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Composition theorems in communication complexity

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Us<strong>in</strong>g the idea of f<strong>in</strong>d<strong>in</strong>g a certificate of the high approximate degree by duality [She09,SZ09b],<br />

we have the follow<strong>in</strong>g fact analogous to Lemma 1.<br />

Lemma 2. For a function f : S → C and a subspace Φ of L C (S), if d(f, span(Φ)) = δ, then<br />

there exists a function h s.t.<br />

ĥ i = 0, ∀ψ i ∈ Φ (1)<br />

∑<br />

|h(g)| ≤ 2, (2)<br />

g∈G<br />

| ∑ g∈G<br />

f(g)h(g)| > δ (3)<br />

Us<strong>in</strong>g the lemma, we can prove the Theorem 11.<br />

Proof. (of Theorem 11) By the regularity property, we know that when (x, y) runs over X×Y ,<br />

g(x, y) runs over S exactly K times where K = MN/|G|. Consider B = 1 K [h(g(x, y))] x,y, h<br />

obta<strong>in</strong>ed by Lemma 2; we want to apply Proposition 1 and Theorem 4 by us<strong>in</strong>g this B. First,<br />

‖B‖ 1 = 1 ∑<br />

|h(g(x, y))| = ∑ |h(g)| ≤ 1. (4)<br />

K<br />

x,y<br />

g∈G<br />

Also,<br />

|〈A, B〉| = 1 K | ∑ x,y<br />

f(g(x, y))h(g(x, y))|<br />

= | ∑ g∈G<br />

f(g)h(g)| > δ<br />

Now we need to compute<br />

Note that<br />

‖B‖ = 1 ∥ [<br />

K<br />

∑<br />

∥<br />

ĥ i χ i (g(x, y))] x,y∥.<br />

χ i∈Hard<br />

[ψ i (g(x, y))] † x,y[ψ j (g(x, y))] x,y = [ ∑ x<br />

ψ i (g(x, y))ψ j (g(x, y ′ ))] y,y ′<br />

and<br />

[ψ i (g(x, y))] x,y [ψ j (g(x, y))] † x,y = [ ∑ y<br />

ψ i (g(x, y))ψ j (g(x ′ , y))] x,x ′.<br />

Thus the orthogonality condition implies that<br />

[ψ i (g(x, y))] † x,y[ψ j (g(x, y))] x,y = [ψ i (g(x, y))] x,y [ψ j (g(x, y))] † x,y = 0<br />

for all i ≠ j. Now as <strong>in</strong> [She09], we can use the max bound<br />

‖B‖ = 1 K<br />

max<br />

∥ ∥ ∥ĥi [ψ i (x, y)] x,y i:ψ i∈Ψ Hard<br />

≤ 1 K<br />

max |ĥi|<br />

i:ψ i∈Hard<br />

≤ 1 max<br />

K|G| ψ i∈Hard<br />

max<br />

ψ i∈Hard<br />

∥<br />

∥ [ψi (x, y)] x,y (<br />

max |ψ i (g)| · ∥∥ ∥) [ψi (x, y)] x,y .<br />

g<br />

where the last <strong>in</strong>equality is due to Eq. (5) and Eq. (4). F<strong>in</strong>ally note K|G| = MN to complete<br />

the proof.

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