# Let $f:\mathbb R \longrightarrow \mathbb R$ be continuous which is also an additive homomorphism

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.

• 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. Jul 29, 2017 at 5:30
• What is q in that? Oh is it $\frac ab$ ? Yes yes I got it :) thanks Jul 29, 2017 at 5:36
• You can use \mathbb R to obtain $\mathbb R$ and \longrightarrow or \to for $\longrightarrow$ or $\to$.
– Pedro
Jul 29, 2017 at 5:40
• 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 $b$. Jul 29, 2017 at 5:40
• 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 $f(1) = \lambda$.
– Our
Jul 29, 2017 at 5:56

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.

• But where did you prove the limit $L$ exists?
– Pedro
Jul 29, 2017 at 7:50
• @PedroTamaroff Step 5. I've rephrased slightly to clarify. Jul 29, 2017 at 8:03
• 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 $t_n\to 0$.
– Pedro
Jul 29, 2017 at 8:11
• I like the standard proof ..the one without actually finding the function ...since it is easy for me...but the other one is interesting... 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,$f(0) = f(0 + 0) = f(0) + f(0), \tag{1}$so$f(0) = 0; \tag{2}$also,$f(a) + f(-a) = f(a + (-a)) = f(0) = 0, \tag{3}$so$f(-a) = -f(a); \tag{4}$(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$f(k) = kf(1), \tag{6}$then$f(k + 1) = f(k) + f(1) = kf(1) + f(1) = (k + 1)f(1); \tag{7}$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$p = q \dfrac{p}{q} = qr; \tag{11}$thus,$f(qr) = f(p) = pf(1); \tag{12}$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):$qf(r) = f(qr) = pf(1), \tag{18}$whence$f(r) = \dfrac{p}{q}f(1) = rf(1) \tag{19}$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!