Affine Function Proofs A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is affine if $f(x+y-z) = f(x)+f(y)-f(z)$ for all $x,y,z\in \mathbb{R}$.  A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is additive if $f(x+y) = f(x)+f(y)$ for all $x,y\in \mathbb{R}$.  
(1) Assume $f$ is an affine function and $m$ is a real number. Define $g:\mathbb{R} \rightarrow \mathbb{R}$ to be $g(x) = f(x)-[mx+f(0)]$. Show that $g$ is an additive function.
(2) Let $m = f(1) -f(0)$ and show that the function defined in item (1) has the property that $g(x+1)=g(x)$.
(3) Suppose $f$ is affine and bounded on $[0,1]$, and $g$ is defined by $g(x) = f(x)-[mx+b]$ where $m = f(1) -f(0)$ and $b=f(0)$. Show that $g$ is bounded on $\mathbb{R}$. 
(4) Show that any bounded, additive function must be the zero function.
(5) Show that if $f$ is affine and continuous, there are real numbers $m$ and $b$ such that $f(x)=mx+b$ for all $x\in \mathbb{R}$. 
 A: 

*$g(x) = f(x) - (mx + b) = f(x) - xf(1) + (x-1)f(0)$. But $|f(x)| \leq M$ on for all $x$ on $(0, 1)$ for some $M$ since $f$ is bounded.

$|f(x) - xf(1) + (x-1)f(0)|<|M| +|M| + |M|= 3M$
by the triangle inequality, and using the fact that $1>x$ and $1>(x-1)$ for all $x \in [0,1]$. It's not a very good estimate, but we don't need it to be good, we just need it to work. That shows that IF $f$ is bounded on $[0, 1]$, then $g(x)$ is bounded on $\mathbb{R}$.


*Prove by induction that $f(m) = f(1 +... + 1) = f(1) + ... + f(1) = mf(1)$. If $|f(1)| > 0$ then...


*My hint for 5 may have been misleading. It's actually much easier because we can use the rest of the problem. I'll let you fill in the gaps:
$f$ is affine and continuous. Since it's continuous, it's bounded on the set $[0, 1]$. Let $g(x) = f(x) - (mx + b)$ as defined in part (3). $g$ is bounded by part (3) and additive by part (1). Thus, g must be the zero function by part (4). So, $f(x) = mx + b$.
EDIT
Interestingly, there is a way to use induction on the reals, but you don't need it here. If you're interested, See Pete Clark's notes. For 4, you can just use normal (positive integer) induction. Like, $f(2) = f(1+1) = f(1) + f(1) = 2f(1)$ And so on, which is good enough. For example, if I want to find some number $m$ so that $f(m) > N$ for any positive $N$, then we can just say $f(m) > n \implies mf(1) > N \implies m > \frac{N}{f(1)}$.
