# Solution of Cauchy functional equation which has an antiderivative

Let $f\colon\mathbb R\to\mathbb R$ be a function such that $$f(x+y)=f(x)+f(y)$$ for any $x,y\in\mathbb R$ i.e., it fulfills Cauchy functional equation.

Additionally, suppose that $F'=f$ for some function $F\colon\mathbb R\to\mathbb R$, i.e., $f$ has a primitive function.

How can I show that every such function must by of the form $f(x)=cx$ for some constant $c\in\mathbb R$?

I have seen an exercise in a book on real analysis, where I would be able to use this fact. I could use the argument that every derivative belongs to the first Baire class and consequently it is measurable. Every measurable solution of Cauchy functional equation is a linear function, nice proof is given, for example, in Herrlich's Axiom of Choice, p.119.

The fact that derivative is Baire function was mentioned in the book before the chapter with this exercise. But measurability is done in this book only later. For this reason (and also out of curiosity) I wonder whether there is a proof not using measurability of $f$.

• Functions of first Baire class have many points of continuity (a comeager $G_\delta$ -- and no direct use of measurability is made when proving that). If you are willing/allowed to use this, then you're done. Commented May 4, 2013 at 20:29
• Oh, I see. And for Cauchy equations, continuity at one point implies continuity on the whole real line. @Martin Could you plaase post your comment as an answer? Commented May 4, 2013 at 20:33

By the functional equation, it suffices to prove that $f$ is continuous at one point.

The fact that $f$ is of first Baire class is very straightforward: $$f(x) = \lim_{n \to \infty} \frac{F(x+1/n)-F(x)}{1/n}$$ is a pointwise limit of continuous functions.

Now a function of first Baire class has a comeager $G_\delta$-set of points of continuity. Done.

Indeed, enumerate the open intervals with rational endpoints as $\langle I_n \mid n \in \omega\rangle$. Then $$f \text{ is discontinuous at }x \iff \exists n\in \omega : x \in f^{-1}[I_n] \setminus \operatorname{int}f^{-1}[I_n]$$ Since $f$ is of first Baire class, $f^{-1}[I_n]$ is an $F_\sigma$ and so is $f^{-1}[I_n] \setminus \operatorname{int} f^{-1}[I_n]$. Therefore we can write $$f^{-1}[I_n] \setminus \operatorname{int} f^{-1}[I_n] = \bigcup_{k \in \omega} F_{k}^{n}$$ for some sequence $\langle F_{k}^n \mid k \in \omega\rangle$ of closed sets. Observe that $F_{k}^n$ has no interior, so the set of points of discontinuity of $f$ is $$\bigcup_{n \in \omega} f^{-1}[I_n] \setminus \operatorname{int}f^{-1}[I_n] = \bigcup_{n\in\omega} \bigcup_{k\in\omega} F_{k}^n,$$ a countable union of closed and nowhere dense sets.

• Thanks a lot! BTW congrats on the functional analysis bronze badge :-) Commented May 5, 2013 at 10:16
• You're welcome. Thanks for the congrats, I seem to have joined a rather select group :-) BTW: The argument for the continuity of measurable solutions to Cauchy's functional equation in Herrlich is very nice. I posted a more traditional argument recently here. Commented May 5, 2013 at 10:21

$$\def \R {\mathbb R}$$ Define the function $$G : \R ^ 2 \to \R$$ by the equation $$G ( x , y ) = F ( x + y ) - F ( x ) - F ( y ) - x f ( y )$$ for all $$x , y \in \R$$. Then, as $$F$$ is differentiable, $$G$$ is differentiable with respect to the first variable. Since $$F ' = f$$ and $$f$$ is additive, we have $$\frac { \partial G ( x , y ) } { \partial x } = F ' ( x + y ) - F ' ( x ) - f ( y ) = f ( x + y ) - f ( x ) - f ( y ) = 0$$ for all $$x , y \in \R$$. Hence, $$G$$ is constant with respect to the first variable; i.e. there is a function $$g : \R \to \R$$ such that $$G ( x , y ) = g ( y )$$ for all $$x , y \in \R$$. Now, using the definition of $$G$$ and taking a look at the expression for $$G ( y , x ) - G ( x , y )$$, we have $$x f ( y ) - y f ( x ) = g ( x ) - g ( y ) \tag {*} \label {eqn}$$ for all $$x , y \in \R$$. By additivity of $$f$$ we have $$f ( 0 ) = 0$$, and thus letting $$y = 0$$ in \eqref{eqn} we get $$g ( x ) = g ( 0 )$$ for all $$x \in \R$$. This, together with \eqref{eqn} itself, proves that $$y f ( x ) = x f ( y )$$ for all $$x , y \in \R$$. In particular, for $$y = 1$$, this gives $$f ( x ) = c x$$ for all $$x \in \R$$, where $$c = f ( 1 )$$.

You can see that there's no need to mention measurability, as you wanted. We didn't even reduce the problem to the case when we have additional assumptions (like continuity) on $$f$$. In the light of the fundamental theorem of calculus, this method is essentially the same as the one here, only avoiding integration.