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This question already has an answer here:

Let $f$ be a real-valued function for which, for every real $x,y$:

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

Does this imply that $f$ is a linear function ($f(x)=a\cdot x$)?

  1. If $f$ is differentiable, I think the answer is yes. Take the derivative of $f(x+y)$ by $x$: $f'(x+y)=f'(x)$. This is true for all $y$, hence $f'$ is a constant function. Additionally, $f(0)=f(x+0)-f(x)=0$. Hence $f(x)=ax$. (is this correct?)

  2. If $f$ is continuous, I think the answer is also yes. First, note that $f(0)=0$. Let $a=f(1)$. By addition, for every integer $x$, $f(x)=a\cdot x$. This also must be true for every rational $x$. Hence, by continuity, it must be true for all $x$. (is this correct?)

  3. What is the answer for general $f$?

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marked as duplicate by Martin Sleziak, Community Aug 31 '15 at 14:58

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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It is quite easy to prove for a continuous function, as you pointed out, and in fact also holds true for measurable functions. This is the most general setting.

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  • $\begingroup$ You mean Lebesgue measurable. $\endgroup$ – Umberto P. Aug 31 '15 at 13:59
  • $\begingroup$ Yes, I mean Lebesgue measurable, thank you :) $\endgroup$ – uniquesolution Aug 31 '15 at 13:59

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