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A question from Introduction to Analysis by Arthur Mattuck:

Suppose $f(x)$ is continuous for all $x$ and $f(a+b)=f(a)+f(b)$ for all $a$ and $b$. Prove that $f(x)=Cx$, where $C=f(1)$, as follows:

(a)prove, in order, that it is true when $x=n, {1\over n}$ and $m\over n$, where $m, n$ are integers, $n\ne 0$;

(b)use the continuity of $f$ to show it is true for all $x$.

I can show the statement is true when $x=n$. As for $x={1\over n},{m\over n}$, I don't know how.


marked as duplicate by jgon, Ross Millikan, clark, Guy Fsone, Did real-analysis Jan 2 '18 at 13:39

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.


$f(1/2)+f(1/2)=f(1)=C$: what is $f(1/2)$?

$f(1/3)+f(1/3)+f(1/3)=f(2/3)+f(1/3)=f(1)=C$: what is $f(1/3)$?


  • $\begingroup$ Could you also please mention how we get to real numbers from the above? $\endgroup$ – user1952500 Jan 2 '18 at 6:34
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    $\begingroup$ @user1952500 Continuity. $\endgroup$ – Lord Shark the Unknown Jan 2 '18 at 6:36
  • $\begingroup$ I have not done much of a course in analysis. I understand that continuity implies that the reals must be satisfied but not able to prove it mentally. If there is a $f(x)$ where x is irrational, do we try to get an approximate rational upper and lower bound and show that the difference in bounds tends to $0$? Sorry if these are basic facts. $\endgroup$ – user1952500 Jan 2 '18 at 6:42
  • $\begingroup$ I have a question:For the case $x = 1/n$, the rule is $f(1/n+1/m)=f(1/n)+f(1/m)$. $f(1/3)+f(1/3)=f(2/3)$ is true, but $f(2/3)+f(1/3)=f(1)$ is not defined? $\endgroup$ – Hongyan Jan 2 '18 at 7:13
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    $\begingroup$ @user1952500 every real number $a$ is the limit of a sequence of rational numbers $(a_n)_{n\in\Bbb N}$. So by continuity $\lim_{n\to\infty}f(a_n)=f(a)$, because $a_n$ is rational we can do the following: $\lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}a_nC=aC$. This is true for all real numbers. $\endgroup$ – Holo Jan 2 '18 at 8:14

For (a):

start with $x=\frac1n$:

$f(1)=f(\sum_{i=1}^n x)=\sum_{i=1}^n f(x)=nf(x)=C\implies f(x)=\frac Cn$

Now do similar thing to $y=\frac mn$:

$f(y)=f(mx)=f(\sum_{i=1}^m x)=\sum_{i=1}^m f(x)=\sum_{i=1}^m \frac Cn=\frac{Cm}n$

For (b) use the fact that all real numbers are the limit of some rational numbers sequence

  • $\begingroup$ That also doesn’t prove for all right? $\endgroup$ – QuIcKmAtHs Jan 2 '18 at 6:29
  • $\begingroup$ @XcoderX right, the problem I had with your answer is that you wrote "always" at the end, except this your way is good, OP didn't asked for all in (a) $\endgroup$ – Holo Jan 2 '18 at 6:31
  • $\begingroup$ Okay @Holo I have edit that already $\endgroup$ – QuIcKmAtHs Jan 2 '18 at 6:31
  • $\begingroup$ @XcoderX okay, I delete my comment $\endgroup$ – Holo Jan 2 '18 at 6:32
  • $\begingroup$ From $f(1/n+1/m)=f(1/n)+f(1/m)$, how do you get $f(1)=nf(1/n)$? Because there are only two numbers:$1/n$ and $1/m$. $\endgroup$ – Hongyan Jan 2 '18 at 8:29

This is true for all: $f(x)=Cx$. For instance, for $f({1\over n})$, you would need $n$ of that term to make C. Hence, it is ${C\over n}$, which is also $C*{1\over n}$. For $f({m\over n})$, it is ${C*m\over n}$. Firstly, ${m\over n}$ is actually just $m*{1\over n}$. Since $f({1\over n})$ = ${C\over n}$, $f({m\over n})$ would be $m$ times of ${C\over n}$, so it would be ${C*m\over n}$, which can be expressed as $C*{m\over n}$. Hence, we have proven that for any $f(x)$, it is always equal to $Cx$(at least for rational numbers).


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