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

  1. Mar 30, 2010 #1
    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. jcsd
  3. Mar 30, 2010 #2

    Dick

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    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.
     
  4. Mar 30, 2010 #3
    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
  5. Mar 30, 2010 #4

    Dick

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    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
  6. Mar 30, 2010 #5
    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)
    ?
     
  7. Mar 30, 2010 #6

    Dick

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    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?
     
  8. Mar 30, 2010 #7
    OMG, you are so right ...

    now it's crystal clear :\

    thanks very much....
     
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