Cauchy's functional equation

Solutions over the rational numbers

A simple argument, involving only elementary algebra, demonstrates that the set of additive maps f \colon V \to W, where V, W are vector spaces over an extension field of \Q, is identical to the set of \Q-linear maps from V to W.

Theorem: Let f \colon V \to W be an additive function. Then f is \Q-linear.

Proof: We want to prove that any solution f \colon V \to W to Cauchy’s functional equation, f(x+y) = f(x) + f(y), satisfies f(q v) = q f(v) for any q \in \Q and v \in V. Let v \in V.

First note f(0) = f(0 + 0) = f(0) + f(0), hence f(0) = 0, and therewith 0 = f(0) = f(v + (-v)) = f(v) + f(-v) from which follows .

Via induction, f(m v) = m f(v) is proved for any m \in \N \cup {0}.

For any negative integer m \in \Z we know , therefore . Thus far we have proved : f(m v) = m f(v) for any m \in \Z.

Let n \in \N, then f(v) = f(n n^{-1} v) = n f(n^{-1} v) and hence .

Finally, any q \in \Q has a representation q = \frac{m}{n} with m \in \Z and , so, putting things together, : f(q v) = f\left( \frac{m}{n} , v \right) = f\left( \frac{1}{n} , (m v) \right) = \frac{1}{n} , f(m v) = \frac{1}{n} , m , f(v) = q f(v), q.e.d.

Properties of nonlinear solutions over the real numbers

We prove below that any other solutions must be highly pathological functions. In particular, it is shown that any other solution must have the property that its graph {(x, f(x)) \vert x \in \R} 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.

It suffices to show that the graph of f is dense in , which is dense in .

Since f is not linear, we have f(a) \neq a for some .

Claim: The lattice defined by L:= {(r_1 + r_2 a, r_1 + r_2 f(a)): r_1, r_2 \in \mathbb Q} is dense in .

Consider the linear transformation A: \R^2 \to \R^2 defined by A(x, y) = \begin{bmatrix} 1 & a\ 1 & f(a) \end{bmatrix} \begin{bmatrix} x\ y \end{bmatrix}

With this transformation, we have .

Since , the transformation is invertible, thus it is bicontinuous. Since \mathbb Q^2 is dense in , so is . \square

Claim: if , and , then .

If , then it is true by additivity. If {{tmath| r_1, r_2

If , we have . Let k be a positive integer large enough such that . Then we have by additivity: f\left(\frac{r_1}k + \frac{r_2 a}k\right) + f\left(\frac{-r_2 a}k\right) = f\left(\frac{r_1}k\right)

That is, \frac{1}{k} f\left(r_1 + r_2 a\right) + \frac{-r_2}k f\left(a\right) =\frac{r_1}k \square

Thus, the graph of f contains , which is dense in . }}

Existence of nonlinear solutions over the real numbers

The linearity proof given above also applies to , where \alpha\Q is a scaled copy of the rationals. This shows that only linear solutions are permitted when the domain of f is restricted to such sets. Thus, in general, we have f(\alpha q)=f(\alpha)q for all \alpha \in \R and . However, as we will demonstrate below, highly pathological solutions can be found for functions f \colon \R \to \R 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 exist models such as the Solovay model where all sets of reals are measurable which are consistent with ZF + DC, and therein all solutions are linear.

To show that solutions other than the ones defined by f(x)=f(1)x exist, we first note that because every vector space has a basis, there is a basis for \R over the field , i.e. a set \mathcal{B} \sub \R with the property that any x\in\R can be expressed uniquely as , where { x_i }_{i \in I} is a finite subset of , and each \lambda_i is in . We note that because no explicit basis for \R over \Q can be written down, the pathological solutions defined below likewise cannot be expressed explicitly.

As argued above, the restriction of f to x_i \Q must be a linear map for each . Moreover, because x_iq\mapsto f(x_i)q for , it is clear that f(x_i) / x_i is the constant of proportionality. In other words, f \colon x_i\Q \to \R is the map . Since any x \in \R can be expressed as a unique (finite) linear combination of the , and f \colon \R \to \R is additive, f(x) is well-defined for all x \in \R and is given by: f(x) = f\Big(\sum_{i \in I} \lambda_i x_i\Big) = \sum_{i \in I} f(x_i\lambda_i) = \sum_{i \in I} f(x_i) \lambda_i.

It is easy to check that f is a solution to Cauchy's functional equation given a definition of f 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 f(x_i) / x_i is constant over all . Thus, in a sense, despite the inability to exhibit a nonlinear solution, "most" (in the sense of cardinality) solutions to the Cauchy functional equation are actually nonlinear and pathological.

References

References

1. Kuczma (2009), p. 130
2. V.G. Boltianskii (1978) "Hilbert's third problem", Halsted Press, Washington
3. 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?".
4. It can easily be shown that {{tmath. f \colon \mathcal{B}\to\R , each of which could be extended to a unique solution of the functional equation. On the other hand, there are only \mathfrak{c} solutions that are linear.