1. ## 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\$ $=
\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?

2. only way I can see is if the integrand is .
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
Now, let and , then...

Suppose again that and let , then....

is closed under scalar multiplication.
Instead, start by saying, let $f$ and $g \in C[0,1]$ and then continue as you did.

3. 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

4. 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.

5. 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 $\int_0^1 g\!\left(x\right)\,dx=0$!!!

6. 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.

7. Noted

Thank you