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This question already has an answer here:

Let $f$ a continuous function defined on $\mathbb R$ such that $\forall x,y \in \mathbb R :f(x+y)=f(x)+f(y)$

Prove that : $$\exists a\in \mathbb R , \forall x \in \mathbb R, f(x)=ax$$

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marked as duplicate by Adam Hughes, Magdiragdag, Ennar, davidlowryduda Oct 12 '16 at 21:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ You need to make a further assumption on $f$. Like $f$ continuous (or measurable) in order to reach the stated conclusion. $\endgroup$ – H. H. Rugh Oct 12 '16 at 21:17
  • $\begingroup$ Not true. For more, see mit.edu/~evanchen/handouts/Monsters/Monsters.pdf $\endgroup$ – jlammy Oct 12 '16 at 21:18
  • $\begingroup$ No i think it's so true $\endgroup$ – user315918 Oct 12 '16 at 21:21
  • $\begingroup$ @jlammy, you need $f$ to be continuous. I can't see how one can prove it otherwise. $\endgroup$ – felasfa Oct 12 '16 at 21:24
  • $\begingroup$ Sorry i've edited the question $\endgroup$ – user315918 Oct 12 '16 at 21:25
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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 $x$.

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