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A Hessian lower bound gives strong convexity and gradient-gap bounds

Statement

Let U⊆Rd be open and convex, let f∈C2(U,R), and fix μ>0. Suppose vTHf(z)v≥μ∥v∥22 for every z∈U and v∈Rd. Then for every x,y∈U,

f(y)≥f(x)+⟨∇f(x),y−x⟩+μ2∥y−x∥22,

and

⟨∇f(y)−∇f(x),y−x⟩≥μ∥y−x∥22.

If x∗∈U is a global minimizer and f∗:=f(x∗), it is unique and every x∈U satisfies

f(x)−f∗≥μ2∥x−x∗∥22,∥∇f(x)∥22≥2μ(f(x)−f∗).

Here Hf is the matrix of second partial derivatives and ∇f 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 U⊆Rd, f∈C2(U,R), μ>0, and vTHf(z)v≥μ∥v∥22 for every z∈U and v∈Rd.

[F1]

Multivariable Taylor formula with a Lagrange remainder along a line segment: for a C2 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.

[F2]

A C2 function is convex exactly when its Hessian is positive semidefinite: a C2 function on an open convex set is convex if and only if its Hessian is positive semidefinite everywhere.

[F3]

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: Hf is the matrix of second partial derivatives; ∇f has the first partial derivatives as its standard coordinates.

[F4]

Directional derivatives and partial derivatives of a map U⊆Rm→Rn: the derivative of t↦f(a+tv) at 0 is the directional derivative Dvf(a); for a standard basis vector it is the corresponding partial derivative.

Proof

1.1F1F2F3given

For any x,y∈U, convexity keeps the segment from x to y in U. Applying [F1] with Taylor order one gives f(y)=f(x)+⟨∇f(x),y−x⟩+12(y−x)THf(x+θ(y−x))(y−x) for some θ∈(0,1). The Hessian hypothesis makes the last term at least μ∥y−x∥22/2, proving the first inequality. Since Hf⪰μI⪰0, [F2] also makes f convex on U.

1.2F3F4given

Let x∗ be an interior minimizer. For each standard basis vector ej, openness gives an interval around 0 on which x∗+tej∈U. The one-variable function t↦f(x∗+tej) has a minimum at 0, so its derivative there is 0; [F4] identifies this derivative with ∂jf(x∗). Thus every coordinate of ∇f(x∗) is zero.

2.1step 1.1algebra

Apply the first inequality from step 1.1 once to (x,y) and once to (y,x), and add. The function-value terms cancel, leaving ⟨∇f(y)−∇f(x),y−x⟩≥μ∥y−x∥22.

2.2step 1.1step 1.2

Apply step 1.1 with x=x∗ and y=x. Using ∇f(x∗)=0 gives f(x)−f∗≥μ∥x−x∗∥22/2. This also proves uniqueness, because equality of the minimum values forces x=x∗.

3.1F5step 1.1algebra∎

Apply step 1.1 with the ordered pair (x,x∗) and write d:=x−x∗. Rearrangement gives f(x)−f∗≤⟨∇f(x),d⟩−μ∥d∥22/2. By [F5] and completing the square, the right-hand side is at most ∥∇f(x)∥22/(2μ), since 0≤(∥∇f(x)∥2−μ∥d∥2)2. This is the gradient-gap inequality.

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