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The following problem is from Golan's linear algebra book. I have posted a solution in the comments.

Problem: Let $f(x):\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function satisfying $f(x+y)=f(x)+f(y)$ for all $x,y\in \mathbb{R}$. Show $f$ is a linear transformation.

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  • $\begingroup$ See also this post. And also here, here and here. $\endgroup$ – Martin Sleziak Dec 21 '15 at 16:17
  • $\begingroup$ Let $T\colon \mathbb{R}^m\to \mathbb{R}^n$, and consider the following properties. (i) $T$ is continuous (ii) $T(cv)=cT(v)$ for any $c\in \mathbb{R}$, $v\in\mathbb{R}^m$ (iii) $T(v+w)=T(v)+T(w)$ for any $v,w\in\mathbb{R}^m$. This question shows (i) and (iii) imply (ii). It's easy to show (ii) and (iii) implies (i). My question: do (i) and (ii) imply (iii)? $\endgroup$ – math.nb Jan 29 '16 at 21:06
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The only property of linear transformations that we still need to verify is that $f(xt)=tf(x)$ for all $x,y\in \mathbb{R}.$ It is enough to establish this result just for rational numbers. If $j$ is irrational and $x\in \mathbb{R}$,, we can find a rational $r$ with $|jx-rx|<\delta$ for any positive real $\delta$. By continuity, for every $\epsilon>0$, we can choose $\delta$ so that $|f(rx)-f(jx)|<\epsilon$. This condition also gives $|r-j|<\delta/|j|$, and choosing $\delta$ to be even smaller if necessary gives $|f(x)-f(r)|<\epsilon$, too. Putting this all together gives

$|jf(x)-f(jx)|<|j|f(x)-f(r)| + |f(jx)-f(rx)|<(|j|+1)\epsilon$

and we can make this arbitrarily small, giving the desired result.

To verify the property for rationals, we first verify it for integers. If $n\in \mathbb{N}$, then

$nf(x)=f(x)+f(x)+...+f(x)=f(nx)$

by hypothesis. Also,

$f(x)=f(x/n)+f(x/n)+\cdots f(x/n)=nf(x/n)$

so $\frac{1}{n}f(x)=f(\frac{x}{n})$. Combining the above shows we have scalar multiplication for all positive rationals.

Noting that $f(0)=f(0)+f(0)$ gives $f(0)=0$, and

$f(0)=f(-x)+f(x)\Rightarrow -f(-x)=f(x)$. Using this allows us to extend scalar multiplication to negative rationals and completes the proof.

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    $\begingroup$ It's fine, but the first part with the irrational is unduly long and cumbersome, imo. Just need to take into account that there's a rational sequence that converges to that irrational and continuity of $f$... $\endgroup$ – DonAntonio Jun 1 '12 at 22:26
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Perhaps a clearer answer is ...

Let $f$a addictive continuous function such that $f:\mathbb{R}\rightarrow\mathbb{R}$.

(a) Note that $f$ is linear in $\mathbb{Q}$:

(i) if $q\in \mathbb{Z}$, $f(q)=f(\sum_{i=1}^{q} (1^i))$, for the additivity of $f$, $$ f(q)=\sum_{i=1}^{q} f(1^i)=q f(1)=q k $$ for some $k\in\mathbb{R}$. So $f$ is linar in $\mathbb{Z}$

(ii) if $q\in \mathbb{Q}\backslash\mathbb{Z}$, $q=\frac{a}{b}$ where $b\neq0$ and $a,b\in \mathbb{Z}$. Note that $f(1)=f\left(\sum_{i=1}^b 1^i/b\right)$, for the additivity of $f$, $$ f(1)=\sum_{i=1}^nf( 1^i/n)=nf(1/n)\Rightarrow \frac{1}{b}f(1)=f(1/b)\Rightarrow f(1/b)=k/b $$ for some $k\in\mathbb{R}$. So $f$ is linear in $\mathbb{Q}$

(b) Let $x\in \mathbb{R}\backslash\mathbb{Q}$ and $\varepsilon>0$

By the continuity of $f$, there is $\delta>0$ such that $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon$.

For the density of $\mathbb{Q}$ in $\mathbb{R}$, there is $j$ in $(x,y)$, such that $|x-j|<\varepsilon$. $$ |f(x)-xf(1)|\leq |f(x) - f(j)| +|f(j) -xf(1)| \leq \varepsilon+|jf(1) -xf(1)| <\varepsilon(1+f(1)) $$ as $\varepsilon$ is an arbitrary positive, we have $f(x)=kx$ for any real numbers, that is it $f$ is linear.

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