Sort the set of functions f:R→R, with R being the set of real numbers. This means you have to find an
order "«" that lets you compare any two pairs of unequal functions f and g; unequal means, f(x)≠g(x) for at least one x.
More precisely, these are the requirements for the order "«" you are challenged to find:
1. If f≠g, either f«g or g«f.
2. If f«g and g«h then f«h
You might be tempted to declare f«g when f(x)<g(x) for all x but that would of course fail because e.g. f(x)=x and g(x)=-x would not be comparable with respect to your order.
For a much, much easier challenge, start by finding an order for all continuous functions.
(In reply to
re(2): For continuous functions... by Eric)
Eric:
I don't think your idea in comment 4 works either.
Consider: Define functions
f(x) = 0 except between d and 2d
= sine(2pi(x-d)/d) between d and 2d
g(x) = -f(x)
The integral of f(x) and g(x) is 0 for any region that includes all of or none
of [d,2d].
The integral of |f(x)| and |g(x)| are equal for any region that includes all of or none
of [d,2d].
These functions is indistinguishable if
the integral is
evaluated between 0 and d, -d, d/2, 2d, -d/2, -2d, d/4, 4d, -d/4, -4d,
d/8, 8d, etc.
re(3): For continuous functions...
