If $f:\Bbb R\to\Bbb R$ is continuous and $\forall x,y\in\Bbb R,\ f(x+y)=f(x)+f(y)$, then $\exists a\in\Bbb R,\forall x\in\Bbb R,\ f(x)=ax$.

Question

Let $f$ be a continuous function defined on $\mathbb R$ such that $$\forall x,y \in \mathbb R , \ f(x+y)=f(x)+f(y) \text.$$ Prove that: $$\exists a\in \mathbb R , \forall x \in \mathbb R, \ f(x)=ax \text.$$

@OP, let me give you the following hint. Could you prove $f(\frac{m}{m}) = a\frac{m}{n}$?, namely for $x\in \mathbb{Q}$, $f(x)=ax$ — felasfa, Oct 12 '16 at 21:25

score 0 · Answer 1 · answered Oct 12 '16 at 21:30

It is the Cauchy functional equation. The question has appeared many times before on MSE.

As mentioned in wiki using the Axiom of Choice you may construct solutions violating the linear relation you stipulate. If continuous it follows from considerations like: $ f(nx)=f(x)+f((n-1)x)=2f(x)+f((n-2)x)=\cdots=nf(x)$ (proof by induction, $n\geq 2$).

In particular: $f(0)=0$, $f(-x)=-f(x)$ and $ qf(p/q)=f(p)=pf(1)$ so that $f(r)=r f(1)$ for all rationals.

Continuity then implies $f(x)=x f(1)$ for all real $x$.

If $f:\Bbb R\to\Bbb R$ is continuous and $\forall x,y\in\Bbb R,\ f(x+y)=f(x)+f(y)$, then $\exists a\in\Bbb R,\forall x\in\Bbb R,\ f(x)=ax$.

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