1. ## Vector Spaces

So here is a 2 part question that I am having a lot of trouble with.

For
each of the following sets determine whether it is a vector space:

(i) $W_1 = \{f(x) \in C(R) | \int_0^{2} x^2 f(x)dx =0\}$

(iii) $W_1 = \{f(x) \in C(R) | \int_0^{2} xf^2 (x)dx =0\}$

Ok so what I know is that in order for it to be a vector space I need to show that all the axioms are true, or to show that it is not a vector space I need to show atleast one axiom fails.

I am kind of confused about how to go about doing this?

For the first one I integrated and got $4G(2) -4R(2)-2H(2)+2H(0)$
where G(x) is the first integral of f(x), R(x) is the second and H(x) is the third. Im not completely sure wether or not this was neccessary, and if so what to do with this.

Any help would be very much appreciated, thanks!

2. Originally Posted by joe909
So here is a 2 part question that I am having a lot of trouble with.

For
each of the following sets determine whether it is a vector space:

(i) $W_1 = \{f(x) \in C(R) | \int_0^{2} x^2 f(x)dx =0\}$

(iii) $W_1 = \{f(x) \in C(R) | \int_0^{2} xf^2 (x)dx =0\}$

Ok so what I know is that in order for it to be a vector space I need to show that all the axioms are true, or to show that it is not a vector space I need to show atleast one axiom fails.

I am kind of confused about how to go about doing this?

For the first one I integrated and got $4G(2) -4R(2)-2H(2)+2H(0)$
where G(x) is the first integral of f(x), R(x) is the second and H(x) is the third. Im not completely sure wether or not this was neccessary, and if so what to do with this.

I think this was unnecessary: what you need to show is $f_1\,,\,f_2\in W_i\Longrightarrow k_1f_1+k_2f_2 \in W_i\,,\,\,i=1,2\,,\,\,k_i\in\mathbb{R}$ . For $W_1$ this follows almost at once from linearity of the Riemann integral, and for $W_2$...well, this is way trickier. Try it yourself or a while

Tonio

Any help would be very much appreciated, thanks!
.

3. Thank you very much for the reply, although I still find myself a little bit confused. Where exactly is $W_2$ comming from?

Just so I make sure Im understanding this properly. Are you using the fact that we know $C(R)$ is a vectorspace and therefore we only need to show that theese are subspaces of $C(R)$?

Thanks!

4. Alright here is what I have can comeone tell me if im atleast on the right track, thanks.

part (i)

$\int_0^{2} x^2 (f+g)(x)=\int_0^{2} x^2 f(x) + \int_0^{2} x^2 g(x)$ Therefore closed on addition.

$\int_0^{2} x^2 (cf(x)) = \int_0^{2} x^2 c(f(x)), c \in R$
therefore closed under multiplication.

Therefore this is a subspace of $\{f(x) \in C(R)\}$ and since $C(R)$ is a vector space, this set is also a vector space.

Im not really sure that I gave enough information as to why theese properties are true in order to really prove this (and am in fact not sure myself why they are true).

Part (ii)

$\int_0^{2} x^2 (f+g)^2(x)=\int_0^{2} x^2 (f^2+2fg+g^2)(x)$ which is clearly not equal to $\int_0^{2} x^2 f(x) + \int_0^{2} x^2 g(x)$. Therefore not closed on vector addition and not a vector space.

If anyone could look this over and let me know where i may be going wrong that would be great, thanks!

5. Originally Posted by joe909
Alright here is what I have can comeone tell me if im atleast on the right track, thanks.

part (i)

$\int_0^{2} x^2 (f+g)(x)=\int_0^{2} x^2 f(x) + \int_0^{2} x^2 g(x)$ Therefore closed on addition.
= 0+ 0= 0, of course! That shows that f+ g also has this property.

$\int_0^{2} x^2 (cf(x)) = \int_0^{2} x^2 c(f(x)), c \in R$
therefore closed under multiplication.
You really want the c outside the integral on the right. And then (c)(0)= 0, showing that cf also has the required property.

Therefore this is a subspace of $\{f(x) \in C(R)\}$ and since $C(R)$ is a vector space, this set is also a vector space.

Im not really sure that I gave enough information as to why theese properties are true in order to really prove this (and am in fact not sure myself why they are true).

Part (ii)

$\int_0^{2} x^2 (f+g)^2(x)=\int_0^{2} x^2 (f^2+2fg+g^2)(x)$ which is clearly not equal to $\int_0^{2} x^2 f(x) + \int_0^{2} x^2 g(x)$. Therefore not closed on vector addition and not a vector space.

If anyone could look this over and let me know where i may be going wrong that would be great, thanks!