Find all real functions that satisfy the functional equations $f(x+y) = f(x) + f(y)$ and $f(xy)=f(x)\,f(y)$

Question

Determine all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy the two functional equations $f(x + y) =f(x)+f(y)$ and $f(xy)=f(x)\,f(y)$.

Answer 1

The idetinty function $f(x)=x$ and the null function $f \equiv 0$ are two obvious solutions.

To see there are no others, take a function $f$ which verifies your conditions and $f(x)\neq 0$ for some $x \in \mathbb{R}$ (that is, $f \not \equiv 0$). We have:

$f(x)=f(1x)=f(1)f(x)$, and since $f(x) \neq 0$ we get $f(1)=1$. By induction and using $f(n+1)=f(n)+f(1)$ we get $f(n)=n$ for all $n \in \mathbb{N}$.

Also note that $f(0)=f(0+0)=f(0)+f(0)$ hence $f(0)=0$.

Therefore $f(0)=f(n-n)=f(n)+f(-n)$ hence $f(-n)=-n$ and we have $f(p)=p$ for all $p \in \mathbb{Z}$.

Now take $r \in \mathbb{Q}$, with $r={p \over q}$, $p,q \in \mathbb{Z}$, $q \neq 0$. Since $rq$ is an integer we have: $rq=f(rq)=f(r)f(q)=f(r)q$ hence $f(r)=r$.

We've now proven that $f$ coincides with the identity in the rational numbers.

If we also know continuity of $f$, take any real $y$ and a sequence of rational numbers $r_n$ which converges to $y$. Since $f$ is continuous:

$$f(y)=\lim_{n \to \infty}f(r_n)=\lim_{n \to \infty}r_n=y$$

So $f$ is the identity function.

Answer 2

$f=0$ is a solution, so assume that $f\not\equiv0$ in what follows.

It is easy to show (see @Reveillark's answer, for instance) that $f(x)=x$ for all $\color{red}{x\in\mathbb Q}$. Now, take $x,y\in\mathbb R,x\leq y$. Then $y-x$ is nonnegative and thus has a root, $z^2=y-x$, say. It follows that $f(y)-f(x)=f(y-x)=f(z^2)=f(z)^2\geq0$ and hence $f(x)\leq f(y)$. Thus, $f$ is monotonically increasing.

Now, fix $x\in\mathbb R$ and take rational sequences $(p_n),(q_n)$ such that both converge to $x$ and $(p_n)$ increases and $(q_n)$ decreases. We then have $p_n\leq x\leq q_n$ and $p_n=f(p_n)\leq f(x)\leq f(q_n)=q_n$. Since both sequences converge to $x$ it follows from the sandwich principle that $f(x)=x$. This shows that $f(x)=x,x\in\mathbb R$ is the only nonconstant function satisfying your assumptions.

Answer 3

If considering continous solutions: That is the intersection of the linear functions $a x$ and the power functions $x^a$ (and the $0$ function), thus $x$ and $0$.