Cauchy's functional equation explained
Cauchy's functional equation is the functional equation:
A function
that solves this equation is called an
additive function. Over the
rational numbers, it can be shown using
elementary algebra that there is a single family of solutions, namely
for any rational constant
Over the
real numbers, the family of
linear maps
now with
an arbitrary real constant, is likewise a family of solutions; however there can exist other solutions not of this form that are extremely complicated. However, any of a number of regularity conditions, some of them quite weak, will preclude the existence of these
pathological solutions. For example, an additive function
is
linear if:
is
continuous (
Cauchy, 1821). In fact, it suffices for
to be continuous at one point (
Darboux, 1875).
is
monotonic on any
interval.
is
bounded on any interval.
is
Lebesgue measurable.On the other hand, if no further conditions are imposed on
then (assuming the
axiom of choice) there are infinitely many other functions that satisfy the equation. This was proved in 1905 by
Georg Hamel using Hamel bases. Such functions are sometimes called
Hamel functions.
[1] The fifth problem on Hilbert's list is a generalisation of this equation. Functions where there exists a real number
such that
are known as Cauchy-Hamel functions and are used in Dehn-Hadwiger invariants which are used in the extension of
Hilbert's third problem from 3D to higher dimensions.
[2]
Cauchy's logarithmic functional equation
and Cauchy's multiplicative functional equation
Solutions over the rational numbers
A simple argument, involving only elementary algebra, demonstrates that the set of additive maps
, where
are vector spaces over an extension field of
, is identical to the set of
-linear maps from
to
.
Theorem: Let
be an additive function. Then
is
-linear.
Proof: We want to prove that any solution
to Cauchy’s functional equation,
, satisfies
for any
and
. Let
.
First note
, hence
, and therewith
0=f(0)=f(v+(-v))=f(v)+f(-v)
from which follows
.
Via induction,
is proved for any
.
For any negative integer
we know
, therefore
f(mv)=f((-m)(-v))=(-m)f(-v)=(-m)(-f(v))=mf(v)
. Thus far we have proved
for any
.
Let
, then
and hence
.
Finally, any
has a representation
with
and
, so, putting things together,
f(qv)=f\left(
v\right)=f\left(
(mv)\right)=
f(mv)=
mf(v)=qf(v)
, q.e.d.
Properties of nonlinear solutions over the real numbers
is
dense in
that is, that any disk in the plane (however small) contains a point from the graph. From this it is easy to prove the various conditions given in the introductory paragraph.
Existence of nonlinear solutions over the real numbers
The linearity proof given above also applies to
where
is a scaled copy of the rationals. This shows that only linear solutions are permitted when the
domain of
is restricted to such sets. Thus, in general, we have
for all
and
However, as we will demonstrate below, highly pathological solutions can be found for functions
based on these linear solutions, by viewing the reals as a
vector space over the
field of rational numbers. Note, however, that this method is nonconstructive, relying as it does on the existence of a
(Hamel) basis for any vector space, a statement proved using
Zorn's lemma. (In fact, the existence of a basis for every vector space is logically equivalent to the
axiom of choice.) There exists models
[3] where all sets of reals are measurable which are consistent with ZF +
DC, and therein all solutions are linear.
[4] To show that solutions other than the ones defined by
exist, we first note that because every vector space has a basis, there is a basis for
over the field
i.e. a set
with the property that any
can be expressed uniquely as
where
is a finite
subset of
and each
is in
We note that because no explicit basis for
over
can be written down, the pathological solutions defined below likewise cannot be expressed explicitly.
As argued above, the restriction of
to
must be a linear map for each
Moreover, because
for
it is clear that
is the constant of proportionality. In other words,
is the map
Since any
can be expressed as a unique (finite) linear combination of the
s, and
is additive,
is well-defined for all
and is given by:
It is easy to check that
is a solution to Cauchy's functional equation given a definition of
on the basis elements,
Moreover, it is clear that every solution is of this form. In particular, the solutions of the functional equation are linear
if and only if
is constant over all
Thus, in a sense, despite the inability to exhibit a nonlinear solution, "most" (in the sense of cardinality
[5]) solutions to the Cauchy functional equation are actually nonlinear and pathological.
References
. Marek Kuczma. An introduction to the theory of functional equations and inequalities. Cauchy's equation and Jensen's inequality. Birkhäuser. Basel. 2009. 9783764387495.
- Hamel. Georg. Eine Basis aller Zahlen und die unstetigen Lösungen der Funktionalgleichung: f(x+y) = f(x) + f(y).. Mathematische Annalen. 1905.
External links
Notes and References
- Kuczma (2009), p.130
- V.G. Boltianskii (1978) "Hilbert's third problem", Halsted Press, Washington
- Solovay . Robert M. . 1970 . A Model of Set-Theory in Which Every Set of Reals is Lebesgue Measurable . Annals of Mathematics . 92 . 1 . 1–56 . 10.2307/1970696 . 0003-486X.
- Web site: E. Caicedo . Andrés . 2011-03-06 . Are there any non-linear solutions of Cauchy's equation $f(x+y)=f(x)+f(y)$ without assuming the Axiom of Choice? . 2024-02-21 . MathOverflow . en.
- It can easily be shown that
; thus there are
functions
each of which could be extended to a unique solution of the functional equation. On the other hand, there are only
solutions that are linear.