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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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One-dimensional and multidimensional ReLU region lower bounds

Statement

For a standard adjacent-layer ReLU network with one input dimension and L≥1 hidden layers of widths n1,…,nL≥3, the maximum number of full-dimensional activation regions is exactly ∏ℓ=1L(nℓ+1). For integers n0,L≥1 and widths n1,…,nL≥3n0, there exists a standard adjacent-layer ReLU network with at least (∏ℓ=1L−1(⌊nℓn0⌋+1)n0)∑j=0n0(nLj) full-dimensional activation regions.

Facts & Assumptions

Given: The layer width conditions in the statement and the fold, region upper-bound, and hyperplane arrangement results below.

[F1]

A width-q ReLU block for q≥3 maps each of its q+1 intervals bijectively onto [0,1] (A width-n ReLU block makes n+1 full folds).

[F2]

For input dimension n0 and widths n1,…,nL, the region count is at most the rank-sensitive sum in Rank-sensitive upper bound for ReLU linear regions.

[F3]

n hyperplanes in general position in dimension d cut that space into ∑j=0d(nj) regions when n≥d (Hyperplane arrangement upper count and general-position count).

Proof

1.1givenF1induction

Compose the width-nℓ blocks from [F1]. Each existing interval is divided into nℓ+1+1 intervals, each mapped onto [0,1], so after all L layers the number is ∏ℓ=1L(nℓ+1).

2.1step 1.1F1F2constructalgebra

For n0=1, each index jℓ in [F2]'s upper sum is 0 or 1: the previous factors leave ni−ji≥2 for every ni≥3. Thus the sum factors as ∏ℓ=1L(1+nℓ), so the count in step 1.1 is exact. For the multidimensional construction, in each of the first L−1 layers allocate qℓ=⌊nℓ/n0⌋≥3 units to each coordinate and implement an independent width-qℓ fold on that coordinate; set any remaining units constantly inactive. Each prior box is thereby subdivided into (qℓ+1)n0 boxes, each mapped bijectively onto [0,1]n0.

3.1step 2.1F3algebrachooseconstruct∎

Take an arrangement of nL hyperplanes in general position in Rn0. It has S=∑j=0n0(nLj) chambers by [F3], and there are finitely many chambers. Choose one point yC in each chamber and set A=max⁡C∥yC∥∞. With c=(1/2,…,1/2) and λ=1/(4(1+A)), all points c+λyC lie in the interior of [0,1]n0. Translate and scale the arrangement by y↦c+λy; it remains in general position and each of its S chambers now meets the interior of the image cube. Use these transformed hyperplanes as the last layer's preactivation zero sets. Each preceding folded box maps affinely and bijectively onto [0,1]n0, so every last-layer chamber has a preimage in the interior of every such box. The earlier layers distinguish different boxes, and the last-layer patterns distinguish the S chambers within each box. Multiplying the number of boxes from step 2.1 by S proves the stated lower bound.

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