global objective-gap contraction Theorem
Dependency tree · two levels
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Sources
- Stephen Boyd and Lieven Vandenberghe, Convex Optimization (standard reference, not scraped)
- Philip E. Gill and Margaret H. Wright, Math 171B Numerical Optimization Class Notes (standard reference, not scraped)