# Linear Algebra - subspaces of f in C[-1,1]

1. Mar 30, 2010

### Roni1985

1. The problem statement, all variables and given/known data
Determine whether the following are subspaces of C[-1,1]:

d) The set of functions f in C[-1,1] such that f(-1)=0 AND f(1)=0

2. Relevant equations

3. The attempt at a solution

I did the question with 'OR', but I don't think I can find the functions.

I am not sure I can use x2 here ...

say f2(x)=f1(x)=x2-1 will give me 0 when x=-1 or x=1

Last edited: Mar 30, 2010
2. Mar 30, 2010

### Dick

C[-1,1] doesn't have anything to with polynomials. You have to show if f(x) and g(x) satisfy your condition then so does f(x)+g(x) and c*f(x). Forget the polynomials.

3. Mar 30, 2010

### Roni1985

I just solved a similar question but with 'OR' instead of 'AND'.

I actually compared my answer with this file:
http://www.math.ohio-state.edu/~chou/Site/Homework_(571)_files/Practice_Final_solution.pdf [Broken]

they used the first polynomial to get the answer.

How would you suggest solving this one ?
I am kind of lost right here....

Last edited by a moderator: May 4, 2017
4. Mar 30, 2010

### Dick

You can use a polynomial to provide give a counterexample in the OR case. Because it's false and the polynomials are contained in C[-1,1]. You can't prove the case of AND just using polynomials because it is true. I already told you. Just show f(x) and g(x) vanish at both endpoints that if f(x)+g(x) and c*f(x) also have that property.

Last edited: Mar 30, 2010
5. Mar 30, 2010

### Roni1985

Ohh, I see....

But, what are my f(x) and g(x) ?
If I can't use the polynomials, what can I use ?
are the simply the 0 functions ?
f(x)=0=g(x)
?

6. Mar 30, 2010

### Dick

f(x) and g(x) are just continuous functions on [-1,1]. All you know about them is that f(-1)=f(1)=0 and g(-1)=g(1)=0, since they are in the set that you are supposed to prove is a subspace. That's all you need to know. Isn't that enough?

7. Mar 30, 2010

### Roni1985

OMG, you are so right ...

now it's crystal clear :\

thanks very much....