Cauchy's functional equation -- additional condition

Question

Consider the function $f:R \to R$ $$f(x+y)=f(x)+f(y)$$ which is known as Cauchy's functional equation. I know that if $f$ is monotonic, continuous at one point, bounded, then the only solutions are $f(x)=cx,~ \forall c$ But once I successfully derived Cauchy's functional equation from a problem, and I know it's an involution, bijective, as well as an odd function. How can I derive (if possible) that $f(x)=x$ is the only solution?

Answer 1

Let $x+y=z$ for some $z$, then $f(z)=f(z-y)+f(z-x)$. If $f(z)=z$ then $$\begin{align}f(z)&=z-y+z-x \\ &=2z-(x+y)\\ &=2z-z =z.\end{align}$$ so we know $f(z)=z$ is a solution. Now, let $f(z)=w\cdot z\,\exists w$. We certainly know that $w=1$ is a solution, but are there other values that satisfy $w$? Let's see: $$\begin{align}f(z)&=w(z-y)+w(z-x)\\ &=w(z-y+z-x)=w\cdot z.\end{align}$$ So, since this is for all $w$, the solution can also be negative; id est, $f(z)=-z$.

Now what if the solutions are not of multiplicative form? We examine this case by letting $f(z)=z$ $+\,z_0\,\exists z_0$ such that $\gcd(z,z_0)=1$; thus, $z$ and $z_0$ must have an opposite parity. Now, $$\begin{align}f(z)&=z-y+z_0+z-x+z_0\\ &=2z+2z_0-x-y \\ &=2(z+z_0)-(x+y) \\ &=2f(z)-z.\end{align}$$ But notice this: $$\begin{align}f(z)&=2f(z)-z \\ \Leftrightarrow f(z)-f(z)&=2f(z)-z-f(z) \\ \Leftrightarrow z&=f(z).\end{align}$$Similarly, $$\begin{align}f(z)&=2f(z)-z \\ \Leftrightarrow f(z)-z_0&=2f(z)-z-z_0 \\ \Leftrightarrow z&=f(z).\end{align}$$ It follows, then, that $z=f(z)$, thus $z_0=0$; it goes to show that $f(z)=w\cdot z$ is the only form. $\;\bigcirc$

Answer 2

There are infinitely many non-linear solutions.

First of all, we of course have $f(0) = 0$, and by plugging $y = -x$ we have that $f$ is odd. We should thus only focus on $f$ being bijective and an involution.

Let $S = \{s_1, s_2, \ldots \}$ be a Hamel basis with the following property:

each real number $r$ can uniquely be expressed as $r = q_1s_1 + q_2s_2 + \ldots + q_ns_n$, for non-zero rationals $q_i$, and where $s_i \in S$ are distinct. Here $n$ is a unique integer (note that $n$ is finite). We can see $S$ as a $\mathbb{Q}$-basis for the real numbers, which is linearly independent. It is well-known that such a set S exists, assuming the axiom of choice (every vector space has a basis). I am not actually sure if there is a non-linear additive function, assuming the axiom of choice is false (there probably is a question about this already).

Now, to create a non-trivial solution we may, for example, define $f$ as follows:

$f(s_i) = (-1)^is_i$

for each $i$, and now in general

$f(q_1s_1 + \ldots + q_ns_n) = q_1f(s_1) + \ldots + q_nf(s_n)$

It is very straight-forward to see that $f$ satisfies the equation $f(x+y) = f(x) + f(y)$.

It is also quite easy to see that $f$ is injective: if

$f(q_1s_1 + \ldots + q_ns_n) = f(t_1s_1' + \ldots + t_ks_k')$,

then due to $S$ being linearly independent in the rational numbers we must have, using the above definition of $f$, $q_1s_1 + \ldots + q_ns_n = t_1s_1' + \ldots + t_ks_k'$.

For surjectivity: for each $y = q_1s_1 + \ldots + q_ns_n'$ we will construct an $x$ such that $f(x) = y$. This works just by putting $x = (-1)^1q_1s_1 + (-1)^2q_2s_2 + \ldots + (-1)^nq_ns_n$

$f$ being an involution is also straightforward to check (in fact, it sufficies to check $f(f(x)) = x$ only for $x \in S$, additivity handles the rest).

Note that the choice $f(s_i) = (-1)^is_i$ was just one example. We may choose the signs arbitrarily, and we could even do $f(s_1) = s_2$, $f(s_2) = s_1$, and stuff like that.

Answer 3

Actually, there are still 'bad' solutions.

consider $$f(x)=x$$ if $x\in \mathbb Q$ and $$f(x+y)=x-y$$ if $x\in \mathbb Q'$(the irrational numbers), where $x \in \mathbb Q$ and $y \in \mathbb Q'$.

This works because $\mathbb Q$ and $\mathbb Q'$ are linear independent, and hence $f(f(x))\in\mathbb Q$ iff $x\in \mathbb Q$.

Hence the involution is resolved. Bijective follows from involution, and you can prove odd with Cauchy.