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Find all functions $f$, defined over real numbers that satisfy $f(x+y) = f(x)+f(y)$ and $f(xy) = f(x)f(y)$.

We can set $x = 0$ and $y = n$ we get $f(n) = f(0)+f(n) \implies f(0) = 0$. Then what I thought of doing was saying $f(x+y)^2 = f(x)^2+2f(xy)+f(y)^2$ and we can get $f(1) = 2f(1)^2 \implies f(1) =1,0$. I am not sure how to proceed from here.

user19405892
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1 Answers1

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It's not necessarily true that $f(1)=1$: the function $f(x)=0$ satisfies both functional equations. However, setting $x=y=1$ in $f(xy)=f(x)f(y)$, we get $f(1)=f(1)^2$, so $f(1)=0$ or $f(1)=1$.

Next, $f(n)=nf(1)$ for all $n\in\mathbb{N}$ since $f(x+y)=f(x)+f(y)$, and similarly $$ 0=f(n-n)=f(n)+f(-n) $$ so $f(-n)=-f(n)=(-n)f(1)$ for all $n\in\mathbb{N}$. And if $r=\frac{m}{n}\in\mathbb{Q}$ with $n>0$, then $$ nf(r)=f(m)=mf(1)$$ so we see that $f(r)=rf(1)$. So either $f=0$ on $\mathbb{Q}$, or $f(r)=r$ for all $r\in\mathbb{Q}$.

Finally, since $f(x^2)=f(x)^2$ and every non-negative real number is a square, it follows that $f$ maps non-negative real numbers to non-negative real numbers. Therefore if $x\leq y$ then $f(x)\leq f(y)$.

So given any $x\in\mathbb{R}$, choose $r,s\in\mathbb{Q}$ with $r<x<s$, then $f(r)\leq f(x)\leq f(s)$. Therefore if $f(1)=0$ then $f$ is identically zero.

If however $f(1)=1$, then we have $r<x<s$ and $r=f(r)\leq f(x)\leq f(s)=s$, so choosing $s-r$ arbitrarily small shows that $f(x)=x$.

carmichael561
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  • Why did you need to check for $n \in \mathbb{N}$ if you showed for rationals? – user19405892 Apr 10 '16 at 16:37
  • I needed to show it for $\mathbb{N}$ first before I could show it for $\mathbb{Q}$. – carmichael561 Apr 10 '16 at 19:20
  • I showed $f(r)=rf(1)$ for $r\in\mathbb{Q}$ by writing $r=\frac{m}{n}$ with $m,n\in\mathbb{Z}$, so $nr=m\in\mathbb{Z}$, hence $nf(r)=f(nr)=f(m)=mf(1)$. The last step isn't possible without already knowing the result for $\mathbb{Z}$. – carmichael561 Apr 10 '16 at 19:33
  • I think there may be a slight issue though in the last step. In particular, since $f$ may or may not be continuous, I don't agree that "choosing $s-r$ arbitrarily small shows that $f(x) = x$." – user19405892 Apr 10 '16 at 19:42
  • I've shown that both $r<x<s$ and $r\leq f(x)\leq s$. If $s-r<\varepsilon$, how big can $|f(x)-x|$ be? – carmichael561 Apr 10 '16 at 19:43
  • Sorry, you are right you just used the squeeze theorem. – user19405892 Apr 10 '16 at 19:45