I think this is false. Consider a continuous function which is negative from to and then positive from to in such a way that .
So it is not helpful to do Instead, start by saying, let and and then continue as you did.
Q: Determine whether the set,
is a subspace of
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 ; thus, is non-empty.
Now, let , then...
is closed under addition.
Suppose again that and let , then....
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?
I think this is false. Consider a continuous function which is negative from to and then positive from to in such a way that .
So it is not helpful to do Instead, start by saying, let and and then continue as you did.
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 ... as I think you did.