This is problem 9 from Shilov's Elementary Real and Complex Analysis. The Hints given seem a bit off the wall and I am curious to see what the community comes up with sans those suggestions. [ f(x) is monotonic on (-inf, inf) and satisfies the functional property\ f(x +y) = f(x) + f(y). Show that f(x) = ax for a constant, real a.]
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It is easy to prove that, for all $x\in\mathbb Q$, $f(x)=ax$, with $a=f(1)$. It follows from this and from the fact that $f$ is monotonic that, for each $y\in\mathbb R$,$$\lim_{x\to y}f(x)=f(y).$$Putting all this together, one gets that $(\forall x\in\mathbb{R}):f(x)=ax$.
José Carlos Santos
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You seem to saying that a monotonic function is continuous everywhere. – Martin Argerami Mar 17 '18 at 22:19
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@MartinArgerami I am using the fact that $f$ is monotonic and that $(\forall x\in\mathbb{Q}):f(x)=ax$. – José Carlos Santos Mar 17 '18 at 22:21
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Makes sense, but it is not as you wrote it. – Martin Argerami Mar 17 '18 at 22:23
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@MartinArgerami You are right. I've edited my answer. Thank you. – José Carlos Santos Mar 17 '18 at 22:26
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Why does being monotonic and the property for all Q imply this for all R? – lalala Mar 17 '18 at 22:41
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@lalala If $x\in\mathbb R$, $q,q'\in\mathbb Q$, and $q\leqslant x\leqslant q'$, then $f(q)\leqslant f(x)\leqslant f(q')$. But this means that $aq\leqslant f(x)\leqslant aq'$. – José Carlos Santos Mar 17 '18 at 22:55
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Since $f$ is monotonic, it is continuous almost everywhere. Let $x_0$ be a point such that $f$ is continuous at $x_0$. Then, noting that $f(0)=0$ and $-f(x_0)=f(-x_0)$, $$ \lim_{h\to0}f(h)=\lim_{h\to0}f(x_0+h-x_0)=\lim_{h\to0} f(x_0+h)-f(x_0)=f(x_0)-f(x_0)=0. $$ So $f$ is continuous at $0$. But then $f$ is continuous at any point: $$ f(x+h)=f(x)+f(h)\xrightarrow[h\to0]{}f(x). $$ Now do the usual $f(n)=nf(1)$, $f(1/n)=f(1)/n$, $f(m/n)=mf(1)/n$. By continuity, $$ f(x)=f(1)x. $$
Martin Argerami
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