Let $f:\mathbb R \longrightarrow \mathbb R$ be continuous which is also an additive homomorphism, that is, $f( x+ y)= f( x )+f(y) $ for all $x,y\in \mathbb R$ then $f( x)= \lambda x$ where $\lambda= f(1)$

I tried like this let $x\in N$

$f(x)=f(\underbrace{1+\cdots+1}_{x \text{ times }})= \underbrace{ f(1)+\cdots+f(1)}_{x \text{ times}} =xf(1)=\lambda x $

Say $\lambda =f(1)$. But it is the case where $x$ is natural number what to do for $x$ real.

  • 2
    $\begingroup$ You can kinda go on like this, showing the statement für $x\in\mathbb{Z}, x\in\mathbb{Q}$ and finally $x\in\mathbb{R}$. The case $x\in\mathbb{Z}$ is done, when you sperate the cases $x\geq 0$ and $x<0$, which is the same as for $x\in\mathbb{N}$. Now take $x\in\mathbb{Q}$. Hence $x=\frac{a}{b}$ with a\in\mathbb{Z}, and $b\in\mathbb{N}\setminus\{0\}$. Then bf(q)=f(bq)=f(a)=a\lambda, hence $f(q)=\frac{a}{b}\lambda$ after division by b. Use that f is continuous and take rational cauchy-sequences with irrational limit and you are done. $\endgroup$
    – Cornman
    Jul 29, 2017 at 5:30
  • $\begingroup$ What is q in that? Oh is it $\frac ab$ ? Yes yes I got it :) thanks $\endgroup$ Jul 29, 2017 at 5:36
  • 1
    $\begingroup$ You can use \mathbb R to obtain $\mathbb R$ and \longrightarrow or \to for $\longrightarrow$ or $\to$. $\endgroup$
    – Pedro
    Jul 29, 2017 at 5:40
  • $\begingroup$ Oh, excuse me, with $q$ i meant $x$. So $bf(x)=f(bx)=f(a)=a\lambda$ and then $f(x)=\frac{a}{b}\lambda$ after division of $b$. $\endgroup$
    – Cornman
    Jul 29, 2017 at 5:40
  • $\begingroup$ In your last equal sign, you can say that, since $\mathbb{R}$ is a field, every non-zero element has an inverse, assuming $x \not = 0$, you can multiply both sides with $x^{-1}$, and you get $f(1) = \lambda$. $\endgroup$
    – Our
    Jul 29, 2017 at 5:56

2 Answers 2


Standard proof:

  • Let $n$ be a positive integer. Then by additivity, $$f(n) = \underbrace{f(1) + \cdots +f(1)}_{n\text{ times}} = n\cdot f(1)$$, which establishes the proof for positive integers.
  • Let $q$ be a positive integer. Then $f(1/q) = \frac{1}{q}f(1).$ To see this, note that $f(1) = f(q\cdot 1/q) = q f(1/q)$ by our previous result with positive integers.
  • Combining these two results, we have that $f(p/q) = \frac{p}{q}f(1)$ whenever $p$ and $q$ are positive integers so that $p/q$ is a positive rational number.
  • We can extend this result to all rational numbers by proving that $f$ is odd: $f(x) = -f(-x)$. To see that $f$ is odd, note that $f(0)$ must be zero because $f(0) = f(0+0) = f(0)+f(0) = 2f(0)$. Then, using additivity, it follows that for any $x$, $0 = f(0) = f(x + (-x)) = f(x) + f(-x)$.
  • Now we know that $f(r) = r\cdot f(1)$ whenever $r$ is any rational number.
  • Finally, we extend this result to all real numbers. Let $x$ be any real number. Then there is a sequence $r_n$ of rational numbers that converges to $x$. Because $f$ is continuous, $f(r_n)$ must converge to $f(x)$. But $f(r_n) = r_n f(1)$ as we have shown, and this sequence evidently converges to $x \cdot f(1)$. Q.E.D.

Here's an alternative proof that shows how this condition affects the derivative of $f$. In particular, we can prove that the derivative of $f$ must be constant—namely, equal to $f(1)$.

  1. Let $n$ be any positive integer. Then $f(1/n) = \frac{1}{n}f(1)$. To see this, note that $f(1) = f(n/n) = \underbrace{f(1/n)+f(1/n)+\cdots + f(1/n)}_{n\text{ times}} = nf(1/n)$. Then divide the first and last terms by $n$.
  2. Suppose the limit $L \equiv \lim_{\epsilon\rightarrow 0}\frac{f(\epsilon)}{\epsilon}$ exists. (We'll prove that it does later.)
  3. The derivative of $f$ exists and is constantly equal to $L$. Indeed, for any $x$, $$\begin{align*}f^\prime(x) &= \lim_{\epsilon\rightarrow 0}\frac{f(x+\epsilon)-f(x)}{\epsilon}\\&= \lim_{\epsilon\rightarrow 0} \frac{f(x) + f(\epsilon) - f(x)}{\epsilon}&\{\text{additivity}\}\\&= \lim_{\epsilon\rightarrow 0} \frac{f(\epsilon)}{\epsilon}\\&= L.\end{align*}$$
  4. Because the derivative of $f$ is constant, we must have by integration that $f(x) = Lx + C$ for some constant $C$. Because $f$ is additive, $C$ must be zero. (Because, for example, $f(2) = 2L+C$ whereas $f(1)+f(1) = 2(L+C)=2L+2C$.) Hence $f(x) = L\cdot x$.
  5. Now we show that the limit $L$ exists. Let $r_n\rightarrow 0$ be a vanishing sequence of rational numbers. We can write the limit $\lim_{\epsilon\rightarrow 0} \frac{f(\epsilon)}{\epsilon}$ as the limit of a sequence $\lim_{n\rightarrow \infty}\frac{f(r_n)}{r_n} = \lim_{n\rightarrow \infty}\frac{f(p_n/q_n)}{p_n/q_n}$. We know from additivity and our first observation above that we can rewrite $f(p_n/q_n) = \frac{p_n}{q_n} f(1)$, yielding $L = \lim_{n\rightarrow \infty}\frac{\frac{p_n}{q_n}f(1)}{p_n/q_n} = \lim_{n\rightarrow\infty}f(1)$. Hence it's the limit of a constant, whose value is $L=f(1)$. Because $f$ is continuous, this result holds for all sequences, not just sequences of rationals, so the limit holds. Q.E.D.

  • $\begingroup$ But where did you prove the limit $L$ exists? $\endgroup$
    – Pedro
    Jul 29, 2017 at 7:50
  • $\begingroup$ @PedroTamaroff Step 5. I've rephrased slightly to clarify. $\endgroup$
    – user326210
    Jul 29, 2017 at 8:03
  • $\begingroup$ You still don't prove the limit exists in that step, only that if it exists, it must be $f(1)$. It is not true that the limit $\lim_{t\to 0}f(t)/t$ exists if it exists for a sequence $t_n\to 0$. $\endgroup$
    – Pedro
    Jul 29, 2017 at 8:11
  • $\begingroup$ I like the standard proof ..the one without actually finding the function ...since it is easy for me...but the other one is interesting... $\endgroup$ Jul 30, 2017 at 18:42

I wrote this one up last night but then fell asleep before I posted; when I awoke, I found the answer of user 326210 and the engaging dialog 'twixt himself and Pedro Tamaroff, but decided to throw my US \$0.02 into the hat anyway. You can think of it as sharing my lecture notes . . .

Well, first off,

$f(0) = f(0 + 0) = f(0) + f(0), \tag{1}$


$f(0) = 0; \tag{2}$


$f(a) + f(-a) = f(a + (-a)) = f(0) = 0, \tag{3}$


$f(-a) = -f(a); \tag{4}$

(4) holds for all $a \in \Bbb R$; now for $n \in \Bbb N$,

$f(n) = nf(1), \tag{5}$

as a simple induction proves: if

$f(k) = kf(1), \tag{6}$


$f(k + 1) = f(k) + f(1) = kf(1) + f(1) = (k + 1)f(1); \tag{7}$

thus (5) binds. Using (2), (4), and (5), we conclude

$f(m) = mf(1) \tag{8}$

for all $m \in \Bbb Z$. So we have (8) for all integers. We move outward in scope and extend the relation

$f(r) = rf(1) \tag{9}$

to all $r \in \Bbb Q$, the rationals. Writing

$r = \dfrac{p}{q}, \tag{10}$

where $p, q \in \Bbb Z$, we have

$p = q \dfrac{p}{q} = qr; \tag{11}$


$f(qr) = f(p) = pf(1); \tag{12}$

we now observe that essentially the same inductive argument which proves (5), (8) may easily be extended to show that

$f(nx) = nf(x) \tag{13}$

where $n \in \Bbb Z$, $x \in \Bbb R$, viz.

$f(x) = f(x), \tag{14}$

$f(2x) = f(x + x) = f(x) + f(x) = 2f(x); \tag{15}$

and assuming

$f(kx) = kf(x), \tag{16}$

we have

$f((k + 1)x) = f(kx + x) = f(kx) + f(x) = kf(x) + f(x) = (k + 1)f(x); \tag{17}$

we now apply (13) to (12):

$qf(r) = f(qr) = pf(1), \tag{18}$


$f(r) = \dfrac{p}{q}f(1) = rf(1) \tag{19}$

holds for all rationals $r \in \Bbb Q$.

Now for real $t \in \Bbb R \setminus \Bbb Q$, we take a sequence $t_i \in \Bbb Q$ with

$\lim_{i \to \infty} t_i = t; \tag{20}$

then since $f$ is continuous by hypothesis,

$f(t) = \lim_{i \to \infty} f(t_i) = \lim_{i \to \infty} (t_i f(1)) = (\lim_{i \to \infty} t_i)f(1) = tf(1), \tag{21}$

and we are done!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.