# Overview of basic facts about Cauchy functional equation

The Cauchy functional equation asks about functions $$f \colon \mathbb R \to \mathbb R$$ such that $$f(x+y)=f(x)+f(y).$$ It is a very well-known functional equation, which appears in various areas of mathematics ranging from exercises in freshman classes to constructing useful counterexamples for some advanced questions. Solutions of this equation are often called additive functions.

Also a few other equations related to this equation are often studied. (Equations which can be easily transformed to Cauchy functional equation or can be solved by using similar methods.)

Is there some overview of basic facts about Cauchy equation and related functional equations - preferably available online?

• This is also done to some extent to make finding duplicate questions easier, since questions of this type appear at MSE quite frequently. See this meta post for the format of the question (and answer) I have used here. Jun 18, 2013 at 10:44
• Wikihammered per OP request. Since this is a sort of an abstract duplicate question I also added the faq tag. Jun 18, 2013 at 11:39
• Aczél J (1966) Lectures on Functional Equations and their Applications, section 2.1 Mar 20, 2020 at 11:24


We are interested in functions $$\Zobr f{\R}{\R}$$ such that $$f(x+y)=f(x)+f(y) \tag{0}\label{0}$$ holds for each $$x,y\in\R$$.

• If $$f$$ is a solution of \eqref{0} and $$q$$ is a rational number, then $$f(qx)=qf(x)$$ holds for each $$x\in\R$$.
• Every continuous solution $$\Zobr f{\R}{\R}$$ of \eqref{0} has the form $$f(x)=cx$$ for some constant $$c\in\R$$.
• If $$\Zobr f{\R}{\R}$$ is a solution of \eqref{0} which is continuous at some point $$x_0$$, then it is continuous everywhere.
• If a solution $$f$$ of \eqref{0} is bounded on some interval, then $$f(x)=cx$$ for some $$c\in\R$$.

One part of this question is about the proof that locally bounded solution of \eqref{0} is necessarily continuous: Real Analysis Proofs: Additive Functions

• Every monotonic solution $$\Zobr f{\R}{\R}$$ of \eqref{0} has the form $$f(x)=cx$$ for some constant $$c\in\R$$.
• If we have the equation \eqref{0} for function $$\Zobr f{[0,\infty)}{[0,\infty)}$$, then every solution is of the form $$f(x)=cx$$.
• There exist non-continuous solutions of \eqref{0}.

NOTE: The proof of this fact needs at least some form of Axiom of Choice - it is not provable in ZF.

For the role of AC in this result, see this MO thread, and theses questions: Cauchy functional equation with non choice, A question concerning on the axiom of choice and Cauchy functional equation, Is there a non-trivial example of a $\mathbb Q-$endomorphism of $\mathbb R$?, What is $\operatorname{Aut}(\mathbb{R},+)$?

• The graph $$G(f)=\big\{\big(x,f(x)\big)\big| x\in\R\big\}$$ is a dense subset of $$\R^2$$ for every non-continuous solution of \eqref{0}.
• Every anti-differentiable solution $$\Zobr f{\R}{\R}$$ of \eqref{0} has the form $$f(x)=cx$$ for some constant $$c\in\R$$.
• Every locally integrable solution $$\Zobr f{\R}{\R}$$ of \eqref{0} has the form $$f(x)=cx$$ for some constant $$c\in\R$$.
• Every Lebesgue measurable solution $$\Zobr f{\R}{\R}$$ of \eqref{0} has the form $$f(x)=cx$$ for some constant $$c\in\R$$.

# Related equations

There are several functional equations that are closely related to Cauchy equation. They can be reduced to it by some methods or solved by very similar methods as the Cauchy equation.

## Cauchy's exponential functional equation

$$f(x+y)=f(x)f(y) \tag{1}\label{1}$$

• If $$f$$ is a solution of \eqref{1}, then either $$f=0$$ (i.e., $$f$$ is constant function equal to zero) or $$f(x)>0$$ for each $$x\in\R$$.
• If $$f$$ is a solution of \eqref{1}, $$x\in\Q$$ and if we denote $$a=f(1)$$, then $$f(x)=a^x$$.

• Every continuous solution of \eqref{1} has the form $$f(x)=a^x$$ for some $$a\ge 0$$.

• If a solution of \eqref{1} is continuous at $$0$$, then it is continuous everywhere.

If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f (x + y) = f (x) f (y)$ and continuous at $0$, then continuous everywhere

• There are non-continuous solutions of \eqref{1}.

They can be obtained from non-continuous solutions of \eqref{1}.

• If a solution of \eqref{1} is differentiable at $$0$$, then it is differentiable everywhere and $$f'(x)=f(x)f'(0)$$.

See Prove that $f'$ exists for all $x$ in $R$ if $f(x+y)=f(x)f(y)$ and $f'(0)$ exists and Differentiability of $f(x+y) = f(x)f(y)$ There are also posts searching for differentiable solutions: Solution for exponential function's functional equation by using a definition of derivative

## Cauchy's logarithmic functional equation

$$f(xy)=f(x)+f(y) \tag{2}\label{2}$$

## Cauchy's multiplicative functional equation

$$f(xy)=f(x)f(y) \tag{3}\label{3}$$

• Every continuous solution of \eqref{3} has the form $$f(x)=x^a$$
• A solution of \eqref{3} which is continuous at $$1$$ is continuous for each $$x>0$$.
• If a function fulfills both \eqref{0} and \eqref{3}, then either $$f(x)=0$$ or $$f(x)=x$$ (i.e., it is either zero function or the identity).

Notice that we do not require continuity here. See Functional equations $f(x+y)= f(x) + f(y)$ and $f(xy)= f(x)f(y)$