Let $\,f:[0,\infty)\to [0,\infty)$ be a function such that $\,f(x+y)=f(x)+f(y),\,$ for all $\,x,y\ge 0$. Prove that $\,f(x)=ax,\,$ for some constant $a$.

My proof :

We have , $\,f(0)=0$. Then , $$\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(h)}{h}=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=f'(0)=a\text{(constant)}.$$

Then, $\,f(x)=ax+b$. As, $\,f(0)=0$ so $b=0$ and $f(x)=ax.$

Is my proof correct?

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    $\begingroup$ It is not correct unless differentiability be an hypothesis $\endgroup$
    – sinbadh
    Feb 15, 2016 at 9:34
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    $\begingroup$ It's also true if $f:\mathbb{Q}\to\mathbb{Q}$, without assuming differentiability $\endgroup$
    – user304329
    Feb 15, 2016 at 9:35
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    $\begingroup$ Assuming AC, without giving that $\;f\;$ is continuous the claim is false. $\endgroup$
    – DonAntonio
    Feb 15, 2016 at 9:42
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    $\begingroup$ Please, note that the question is not about a function defined over $\mathbb{R}$, but a function defined on $[0,\infty)$ with values in $[0,\infty)$, which makes this a different problem from Cauchy functional equation. $\endgroup$
    – egreg
    Feb 15, 2016 at 9:54
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    $\begingroup$ I have added to the title domain and codomain, to stress the difference pointed out by @egreg. (And I have also retraced my close vote. I should have read the question more carefully.) $\endgroup$ Feb 15, 2016 at 11:34

4 Answers 4


In your proof you assume that $f$ is differentiable, which is not given.

Let me suggest how to obtain the formula of $f$:

Step I. Show that $\,f(px)=p\,f(x),\,$ when $p$ is a positive rational and $x$ a non-negative real. (At first show this for $p$ integer.) We obtain also that, $\,f(0)=0$.

Step II. Observe that $f$ is increasing, since, for $y>x$, we have $$ f(y)=f(x)+f(y-x)\ge f(x). $$

Step III. Since $f$ is increasing, then the limit $\,\lim_{x\to 0^+}f(x)\,$ exists. However $$ \lim_{x\to 0^+}f(x)=\lim_{n\to\infty}f\Big(\frac{1}{n}\Big) =\lim_{n\to\infty}\frac{1}{n}\,f(1)=0. $$

Step IV. Pick an arbitrary $x\in(0,\infty)$, and a decreasing sequence $\{q_n\}\subset\mathbb Q$ tending to $x$. Then $$ f(q_n)=q_n\,f(1) $$ and $$ x\,f(1)\longleftarrow q_n\,f(1)=f(q_n)=f(x)+f(q_n-x)\longrightarrow f(x), $$ since $\,\,q_n-x\to 0^+$, and thus $\,\,\lim_{n\to\infty}f(q_n-x)=0$.

Therefore, $\,f(x)=x\,f(1),\,$ for all $x\in\mathbb [0,\infty)$, and hence $\,f'(x)=f(1)$.

  • $\begingroup$ Why $\;q_nf(0)=f(x)+f(q_n-x)=f(q_n)\;$ ? Shouldn't it be $\;f(q_n)=q_nf(1)\;$ ? Also, why $\;q_n-x\,\to\, \implies\,f(q_n-x)\,\to f(0)\;?$ Isn't this assuming continuity? $\endgroup$
    – DonAntonio
    Feb 15, 2016 at 10:27
  • $\begingroup$ You are absolutely right! I am correcting it! $\endgroup$ Feb 15, 2016 at 10:28
  • $\begingroup$ For Step III: by Monotone Limit Theorem, $\ell=\lim\limits_{x\to0^+}f(x)$ exists; by composition of limits: $\ell=\lim\limits_{n\to+\infty}f(1/n)=\lim\limits_{n\to+\infty}f(1)/n$ (by Step I) hence $\ell=0$. I guess you don't need to use a reasoning by contradiction. $\endgroup$ Feb 15, 2016 at 13:44
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    $\begingroup$ @gniourf_gniourf I simplified my answer according to your comment. Thanx! $\endgroup$ Feb 15, 2016 at 15:24

Note that $f$ is monotonic: if $y>0$, then $f(y)\ge0$ so $f(x+y)=f(x)+f(y)\ge f(x)$.

In particular, the function is continuous over $[0,\infty)$ except for an at most countable set.

You can extend $f$ to $f_e\colon\mathbb{R}\to\mathbb{R}$ by setting $f_e(x)=-f(-x)$, for $x<0$. Show that this function still has the property that $f_e(x+y)=f_e(x)+f_e(y)$. Then continuity at a point implies continuity at $0$.

The result now follows from methods in Overview of basic facts about Cauchy functional equation

Your attempt is not good, I'm afraid: you're assuming differentiability at $0$, which is not among the hypotheses.


By induction, $f(nx)=nf(x)$ for an integer $n$.

Now take any real $x$. From

$$\lfloor nx\rfloor\le nx\le\lceil nx\rceil,$$applying the non-decreasing function $f$, we deduce $$f\left(\lfloor nx\rfloor\right)\le f(nx)\le f\left(\lceil nx\rceil\right).$$

By the above induction property, $$\lfloor nx\rfloor f(1)\le nf(x)\le \lceil nx\rceil f(1),$$ and $$\frac{\lfloor nx\rfloor}nf(1)\le f(x)\le \frac{\lceil nx\rceil}nf(1).$$

As $n$ can be arbitrarily large, by squeezing


  • $\begingroup$ I don't think $\;f\;$ must be continuous: choosing a Hamel basis for linear space $\;\Bbb R_{\Bbb Q}\;$ any linear operator will do the trick, and not all are of the wanted form. $\endgroup$
    – DonAntonio
    Feb 15, 2016 at 11:07
  • $\begingroup$ @Joanpemo Monotone functions can only have at most countably many discontinuities in an interval, so $f$ is continuous somewhere, hence everywhere. $\endgroup$
    – Ningxin
    Feb 15, 2016 at 11:13
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    $\begingroup$ Thank you@QiyuWen . Apparently the domain and codomain being the positive reals make the trick for monotony. $\endgroup$
    – DonAntonio
    Feb 15, 2016 at 11:17
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    $\begingroup$ @Joanpemo: I have completely rewritten, with a much simpler argument. $\endgroup$
    – user65203
    Feb 15, 2016 at 11:50
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    $\begingroup$ Only answer not needing a sledgehammer to crack a nut. $\endgroup$
    – Bananach
    Feb 22, 2016 at 8:48

For each $\,m,n\geqslant1\,,$

$f(m)=f\left(n\!\cdot\!\dfrac mn\right)=f\left(\dfrac mn+\dfrac mn+\ldots+\dfrac mn\right)=nf\left(\dfrac mn\right).$

So $\,f\left(\dfrac mn\right)=\dfrac{f(m)}n\,.\;$

Next $\,f(m)=f(1+1+···+1)=mf(1)=ma\,.$

So $\,f\left(\dfrac mn\right)=a\!\cdot\!\dfrac mn\;.$

Also $\,f\left(−\dfrac mn\right)=−f\left(\dfrac mn\right)=a\left(−\dfrac mn\right).$

It follows that for each nonzero $\,x\in\mathbb Q\,,\,f(x)=ax\;$ since $f(0) = 0\,.$


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