Subspace question

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July 10th 2009, 03:01 PM
Danneedshelp

Subspace question

Q: Determine whether the set,

$S=\{f\in\\C[0,1]:\int_{0}^{1}\\f(x)dx=0\}$

is a subspace of $C[0,1]$

A: I said yes, because...

Firstly, the set contains the zero vector, since the set consists of all functions with zero area on the interval $C[0,1]$ ; thus, $S$ is non-empty.

Now, let $f,g\in{C[0,1]}$ , then...

$\int_{0}^{1}\\(f+g)(x)dx\$ $=\int_{0}^{1}\\(f(x)+g(x))dx\$ $=<br /> \int_{0}^{1}\\f(x)dx+\int_{0}^{1}\\g(x)dx=0\$

$\therefore$ $S$ is closed under addition.

Suppose again that $f\in{C[0,1]}$ and let $k\in\mathbb{R}$ , then....

$k\int_{0}^{1}\\f(x)=\int_{0}^{1}\\kf(x)dx=0$ $(\forall{k}\in\mathbb{R})$

$\therefore$ $S$ is closed under scalar multiplication.

In general, is it true that a set made some function equal to zero will automatically contain the zero vector and be nonempty?
July 10th 2009, 03:23 PM
arbolis

Quote:

only way I can see http://www.mathhelpforum.com/math-he...bc0c5f15-1.gif is if the integrand is http://www.mathhelpforum.com/math-he...f98764da-1.gif.

I think this is false. Consider a continuous function which is negative from $0$ to $\frac{1}{2}$ and then positive from $\frac{1}{2}$ to $1$ in such a way that $\int_0^1 f(x)dx=0$ .

So it is not helpful to do

Quote:

Now, let http://www.mathhelpforum.com/math-he...bd8576bc-1.gif and http://www.mathhelpforum.com/math-he...3aa481ac-1.gif, then...

http://www.mathhelpforum.com/math-he...04538411-1.gif http://www.mathhelpforum.com/math-he...b0c1d457-1.gif http://www.mathhelpforum.com/math-he...2a2bc8a6-1.gif

http://www.mathhelpforum.com/math-he...e977e4ec-1.gif http://www.mathhelpforum.com/math-he...1a47546e-1.gif is closed under addition.

Suppose again that http://www.mathhelpforum.com/math-he...bd8576bc-1.gif and let http://www.mathhelpforum.com/math-he...0b3babe7-1.gif, then....

http://www.mathhelpforum.com/math-he...8343bc4a-1.gif http://www.mathhelpforum.com/math-he...29e7ba1e-1.gif

http://www.mathhelpforum.com/math-he...e977e4ec-1.gif http://www.mathhelpforum.com/math-he...1a47546e-1.gif is closed under scalar multiplication.

Instead, start by saying, let $f$ and $g \in C[0,1]$ and then continue as you did.
July 10th 2009, 04:24 PM
Danneedshelp

Thanks alot for the response. I totally looked right past that, thanks for pointing that out! So, with my above changes, does the "proof" look alright? I am still not sure if my reasoning at the first step (showing the subspace contains the zero vector and is nonempty) is correct.

Thank you
July 10th 2009, 04:50 PM
arbolis

Quote:

Originally Posted by Danneedshelp

Thanks alot for the response. I totally looked right past that, thanks for pointing that out! So, with my above changes, does the "proof" look alright? I am still not sure if my reasoning at the first step (showing the subspace contains the zero vector and is nonempty) is correct.

Thank you

You're welcome.
I think the proof looks alright if you make the change I suggested.
If you show that the zero vector is in the set S (you cannot say it is in the subspace yet because you are not yet sure it is a vector space), then it is non empty.
And of course it is very important for the zero vector to be in the set, otherwise the set would not be a vector space, hence not a subspace.
The zero vector is indeed in the set : take $f(x)=0$ ... as I think you did.
July 10th 2009, 07:46 PM
Chris L T521

Quote:

Originally Posted by arbolis

[snip]
Instead, start by saying, let $f$ and $g \in C[0,1]$ and then continue as you did.
[/snip]

I would suggest saying let $f,g\in{\color{red}S}$ . If you say $f,g\in\color{red}C\left[0,1\right]$ , then it may not necessarily be true that $\int_0^1 f\!\left(x\right)\,dx=0$ and $\int_0^1 g\!\left(x\right)\,dx=0$ !!!
July 10th 2009, 08:16 PM
arbolis

Quote:

Originally Posted by Chris L T521

I would suggest saying let $f,g\in{\color{red}S}$ . If you say $f,g\in\color{red}C\left[0,1\right]$ , then it may not necessarily be true that $\int_0^1 f\!\left(x\right)\,dx=0$ and $\int_0^1 g\!\left(x\right)\,dx=0$ !!!

Oops, you're absolutely right! Sorry about that.
Dan take note of this correction. (Itwasntme)
July 11th 2009, 01:15 PM
Danneedshelp

Noted

Thank you