# Subspace of even and odd functions

1. Sep 24, 2009

Hey guys. I came upon this problem in the professor's recommended problems, and I have no idea how to solve it. After spending an hour on it, I've got nothing. Any suggestions would be much appreciated!

A function f : R -> R is called even just in case f(-r) = f(r) for every
real number r. f : R - > R is called odd just in case f(-r) = -f(r) for
every real number r. Let V be the real vector space of all functions from
R to R, U be the set of even functions, and W be the set of odd functions.
(a) Show that U and W are subspaces of V .
(b) Show that U + W = V and U $$\bigcap$$ W = {0}, where here 0 means the
constant function 0. (In such a situation, as will be pointed out in
class, we will say that V is the direct sum of U and W and write
V = U $$\otimes$$W.)

2. Sep 24, 2009

### aPhilosopher

Ok, for (a) let's work in U. if $$f, g \in U$$ is $$cf \in U$$ where $$c \in R$$? is $$f + g \in U$$?

To prove the above, let x be an arbitrary real number and see how cf(-x) and f(-x) + g(-x) behave with respect to being even.

If you want to be ambitious, you can just try to prove that cf(-x) + dg(-x) are even for c and d both real.

Last edited: Sep 24, 2009
3. Sep 24, 2009

### aPhilosopher

Sorry, I messed up my earlier post in that it wasn't as helpful as I meant it to be! I'll edit it.

4. Sep 24, 2009

Thanks for the rapid reply. I mean that's the definition of a subspace, so it just seemed to simple to me to say "f(-x) + g(-x) = f(x) + g(x) therefore U is a subspace." Is that really all I'd need to say (along with being closed under scalar multiplication?

And any idea about how to prove part b?

Regards,

5. Sep 24, 2009

### aPhilosopher

well if you wanted to dress it up some, you could do something like (f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x) but I just wanted to help you get the basic idea. Normally, proving something is a sub-space is pretty easy ;)

As for part b, suppose that you are given a function f.

What can you say about the function g(x) = f(x) - f(-x)?

6. Sep 24, 2009

That depends, is f(x) even or odd? If even, then g(x) = 0, if odd, then g(x) = 2f(x). What's the next logical step that you're hinting at?

7. Sep 24, 2009

### aPhilosopher

Sorry, f is just a general function on R. What happens when you plug -x in to g?

8. Sep 24, 2009

It gives the sum of both even and odd functions, and hence all of V? I sense that's where you're leading me, but I still don't fully see it.

9. Sep 24, 2009

### aPhilosopher

Ok, is g an even function or an odd function?

10. Sep 24, 2009

It's neither since it's composed of the sum of both an even and odd function?

11. Sep 24, 2009

### aPhilosopher

I think we got confused. f(x) is neither even nor odd because we assumed that it was a general function on R, right?

So then f(-x) is neither even nor odd either because it's just f(x) reflected about the y axis when you graph it, correct?

Substitute -x for x in the definition of g(x) and see what happens.

12. Sep 24, 2009

g(-x) = f(-x) - f(x), right? So g(x) + g(-x) = 0; therefore, g encompases all of the vector space? Couldn't we just say since even functions map all of R to positive R's, and odd functions map all of R to negative R's, the sum of both functions will allow us to produce any function?

I still don't fully get it :(. I feel like I'm missing something about the entire question, and that's preventing me from making the next step. Thanks for your continual help though :)

13. Sep 24, 2009

### aPhilosopher

What do you mean g encompasses all the vector space? g is just one vector in the space!

you have g(x) = f(x) - f(-x) and g(-x) = f(-x) - f(x).

Now stop thinking and answer my question! Is g even or odd? ;)

Stick with it. This stuff is fun.

14. Sep 24, 2009

Ah, right, it's an odd function! While h(x) = f(-x) is an even function, correct? so h(x) + g(x) = f(x)... So does that mean that any even function added to any odd function can give any function in the space? Or did I skip a step?

15. Sep 24, 2009

### aPhilosopher

Correct, it's an odd function. Can you prove that h(x) = f(-x) is even? (hint: no because it's not true!)

Try to cook up an even function from f though. It's pretty similar to g. think about what makes g an odd function and then do the opposite of that. Just as a hint, i'll tell you that when you find it, h + g does not equal f. You have to take a linear combination of them.

Then we can get on with the second part of part (b)

16. Sep 24, 2009

okay, so then h(x) = f(x) + f(-x).

In order to map all the functions in V, we need to be able to go from any x to any y, i.e. f(x) - f(x) + f(-x) - f(-x). Is this the right way of thinking about it? What I'm not sure about is whether functions can be manipulated in this way? i.e. would f(x) - f(x) just cancel out, or can I use it the way I intended?

17. Sep 25, 2009

### aPhilosopher

Good job finding h!

No, you don't need to be able to go to any y from any x! Some functions won't do that. think of $$\frac{x^{2}}{x^{2} + 1}$$ for instance. x and y really have nothing to do with this. See below.

In response to your question about f(x) being canceled by -f(x), you can do that. Think of sin(x) and -sin(x) for instance. But I don't see what you are getting at with that sum. Remember, we are dealing with vectors here. That's the name of the game. We are only dealing with x because these vectors happen to be functions of x but all the rules for vectors apply. If you can do it with a vector, you can do it with a function!

So what is h + g? You can probably think of two scalars c and d (not necessarily different) so that cg + dh = f.

18. Sep 25, 2009