Let $f:\mathbb{R}\to \mathbb{R}$ is a function such that for all real $x$ and $y$, $f(x+y)= f(x) + f(y)$ and $f(xy)= f(x)f(y)$, then prove that $f$ must be one of the two following functions:

  • $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=0$ for all real $x$


  • $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x$ for all real $x$

I got to the point where putting the two equations together, you get $f(x+y)f(x)= f(xy) + f(x)^2$ and plugging in $f(x)=x$ checks with it. So am I going in the right direction or am I just doing some guess work? Is there a more elegant way of doing it?



Easier way from the first identity: $$ f(x) = f(x+0) = f(x) + f(0) $$ so $f(0) = 0$ and similarly $$ f(x) = f(x \cdot 1) = f(x) f(1) $$ so $f(x) = 0$ or $f(1) = 1$.

In the first choice, you are done. Suffices to prove that if $f(1)=1$ then $f(x) = x$. Can you take it from here?

EDIT Another hint: Note that if $f(1)=1$, $$ f(n) = f(1 + 1 \ldots + 1) = n f(1) = n $$ for all integer $n$...

EDIT 2 Another hint... to do rational numbers, $$f(1) = f(1/n) + f(1/n) + \ldots + f(1/n) = n f(1/n),$$ so $f(1/n) = f(1)/n = 1/n$ and similarly $f(a/b) = a/b$.

You can use a similar technique to show $f(\sqrt[b]{a}) = \sqrt[b]{f(a)}$.

There must be a more direct way to prove $f(ax) = af(x)$ for all real $a$...

  • $\begingroup$ Would you please elaborate how you concluded that f(x)=0? And no I can't seem to figure out how to proceed with the second part. $\endgroup$ – user130113 Feb 20 '14 at 22:28
  • $\begingroup$ @user130113 included another hint for you. As for $f(0)$, note we proved $f(x) = f(x) + f(0)$ so subtract $f(x)$ from both sides. $\endgroup$ – gt6989b Feb 20 '14 at 22:29
  • $\begingroup$ Yes, I understood how f(0)= 0 is true, but how did you conclude f(x)= 0? $\endgroup$ – user130113 Feb 20 '14 at 22:33
  • $\begingroup$ @user130113 So you have $f(x) = f(x) \cdot f(1)$. Thus, either $f(1) =1$ or $f(x) = 0$ - it's the only other possible solution to the equation... $\endgroup$ – gt6989b Feb 20 '14 at 22:34
  • $\begingroup$ Oh I see, that explains. As for the 2nd hint, are you suggesting I should prove it separately for integers, rationals and irrationals? A hint was given alongwith the question that at some point the fact that every x>0 can be expressed as x=t^2 will be essential. $\endgroup$ – user130113 Feb 20 '14 at 22:45

I am providing another answer without the use of the Axiom of Choice this time:

As already observed $f(x)=0$ is a solution of this functional equation.

Assuming that $f$ is not the identically zero solution we obtain that $f(1)=1$ since for every $x$: $$ f(x)=f(x\cdot 1)=f(x)f(1). $$ Next, obtain that $$f(n)=\underbrace{f(1)+\cdots+f(1)}_{n\,\,\text{times}}=n,$$ for every $n\in\mathbb N$. Also $f(0)=f(0+0)=f(0)+f(0)$, and thus $f(0)=0$, and $$0=f(n-n)=f(n)+f(-n),$$ and thus $f(-n)=-f(n)$, and hence $f(k)=k$, for all $k\in\mathbb Z$. Finally $$f(1)=\underbrace{f(1/q)+\cdots+f(1/q)}_{q\,\,\text{times}},$$ and thus $f(1/q)=1/q$ similarly $f(p/q)=p/q$ and $$ f(r)=r\quad \text{for every}\quad r\in\mathbb Q.\tag{1} $$

Next, observe that, if $x\ge 0$, then $$f(x)=f\big(\sqrt{x}\big)\,f\big(\sqrt{x}\big),$$ and hence $$ x\ge 0\quad\Longrightarrow f(x)\ge 0. $$ In particular, $$ \text{If}\,\,\,\,\, y \ge x\,\,\,\,\, \text{then}\,\,\,\,\, f(y)=f(x)+f(y-x)\ge f(x). \tag{2} $$

Finally, we shall show that $f(x)=x$ for all $x\in\mathbb R$. Suppose that for some $x\in\mathbb R$: $$f(x)<x.$$ Then there is a rational $r$, such that $$f(x)<r<x.\tag{3}$$ But $(1)$ and $(2)$ imply that $$ f(x)\le f(r)=r, $$ which contradicts $(3)$. The case $f(x)>x$ is dealt with similarly.


If $f(x+y)=f(x)+f(y)$, then $f(px+qy)=pf(x)+qf(y)$, for all $p,q\in\mathbb Q$, and thus $f$ is a linear functional on the vector space $\mathbb R$ over $\mathbb Q$.

Therefore, $f$ is fully determined once its values are known on a Hamel basis of $\mathbb R$ over $\mathbb Q$. A Hamel basis in this case is a set $B\subset\mathbb R$, such that every $x\in\mathbb R$ can be written uniquely as a linear combination of elements of this $B$ with rational coefficients. (Such basis exists as a consequence of Zorn's Lemma.)

But for every $x\ge 0$ $$f(x)=f\big(\sqrt{x}\big)\,f\big(\sqrt{x}\big),$$ which means that $$f(x)\ge 0\quad\text{whenever} \quad x\ge 0.\tag{1}$$ In the case $f$ is not a continuous linear functional on $\mathbb R$ over $\mathbb Q$, there exist $a,b\in\mathbb R$, linearly independent over $\mathbb Q$, such that the values $$ f(a),\, f(b), $$ are not proportional to the values $$ a,\,b. $$ In no such pair existed, then $f$ would be $f(x)=cx$. Using these values we can create a linear combination of them $$ pa+qb>0, $$ such that $$ f(pa+qb)<0, $$ which contradicts $(1)$.

Thus $f$ is continuous, which implies that $f(x)=cx$, for some $c\ge 0$, and since $$ cxy=f(xy)=f(x)f(y)=c^2xy, $$ then the only acceptable values of $c$ are $0$ and $1$. Thus $$ f(x)=x \quad\text{or}\quad f(x)=0. $$


These functional equations are rather famous and you can read more here. Actually:

$$f(x+y) = f(x) + f(y)$$

is called Cauchy functional equation and its solutions are $f(x) = cx$; $\forall c \in R$

Now substitute into the second functional equations and we have:

$$f(xy) = f(x)f(y) \implies cxy = c^2xy$$

So we have two cases: $c = 0$ or $c=1$. So we have $f(x) = 0$ and $f(x) = x$ as solutions.

Hence the proof.

  • 5
    $\begingroup$ Only if you assume continuity $f(x+y)=f(x)+f(y)$, implies that $f(x)=cx$. $\endgroup$ – Yiorgos S. Smyrlis Feb 20 '14 at 22:20

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