Results 1 to 7 of 7

Math Help - Subspace question

  1. #1
    Senior Member Danneedshelp's Avatar
    Joined
    Apr 2009
    Posts
    303

    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\ =<br />
\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?
    Last edited by Danneedshelp; July 10th 2009 at 05:06 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    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...



    is closed under addition.



    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Danneedshelp's Avatar
    Joined
    Apr 2009
    Posts
    303
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Danneedshelp View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by arbolis View Post
    [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!!!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Chris L T521 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member Danneedshelp's Avatar
    Joined
    Apr 2009
    Posts
    303
    Noted

    Thank you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. R^4 subspace question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: June 25th 2011, 12:57 AM
  2. Replies: 7
    Last Post: May 17th 2011, 02:31 PM
  3. question about subspace
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: March 1st 2010, 10:25 PM
  4. Subspace question...
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: July 14th 2009, 10:59 PM
  5. Vector subspace question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 13th 2008, 01:23 PM

Search Tags


/mathhelpforum @mathhelpforum