Linear Algebra - Linear Spaces

[Mano Jow]

Hello there,
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 $$\leq$$ 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 x₁ < x₂, then f(x₁) < f(x₂). The same applies to a increasing function g.
So, adding g(x₁) to both sides of the f(x) equation and f(x₂) to both sides of the g(x) equation, we have:

(1) f(x₁) + g(x₁) < f(x₂) + g(x₁)
(2) g(x₁) + f(x₂) < g(x₂) + f(x₂)
$$\Rightarrow$$ f(x₁) + g(x₁) < g(x₂) + f(x₂)
$$\Rightarrow$$ (f+g)(x₁) < (f+g)(x₂)

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(x₁) < f(x₂) $$\Rightarrow$$ af(x₁) < af(x₂) $$\Rightarrow$$ (af)(x₁) < (af)(x₂). But I don't think I got anything useful.

I'd be very grateful if anyone could help.
Thanks in advance.

 

[Mark44]

Hello there,
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 $$\leq$$ 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 x₁ < x₂, then f(x₁) < f(x₂). The same applies to a increasing function g.
So, adding g(x₁) to both sides of the f(x) equation and f(x₂) to both sides of the g(x) equation, we have:

(1) f(x₁) + g(x₁) < f(x₂) + g(x₁)

Since g(x₁) < g(x₂), you can continue the inequality above with < f(x₂) + g(x₂).

This shows that f(x₁) + g(x₁) < f(x₂) + g(x₂), or
(f + g)(x₁) < (f + g)(x₂), and you're done with that part.

(2) g(x₁) + f(x₂) < g(x₂) + f(x₂)
$$\Rightarrow$$ f(x₁) + g(x₁) < g(x₂) + f(x₂)
$$\Rightarrow$$ (f+g)(x₁) < (f+g)(x₂)

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?

This is enough to show that the set of increasing functions is not closed under scalar multiplication.

I have tried this (let a be a real number):
f(x₁) < f(x₂) $$\Rightarrow$$ af(x₁) < af(x₂) $$\Rightarrow$$ (af)(x₁) < (af)(x₂). But I don't think I got anything useful.

I'd be very grateful if anyone could help.
Thanks in advance.

 

[Mano Jow]

Thanks a lot for the fast reply!