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.
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