A function f : R -> R is affine if f(x + y - z) = f(x) + f(y) - f(z) for all x, y, z in R. In
this project, you will prove that continuous affine functions have a special form.
Problems
A function f : R -> R is additive if f(x + y) = f(x) + f(y) for all x, y in R.
1. Assume f is an affine function and m is a real number. Define g : R -> R by
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 above 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 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 R.