ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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FISTA can increase the objective between consecutive iterates

Example

Take $f (x) = x^{2} / 4$ on $R$ , $g = 0$ , $L = 1$ , and $x_{0} = 1$ . The gradient has Lipschitz constant $1 / 2$ , so $L = 1$ is valid. Standard FISTA gives $∣ x_{5} ∣ > ∣ x_{4} ∣$ , hence $f (x_{5}) > f (x_{4})$ ; consecutive accelerated objective values need not decrease.

Facts & Assumptions

Given: The scalar objective, smoothness constant and initial point in the Example.

[F1]

Proximal-gradient, FISTA, and fixed-period restarted FISTA: $x_{k} = T_{L} (y_{k})$ , $t_{k + 1} = (1 + \sqrt{1 + 4 t_{k}^{2}}) / 2$ , and $y_{k + 1} = x_{k} + ((t_{k} - 1) / t_{k + 1}) (x_{k} - x_{k - 1})$ .

[F2]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : identify a scalar $s$ with the sole coordinate $s$ of its corresponding vector in $R^{1}$ .

[F3]

Proper closed convex extended-real functions on $R^{n}$ : properness is a nonempty effective domain, convexity is the Jensen inequality, and closedness is closed epigraph.

Verification

1.1F1F2F3givenalgebra

Under the scalar/one-coordinate identification [F2], $f (x) = x^{2} / 4$ is convex and continuously differentiable, since $f ((1 - α) x + α z) = (1 - α) f (x) + α f (z) - \frac{α (1 - α)}{4} (x - z)^{2} (0 \leq α \leq 1) .$ Also $f^{'} (x) = x / 2$ , so its gradient is $1 / 2$ -Lipschitz and $L = 1$ is valid. The zero function $g$ has nonempty domain, is convex, and has epigraph $R^{1} \times [0, + \infty)$ , which is closed; hence it is proper, closed and convex by [F3]. Its proximal map is the identity, so $T_{1} (y) = y - \nabla f (y) = y / 2$ . Starting from $t_{1} = 1, y_{1} = x_{0} = 1$ , recurrence [F1] gives $x_{1} = 1 / 2$ , $x_{2} = 1 / 4$ , and $t_{2} = \frac{1 + \sqrt{5}}{2}, t_{3} = \frac{1 + \sqrt{1 + 4 t_{2}^{2}}}{2}, t_{4} = \frac{1 + \sqrt{1 + 4 t_{3}^{2}}}{2}, t_{5} = \frac{1 + \sqrt{1 + 4 t_{4}^{2}}}{2} .$ The first interval follows from ${2.2360}^{2} < 5 < {2.2361}^{2}$ . For the next three, substituting the already established intervals gives radicands $1 + 4 t_{2}^{2}$ , $1 + 4 t_{3}^{2}$ , and $1 + 4 t_{4}^{2}$ in $(11.471696, 11.472991)$ , $(20.244014, 20.249279)$ , and $(31.228004, 31.272004)$ respectively. The endpoint squares ${3.3868}^{2} < 11.471696$ , ${3.3874}^{2} > 11.472991$ , ${4.498}^{2} < 20.244014$ , ${4.502}^{2} > 20.249279$ , ${5.586}^{2} < 31.228004$ , and ${5.594}^{2} > 31.272004$ then give $1.6180 < t_{2} < 1.6181$ , $2.1934 < t_{3} < 2.1937$ , $2.749 < t_{4} < 2.751$ , and $3.293 < t_{5} < 3.297$ .

2.1F1step 1.1algebra

The momentum coefficients satisfy $0.281 < (t_{2} - 1) / t_{3} < 0.283$ , $0.433 < (t_{3} - 1) / t_{4} < 0.435$ , and $0.529 < (t_{4} - 1) / t_{5} < 0.533$ : for example, the bounding cross-products are $0.281 (2.1937) = 0.6164297 < 0.6180$ , $0.283 (2.1934) = 0.6207322 > 0.6181$ , $0.433 (2.751) = 1.191183 < 1.1934$ , $0.435 (2.749) = 1.195815 > 1.1937$ , $0.529 (3.297) = 1.744113 < 1.749$ , and $0.533 (3.293) = 1.755169 > 1.751$ . Substitution into $x_{3} = \frac{1}{8} (1 - (t_{2} - 1) / t_{3})$ gives $0.0896 < x_{3} < 0.0899$ . Then $x_{4} = \frac{1}{2} [x_{3} + \frac{t_{3} - 1}{t_{4}} (x_{3} - 1 / 4)]$ lies between $\frac{1}{2} [(1 + 0.435) (0.0896) - 0.435 / 4] = 0.009913$ and $\frac{1}{2} [(1 + 0.433) (0.0899) - 0.433 / 4] = 0.01028835$ , so $0.0099 < x_{4} < 0.0103$ . Finally $x_{5} = \frac{1}{2} [x_{4} + \frac{t_{4} - 1}{t_{5}} (x_{4} - x_{3})]$ and endpoint monotonicity gives $\frac{1}{2} [(1 + 0.533) (0.0099) - 0.533 (0.0899)] = - 0.01637 < x_{5} < \frac{1}{2} [(1 + 0.529) (0.0103) - 0.529 (0.0896)] = - 0.01582485.$ Thus $- 0.0164 < x_{5} < - 0.0158$ .

3.1step 2.1givenalgebra∎

The intervals imply $∣ x_{5} ∣ > 0.0158 > 0.0103 > ∣ x_{4} ∣$ . Since $f (x) = x^{2} / 4$ is strictly increasing as a function of $∣ x ∣$ , $f (x_{5}) > f (x_{4})$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

Amir Beck and Marc Teboulle, A Fast Iterative Shrinkage-Thresholding Algorithm for Linear Inverse Problems, §4 (standard reference, not scraped)