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The projected OGD bound does not cover infeasible comparators

Statement refuted

Refuted claim: For integers $d, T \geq 1$ , let $K \subseteq R^{d}$ be nonempty, closed, and convex with diameter at most a finite $D > 0$ . Let each $f_{t} : R^{d} \to R$ be differentiable and convex on an open convex neighborhood of $K$ , and suppose the projected online-gradient algorithm starts at $x_{1} \in K$ , uses $η_{t} = D / (G \sqrt{t})$ for finite $G > 0$ , and has $∥ \nabla f_{t} (x_{t}) ∥_{2} \leq G$ on every round. Assume Countable Choice for the metric projections. The refuted claim is that

\sum_{t = 1}^{T} (f_{t} (x_{t}) - f_{t} (u)) \leq \frac{3}{2} D G \sqrt{T}

holds for every comparator $u \in R^{d}$ , including points outside $K$ .

Statement references

Projected online gradient descent has static regret at most three-halves of DG square-root T: its stated comparator belongs to $K$ .

Facts & Assumptions

Given: The projected OGD algorithm, $K = [0, 1]$ , diameter bound $D = 1$ , horizon $T = 2$ , losses $f_{t} : R \to R$ defined by $f_{t} (x) = - x$ for $t = 1, 2$ , initial point $x_{1} = 0$ , gradient bound $G = 1$ , and infeasible comparator $u = 2$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is assumed for the nondegenerate projection updates.

[F1]

Projected online gradient descent: $η_{t} = D / (G \sqrt{t})$ and $x_{t + 1} = P_{K} (x_{t} - η_{t} g_{t})$ when $D, G > 0$ .

[F2]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : in $R$ , $∥ x ∥_{2} = ∣ x ∣$ .

[F3]

Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length and A convex subset of $R^{m}$ contains every line segment between two of its points: $[0, 1]$ is a nonempty convex set.

[F4]

The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement: a set is closed when its complement is open.

[F5]

The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case: the gradient of a scalar-valued differentiable function on $R$ is its ordinary derivative.

[F6]

Metric projection onto a closed convex set satisfies the variational inequality: $p = P_{K} (z)$ exactly when $(z - p) (v - p) \leq 0$ for every $v \in K$ .

[F7]

Convex and strictly convex functions on Euclidean convex sets: a function is convex when it lies below the chord between any two points of its convex domain.

[F8]

Projection onto a nonempty closed convex set: under Countable Choice, the metric projection onto $K$ exists and is unique.

Counterexample

technique · direct

1.1F2F3F4givenalgebra

The interval $K = [0, 1]$ is nonempty. For $v, w \in K$ and $λ \in [0, 1]$ , $0 \leq (1 - λ) v + λ w \leq (1 - λ) + λ = 1$ , so [F3] makes it convex. Its complement is $(- \infty, 0) \cup (1, \infty)$ . If $z < 0$ , the ball of radius $- z / 2$ about $z$ lies in $(- \infty, 0)$ ; if $z > 1$ , the ball of radius $(z - 1) / 2$ about $z$ lies in $(1, \infty)$ . Thus [F4] makes $K$ closed. Its diameter is $D = 1$ because $∣ v - w ∣ \leq 1$ for $v, w \in [0, 1]$ and the endpoints attain distance $1$ .

1.2F2F5F7givenalgebra

For each $t$ and $h \neq 0$ , $(f_{t} (x + h) - f_{t} (x)) / h = (- (x + h) + x) / h = - 1$ , so the derivative is $g_{t} = - 1$ by [F5]; the affine identity $f_{t} ((1 - λ) x + λ y) = (1 - λ) f_{t} (x) + λ f_{t} (y)$ and [F7] show convexity. Also $∣ g_{t} ∣ = 1 = G$ .

2.1A1F1F6F8step 1.1step 1.2algebra

By [A1] and [F8], both metric projections exist uniquely. Since $D = G = 1$ , $η_{1} = 1$ and $η_{2} = 1 / \sqrt{2}$ . The first unprojected point is $x_{1} - η_{1} g_{1} = 1$ ; for every $v \in [0, 1]$ , $(1 - 1) (v - 1) = 0$ , so [F6] gives $x_{2} = P_{K} (1) = 1$ . The next unprojected point is $x_{2} - η_{2} g_{2} = 1 + 1 / \sqrt{2} > 1$ ; for every $v \in [0, 1]$ , $(1 + 1 / \sqrt{2} - 1) (v - 1) \leq 0$ , so [F6] gives $x_{3} = P_{K} (1 + 1 / \sqrt{2}) = 1$ .

3.1givenstep 1.1step 2.1algebra

The learner's two played losses sum to $f_{1} (x_{1}) + f_{2} (x_{2}) = 0 - 1 = - 1$ , whereas the infeasible comparator has $\sum_{t = 1}^{2} f_{t} (2) = - 4$ . Its regret is therefore $3$ .

4.1givenstep 3.1algebra∎

The proposed right-hand side is $\frac{3}{2} \cdot 1 \cdot 1 \cdot \sqrt{2} = \frac{3}{2} \sqrt{2} < 3$ , since $\sqrt{2} < 2$ . Thus the regret inequality fails for $u = 2 \notin K$ .

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Sources

Elad Hazan, Introduction to Online Convex Optimization, second edition (standard reference, not scraped)
Tengyu Ma, Lecture Notes for Machine Learning Theory, Chapter 11 (standard reference, not scraped)