REMARKS ON THE CAUCHY FUNCTIONAL EQUATION AND
VARIATIONS OF IT

arXiv:1002.3721v7 [math.CA] 8 Feb 2017

DANIEL REEM

Abstract. This paper examines various aspects related to the Cauchy functional
equation f (x + y) = f (x) + f (y), a fundamental equation in the theory of functional equations. In particular, it considers its solvability and its stability relative
to subsets of multi-dimensional Euclidean spaces and tori. Several new types of
regularity conditions are introduced, such as a one in which a complex exponent
of the unknown function is locally measurable. An initial value approach to analyzing this equation is considered too and it yields a few by-products, such as
the existence of a non-constant real function having an uncountable set of periods
which are linearly independent over the rationals. The analysis is extended to related equations such as the Jensen equation, the multiplicative Cauchy equation,
and the Pexider equation. The paper also includes a rather comprehensive survey
of the history of the Cauchy equation.

1. Introduction
1.1. History. A well-known and fundamental equation in the theory of functional
equations is the Cauchy functional equation
(1)

f (x + y) = f (x) + f (y).

This equation (as well as its three sisters f (x + y) = f (x)f (y), f (xy) = f (x) + f (y),
f (xy) = f (x)f (y)), was introduced by Cauchy in his book [19, p. 103] from 1821.
Cauchy rigorously analyzed (1) under the assumptions that the unknown function
f is a continuous function from R to R and the variables x and y can be arbitrary
real numbers. (Gauss [42, p. 212] also considered a certain version of (1) in
his book from 1809, but the consideration was neither rigorous nor very explicit;
see also [1, pp. 47, 106–109]. Going back even further, to 1794, one can find in
the book of Legendre [71], in the section devoted to the consideration of ratios of
areas of rectangles, both a use and an analysis of (1), but they are also neither
rigorous nor explicit; see also [3, pp. 369–370]) Cauchy’s equation (1) has many
applications inside the theory of functional equations and in other mathematical and
scientific fields including geometry, real and complex analysis, probability, functional
analysis, dynamical systems, partial differential equations, classical and statistical
mechanics, and economics. It is therefore nothing but natural that it has attracted
Date: February 8, 2017.
2010 Mathematics Subject Classification. 39B52, 43A22, 39B22, 39B05, 26B99, 22B99.
Key words and phrases. Complex exponent, initial value, locally measurable, periodic, regularity condition, restricted domain, stability, the Cauchy functional equation.
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DANIEL REEM

the attention of many authors for a long period of time. Formulating this fact in
more romantic terms [63, p. 2], Kannappan wrote: “Researchers fell in love with
these equations [Cauchy’s equation (1) and its sisters mentioned above], and the
romance will continue and will result in many more interesting results”.
A common path of investigation of (1) is to impose various types of “regularity”
conditions on the unknown function. It turns out that in the specific case where
f : R → R, each of these conditions implies the existence of some c ∈ R such that
f (x) = cx for all x ∈ R, and this fact has been proved in various ways. For instance,
Cauchy [19, pp. 103–106] assumed that f is continuous, Darboux showed that f
can be assumed to be either monotone [28] or bounded on an interval [29], Fréchet
[39], Blumberg [15], Banach [13], Sierpiński [96, 97], Kac [61], Alexiewicz-Orlicz
[6], and Figiel [35] assumed that f is Lebesgue measurable, Kormes [65] assumed
that f is bounded on a measurable set of positive measure, Ostrowski [81] and
Kestelman [64] assumed that f is bounded from one side on a measurable set of
positive measure, and Mehdi [74] assumed that f is bounded above on a second
category Baire set. On the other hand, in 1905 Hamel [47] investigated (1) without
any further condition of f . By using the Hamel basis, which was introduced for this
purpose, he showed that there are nonlinear solutions to (1) and he found all of
them (see also Remark 3.4 below).
The functional equation (1) has been generalized or modified in many other directions. A typical path in this spirit is to take the domain and range of f to be groups
of a certain kind, e.g., (locally compact) Polish groups, and to prove that if f satisfies a certain kind of measurability assumption (say Baire, Haar, or Christensen),
and maybe additional assumptions, then it must be continuous (and, depending
on the setting, possibly linear [3, pp. 36–38]). See, among others, Banach [14,
p. 23], Pettis [85], Hewitt-Ross [49, p. 346], Monroe [77], Figiel [35], Itzkowitz
[55], Baker [12], Brzdek [18], Christensen [21], Gajda [40], Grosse-Erdmann [45],
Járai-Székelyhidi [57], Neeb [80], and Rosendal [91]. In some of these generalizations the form of the functional equation is more general than (1). (Remark: the
terminology used in some of the references mentioned above and below is sometimes
different from that used here, and, in particular, instead of speaking on the Cauchy
functional equation, from time to time one speaks on homomorphism or additive
functions; below we will sometimes also call solutions of (1) “additive functions”.)
A different path is to impose assumptions different from measurability, e.g., algebraic ones. See the book of Dales [25] for an extensive discussion on this issue. Yet
another path of generalization/modification is to change the domain of definition of
f so that it will no longer have a necessarily nice algebraic structure but rather it
will merely be a certain subset of the original domain, e.g., an orthant, a convex set,
a complement of a measure zero set, and so forth. A variation of this is to change
the domain of validity of the equation, for instance, to assume that f satisfies (1)
only for pairs (x, y) belonging to a subset of R2n , e.g., a manifold (and f may be
defined on the whole space or on a subset of it). In both cases one may conclude
the existence of nonlinear solutions to (1) (even under “strong” regularity conditions) or that (in the case where f is assumed to be defined on the whole space) f

REMARKS ON THE CAUCHY FUNCTIONAL EQUATION

3

must satisfy (1) for all possible pairs (x, y). For a very partial list of related works
see Pisot-Schoenberg [87], Aczel-Erdös [4], Jurkat [60], de Bruijn [30], Zdun [105],
Kuczma [66], Dhombres-Ger [32], Dhombres [31], Forti [37], Matkowski [73], Sablik [93], Szabo [99], Choczewski et al. [20], Alsina and Garc Roig [7], Ger-Sikorska
[44], Skof [98], Paneah [82, 83], Shalit [94, 95], Laohakosol-Pimsert [70] and the
references therein.
More information related to (1), including many additional references, can be
found, for instance, in the books of Aczél [1], Aczél-Dhombres [3] (both books also
mention applications of (1) in various scientific areas), Czerwik [24] (the book contains also a treatment of the set valued version of Cauchy’s equation), Einchorn [33]
(for applications in economics), Járai [56] (various regularity conditions), Kannappan [63], Kuczma [68], and the surveys of Kannappan [62] and Wilansky [103].
See also Section 10 below regarding a short survey and several references related to
stability in the context of the Cauchy equation.

1.2. Contribution and paper layout. This paper examines various aspects related to (1), including its solvability and its stability under new kind of assumptions,
and presents new methods to analyze it. The first result that is proved (Section 2
below) is the following: given n ∈ N, if f : Rn → R satisfies (1) for every x, y ∈ Rn
and if a complex exponent of f is locally measurable, that is, eif is Lebesgue measurable on a certain open ball (and hence on a certain non-degenerate and compact
hypercube contained in that ball), then f must be linear, namely, there exists c ∈ Rn
such that f (x) = c · x for all x ∈ Rn (here, of course, c · x denotes the
P inner product
between the vectors c = (ck )nk=1 and x = (xk )nk=1, that is, c · x := nk=1 ck xk ). This
regularity condition is strictly weaker than measurability (Remark 2.2 below) and
although it seems to be new, traces of it can be detected in the paper of Kac [61] (Remark 2.3 below). In Section 3 the same result is proved using a different approach
by considering (1) as an initial value problem and utilizing periodicity properties
which arise naturally. Two by-products of this approach are the observation that
there exist nonlinear solutions of (1) having an uncountable dense set of periods
which are (the periods) linearly independent over the rationals (Remark 3.4 below)
and the observation that any solution to (1) defined on a multidimensional topological (flat) torus must vanish identically (Section 4 below). The previous analysis
is extended to other settings: to Jensen’s equation (Section 5 below), to Cauchy’s
equation on restricted domains satisfying a certain abstract algebraic condition and
related conditions (Section 6 below), to multiplicative Cauchy’s equation (Section
7 below), to an alternative version of Cauchy’s equation (Section 8 below), to Pexider’s equation (Section 9 below), and to the stability of Cauchy’s equation (Section
10 below). We conclude the paper in Section 11 with a few remarks regarding possible additional extensions, such as to a setting in which the regularity condition is
abstract (Remark 11.3 below). The paper also has an appendix (Section 12 below)
which contains proofs of several assertions.

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DANIEL REEM

2. The case where eif is locally measurable, Rn
This section contains the main theorem of the paper (Theorem 2.1 below), as
well as several remarks related to it (Remarks 2.2–2.3 below).
Theorem 2.1. If f : Rn → R satisfies (1) for every x, y ∈ Rn and if eif is locally
Lebesgue measurable, then there exists some c ∈ Rn such that f (x) = c · x for all
x ∈ Rn .
Proof. By applying the function t 7→ eit , t ∈ R on both sides of (1) we have
eif (x+y) = eif (x) eif (y) for all x, y ∈ Rn . Let I be any hypercube on which eif is
known to be measurable. By taking an arbitrary y ∈ Rn and x ∈ I and using the
fact that the function x 7→ eif (x) eif (y) , x ∈ I is measurable on I, we see that eif
is measurable on any translated copy of I. Because Rn can be represented as a
countable union of translated copies of I, we conclude that eif is measurable on Rn .
Now, from a well-known theorem about the continuity of measurable homomorphisms acting between certain groups (in particular, between a locally compact
group and the circle group: see [49, p. 346] for an even more general statement),
it follows that the measurable homomorphism eif is continuous. Since its image is
obviously contained in the unit circle of the complex plane C, it follows that eif is
a character (namely, a continuous homomorphism from Rn to the unit circle in C).
By a well known representation result regarding the character group of Rn (see, e.g.,
[49, pp. 366–368]) it follows that there exists c ∈ Rn such that
(2)

eif (x) = eic·x ,

∀x ∈ Rn .

Alternatively, one can deduce (2) without referring to the theory presented in [49]
by using an approach based on Fourier series [90].
Anyway, because of (2) there exists a function k from Rn to the set of integers such
that f (x) = c·x+2πk(x) for all x ∈ Rn . Since f satisfies (1) for every x, y ∈ Rn and
since (1) is linear, we conclude that k also satisfies (1) for all x, y ∈ Rn . It remains
to show that k vanishes identically. Assume that this is not true, namely k(x) 6= 0
for some fixed x. Then by (1) (with k instead of f ) we have k(x/m) = k(x)/m for
any positive integer m. In particular this is true for m > 2|k(x)|, a contradiction
since 0 < |k(x)/m| < 0.5 < 1 but k(x/m) is an integer. Thus k ≡ 0 and f (x) = c · x
for each x ∈ Rn . Finally, an immediate verification shows that f , having this linear
form, still satisfies (1) for every x, y ∈ Rn .

Remark 2.2. The regularity condition imposed on f is strictly weaker than merely
measurability. Indeed, if f is Lebesgue measurable, then so is eif as a composition
of f with the complex exponent function t 7→ eit , t ∈ R which is continuous. On the
other hand, even if eif is a continuous (multiplicative) homomorphism, then f itself
need not be measurable as the following simple example shows: for each integer m
let Am be a non-measurable set of [m, m + 1), and let f : R → R be defined by
f (x) := 2πm if x ∈ Am , and f (x) := −2π(m + 1) if x ∈ [m, m + 1)\Am .
Another example, perhaps more interesting (see also Remark 10.2 below), is
f (x) := 2π[g(x)], x ∈ R where [t] is the greatest integer not exceeding the real

REMARKS ON THE CAUCHY FUNCTIONAL EQUATION

5

number t and g is an arbitrary nonlinear solution of (1). Indeed, eif ≡ 1 but there
can be no set of positive measure on which f is bounded, since if A is such a set,
then also g is bounded on A and by the theorem of Kormes [65, Theorem I] it
follows that g is linear, a contradiction. Since any measurable function h : R → R is
bounded on some set of positive measure (simply on Am := {x : h(x) ∈ [m, m + 1)}
for some
S integer m because a measurable function is finite almost everywhere and
hence m∈Z Am is, up to a set of measure 0, the whole space) it follows that f is
not measurable.

Remark 2.3. The regularity condition that eif is measurable has traces in the 1937
paper of Kac [61]. Indeed, while he assumed that f : R → R was measurable, in
the proof he implicitly used only the measurability of eif . Modifications of Kac’s
proof can be found in other places, e.g., in [3, pp. 54–56], [24, pp. 39–40] (in these
places, however, f is from R to C and it satisfies the multiplicative Cauchy equation
(8) below rather than the additive one (1) above). Anyway, in all of these places
(including in Kac’s paper) the proofs seem to be restricted to the one-dimensional
case and it is explicitly assumed that f is measurable and not that eif is measurable.
3. The initial value approach

In this section a different proof of Theorem 2.1 is presented. The approach is
to consider (1) as an initial value problem. The inspiration comes from differential
equations. The derivation is based on two lemmas. The first (Lemma 3.1 below)
is known. For the sake of completeness and because it is not easy to find in the
literature a proof of this lemma for the case n > 1, a proof is provided in the
appendix (Section 12 below). The second lemma (Lemma 3.2 below) contains a
key technical assertion.
In Lemma 3.1 below (and elsewhere) we use the following terminology and observations.
P A (closed) hyper-parallelepiped spanned by some basis is the set I :=
{u0 + nk=1 tk uk : (tk )nk=1 ∈ [0, 1]n } where U := {u1, . . . , un } is a given basis of
Rn and u0 P
∈ Rn is a given point. A right-semi-open hyper-parallelepiped is the set
I := {u0 + nk=1 tk uk : (tk )nk=1 ∈ [0, 1)n }, namely a hyper-parallelepiped from which
the right facets have been removed. Of course, a (closed or right-semi-open) hypercube is a (closed or right-semi-open, respectively) hyper-parallelepiped spanned by
an orthogonal basis, where the lengths of all the basis elements are equal. We say
that a function h : Rn → C is U-periodic (or periodic with respect to the basis U)
if it satisfies h(x + uk ) = h(x) for each x ∈ Rn and each k ∈ {1, . . . , n}.
Lemma 3.1. Let I ⊂ Rn be a given (closed or semi-open) hyper-parallelepiped
spanned by some basis U := {u1 , . . . , un } of Rn . Suppose that h : Rn → C is Uperiodic. If h is Lebesgue integrable on I, then for each y ∈ Rn the translation
n
n
hy : R
R → R defined
R by hy (x) := h(x + y) for all x ∈ R is Lebesgue integrable on I
and I hy (x)dx = I h(x)dx.
Lemma 3.2. Let g : Rn → R. Suppose that I ⊂ Rn is a hypercube spanned by some
basis U := {u1 , . . . , un }. Assume that eig is Lebesgue measurable on I and that g is
U-periodic. If g satisfies (1) for every x, y ∈ Rn , then g vanishes identically.

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DANIEL REEM

Proof. Since g satisfies (1) for every x, y ∈ Rn and since eig is measurable on some
hypercube I, we can use the same reasoning given in Section 2 above (the beginning of the first proof of Theorem 2.1) to conclude that eig is measurable on Rn
and hence on any hypercube contained in Rn . Given any rational number α, let
hα : Rn → C be defined by hα (x) := eig(αx) for every x ∈ Rn . The change of
variables formula for scaling [58, pp. 170–171] and the previous discussion on the
measurability of eig imply that hα is measurable on Rn , and, being bounded, it
is Lebesgue
integrable over any hypercube. In particular hα is measurable over
P
I0 := { nk=1 tk uk : (tk )nk=1 ∈ [0, 1)n }. In addition, we have
hα (x) = eiαg(x) ,

(3)

∀x ∈ Rn , ∀ α rational

because g(αx) = αg(x) for each rational α and each x ∈ Rn (this identity is, of
course, a well known fact which easily follows from the assumption that g satisfies
(1) for every x, y ∈ Rn : see [68, Theorem 5.2.1,
pp. 128–129]).
R
Let α > 0 be some rational satisfying I0 hα (x)dx 6= 0. It will be shown in a
moment that such an α does exist. Since g is U-periodic, the identity (3) implies that
hα is also U-periodic. This fact, the fact that I0 is also spanned by U,
R and Lemma
3.1 (with hα and I0 instead of h and I, respectively), imply that I0 hα (x)dx =
R
R
hα (x + y)dx for every y ∈ Rn . But (3) and (1) imply that I0 hα (x + y)dx =
I
0
R iαg(x+y)
R
R
e
dx = I0 eiαg(x) eiαg(y) dx = eiαg(y) I0 hα (x)dx for all y ∈ Rn . We conclude
I0
R
R
that I0 hα (x)dx = eiαg(y) I0 hα (x)dx for all y ∈ Rn . As a result of this equality
R
and the assumption that I0 hα (x)dx 6= 0 it follows that 1 = eiαg(y) for all y ∈ Rn .
Therefore αg(y) ∈ {2πk : k ∈ Z} for any y ∈ Rn . Assume for a contradiction that
αg(y0) 6= 0 for some y0 . We can write 2πk0 = αg(y0) for some 0 6= k0 ∈ Z. Since
k0 6= 0 and because g(qy0) = qg(y0) for each rational q, we have αg(y0/(7k0 )) =
α(1/(7k0))g(y0) = 2πk0 /(7k0 ) = 2π/7. Thus αg(y) ∈
/ {2πk : k ∈ Z} for y :=
y0 /(7k0 ), a contradiction. Hence
g
≡
0.
R
It remains to prove that I0 eiαg(x) dx 6= 0 for some positive rational α. If this is
R
not true, then in particular I0 eiαg(x) dx = 0 for all rationals α ∈ (0, 1]. As a result
Z
Z
Z
Z
ig(w)
n
ig(αx)
iαg(x)
eig(w) dw
e
dw/α =
e
dx =
e
dx =
0=
I0

αI0

αI0

I0

for each rational α ∈ (0, 1] by the change of variables formula for scaling. Therefore,
given y ∈ Rn and a rational number α ∈ (0, 1], we have
Z
Z
Z
ig(z+y)
ig(y)
ig(x)
eig(z) dz = 0.
e
dz = e
e
dx =
αI0

y+αI0

R

ig(x)

αI0

It follows in particular that S e
dx = 0 whenever S is a dyadic copy of I0 ,
namely, S is obtained from I0 by scaling it byP
2−m for some nonnegative integer m
and translating it by a vector of the form y = nk=1 (mk /2m )uk , mk ∈ Z. The above
is true even if we remove from, or add to, these dyadic copies of I0 several parts of
their (measure 0) boundaries. It is well-known that any nonempty open set S of Rn
is a countable union of pairwise disjoint dyadic copies of I0 , namely S = ∪∞
m=1 Sm

REMARKS ON THE CAUCHY FUNCTIONAL EQUATION

7

where Sm is a dyadic copy of I0 for each m ∈ N, and Sm ∩Sp = ∅ whenever m 6= p (see
[92, p. 50]; the proof given there is for I0 = [0, 1)n , namely U is the canonical basis,
but the same proof holds for an arbitrary hypercube, or, more generally, an arbitrary
hyper-parallelepiped, simply by representing the vectors in Rn using the given basis
U). This fact, combined with the countable
implies that
R integral,
R ig(w)additivity
P
P∞of the
ig(w)
= ∞
for each nonempty open set S we have S e
dw = m=1 Sm e
m=1 0 = 0.
Now, given a measurable subset S 6= ∅ of I0 , a simple consequence of the (upper)
regularity of the Lebesgue measure µ (see [58, Property M9, p. 50]) implies that
n
there exists a decreasing sequence (Um )∞
m=1 of open subsets of R satisfying S ⊆ Um
n
and µ(Um \S) < 1/m for each m ∈ N. For each A ⊆ R let 1A be the characteristic
function of A, namely 1A (x) := 1 if x ∈ A and 1A (x) := 0 otherwise. It must
/ (∩∞
be that limm→∞ 1Um (x) = 1S (x) for each x ∈
m=1 Um )\S. Indeed, if x ∈ S,
/ ∩∞
then both sides are equal to 1 since 1Um (x) = 1S (x) = 1, and if x ∈
m=1 Um ,
then x ∈
/ Um0 for some m0 ∈ N, so x ∈
/ Um for all m > m0 because (Um )∞
m=1 is
decreasing. This fact and the inclusion S ⊆ Um for all m ∈ N imply that both sides of
limm→∞ 1Um (x) = 1S (x) equal 0. But (∩∞
m=1 Um )\S is a measurable subset satisfying
∞
µ((∩m=1 Um )\S) ≤ µ(Um \S) < 1/m for each m ∈ N. Thus µ((∩∞
m=1 Um )\S) = 0. It
follows that limm→∞ 1Um (x) = 1S (x) and hence limm→∞ (1Um (x)eig(x) ) = 1S (x)eig(x)
for almost all x ∈ Rn . Since (Um )∞
m=1 is a decreasing sequence of subsets which
contain S we have 1S (x) = |1S (x)eig(x) | ≤ |1Um (x)eig(x) | = 1Um (x) ≤ 1U1 (x) for all
x ∈ Rn and all m ∈ N. Since 1U1 is Lebesgue integrable on Rn (because µ(U1 ) ≤
µ(U1 \S) + µ(S) < 1 + µ(I0 ) < ∞), we can conclude the following equality from the
dominated convergence theorem and the fact established earlier that the integral of
eig vanishes on every nonempty open subset of Rn :
Z
Z
Z
Z
ig(w)
ig(w)
ig(w)
eig(w) dw = 0.
dw = lim
1Um e
e
dw =
1S (w)e
dw = lim
S

Rn

m→∞

Rn

m→∞

Um

Because S was an arbitrary measurable subset of I0 we can conclude that eig(w) = 0
for almost every w ∈ I0 (see, e.g., [69, p. 105]). On the other hand, we obviously
have |eig(w) | = 1 for every w ∈ I0 since the rangeRof g is R. This contradiction shows
the existence of a positive rational α such that I0 eiαg(x) dx 6= 0, as claimed.


In Lemma 3.2 above instead of a hypercube we can take any hyper-parallelepiped
spanned by some basis (the proof is the same), but we have found no real value in
working with a general basis since any hyper-parallelepiped contains a hypercube
and we are interested only in local measurability of eig . On the other hand, in the
abstract regularity condition mentioned in Remark 11.3 below, we preferred to use
a more general statement using a general basis.
Proof of Theorem 2.1. Let I be a hypercube on which eif is measurable and
suppose that U := {u1 , . . . , un } is the (orthogonal) basis of Rn which spans I.
Consider the n × n matrix whose k-th raw is uk , k ∈ {1, . . . , n}. This matrix is
invertible since {u1 , . . . , un } is a basis. Hence there exists a unique c = (ck )nk=1 ∈ Rn

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DANIEL REEM

which satisfies the linear system
(4)

uk · c = f (uk ),

k ∈ {1, . . . , n}.

One solution to (1) is g1 := f and another one is the linear function g2 (x) := c · x,
x ∈ Rn . Moreover, by the choice of c we have g1 (uk ) = g2 (uk ) for all k ∈ {1, . . . , n}.
This fact and the linearity of (1) imply that g := g1 − g2 satisfies the following
initial value problem
(5a)
(5b)

g(x + y) = g(x) + g(y) ∀x, y ∈ Rn ,
g(uk ) = 0,

k = 1, . . . , n.

We claim that (5) has g ≡ 0 as its unique solution in the class of functions g :
Rn → R having the property that eig is measurable on I. Once we prove this
claim we can conclude that g1 = g2 because g := g1 − g2 solves (5) and it satisfies
eig(x) = eif (x) e−ic·x for all x ∈ Rn , namely it is a multiplication of two functions which
are measurable on I and hence it is also measurable on I. To see that the above
mentioned claim holds, we observe that the zero function obviously solves (5) and it
belongs to the considered class of functions. On the other hand, given any solution g
to (5) in the considered class of functions, we have g(x+uk ) = g(x)+g(uk ) = g(x) for
all x ∈ Rn and all k ∈ {1, . . . , n}, that is, g is U-periodic. Since eig is measurable on
I we can use Lemma 3.2 to deduce that indeed g ≡ 0, i.e., we also have uniqueness,
as required.

Remark 3.3. Under the stronger assumption that f : Rn → R satisfies (1) for
every x, y ∈ Rn and is locally integrable, the approach used in this section allows
one to give a simple proof that f must be linear. Indeed, since f is locally integrable,
there exists a hypercube I on which f is integrable. Let U := {u1, . . . , un } be the
basis which spans I. The linearity of (1) implies, as in the second proof of Theorem
2.1 given above, that it is enough to show that g ≡ 0 is the unique solution to (5)
in the class of functions which are integrable on I. By substituting y := uk in (1)
we see that g(x + uk ) = g(x) + 0 for all x ∈ Rn and all k ∈ {1, . . . , n}, i.e., g is
U-periodic. Since we assume that g is integrable on I, its periodicity implies (by
Lemma 3.1) that it is integrable on any translated copy of I. Now, by fixing y ∈ Rn
in (1) and integrating over I with respect to x, we see that
Z
Z
Z
g(x + y)dx = g(x)dx + g(y) dx.
I
I
I
R
R
But I g(x + y)dx = I g(x)dx according to Lemma 3.1. Hence g(y)µ(I) = 0 for
each y ∈ R, that is, g ≡ 0.
Remark 3.4. The nonlinear solutions of (1) are examples of exotic functions, not
only because they are not measurable (or, as Theorem 2.1 shows, even a complex
exponent of them cannot be measurable), but, for instance, because their graphs
are dense in R2 [3, p. 14],[47],[103]. (But see, e.g., [59, 103] for “nice” properties
that some of them satisfy.) The approach presented in this section suggests an idea
for creating functions which are even more exotic.

REMARKS ON THE CAUCHY FUNCTIONAL EQUATION

9

Indeed, assume for simplicity that f : R → R and let H := {xγ : γ ∈ Γ} be
a Hamel basis, that is, a basis of R where R is considered as a vector space over
the field of rationals Q. It is well-known that the cardinality of H is the continuum
cardinality, that is, the same as the cardinality of R (see [68, Theorem 4.2.3, p. 82]).
As is well-known, and was observed by Hamel [47] (see also [1, pp. 35–36],[3, pp. 18–
19]), the general solution f : R → R which satisfies (1) for all x, y ∈ R is obtained by
associating with each γ ∈ Γ an arbitrary yγ ∈ R, settingP
f (xγ ) := yγ , and extending
f to R in a linear manner (that is, given y ∈ R, since y = m
k=1
P qk xγk for some m ∈ N,
q1 , . . . , qm ∈ Q, and xγ1 , . . . , xγm ∈ H, we define f (y) := m
k=1 qk yγk ). Now fix an
arbitrary nonempty finite or countable set J of basis elements, define f (xγ ) to be
any nonzero real number for all γ ∈ J, define f (xγ ) := 0 for the other basis elements
xγ , γ ∈
/ J, and extend f to R in a linear manner as explained above (in particular,
when card(J) = ℵ0 , then there are at least 2ℵ0 = card(R) such functions f ). The
exotic property that f has is that it is a non-constant solution to (1) which has a set
of linearly independent (over Q) periods whose cardinality is card(R). Indeed, the
equality f (xγ ) = 0 for each γ ∈ Γ\J implies that f (x + xγ ) = f (x) + f (xγ ) = f (x)
for each x ∈ R, i.e., xγ is a period of f for every γ ∈ Γ\J, and it remains to observe
that card(Γ\J) = card(Γ) = card(H) = card(R).
Actually, it is possible to make the set of periods of f to be dense in R (this, in
particular, will also show that f is so-called microperiodic [68, p. 332], that is, it
has arbitrary small positive periods). Indeed, since the set of periods of f coincides
with the Hamel basis that we fixed in the beginning, it is sufficient to show that
there exists a Hamel basis which is dense in R and then to define f with respect to
that basis as done above. The existence of a dense Hamel basis is known (the main
arguments below are based on ideas of Zoltan Boros [16]): roughly speaking, one
starts with an arbitrary Hamel basis H and then rescales its elements by appropriate
rational coefficients so the resulting basis will be dense. More precisely, one first
observes using the Cantor-Schroeder-Bernstein theorem that the cardinality of the
set K := {(a, b) : a, b ∈ R, a < b} of all bounded and open intervals in R is at
most card(R2 ) = card(R) and at least card(R), and hence it is card(R). Therefore
there exists a bijection ψ from K to H. Now, given I ∈ K , since ψ(I) ∈ H
we have ψ(I) 6= 0 because the elements of a Hamel basis do not vanish. Since
for all x ∈ R, x 6= 0 the set {qx : q ∈ Q\{0}} is dense in R, it follows that
{qψ(I) : q ∈ Q\{0}} is dense in R. From this result and the fact that I is an
open interval it follows that there exists a rational number qI such that qI ψ(I) ∈ I.
Denote H̃ := {qI ψ(I) : I ∈ K }. Then H̃ is a weighted rescaling of H by non-zero
rationals and hence H̃ is also a Hamel basis. Since from the construction of H̃ for
all I ∈ K we can find an element of H̃ which belongs to I, it follows that H̃ is
dense in R, as required.
4. The case where eif is measurable, topological torus
Now f is assumed to be a real function defined on the n-dimensional topological
(flat) torus Tn := Rn /Zn and having the property that eif is (Haar) measurable.

10

DANIEL REEM

We need the following known lemma which is a variation of Lemma 3.1. For the
sake of completeness and since it is not very easy to find a full proof of this Lemma
in the literature, we decided to include a proof in the appendix (Section 12 below).
Lemma 4.1. Given any y ∈ Tn and h : Tn → C, if h ∈ L1 (Tn ), then the
n
Rtranslation function
R hy (x) := h(x + y), x ∈ T is (Haar) integrable and we have
h (x)dµ(x) = Tn h(x)dµ(x).
Tn y

Theorem 4.2. Given an additive homomorphism f from the finite dimensional
topological torus Tn to R, if eif is (Haar) measurable, then f ≡ 0.
Proof. The proof is almost word for word as in the proof of Lemma 3.2. Here we
use the periodicity of f with respect to the basis which spans the torus, and Lemma
4.1 instead of Lemma 3.1. We note that we need the change of variables formula
for scaling only for small enough (at most 1) positive rational α, and in this case
this formula is true for the topological torus.

5. Jensen’s equation
This is the functional equation


x+y
f (x) + f (y)
(6)
f
=
2
2

where x and y belong to, say, Rn or a subset of Rn and f is a real-valued function.
See [1, pp. 43–48],[3, pp. 242–243],[68, pp. 351–354] for more details and related
references. The next theorem extends [1, Theorem 1, p. 46],[3, Proposition 3, p.
243],[68, Theorem 13.2.3, p. 354].
Theorem 5.1. Let S be a convex subset of Rn and assume that its interior is
nonempty. Suppose that f : S → R satisfies (6) for all (x, y) ∈ S 2 . If eif is
measurable on S, then there exist c ∈ Rn and b ∈ R such that f (x) = c · x + b for
each x ∈ S.
Proof. Because S is convex, (x + y)/2 ∈ S, so (1) is well defined. By [68, Theorem
13.2.1, p. 353] there exists a function g : Rn → R satisfying (1) for every x, y ∈ Rn
and a constant b ∈ R such that f (x) = g(x) + b for all x ∈ S. Since eif is measurable
on S and the interior of S is nonempty, eif is measurable on a hypercube contained
in S. Therefore eig = e−ib eif is measurable on this hypercube, and by Theorem 2.1
there exists c ∈ Rn such that g(x) = c · x for each x ∈ Rn . Thus f (x) = c · x + b for
each x ∈ S, and an immediate check shows that this function does satisfy (6) for
all (x, y) ∈ S 2 .

6. Cauchy’s equation on restricted domains
Now we consider the Cauchy functional equation (1) assuming f is real-valued
function defined on a subset of Rn (and hence (x, y) should belong to a subset of R2n ).
Sometimes (1) with these additional conditions is also referred to as “conditional
Cauchy’s equation” or “the restricted Cauchy equation”. More details about this

REMARKS ON THE CAUCHY FUNCTIONAL EQUATION

11

(quite studied) issue can be found in [1, pp. 44–49],[3, pp. 12, 18, 73, 84–92],
[31],[68, pp. 369–374].
In what follows we will derive a few theorems. The first one extends [3, Corollary
9, p. 18] in which S = [0, ∞). It is based on a certain definition (see Remark 6.4
below for related examples) and several lemmas (the first lemma generalizes the
discussion in [3, p. 12] and is also related to [2]).
Definition 6.1. Let (G, +) be a group and let S ⊆ G.
(a) S is said to subtractively span G if G = S − S := {s − t : (s, t) ∈ S 2 }.
(b) S is said to strongly subtractively span G if for all x1 , x2 ∈ G there exist
s1 , s2 , t1 , t2 ∈ S such that xi = si − ti , i = 1, 2 and such that s1 + s2 ∈ S,
t1 + t2 ∈ S.
Lemma 6.2. Let (G, +) and (H, +) be two commutative groups. Suppose that S ⊆
G strongly subtractively spans G. Let S + S := {a + b : (a, b) ∈ S 2 }. Let A ⊆ G
satisfy S ∪ (S + S) ⊆ A. If f : A → H satisfies
(7)

f (x + y) = f (x) + f (y),

∀(x, y) ∈ S 2 ,

then there exists an additive homomorphism F : G → H which coincides with f on
S.
Proof. Since S strongly subtractively spans G it subtractively spans it as an immediate verification shows. Hence, given x ∈ G, there exists (s, t) ∈ S 2 such that
x = s − t. Define
F (x) := f (s) − f (t).
To see that F is well defined, suppose that we also have x = s′ − t′ for another pair
(s′ , t′ ) ∈ S 2 . Then s + t′ = s′ + t and f (s) + f (t′ ) = f (s + t′ ) = f (s′ + t) = f (s′ ) + f (t)
because of (7). It follows that f (s) − f (t) = f (s′ ) − f (t′ ) and hence F is indeed well
defined. To see that F coincides with f on S let x ∈ S be given. As explained above,
because S subtractively spans G there exists (s, t) ∈ S 2 such that x = s − t. By (7)
it follows that f (s) = f (x + t) = f (x) + f (t). This and the definition of F imply the
equality F (x) = f (s)−f (t) = f (x). Finally, to see that F is additive let (x1 , x2 ) ∈ G2
be arbitrary. Since S strongly subtractively spans G there exist s1 , s2 , t1 , t2 , ∈ S
which satisfy the relations x1 = s1 − t1 , x2 = s2 − t2 , s1 + s2 ∈ S, t1 + t2 ∈ S. This
and the commutativity of G imply that x1 + x2 = (s1 + s2 ) − (t1 + t2 ) ∈ S − S. This,
the definition of F , the commutativity of H, and (7) imply the required assertion:
F (x1 + x2 ) = f (s1 + s2 ) − f (t1 + t2 ) = f (s1 ) − f (t1 ) + f (s2 ) − f (t2 ) = F (x1 ) + F (x2 ).

Theorem 6.3. Suppose that S ⊆ Rn strongly subtractively spans Rn . Let A ⊆ Rn
satisfy S ∪ (S + S) ⊆ A. Let f : A → R and assume that f satisfies (7). If S
contains a hypercube I on which eif is measurable, then there exists c ∈ Rn such
that f (x) = c · x for each x ∈ S.

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DANIEL REEM

Proof. By Lemma 6.2 there exists an additive function F : Rn → R satisfying
F (x) = f (x) for all x ∈ S. Since F is additive and eiF = eif is measurable on I,
it follows from Theorem 2.1 that there exists c ∈ Rn such that F (x) = c · x for all
x ∈ Rn . The assertion follows since f (x) = F (x) for all x ∈ S.

Remark 6.4. Examples of subsets S of Rn which strongly subtractively span Rn
and containing a hypercube are orthants (with or without the boundary), halfspaces, and any set containing a subset which strongly subtractively spans Rn and
has a nonempty interior. In addition, any semigroup S ⊆ Rn (that is, S + S ⊆
n
n
S) which
S∞ subtractively spans R strongly subtractively spans R . In particular,
S := m=1 [2m, 3m] strongly subtractively spans R. There are also examples of
subsets S with a nonempty interior which strongly subtractively span Rn but are
not semigroups. A simple example is a translated copy of an orthant by a vector
which does
But there are more exotic examples such
S not belong to the orthant.
m
as S := ∞
S
where
S
:=
[10
,
5
· 10m )n for each m ∈ N (because given
m
m=1 m
n
x, y ∈ R , let 2 ≤ m ∈ N be such that kxk + kyk < 10m−1 ; let pm := (2 · 10m )nk=1
be the vector whose components are 2 · 10m ; then x = (pm + x) − pm ∈ Sm − Sm ,
y = (pm + y) − pm ∈ Sm − Sm , and x + y = (2pm + x + y) − 2pm ∈ Sm − Sm as
claimed).
The same reasoning of the proof of Theorem 6.3 can be used for easily extending
it much further, as shown in the next theorem.
Theorem 6.5. Let S ⊆ Rn . Let A ⊆ Rn satisfy S ∪ (S + S) ⊆ A. Suppose
that f : A → R satisfies (7). Suppose also that S contains a hypercube I. If eif
is measurable on I and if there exists a (not necessarily unique) additive function
F : Rn → R such that F (x) = f (x) for all x ∈ S, then there exists c ∈ Rn such that
f (x) = c · x for each x ∈ S.
Now the theory developed in the literature can be used, so for instance, we may
take S to be any interval in R having 0 in its closure [68, Theorem 13.5.2, p. 368]
and A := S + S. We can also take S to be a multidimensional ball having the origin
as its center [68, Corollary 13.6.3, p. 373] (and A := S + S) but this actually follows
from the next theorem (which generalizes the case of a continuous additive function
defined on a compact interval [1, pp. 45–46]).
Theorem 6.6. Let S ⊆ Rn be a convex subset having a nonempty interior. Let
A ⊆ Rn satisfy S ∪ (S + S) ⊆ A. Suppose that f : A → R satisfies (7). If eif is
measurable on S, then there exists c ∈ Rn such that f (x) = c · x for each x ∈ S.
Proof. Let u ∈ S + S be arbitrary. Then u = s + t for some s, t ∈ S and hence
u/2 = (s + t)/2 ∈ S because S is convex. By putting x = y = u/2 in (7) it follows
that f (u)/2 = f (u/2) for all u ∈ S + S. In particular this is true for u := x + y
where x, y ∈ S are given. Thus f ((x + y)/2) = f (x + y)/2 = (f (x) + f (y))/2 where
the last equality follows from (7). Hence f satisfies the Jensen equation (6) for all
(x, y) ∈ S 2 , and from Theorem 5.1 it follows that there exist c ∈ Rn and b ∈ R such
that f (x) = c · x + b for each x ∈ S. By plugging this expression in (7) it follows
that b = 0.


REMARKS ON THE CAUCHY FUNCTIONAL EQUATION

13

7. The multiplicative Cauchy equation
This is the functional equation
(8)

f (x + y) = f (x)f (y)

where the pairs (x, y) belong to a subset of R2n and f is a real-valued function.
See [1, pp. 37–39],[3, pp. 28–29],[68, pp. 343-344, 349–350] for related details
and related references. An obvious solution to (8) is f ≡ 0. As is well known,
when the whole space is considered, if f (x0 ) = 0 at some point x0 ∈ Rn , then
f (x) = f (x − x0 )f (x0 ) = 0 for each x ∈ Rn . Since in addition f (x) = (f (x/2))2 for
each x ∈ Rn , it follows that when f is not the constant zero function, then f (x) > 0
for all x. However, when the pairs (x, y) belong only to a subset of R2n , then it is
not clear why a solution to (8) must be positive. Hence, in the next theorem (which
extends [1, Theorem 1, pp. 38–39], [3, Theorem 5, p. 29], [68, Theorem 13.1.4, p.
349], [68, Theorem 13.1.7, p. 350]) the positivity of f is assumed in advance.
Theorem 7.1. Let S ⊆ Rn . Let A ⊆ Rn satisfy S ∪ (S + S) ⊆ A. Suppose that
f : A → R is a positive function satisfying (8) for all (x, y) ∈ S 2 . Assume that
either S is a convex subset having a nonempty interior and f i is measurable on S,
or that S strongly subtractively spans Rn and it contains a hypercube I on which f i
is measurable. Then there exists c ∈ Rn such that f (x) = ec·x for each x ∈ S.
Proof. Since f is positive, the function g : A → R defined by g := ln(f ) is welldefined on A and by taking logarithm on (8) we see that g is additive. Since
f i = exp(i ln(f )) = eig is assumed to be measurable on either S or I, Theorem 6.6
or Theorem 6.3 respectively imply the existence of c ∈ Rn such that g(x) = c · x
for each x ∈ Rn and hence f (x) = ec·x for all x ∈ S. This f indeed satisfies (8) and
the assertion follows.

8. An alternative Cauchy’s equation
This is the functional equation
(9)

(f (x + y))2 = (f (x) + f (y))2

where x, y ∈ Rn are arbitrary and f is a real-valued function. It seems to appear
first in the context of certain problems related to physics [51]. See [36],[50],[67],[68,
pp. 380–382] for more details and related references. Obviously, any solution to the
Cauchy functional equation (1) solves (9), but the converse is less trivial because
at least at first glance one may only deduce the relation f (x + y) = ±(f (x) + f (y))
(where the sign depends on the pair (x, y)). However, as follows from [68, Theorem
13.9.2, p. 382], any solution to (9) defined on a semigroup S of Rn must be additive.
This, Theorem 6.3, and Remark 6.4, imply the following theorem.
Theorem 8.1. Suppose that S is a semigroup of Rn which subtractively spans Rn .
Assume that f : S → R satisfies (9) for every x, y ∈ S. If there exists a hypercube
I ⊂ S on which eif is measurable, then there exist c ∈ Rn such that f (x) = c · x for
each x ∈ S.

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DANIEL REEM

9. Pexider’s equation
This is a functional equation generalizing the Cauchy functional equation and
involving three unknown functions:
(10)

f (x + y) = g(x) + h(y)

where (x, y) belongs to R2n or a subset of R2n and f ,g, and h are all real-valued
functions. See [1, pp. 141–142],[3, pp. 42–43],[68, pp. 355–363],[86] for more
details and additional references.
The next theorem extends well-known results regarding the solvability of (10)
(see, e.g., [1, Corollary, p. 142], [68, Theorem 13.3.9, p. 363]).
Theorem 9.1. Let S ⊆ Rn be a semigroup satisfying 0 ∈ S. Assume that f :
S → R, g : S → R, h : S → R satisfy (10) for all (x, y) ∈ S 2 . Suppose that S
subtractively spans Rn and it contains a hypercube I on which a complex exponent
of one of the given functions is measurable. Then there exist c ∈ Rn and constants
a, b ∈ R such that f (x) = c · x + a + b, g(x) = c · x + a, and h(x) = c · x + b for all
x ∈ S.
Proof. Let a := g(0), b := h(0) and let p : S → R be defined by p(x) := f (x) − a − b
for all x ∈ S. An immediate verification shows that f (x) = p(x) + a + b, g(x) =
p(x) + a, h(x) = p(x) + b for all x ∈ S and that p is additive. Since a complex
exponent of one of the functions f , g, or h is assumed to be measurable on a certain
hypercube, the above expressions imply that eip is measurable on this hypercube.
Since a semigroup which subtractively spans a commutative group actually strongly
subtractively spans it (this is an immediate consequence of Definition 6.1) and since
S + S ⊆ S, Theorem 6.3 (with A := S) implies that there exists c ∈ Rn such that
p(x) = c · x for each x ∈ S. The assertion follows after easily checking that the
obtained triplet (f, g, h) does solve (10).

10. Stability
Consider the following approximate version of (1):
(11)

|f (x + y) − f (x) − f (y)| ≤ ǫ

for all x, y ∈ Rn where f is a real-valued function. Here ǫ is a given positive number.
A function f : Rn → R satisfying (11) is called an approximately additive function
or an ǫ-additive function. Inequality (11) is a perturbed version of (1) and one may
be interested in knowing whether any ǫ-additive function is a small perturbation of
a pure additive function. This issue, which was raised by Ulam in 1940 [52, p. 222],
[102, p. 64] (and traces of it can be found in the 1925 book of Pólya and Szegö
[88, Problem 99, p. 17]), is, in some sense, related to the question of the stability
of solutions to differential equations. Anyhow, it was proved by Hyers [52] that the
answer to the above question is positive: there exists a unique function g : Rn → R
satisfying (1) and |f (x) − g(x)| ≤ ǫ for all x ∈ Rn . The additive function g satisfies
f (mx)
(12)
g(x) = lim
m→∞
m

REMARKS ON THE CAUCHY FUNCTIONAL EQUATION

15

for all x ∈ Rn (a proof can be found in [68, p. 484]; the more common expression
is g(x) = limm→∞ f (2m x)/2m ). As a matter of fact, Hyers proved his theorem for
functions acting between Banach spaces. As is well-known, and follows e.g., from
the proof of [68, Theorem 17.1.1, pp. 483–484] (which is a minor variation of Hyers’
proof), the above holds whenever f : S → X satisfies (11) for all x, y ∈ S where X
is a Banach space and S is a semigroup.
The issue of stability of functional equations (and in particular, of the Cauchy
functional equation) and various problems related to it have become quite popular
during the last decades. This is somewhat illustrated in the following very short
(and semi random) list of related works: [8, 9, 10, 11, 17, 22, 26, 41, 43, 44,
72, 75, 76, 78, 79, 84, 89, 101]. A few reviews related to the issue of stability
are [38, 53, 54, 100]. Here we will concentrate only on extending [68, Theorem
17.1.2, p. 485], saying that under familiar regularity conditions (e.g., f is measurable
or bounded above on a set of positive measure) the associated additive function g
must be continuous. In a sense, this result is related to the stability notion involving
integrability conditions mentioned in [5, 34] because of the measurability condition
which is involved. However, the results and proofs mentioned there (in particular,
the stability condition on f ) are different from the one given below and also the
setting is different (the functions are from the positive real line to itself).
Theorem 10.1. Let S ⊆ Rn be a semigroup which subtractively spans Rn and
containing a hypercube I. If f : S → R satisfies (11) for all x, y ∈ S and if
there exists an infinite sequence (mk )∞
k=1 of natural numbers such that each of the
functions hk : S → C defined by hk (x) := eif (mk x)/mk for every x ∈ S is measurable
on I, then there exists c ∈ Rn such that |f (x) − c · x| ≤ ǫ for all x ∈ S.
Proof. As mentioned near (12), there exists an additive function g : S → R satisfying |f (x) − g(x)| ≤ ǫ for all x ∈ S. From (12) we have g(x) = limk→∞ f (mk x)/mk
for all k ∈ N and x ∈ S. Of course, f (mx) is well-defined for all x ∈ S and m ∈ N
because S + S ⊆ S. By the continuity of the exponential function we have
h(x) := exp(ig(x)) = lim exp(if (mk x)/mk ) = lim hk (x)
k→∞

k→∞

for all x ∈ S. Thus the restriction of h to I is a pointwise limit of measurable
functions and hence measurable [92, p. 15]. Since g satisfies (1) for every x, y ∈ S
and since a semigroup which subtractively spans a commutative group actually
strongly subtractively spans it (as follows from Definition 6.1), we conclude from
Theorem 6.3 (with A := S and with g instead of f ) the existence of c ∈ Rn such
that g(x) = c · x for all x ∈ S and the assertion follows.

Remark 10.2. If f is locally measurable, then x 7→ eif (mx)/m is locally measurable
for all m ∈ N and hence the condition on the functions hk from Theorem 10.1 is
satisfied and therefore g is measurable. On the other hand, if we only know that eif
is measurable, then it is not necessarily true that g is measurable. Indeed, consider
as in Remark 2.2 the function f (x) := 2π[g(x)], x ∈ R where g is an arbitrary
nonlinear solution of (1). Then eif is even continuous since it equals the constant

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DANIEL REEM

function 1. Since |t − [t]| ≤ 1 for all t ∈ R we have |f (x) − (2π)g(x)| ≤ 2π for each
x ∈ R. This inequality, the triangle inequality, and the additivity of g, imply that
f satisfies (11) for every x, y ∈ R with ǫ := 6π. Since 2πg satisfies (1) for every
x, y ∈ R, Hyers’ theorem implies that this is the unique additive function which
ǫ-approximates f and also that 2πg(x) = limm→∞ f (mx)/m for all x ∈ R (this limit
can also be computed directly). But g is non-measurable. We can actually say more:
although eif is measurable, Theorem 10.1 implies that among the infinitely many
functions hm (x) := eif (mx)/m , x ∈ Rn , m ∈ N, only finitely many of them can be
measurable.
11. Concluding Remarks
We conclude this paper with the following remarks.
Remark 11.1. Systems of equations: Theorem 2.1 and other results regarding
variations of the Cauchy equation (1) can be generalized to systems of equations.
For instance, given m, n ∈ N and a function f : Rn → Rm satisfying (1) for all
x, y ∈ Rn , if f = (f1 , . . . , fm ) and eifk is (locally) measurable for each k ∈ {1, . . . , m},
then there exists an m × n matrix C having real entries such that f (x) = Cx for all
x ∈ Rn . This is a simple consequence of Theorem 2.1, because fk : Rn → R satisfies
(1) for every x, y ∈ Rn and each k.
Remark 11.2. Infinite dimensional spaces: It is interesting whether it is possible to generalize the results of this paper to infinite dimensional spaces on which a
measure can be defined, e.g., the ones described in [27], [46, pp. 154–160],[48, pp.
157–166],[104].
Remark 11.3. An abstract regularity condition: An examination of the proof
of Theorem 2.1 using the initial value approach presented in Section 3 suggests
that the regularity condition can be further generalized by replacing the measure
theoretic aspect in it by a more abstract one. In what follows, we say that a triplet
(A, B, F ) has the periodic integral property if the following conditions hold:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)

(viii)

A is a set of real functions defined on Rn ;
B is a set of functions from Rn to C and B contains {eig : g ∈ A};
F : B → C is a functional;
for all β ∈ C satisfying |β| = 1 and for all h ∈ B, we have βh ∈ B and
F (βh) = βF (h);
the functions x 7→ c · x are in A for all c ∈ Rn ;
the set A is closed under addition and under multiplication by positive rationals;
there exists a basis {u1, . . . , un } of Rn such that for all g ∈ A satisfying the
relation g(x + uk ) = g(x) for all x ∈ Rn and k ∈ {1, . . . , n}, the function gy
defined by gy (x) := g(x + y) for each x ∈ Rn is in A for each y ∈ Rn and we
have F (eigy ) = F (eig );
for each g ∈ A there exists some rational α > 0 such that F (eiαg ) 6= 0.

REMARKS ON THE CAUCHY FUNCTIONAL EQUATION

17

(A, B, F ) is A := {f : Rn →
As follows from Section 3, an example of such a triplet
R
R : eif is measurable}, B := {eig : g ∈ A}, F (v) := I v(x)dx for all v ∈ A, where
I ⊂ Rn is an arbitrary hypercube which is spanned by the basis {u1 , . . . , un }.
Suppose now that we are given such a triplet (A, B, F ) satisfying the periodic
integral property. We claim that in this case, if f ∈ A and if f satisfies (1) for
every x, y ∈ Rn , then there exists c ∈ Rn such that f (x) = c · x for each x ∈ Rn .
Indeed, the proof is almost word for word the proof given in Section 3, where there
are two main differences. First, in Lemma 3.2 we need to follow only the first two
paragraphs. Second, after applying the function t 7→ eiαt , t ∈ R on (5a) (where α
is the positive rational number from Property (viii) above) we apply the functional
F on both sides of the resulting equation eiαg(x+y) = eiαg(x) eiαg(y) Rand use Property
(vii) above instead of applying the integral operator G(v) := I0 v(x)dx on this
equation (where I0 is the hyper-parallelepiped spanned by the basis {u1 , . . . , un }
and defined in the proof of Lemma 3.2, near (3)) and using Lemma 3.1, as done
there.
It is very interesting to find a triplet (A, B, F ) satisfying the periodic integral
property such that A 6= {f : Rn → R : eif is measurable}
and F is not an integral
R
operator (that is, F does not coincide with v 7→ I0 v(x)dx or a minor modification of
this operator) or to show that this is impossible. It is of a considerable interest to find
at least one triplet (A, B, F ) satisfying the periodic integral property in an infinite
dimensional setting or to show that no such a triplet can exist. We may also try to
replace F by a family of functions Fj : B → C. Anyway, because there are nonlinear
solutions to (1), these solutions cannot belong to A. Thus no matter what A and F
are, A cannot be the set of all real functions. A corresponding modification of what
written in the previous lines and paragraphs holds for the topological torus. Finally,
it will be valuable to combine the above mentioned abstract regularity condition with
some existing theories of regularity conditions of functional equations, such as the
ones surveyed and developed in [56, 57].

12. Appendix
Proof of Lemma 3.1. The proof is not conceptually difficult but it is somewhat
technical. The main ideas behind it are illustrated in Figure 1 below. For the sake
of convenience, the proof is divided into steps. In what follows y ∈ Rn is fixed.
Step 0: It is sufficient to assume that I isPright semi-open because the (right)
facets of I (i.e., the sets Ij := {u0 + uj + k∈{1,...,n}\{j} tk uk : tk ∈ [0, 1] ∀k ∈
{1, . . . , n}\{j}} where j ∈ {1, . . . , n}) are of measure 0 and hence adding them to
or removing them from I contributes nothing to the considered integrals. Moreover,
it suffices to assume that u0 = 0, because if we know that the
P assertion holds for
u0 = 0 (equivalently, with the integration done over I(0) := { nk=1 tk uk : (tk )nk=1 ∈
[0, 1)n }) and arbitrary y, then a few applications of this assertion (below: first with
y + u0 and second with u0 , both instead of y) together with a few applications of

18

DANIEL REEM
u2 axis
Iy
R(0,1)
I

R'(1,0)
R'(1,1)

R'(0,0)=R(0,0)

R(1,1)
R(1,0)

R'(0,1)

u1 axis

Figure 1. A schematic description of the proof of Lemma 3.1 for
n = 2. Both Iy and I are decomposed into unions of disjoint subparallelepipeds: Iy = ∪s∈{0,1}n Rs and I = ∪s∈{0,1}n Rs′ . The Uperiodicity Rof h and the Rchange of variable formula for translations
imply that Rs h(z)dz = R′ h(x)dx for all s ∈ {0, 1}n . Hence (using
s
R
R
R
the additivity of the integral) I hy (x)dx = Iy h(z)dz = I h(x)dx.
the change of variables formula for translations, imply the desired conclusion:
Z
Z
Z
Z
h(x + y)dx =
h(w + u0 + y)dw
hy (x)dx = h(x + y)dx =
I(0)
I
I
I(0)+u0
Z
Z
Z
=
hu0 +y (w)dw =
h(w)dw =
hu0 (w)dw
I(0)
I(0)
I(0)
Z
Z
Z
h(x)dx = h(x)dx.
=
h(w + u0 )dw =
I(0)

I(0)+u0

I

Pn

Step 1: Since U is a basis, we have y =
k=1 yk uk where yk ∈ R for each
k ∈ {1, . . . , n} are uniquely determined.
P Given z ∈ Iy :=PI + y, we have z = x + y
for some x ∈ I. We can write x = nk=1 xk uk and z = nk=1 zk uk where (because
u0 = 0 by our assumption) xk ∈ [0, 1] and zk ∈ R are uniquely determined for all
k ∈ {1, . . . , n}. Suppose first that yk ∈ [0, 1) for each k ∈ {1, . . . , n}. The general
case is a simple consequence of this special case (Step 9 below).
Step 2: Since zk = xk + yk , it follows that zk ∈ [0, 2) and hence either zk ∈ [0, 1) or
zk ∈ [1, 2) for all k ∈ {1, . . . , n}. Consider the greatest integer function [·] : R → Z
which assigns to each t ∈ R the greatest integer [t] not exceeding t. In particular,
[t] = 0 when t ∈ [0, 1) and [t] = 1 when t ∈ [1, 2). Define φ : Iy → {0, 1}n by
φ(z) := ([zk ])nk=1 for every z ∈ Iy , where here we used the coordinates of z with
respect to the basis U. For each s ∈ {0, 1}n let Rs := φ−1 (s). Since φ is a measurable function, Rs is measurable for every s ∈ {0, 1}n . In fact, each Rs is a right
semi-open hyper-parallelepiped, possibly empty (if yk = 0 for some k ∈ {1, . . . , n},
then Rs = ∅ for all s ∈ {0, 1}n whose k-th component is 1). For instance, if n = 2,
then R(0,0) = {z1 u1 + z2 u2 : z1 ∈ [y1 , 1), z2 ∈ [y2 , 1)}, R(0,1) = {z1 u1 + z2 u2 :

REMARKS ON THE CAUCHY FUNCTIONAL EQUATION

19

z1 ∈ [y1 , 1), z2 ∈ [1, 1 + y2 )}, R(1,0) = {z1 u1 + z2 u2 : z1 ∈ [1, 1 + y1 ), z2 ∈ [y2 , 1)},
R(1,1) = {z1 u1 + z2 u2 : z1 ∈ [1, 1 + y1 ), z2 ∈ [1, 1 + y2 )}. See Figure 1.
Step 3: The above construction implies that Iy = ∪s∈{0,1}n Rs and Rs ∩ Rp = ∅
whenever s 6= p, s, p ∈ {0, 1}n . Now we associate with each Rs a certain translated copy of it which we denote by Rs′ . The correspondence
is as follows: given
P
s ∈ {0, 1}n , let Rs′ := Rs − s · U where s · U := nk=1 sk uk ∈ Rn . In the next steps
we will see that the sets Rs′ form a disjoint decomposition of I.
Step 4: We first show that Rs′ is contained in I for each s ∈ {0, 1}n . Indeed,
given x ∈ Rs′ we have x = z − s · U for some z ∈ Rs . By the definition of Rs
we have [zk ] = sk for all k ∈ {1, . . . , n}. Fix an index k ∈ {1, . . . , n}. If sk = 0,
then according to the definition of [·] we have zk ∈ [0, 1). By comparing coordinates
with respect to the basis U we see that xk = zk − sk = zk ∈ [0, 1). If sk = 1, then
according to the definition of [·] we have zk ∈ [1, 2). Thus again xk = zk −sk ∈ [0, 1).
Because k was arbitrary we conclude that indeed x ∈ I.
Step 5: Here we show that Rs′ ∩ Rp′ = ∅ whenever s, p ∈ {0, 1}n , s 6= p. Indeed,
suppose for a contradiction that there exists x ∈ Rs′ ∩ Rp′ for two different points
p and s in {0, 1}n . Since s 6= p there exists an index k ∈ {1, . . . , n} such that
sk 6= pk . Because x ∈ Rs′ we have x = z − s · U for some z ∈ Rs . Because x ∈ Rp′
we have x = z̃ − p · U for some z̃ ∈ Rp . In particular zk − sk = xk = z̃k − pk .
By the definition Rs and Rp we have [zk ] = sk and [z̃k ] = pk . Suppose first that
sk = 0. Then xk = zk − 0. In addition, becausePz ∈ Rs ⊆ Iy there is v ∈ I such
that z = v + y. Since v ∈ I we can write v = nj=1 vj uj where vj ∈ [0, 1) for all
j ∈ {1 . . . , n}. Hence vk ≥ 0 and
(13)

xk = zk = vk + yk ≥ yk .

Now, since pk 6= sk we have pk = 1. This and the fact that z̃ ∈ Rp imply, according to the definition of [·], that z̃k ∈ [1, 2). In addition, because z̃ ∈PRp ⊆ Iy
n
there is ṽ ∈ I such that z̃ = ṽ + y. Since ṽ ∈ I we can write ṽ =
j=1 ṽj uj
where ṽj ∈ [0, 1) for each j ∈ {1, . . . , n}. Thus ṽk < 1 and z̃k < 1 + yk . Hence
xk = z̃k − pk < 1 + yk − 1 = yk . This is a contradiction to (13). The case sk = 1,
namely pk = 0, is treated similarly and yields a similar contradiction. We conclude
that Rs′ ∩ Rp′ = ∅, as claimed.
Step 6: Here we show that ∪s∈{0,1}n Rs′ = I. Indeed, ∪s∈{0,1}n Rs′ ⊆ I by Step 4. On
the other hand, let x ∈ I be arbitrary. Then xk ∈ [0, 1) for all k ∈ {1, . . . , n}. Define
s = s(x) ∈ {0, 1}n as follows: given k ∈ {1, . . . , n}, if xk ∈ [0, yk ), then sk := 1. If
xk ∈ [yk , 1), then sk := 0. We claim that x ∈ Rs′ . Since Rs′ = Rs − s · U it suffices
to prove that x + s · U ∈ Rs . To see this we recall that by the definition of Rs and
of [·] (Step 2) we need to show that [xk + sk ] = sk for each k ∈ {1, . . . , n}. Indeed,
given k ∈ {1, . . . , n}, if xk ∈ [0, yk ), then sk = 1 and hence 1 ≤ xk + sk < yk + 1 < 2,
so the definition of [·] implies that [xk + sk ] = 1 = sk . If xk ∈ [yk , 1), then sk = 0.

20

DANIEL REEM

Hence xk + sk ∈ [yk , 1) ⊆ [0, 1) and therefore [xk + sk ] = 0 = sk again.
Step 7: A useful observation: h(z) = h(z − s · U) for all z ∈ Rn . Indeed, this is
just an immediate consequence of the periodicity of h with respect to the basis U.
Step 8: Here we show that the desired integrals are equal when yk ∈ [0, 1) for each
k ∈ {1, . . . , n}. First we observe that since h is defined on Rn and integrable on
I, the function 1I h is integrable on Rn . Hence the change of variables formula for
translations [58, p. 171] (applied to Re(h) and Im(h) and hence to h = Re(h) +
iIm(h)) implies that given w
Rn the function x 7→ 1I (x − w)h(x − w), x ∈ Rn , is
P∈
n
n
integrable on R . Let w = k=1 mk uk where mk is an integer for all k ∈ {1 . . . , n}.
The periodicity of h with respect to U implies that h(z − w) = h(z) for all z ∈ Rn .
Hence by the change of variables formula for translations
Z
Z
Z
h(x)dx =
1I (x)h(x)dx =
1I (z − w)h(z − w)dz
I
Rn
Rn
Z
Z
=
h(z − w)dz =
h(z)dz.
Iw

Iw

Therefore h is integrable (in particular, measurable) over any translated copy of
I where the translation vector is a linear combination of the elements in U with
integer coefficients. Since Rn can be represented as a countable union of translated
copies of I it follows that h is measurable on Rn and (because of the additivity of
the integral) integrable on any measurable subset of Rn contained in a finite union
of
S translated copies of nI. Because Iy is a measurable set contained in the union
s∈{0,1}n (I + s · U) of 2 translated copies of I, it follows that h is integrable on Iy .
Now, by combining this with the previous steps, with the additivity of the integral,
with the periodicity of h with respect to U, and with the change of variables formula
for translations, we have
Z
Z
Z
Z
h(z)dz
h(z)dz = S
hy (x)dx = h(x + y)dx =
I

I

=

X Z

s∈{0,1}n

Iy

h(z)dz =
Rs

s∈{0,1}n

X Z

s∈{0,1}n

h(z − s · U)dz =

Rs

=

Rs

X Z

s∈{0,1}n

X Z

s∈{0,1}n

h(v)dv =
R′s

Z

I

h(v)dv

Rs −s·U

h(v)dv =

Z

h(x)dx.
I

Step 9: Here we finish the proof by considering the
arbitrary y ∈ Rn .
Pncase of an P
Using the greatest integer function we can write y = k=1 yk uk = nk=1 ([yk ]+ ŷk )uk ,
where
Pn ŷk = yk − [yk ]P∈n [0, 1) for all k ∈ {1, . . . , n}. Hence, by denoting ŷ :=
k=1 ŷk uk and [y] :=
k=1 [yk ]uk and using the fact that h is periodic with respect
to the basis U we have h(x + y) = h(x + ŷ + [y]) = h(x + ŷ) for each x ∈ Rn .
Thus hy = hŷ and hence,
of y), we conclude that hy is
R from StepR 8 (with ŷ instead
R
integrable over I and I hy (x)dx = I hŷ (x)dx = I h(x)dx.


REMARKS ON THE CAUCHY FUNCTIONAL EQUATION

21

Proof of Lemma 4.1. Suppose first that h is a characteristic function of some
measurable subset S of Tn , namely h = 1S . Given x ∈ Tn we have x + y ∈ S
if and only if x ∈ S − y := {s − y : s ∈ S} for each x ∈ Tn . Thus hy (x) =
n
h(x
R + y) = 1S (x +
R y) = 1S−y (x) for all x ∈ T . Therefore hy is measurable and
h (x)dµ(x) = Tn 1S−y (x)dµ(x) = µ(S − y). This equality and the fact that the
Tn y
Haar measure
µ is invariant
R
R under translations R[23, p. 285] implyR that µ(S − y) =
µ(S) = Tn 1S (x)dµ(x) = Tn h(x)dµ(x). Hence Tn hy (x)dµ(x) = Tn h(x)dµ(x).
Suppose now that h is a simple real valued function, namely, there exists m ∈ N,
n
mutually disjoint
Pmmeasurable subsets S1 , . . . , Sm of T , and real numbers
Pm β1 , . . . , βm
such that
paragraph h(x+y) = k=1 βk 1Sk (x+
P h = k=1 βk 1Sk . From the previous
n
y) = m
β
1
(x)
for
all
x,
y
∈
T
.
Hence
for every y ∈ Tn the function hy is
k=1 k Sk −y
a finite sum of measurable functions and therefore it is measurable. Moreover, the
linearity of the integral and its translation invariance with respect to characteristic
functions established in the previous paragraph imply the desired equality:
Z

Tn

hy (x)dµ(x) =

m
X
k=1

βk

Z

Tn

1Sk (x + y)dµ(x) =

m
X
k=1

=

Z

Tn

m
X

βk

Z

Tn

!

1Sk (x)dµ(x)

βk 1Sk (x) dµ(x) =

k=1

Z

h(x)dµ(x).

Tn

Now assume that h is an arbitrary nonnegative measurable function. The construction of the abstract Lebesgue integral (in particular, the construction of the Haar
integral) [23, Chapter 2] shows that there exists an increasing sequence (hk )∞
k=1 of
nonnegative simple functions which converges pointwise to h. Hence the sequence
n
(hk (x + y))∞
k=1 of nonnegative numbers increases to h(x + y) for all x, y ∈ T , i.e.,
n
for each y ∈ T the function hy is the increasing limit of the sequence (hk,y )∞
k=1
of nonnegative and measurable functions (here hk,y (x) := (hk )y (x) = hk (x + y) for
all x, y ∈ Tn and all k ∈ N). Thus hy is measurable. These facts, together with
the monotone convergence theorem and the translation invariance of the integral of
simple functions established earlier, all imply that for each y ∈ Tn
Z
Z
Z
Z
hk,y (x)dµ(x) =
hy (x)dµ(x).
hk (x)dµ(x) = lim
h(x)dµ(x) = lim
Tn

k→∞

Tn

k→∞

Tn

Tn

Finally, let h ∈ L1 (Tn ) be arbitrary. Then h = Re(h)+iIm(h) = (Re(h)+ −Re(h)− )+
i(Im(h)+ − Im(h)− ) where, as usual, g+ := max{0, g} and g− := − min{0, g} for
each function g : Tn → R and Re(h)+ , Re(h)− , Im(h)+ , Im(h)− are all nonnegative
and integrable over Tn . By what established in the previous paragraph
and the
R
linearity
of the integral we conclude that hy is measurable and Tn hy (x)dµ(x) =
R
h(x)dµ(x) for all y ∈ Tn .

Tn
An examination of the proof of Lemma 4.1 given above shows that it can be
generalized almost word for word to any locally compact topological group (instead
of Tn ).

22

DANIEL REEM

Acknowledgments
I would like to use this opportunity to thank several people for either helpful remarks
and/or for reading the paper and/or for sending me copies of some papers, especially
to Orr Shalit, Eytan Paldi, Tadeusz Figiel, Roman Ger, Maciej Sablik, Zoltan Boros,
and Jolanta Olko. I also thank the referee for valuable remarks. Parts of this work
were done in various years when I have been associated with various institutes:
The Technion - Israel Institute of Technology, Haifa, Israel (2010, 2016), IMPA The National Institute of Pure and Applied Mathematics, Rio de Janeiro, Brazil
(2013), ICMC - Institute of Mathematical and Computer Sciences, University of
São Paulo, São Carlos, Brazil (2014-2015), and it is an opportunity to me to thank
the Gurwin Foundation, a special postdoc fellowship from IMPA (“Pós-doutorado
de Excelência”), and FAPESP.
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Department of Mathematics, The Technion - Israel Institute of Technology,
Haifa 3200003, Israel. Part of this work was done while the author was at the National Institute of Pure and Applied Mathematics (IMPA), Rio de Janeiro, Brazil,
and at the Institute of Mathematical and Computer Sciences (ICMC), University
of São Paulo, São Carlos, Brazil.
E-mail address: dream@tx.technion.ac.il

