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

Question

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.

Answer 1

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.

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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.

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Answer 2

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!