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?

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    $\begingroup$ how can f be a bounded function except the case when c is zero? $\endgroup$ – Ben May 22 '18 at 0:04
  • $\begingroup$ Sorry... Bounded on some interval. $\endgroup$ – abc... May 22 '18 at 0:34
  • $\begingroup$ $f(x)$ is not the only solution. Why do you believe it must be the only solution? $\endgroup$ – Mr Pie May 26 '18 at 15:01

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.


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. $\;\bigcirc$

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    $\begingroup$ -1 You seem to be assuming that we are working over the integers (the talk about gcd and parity), whereas that need not be the case. It's not clear how examining only he case when gcd$(z, z_0) =1$ is helpful. (Not that it doesn't imply that they must have opposite parity, since gcd$(3,5)=1$.) $\endgroup$ – Calvin Lin Jun 10 '19 at 20:25

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.

  • $\begingroup$ Your example unfortunately doesn't work. We have $-\sqrt{2}-1=f(\sqrt{2}+1)=f(\sqrt{2})+f(1)=-\sqrt{2}+1$ - a contradiction. $\endgroup$ – Bartek Jun 9 '19 at 12:06
  • $\begingroup$ @Bartek I changed it. See the new function. $\endgroup$ – abc... Jun 9 '19 at 13:35
  • $\begingroup$ But still $1-\sqrt{2}=f(\sqrt{2}+1)=f((\sqrt{2}+1)+0)=-\sqrt{2}-1$, so your function is not well defined (and btw in your definition $x$ is both rational and irrational). $\endgroup$ – Bartek Jun 9 '19 at 13:42
  • $\begingroup$ @Bartek Ohh I mean y is 'pure irrational' idk if that is a thing. Or we could say 'consider the Hamel Basis and take the $\mathbb Q$ value as $x$' or something. $\endgroup$ – abc... Jun 9 '19 at 13:47
  • $\begingroup$ I'm afraid that Hamel Basis is necessary here - I haven't seen any other construction of nonlinear solutions (obviously all solutions can be described using a Hamel basis but I've never seen any simpler representation). $\endgroup$ – Bartek Jun 9 '19 at 13:51

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