Suppose that $f:R\to R$ is a function such that $f(x+y)=f(x)+f(y)$ for all $x,y∈R$.

Assume that $f$ has a limit at $0$, $f(1)=1$. Prove that $f(x)=x$ for all $x \in R$

Hint: Show first that $f$ is continuous at any point $c ∈ R$. Prove then that for any rational number $r$, $f(r) = r$, and deduce the statement using the continuity of $f$.

My attempt:

To show continuity at any point $c\in R$

$$\lim_{x \to c}f(x)=\lim_{ h\to0}f(c+h)=\lim_{h\to0}f(h)+f(c)=f(c)$$

Suppose $\lim_{x\to0}f(x)$ exists (as given). We need to show that for $c∈R$ arbitrary, also $\lim_{x\to c}f(x)$ exists.

We have $$f(x)=f(x−c+c)=f(x−c)+f(c)$$

Letting $x→c$ and setting $y=x−c$ we see that $y→0$, so the limit in question exists by assumption and equals


which implies $f(x)=x$ QED

My question is how to use $f(1)=1$

Then how to prove continuity for any rational number?

  • 1
    $\begingroup$ I don't quite understand the part where you write the limit and then 'which implies $f(x)=x$. As far as I can tell, or at least what the 'usual' approach would be, is to show continuity, then prove by induction $f(n)=n \forall n\in\mathbb{N}$, do the same for $\mathbb{Q}$, and since $\mathbb{Q}$ is dense in $\mathbb{R}$ we are done. But you seem to have somehow skipped that which is great if it works - if it works. $\endgroup$ – Some Math Student Oct 28 '14 at 21:26
  • $\begingroup$ Just to make sure I get your points: Induction: the base case f(1)=1, then assume f(n)=n, f(n+1)=f(n)+f(1)=n+1? $\endgroup$ – Linh Phan Oct 29 '14 at 1:34
  • $\begingroup$ Can you be more specific about proof for Q? $\endgroup$ – Linh Phan Oct 29 '14 at 1:36

The posted question assumes $\lim_{x\to0}f(x)=0$, but that was not given. What was given is that $\lim_{x\to0}f(x)$ exists. That the limit is $0$ must be proved.

The functional equation $f(x+y)=f(x)+f(y)$ entails that $f(0)$ $=f(x+(-x))$ $=f(x)+f(-x)$; hence we must have $f(-x)=-f(x)$, i.e. this is an odd function. If an odd function $f$ has a limit $L$ at $0$, then $$ \lim_{x\downarrow0}f(x) =\lim_{x\to0} f(x) = \lim_{x\uparrow0} f(x), $$ but the last one is $$ \lim_{x\uparrow0} f(x) = \lim_{x\downarrow0} f(-x) = -\lim_{x\downarrow0} f(x) = -L. $$ So $L=-L$, and therefore $L=0$.

As to using $f(1)=1$, notice that the function $g(x)=3x$ also satisfies the equation $g(x+y)=g(x)+g(y)$ (and similarly if we had used any other number besides $3$), so certainly that condition must be used.

What the two equalities $f(x+y)=f(x)+f(y)$ and $f(1)=1$ imply is that $f(x)=x$ whenever $x$ is rational. Let's see that first for positive rational numbers.

\begin{align} 1 =f(1) & = f\left(\frac15+\frac15+\frac15+\frac15+\frac15\right) \\[10pt] & = f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) = 5f\left(\frac 15 \right) \end{align} so $f\left(\dfrac15\right)= \dfrac15$. And $$ f\left(\frac35\right)=f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) = \frac15+\frac15+\frac15 = \frac 3 5 $$ and similarly for all other positive rational numbers.

If $x$ is irrational, look at a sequence of rational numbers approaching $x$ and then use continuity.

  • $\begingroup$ Regarding the use of f(1), why do you have to split f(1) in to 5* (1/5). Do you mean that we can always find x, y st x+y=1 and hence, x,y $\in Q$ $\endgroup$ – Linh Phan Oct 29 '14 at 1:41
  • $\begingroup$ How would you formulate a general proof for ANY rational number? $\endgroup$ – Linh Phan Oct 29 '14 at 1:45
  • $\begingroup$ @LinhPhan : I split it up that way in order to show that $f(1/5)=1/5$. Then I used that to show that $f(3/5)=3/5$. The same thing works with every positive rational number. For $m/n$, first show that $f(1/n)=1/n$ by writing $1=f(1)=f(1/n+\cdots+1/n)=f(1/n)+\cdots+f(1/n)=nf(1/n)$, and then divide both sides by $n$ to get $1/n=f(1/n)$. Then you can write $f(m/n)=f(\,\underbrace{1/n+\cdots+1/n}_{m\text{ terms}}\,) = \underbrace{f(1/n)+\cdots+f(1/n)}_{m\text{ terms}} = mf(1/n) = m\cdot(1/n) = m/n$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 29 '14 at 1:50
  • $\begingroup$ Do you have to prove a separate case for the negative ie m/n<0 since you want f(x)=x $\in R$ which means x is possibly negative? $\endgroup$ – Linh Phan Oct 29 '14 at 2:14
  • $\begingroup$ Once you've done it for positive numbers, you can just rely on the fact that it's an odd function to show that it works for negative numbers. $\endgroup$ – Michael Hardy Oct 29 '14 at 2:21

We see that $f$ is an automorphism on the group of real numbers with addition.

It will suffice to show that only the identity function will be the only continuous automorphism that leaves $f(1) = 1$. If we didn't assume $f(1)=1$, we would have a lot more potential automorphisms.

  • $\begingroup$ There are other automorphisms that map $1$ to itself. But none of the others are continuous (except at $0$). $\endgroup$ – Michael Hardy Oct 29 '14 at 2:22
  • $\begingroup$ Ah, yes a nice subtlety in the premise that $f$ has a limit at 0. But thinking of $f$ as an automorphism makes the proof that $f(0) = 0$ much simpler :) $\endgroup$ – Jonny Oct 29 '14 at 2:37
  • $\begingroup$ The proof that $f(0)=0$ is trivial regardless of whether you know the word "automorphism" or not. And you don't need continuity to do that. $\endgroup$ – Michael Hardy Oct 29 '14 at 2:39
  • $\begingroup$ Oh: I should add: None of the others are continuous anywhere. All are everywhere discontinuous, even at $0$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 29 '14 at 2:40
  • $\begingroup$ So together with $f(0) = 0$ and the limit at that point also being 0 we have that $f$ is continuous somewhere, leaving only the identity automorphism. $\endgroup$ – Jonny Oct 29 '14 at 2:43
  1. You only proved that $f$ is continuous at every point yet, that is $$\lim_{x\to c}f(x)=f(c)$$ at every $c\in\Bbb R$. (So, your first line was enough here.)
  2. First prove that $f(1/n)=1/n$ for all $n\in\Bbb N$. Here you will have to use $f(1)=1$.
  3. Then it follows that $f(r)=r$ for all $r\in\Bbb Q$.
  4. Finally use continuity to prove $f(x)=x$ for all $x\in\Bbb R$.
  • $\begingroup$ $-1$ because the proof given in the original question is incorrect. It just assumed the limit at $0$ is $0$, but that was not given and needs to be proved. $\endgroup$ – Michael Hardy Oct 28 '14 at 21:48

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