Continuous and additive implies linear

Question

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.

Answer 1

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.

Answer 2

Let $x \in \mathbb{R}$. We wish to show that $f(x\lambda) = f(x)\lambda$ for any $\lambda \in \mathbb{R}$. We will split the proof into cases.

Case 1: Integers

We will proceed by induction. The base case is quick: $f(x1) = f(x)1$ since both equal $f(x)$.

Now, assume that $f(xn) = f(x)n$ for some $n \in \mathbb{Z}$. We will show the result holds for both $n+1$ and $n-1$, abbreviated as $n \pm 1$:

\begin{align} f\big( x(n \pm 1) \big) &= f(xn \pm x1)\\ &= f(xn) \pm f(x1)\\ &= f(x)n \pm f(x)1 \text{ (by induction hypothesis)}\\ &= f(x)(n \pm 1). \end{align}

Case 2: Rationals

Let $\frac{n}{d}$ be a rational number, where $n, d \in \mathbb{Z}$ and $d \neq 0$. Set $$w = f\left(x \cdot \frac{n}{d}\right).$$ Multiplying both sides by $d$ yields $$wd = f\left(x \cdot \frac{n}{d}\right) d.$$ Since $d$ is an integer, we may apply case 1 twice: \begin{align} wd &= f\left(x \cdot \frac{n}{d} \cdot d\right)\\ &= f(xn)\\ &= f(x)n. \end{align} Dividing both sides by $d$ and expanding the definition of $w$ yields $$ f\left(x \cdot \frac{n}{d}\right) = f(x) \cdot \frac{n}{d},$$ which is the desired result.

Case 3: Reals

Until now, we have not used the continuity of $f$. That is about to change.

Let $\lambda$ be any real number. By the density of $\mathbb{Q}$ in $\mathbb{R}$, there is a sequence of rationals $(\lambda_k : k \in \mathbb{N})$ converging to $\lambda$.

By case 2, for each $k \in \mathbb{N}$ we have:

\begin{align} f(x \lambda_k) &= f(x) \lambda_k.\\ \end{align}

Passing to the limit yields $$\lim_{k}{f(x \lambda_k)} = \lim_{k}{f(x)\lambda_k}.$$

The continuity of $f$ (and of multiplication) allows us to interchange the limit:

\begin{align} f\left(\lim_k{x \lambda_k}\right) &= \lim_{k}{f(x)} \lim_{k}{\lambda_k};\\ f(x \lambda) &= f(x) \lambda. \end{align}

This completes the proof.