- #1

Mano Jow

- 18

- 0

I'm having some problems with a proof-like exercise on linear algebra. Here's what I'm supposed to do:

## Homework Statement

Determine whether each of the given sets is a real linear space, if addition and multiplication by real scalars are defined in the usual way. For those that are not, tell which axioms fail to hold.

## Homework Equations

All increasing functions.

(there are others, like rational functions, all Taylor polynomials of degree [tex]\leq[/tex] n, etc. but I guess my doubts are related to all of them).

## The Attempt at a Solution

Well, after a couple of attempts my teacher told me that once I prove that increasing functions are a subset of the functions linear space, I just need to prove the closure axioms. This can be done just by saying "the increasing fuctions are contained in the set of functions". Ok, then I started to prove the closure under addtion.

Let f be an increasing funcion, if x

_{1}< x

_{2}, then f(x

_{1}) < f(x

_{2}). The same applies to a increasing function g.

So, adding g(x

_{1}) to both sides of the f(x) equation and f(x

_{2}) to both sides of the g(x) equation, we have:

(1) f(x

_{1}) + g(x

_{1}) < f(x

_{2}) + g(x

_{1})

(2) g(x

_{1}) + f(x

_{2}) < g(x

_{2}) + f(x

_{2})

[tex]\Rightarrow[/tex] f(x

_{1}) + g(x

_{1}) < g(x

_{2}) + f(x

_{2})

[tex]\Rightarrow[/tex] (f+g)(x

_{1}) < (f+g)(x

_{2})

And I guess it's proven that it's closed under addition. Is that right?

Now, I know it's not closed under multiplication because if I multiplicate f by -1 it won't be an increasing function anymore. But can I just say that and it's ok or is there another way to show it's not closed under multiplication?

I have tried this (let a be a real number):

f(x

_{1}) < f(x

_{2}) [tex]\Rightarrow[/tex] af(x

_{1}) < af(x

_{2}) [tex]\Rightarrow[/tex] (af)(x

_{1}) < (af)(x

_{2}). But I don't think I got anything useful.

I'd be very grateful if anyone could help.

Thanks in advance.