Continuous and additive implies linear

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 $\require{enclose} \enclose{horizontalstrike}{|r-j|<\delta/|j|}$, and choosing $\delta$ to be even smaller if necessary gives $|f(x)-f(r)|<\epsilon$, too. Putting this all together gives

$\require{enclose} \enclose{horizontalstrike}{|jf(x)-f(jx)|<|j|f(x)-f(r)| + |f(jx)-f(rx)|<(|j|+1)\epsilon}$

**The above line is wrong, especially you can't find an $r \in \Bbb Q$ s.t $|rx-jx|<\delta$ as well as $|x-r|<\delta$ in stead of this you can find an $r \in \Bbb Q$ s.t $|rx-jx|<\delta$ as well as $|j-r|<\delta$

The line should be $|jf(x)-f(jx)|<|jf(x)-f(rx)+f(rx)-f(jx)| \leq |f(jx)-f(rx)|+|jf(x)-rf(x)|\leq 2\epsilon$

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.

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.