Solution of Cauchy functional equation which has an antiderivative

Question

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 $c\in\mathbb R$?

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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 $f$.

Answer 1

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.

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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.

Answer 2

$ \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.