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Networks approximate smooth convolutions of the activation

Statement

Assume Countable Choice. Let $σ \in M_{p t}$ , let $φ \in C_{c}^{\infty} (R; C)$ , and let $K \subseteq R$ be compact. Define the convolution

$g (x) = \int_{R} σ (x - t) φ (t) d t .$

Then $Re g$ and $Im g$ are each uniform limits on $K$ of finite sums $x \mapsto c + \sum_{j = 1}^{m} a_{j} σ (x + b_{j})$ with real coefficients, hence of single-hidden-layer networks with activation $σ$ . In particular, if $φ$ is real-valued, then $g$ itself is such a uniform limit.

Statement references

Leshno et al., proof of Theorem 1, convolution and Riemann-sum step

Moreau, Chapter 2 §2.3, convolution reduction

Facts & Assumptions

Given: Countable Choice, $σ \in M_{p t}$ , a test function $φ$ , and a compact interval containing $K$ .

[F1]

The activation is pointwise bounded on compact intervals and its discontinuity set lies in a closed null set (Locally essentially bounded activations with null discontinuity closure).

[F2]

A bounded function on a compact interval whose discontinuity set is null is Riemann integrable under Countable Choice, and its Riemann and Lebesgue integrals agree (A bounded function on a closed bounded interval, or on a closed nondegenerate rectangle, is Riemann integrable exactly when its discontinuity set has Lebesgue measure zero, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral).

[F3]

A compact null set has open covers of arbitrarily small measure, and Lebesgue measure is translation invariant (Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of $R^{n}$ is the infimum of the measures of the open sets containing it, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[F4]

A finite sum $c + \sum_{j} a_{j} σ (x + b_{j})$ is a network from [F1] with input weight $1$ in each unit (Single-hidden-layer network realization).

[F5]

Every continuous real function on a compact subset of $R$ is uniformly continuous (Heine-Cantor in $R$ : a continuous real function on a compact subset of $R$ is uniformly continuous).

[F6]

Compact subsets of $R$ are bounded, so a compact interval can contain $K - [- R, R]$ (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[F7]

Reciprocals $1 / n$ tend to zero, and for every positive mesh tolerance $η$ there is an integer $m$ with $2 R / m < η$ , so $[- R, R]$ has uniform partitions of arbitrarily small mesh (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[A1]

Countable Choice means every sequence of nonempty sets has a function choosing one element from each (The Axiom of Countable Choice ( ${A C}_{ω}$ )); it is used by [F2] for Lebesgue integration.

[F8]

Under Countable Choice, Lebesgue measure is complete: every subset of a measurable null set is measurable and null (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

[F9]

The invertible linear map $t \mapsto - t$ has absolute determinant one, so it preserves Lebesgue measurability and measure under Countable Choice (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

Proof

1.1givenF1F2F3F6F8F9A1

Write $φ = φ_{R} + i φ_{I}$ with real-valued smooth compactly supported components, and put $g_{R} = σ * φ_{R}$ and $g_{I} = σ * φ_{I}$ , so $g = g_{R} + i g_{I}$ . Choose $R > 0$ with $supp φ \subseteq [- R, R]$ and a compact interval $J$ containing $K - [- R, R]$ by [F6]. Put $C = \overline{D_{σ}} \cap J$ , where $D_{σ}$ is the discontinuity set of $σ$ . Then $C$ is compact and null. Reflection [F9] makes $- C$ measurable and null, and translation invariance [F3] makes $x - C = x + (- C)$ measurable and null for every $x \in K$ . The discontinuities on $[- R, R]$ of each bounded real function $t \mapsto σ (x - t) φ_{α} (t)$ , $α \in {R, I}$ , lie in $(x - D_{σ}) \cap [- R, R] \subseteq x - C$ and hence are measurable and null by completeness [F8]. By [F2], its Riemann integral equals the Lebesgue integral defining $g_{α} (x)$ .

2.1step 1.1F1F3F5A1construct

Fix $δ > 0$ . By [F3] and compactness, cover $C$ by a finite union $U$ of open intervals with total length less than $δ$ . On the compact set $J ∖ U$ , $σ$ is continuous and therefore uniformly continuous by [F5]; it is also bounded on $J$ by [F1]. Both $φ_{R}$ and $φ_{I}$ are uniformly continuous by [F5]. Hence the two products $σ (x - t) φ_{α} (t)$ , $α \in {R, I}$ , have a common modulus of continuity on every grid cell for which $x - t \notin U$ , uniformly over $x \in K$ .

3.1step 1.1step 2.1F3F7A1choosealgebra

For each integer $n \geq 1$ , use [F3] and compactness to cover $C$ by a finite union $U_{n}$ of open intervals with $∣ U_{n} ∣ < 1 / n$ ; Countable Choice selects this sequence of covers. Let $N_{n}$ be the number of component intervals. For each $α \in {R, I}$ , use [F7] to choose a uniform partition of $[- R, R]$ with mesh $h_{n}$ , midpoint tags $t_{i}$ , and define $S_{n, α} (x) = \sum_{i} h_{i} φ_{α} (t_{i}) σ (x - t_{i})$ . For fixed $x$ , cells whose translate meets $U_{n}$ have total length at most $∣ U_{n} ∣ + 2 N_{n} h_{n}$ . Their total integral and sum error is at most twice this length times $\sup_{J} ∣ σ ∣ \max_{α} \sup ∣ φ_{α} ∣$ . On all other cells the common modulus from step 2.1 bounds the total error by $2 R$ times that modulus at $h_{n}$ . Choose the mesh sufficiently small that the terms involving $N_{n} h_{n}$ and the modulus are below $1 / n$ . Thus for both components the uniform error is bounded by a constant times $1 / n$ , and tends to zero. Each $S_{n, α}$ is a finite sum of translates.

4.1step 3.1F4∎

Each term $h_{i} φ_{α} (t_{i}) σ (x - t_{i})$ is a hidden unit with bias $- t_{i}$ and real output weight $h_{i} φ_{α} (t_{i})$ . Hence every $S_{n, α}$ is a network by [F4], proving the claim for both components; for real $φ$ , $g = g_{R}$ .

Depends on

Used by

Rough nonpolynomial activations are dense on compact sets Lemma

Dependency tree · two levels

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Sources

Leshno, Lin, Pinkus, and Schocken, Multilayer Feedforward Networks With a Non-Polynomial Activation Function Can Approximate Any Function, proof of Theorem 1, convolution and Riemann-sum step (standard reference, not scraped)
Yves Moreau, Dynamical Neural Networks and Composition Networks for System Modeling, Chapter 2 §2.3 (standard reference, not scraped)