52 1. Higher order Fourier analysis

thus φ3 vanishes identically; we can integrate this twice to conclude that φ3

is a linear function of its input,

φ3(x) = a3x + b3

for some constants a3,b3 ∈ R. A similar argument (using the partial deriva-

tive operator ∂x −∂y to eliminate φ3, or by applying change of variables such

as (x, z) := (x, x + y)) shows that φ1(x) = a1x + b1 and φ2(x) = a2x + b2

for some additional constants a1,b1,a2,b2. Finally, by returning to (1.28)

and comparing coeﬃcients we obtain the additional compatibility condition

a3 = −a1 = −a2, which one then easily verifies to completely describe all

possible solutions to this equation in the case of smooth functions on R.

Returning now to the discrete world, we mimic the continuous operation

of a partial derivative by introducing difference operators

∂hφ(x) := φ(x + h) − φ(x)

for h ∈ G. If we take differences in (1.28) with respect to the x variable by

an arbitrary shift h ∈ G by replacing x by x + h and then subtracting, we

eliminate φ2 and obtain

(∂hφ1)(x) + (∂hφ3)(x + y) = 0;

if we then takes differences with respect to the y variable by a second arbi-

trary shift k ∈ G, one obtains

(∂k∂hφ3)(x + y) = 0

for all x, y, h, k ∈ G; in particular, ∂k∂hφ3 ≡ 0 for all k, h ∈ G. Such

functions are aﬃne-linear:

Exercise 1.3.8. Let φ: G → R/Z be a function. Show that ∂k∂hφ = 0 if

and only if one has φ(x) = a(x) + b for some b ∈ G and some homomor-

phism a: G → R/Z. Conclude that the solutions to (1.28) are given by the

form φi(x) = ai(x) + bi, where b1,b2,b3 ∈ G and a1,a2,a3 : G → R/Z are

homomorphisms with a1 = −a2 = −a3.

Having solved the functional equation (1.28), let us now look at an equa-

tion related to four term arithmetic progressions, namely

(1.29) φ1(x) + φ2(x + y) + φ3(x + 2y) + φ4(x + 3y) = c

for all x, y ∈ G, some constant c ∈ G, and some functions φ1,φ2,φ3,φ4 : G →

R/Z. We will try to isolate φ4 by using discrete derivatives as before to

eliminate the other functions. First, we differentiate in the y direction by

an arbitrary shift h ∈ G, leading to

(∂hφ2)(x + y) + (∂2hφ3)(x + 2y) + (∂3hφ4)(x + 3y) = 0.