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Let $f: \mathbb{R} \to \mathbb{R} $ be a continuous function with $f(x+y) = f(x) +f(y),\forall x,y\in \mathbb{R}.$

Find $\frac {df} {dx} $, if it is exist?

Martin Sleziak
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Jbrail Fars Snjawy
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    I am not sure this question should have been closed as duplicate. Looking for all differentiable solutions of Cauchy functional equation is easier task than looking for all continuous solutions. – Martin Sleziak Jan 08 '15 at 10:04

2 Answers2

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This is called Cauchy's functional equation. You can show that $f(0)=0$, and if you let $c=f(1)$ then you can deduce that $$f(2)=f(1+1)=f(1)+f(1)=2c\qquad \text{ and } \qquad f\left(\frac{1}{2}\right)=\frac{f(1)}{2}= \frac{c}{2} $$

This extends similarly to

$$f(x)=cx$$ for all rational $x$, but because $f$ is continuous, $f(x)=cx$ for all real $x$. You should be able to show that the derivative of this linear function exists.

Zubin Mukerjee
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  • please prove it completely because it is very important for me – Jbrail Fars Snjawy Jan 08 '15 at 08:59
  • I've intentionally left a few gaps for you to fill in; the point is that you'll gain more by fleshing it out yourself. If you have any specific troubles with any of the steps in the proof, feel free to ask :) – Zubin Mukerjee Jan 08 '15 at 09:03
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If a function $f$ fulfills $f(x+y)=f(x)+f(y)$ for any $x,y\in\mathbb R$, then by differentiating by $x$ you get $$f'(x+y)=f'(x).$$ (Here we use the assumption that $f$ is differentiable.) This equality is true, again, for any $x,y\in\mathbb R$. So we see that $f'(x)$ is constant.

Let $f'(x)=a$. Then we get $f(x)=ax+b$.

It is relatively easy to show that $f(0)=0$, from which we get $b=0$.

Therefore $f(x)=ax$ for some $a\in\mathbb R$. We get that $f'(x)=a=f(1)$.

It is worth mentioning that the equation $f(x+y)=f(x)+f(y)$ is called Cauchy's functional equation. Solutions of this equation are called additive functions.

We saw that if $f$ is differentiable, then $f(x)=ax$. In fact, we get only solutions of this form even with much weaker conditions on $f$. But if we do not have any restrictions on $f$, then there are also different solutions. See Overview of basic facts about Cauchy functional equation.

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Martin Sleziak
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  • İs this proof is complete ? because İ have take home exam . – Jbrail Fars Snjawy Jan 08 '15 at 10:00
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    Well, if you have take-home-exam, then you are supposed to do the work. (What is policy of your teacher on using internet and other materials?) If you found a solution in an answer here on MSE or elsewhere, the least you should do is to check the solution and fill in the gaps if needed. In my opinion, the question whether the solution is complete is somewhat subjective. If there are steps you do not understand, you should try to think about them and fill all the details. – Martin Sleziak Jan 08 '15 at 10:03
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    @JbrailFarsSnjawy I should add this: I have understood your question as follows: Assume that $f'(x)$ exists for each $x$ and that $f(x+y)=f(x)+f(y)$ for each $x$ and $y$. Then find the derivative. If you're question is supposed to ask about the derivative under the assumption that the derivative $f'(x)$ exists only in one point $x$, then it is more difficult. In such case, you should look at the other answer or at the answers in the linked threads. – Martin Sleziak Jan 08 '15 at 14:20