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Subspace of a Function?

  1. Feb 3, 2009 #1
    Subspace of a Function?!?

    1. The problem statement, all variables and given/known data

    {f [tex]\in[/tex] C([0, 1]): f(1/2) = 0}

    Is this subset of C([0,1]) a subspace?

    2. Relevant equations

    C[0,1] be the set of all functions that are continuous on [0, 1].

    (f + g)(x) = f(x) + g(x)

    (af)(x) = a*f(x)

    3. The attempt at a solution

    Okay, so if f is in the subspace of C than any linear combination of functions in the set should also be in the set. I understand that f is just the set of all functions for which f(1/2) = 0, but how am I supposed to answer the question formally?

    Intuitively I understand that you might have two functions in the set, like:

    f(x) = x - 1/2

    g(x) = x2 - 1/4

    And I know that:

    (f + g)(x) = x2 + x - 3/4

    (f + g)(1/2) = 1/4 + 2/4 - 3/4

    (f + g)(1/2) = 0

    Showing that it is a subspace if I just use those two functions. But how do I generalize my results to include any functions that could be in the set?
     
  2. jcsd
  3. Feb 3, 2009 #2

    Mark44

    Staff: Mentor

    Re: Subspace of a Function?!?

    Take any two functions f and g that are continuous on [0, 1]. It's given that f(1/2) = 0 = g(1/2). What about their sum? What willl (f + g)(1/2) be? What will a*f(1/2) be for any scalar a?
     
  4. Feb 4, 2009 #3
    Re: Subspace of a Function?!?

    I definitely understand the logic of what you're saying. If f(1/2) and g(1/2) are both 0, then (f + g)(1/2) will also be zero.

    But I don't know how to say it formally.
     
  5. Feb 4, 2009 #4

    Mark44

    Staff: Mentor

    Re: Subspace of a Function?!?

    Let's give the subset a name:
    W = {f [itex]\in[/itex] C([0, 1]): f(1/2) = 0}

    If f and g are in W, what does that mean, and what can you say about (f + g)?
    If f is in W, what can you say about (af), where a is a constant?
     
  6. Feb 4, 2009 #5
    Re: Subspace of a Function?!?

    The way I did it was:

    W = {(f in C[0, 1]): f(1/2) = 0}

    f(x) and g(x) are in the subset W

    if W is a subspace then h(1/2) = 0 for h(x) = a*f(x) + b*g(x)

    h(1/2) = a*f(1/2) + b*g(1/2)
    h(1/2) = a*0 + b*0
    h(1/2) = 0

    So W is a subspace.

    --

    Does that seem about right?
     
  7. Feb 4, 2009 #6

    Office_Shredder

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    Re: Subspace of a Function?!?

    That's nearly textbook perfect. All you missed is

    This is true, except that's not what you want to prove. What you should have written is the converse: If h(1/2)=0 for h(x) = a*f(x) + b*g(x), then W is a subspace.
     
  8. Feb 4, 2009 #7
    Re: Subspace of a Function?!?

    Thanks. Can I ask if I did another question correctly?

    The question asks:

    Show that if S and T are subspaces of a vector space V, then S [tex]\cap[/tex] T is also a subspace.

    My Solution thus far

    S [tex]\cap[/tex] T [tex]\subset[/tex] S and S is a subspace, so S [tex]\cap[/tex] T is also a subspace.

    That almost seems too easy though, but it seems obvious that if a set is a subspace, then any part of that set must also be a subspace.
     
  9. Feb 5, 2009 #8

    Mark44

    Staff: Mentor

    Re: Subspace of a Function?!?

    The trouble is that not everything that belongs to a vector space (or subspace) is a subspace of that vector space (subspace). For example, R^2 with the usual rules for vector addition and scalar multiplication is a vector space, and the line x = 1 is in that space, but that line is not a subspace.

    You have to show that if u and v are elements of S [tex]\bigcap[/tex] T, then u + v is in S [tex]\bigcap[/tex] T, and so is au, where a is any scalar (which is one way to show that 0 is in S [tex]\bigcap[/tex] T).
     
  10. Feb 5, 2009 #9
    Re: Subspace of a Function?!?

    Okay, here's how I approached the problem.

    u is in S [tex]\cap[/tex] T
    v is in S [tex]\cap[/tex] T
    a, b are scalars

    If S [tex]\cap[/tex] T is a subspace, then au + bv is in S [tex]\cap[/tex] T.
    If au + bv is in S [tex]\cap[/tex] T, then au + bv is in both S and T.

    au + bv is in S because S is a subspace and au + bv is in T because T is a subspace.

    There S [tex]\cap[/tex] T is a subspace.

    Does that seem about right?
     
  11. Feb 5, 2009 #10

    Office_Shredder

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    Re: Subspace of a Function?!?

    You got your if statements backwards. While

    Is true, what you did is prove au+bv is in both S and T, so what you should write is


    S [tex]\cap[/tex] T is a subspace only if au + bv is in S [tex]\cap[/tex] T.
    au + bv is in S [tex]\cap[/tex] T, only ifau + bv is in both S and T.
     
  12. Feb 5, 2009 #11

    Mark44

    Staff: Mentor

    Re: Subspace of a Function?!?

    To concur with Office_Shredder, but saying it a little differently, don't start off by saying "If <what you're trying to prove>". Instead, that should go at the end of your logical progression.

    IOW,
    if u in S [itex]\cap[/itex] T, v in S [itex]\cap[/itex] T, a, b are scalars, where S and T are subspaces of a vector space then <your argument here> ==> S [itex]\cap[/itex] T is a subspace.

    Do you get what I'm saying?
     
  13. Feb 5, 2009 #12
    Re: Subspace of a Function?!?

    Yeah. This is like my 2nd or 3rd proof so admittedly I'm a little rusty on how to formally write out the proofs.
     
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