LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Populated sampling cells control the uniform PINN residual

Statement

Let $X \subset R^{D}$ carry a probability measure $μ$ , let $s, c > 0$ , and fix an integer $m \geq 1$ . Put $ℓ = (c \sqrt{m})^{- 1 / s}$ . Suppose a specified finite measurable partition $X_{1}, \dots, X_{K}$ of $X$ has each cell contained in a cube of side $ℓ$ and of mass at least $c ℓ^{s}$ . Let $Y_{1}, \dots, Y_{m}$ be independent points of law $μ$ , and put $δ = \sqrt{D} ℓ$ . With probability at least $\max {0, 1 - \sqrt{m} \exp (- \sqrt{m} / 8)}$ every cell contains at least $\sqrt{m} / 2$ samples, counted with multiplicity. On this event, simultaneously for all real functions $v, g$ with finite $α$ -Hölder seminorms on $X$ , $∥ v - g ∥_{\infty; X}^{2} \leq 6 \sqrt{m} (\frac{1}{m} \sum_{i = 1}^{m} ∣ v (Y_{i}) - g (Y_{i}) ∣^{2}) + 3 δ^{2 α} ([v]_{α; X}^{2} + [g]_{α; X}^{2}) .$ No independence between the functions and the samples is required.

Statement references

Independent random elements

Shin et al., Appendix B: the underlying cell construction

Facts & Assumptions

Given: The specified partition, iid samples, and functions in the Statement.

[F1]

A probability measure assigns total mass one; independence gives product probabilities for finite sample tuples (Probability measures and probability spaces, Independent random elements).

[F2]

The probability of a finite union of events is bounded by the sum of their probabilities (Finite and countable subadditivity of measures).

Proof

1.1givenF1algebra

Write $p_{j} = μ (X_{j})$ and $N_{j} = \sum_{i = 1}^{m} 1_{{Y_{i} \in X_{j}}}$ . Since $p_{j} \geq c ℓ^{s} = m^{- 1 / 2}$ and $\sum_{j} p_{j} = 1$ , we have $K \leq \sqrt{m}$ . Independence gives $E e^{- N_{j} / 2} = (1 - p_{j} + p_{j} e^{- 1 / 2})^{m} \leq \exp (- m p_{j} (1 - e^{- 1 / 2}))$ . For a nonnegative random variable $Z$ , $a P (Z > a) \leq E Z$ , by integrating $Z \geq a 1_{Z > a}$ . Apply this with $Z = e^{- N_{j} / 2}$ and $a = e^{- m p_{j} / 4}$ to get $P (N_{j} < m p_{j} / 2) \leq \exp (m p_{j} / 4 - m p_{j} (1 - e^{- 1 / 2})) \leq e^{- m p_{j} / 8}$ . Here $e^{1 / 2} \geq 1 + 1 / 2 + 1 / 8 = 13 / 8 > 8 / 5$ implies $e^{- 1 / 2} < 5 / 8$ , and $1 - x \leq e^{- x}$ follows by differentiating $e^{- x} - 1 + x$ .

2.1step 1.1F2algebra

The event $N_{j} < \sqrt{m} / 2$ is contained in $N_{j} < m p_{j} / 2$ . By step 1.1 and [F2], its union over cells has probability at most $K e^{- \sqrt{m} / 8} \leq \sqrt{m} e^{- \sqrt{m} / 8}$ . This proves the probability assertion, with the trivial lower bound zero when necessary.

3.1step 2.1algebra∎

On the complementary event fix $x \in X_{j}$ . Every sample $Y_{i} \in X_{j}$ is within $δ$ of $x$ . The three-term square inequality gives $∣ v (x) - g (x) ∣^{2} \leq 3 ∣ v (Y_{i}) - g (Y_{i}) ∣^{2} + 3 δ^{2 α} ([v]_{α; X}^{2} + [g]_{α; X}^{2})$ . Average over the $N_{j} \geq \sqrt{m} / 2$ samples in this cell. Its first term is at most $(3 / N_{j}) \sum_{i = 1}^{m} ∣ v (Y_{i}) - g (Y_{i}) ∣^{2} \leq 6 \sqrt{m} m^{- 1} \sum_{i} ∣ v (Y_{i}) - g (Y_{i}) ∣^{2}$ . Taking the supremum over $x$ proves the inequality. This is a deterministic inequality on the occupancy event, so it holds for sample-dependent trial functions as well.

Depends on

Used by

Regularized PINN minimizers have vanishing residuals under sample refinement Theorem

Dependency tree · two levels

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Sources

Shin, Darbon, and Karniadakis, On the Convergence of Physics-Informed Neural Networks for Linear Second-Order Elliptic and Parabolic Type PDEs (standard reference, not scraped)