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Homework Help: Is that subset of the set of continuous differential functions closed?

  1. Oct 20, 2011 #1
    Hi! I have used the physics forum a lot of times to deal with several tasks that I had and now its the time to introduce my own query! So please bear with me :-)

    1. The problem statement, all variables and given/known data
    Equip the set [itex] C^1_{[0,1]} [/itex] with the inner product:
    \left\langle f,g \right\rangle= \int_{0}^{1} f(x)\bar{g(x)} + \int_{0}^{1} f'(x)\bar{g'(x)}dx
    (the bar above the [itex] g [/itex] function is the conjugate symbol)
    I need to show that the subspace:

    W = \{f\in C^1_{[0,1]} | f(1)=0\}

    is a closed subspace of [itex] C^1_{[0,1]} [/itex].

    2. Relevant equations

    [itex] \left\langle f,cosh \right\rangle = f(1)sinh(1) [/itex].

    The Cauchy inequality: [itex] |\left\langle f,g \right\rangle | \le \|f\|\|g\|[/itex],
    the Pythagoras theorem: [itex] \|f+g\|^2 = \|f\|^2 + \|g\|^2 [/itex],
    the parallelogram law: [itex] \|f+g\|^2 + \|f-g\|^2 = 2(\|f\|^2 + \|g\|^2)[/itex],
    the triangular inequality: [itex] \|f+g\| \le \|f\| + \|g\|[/itex]

    3. The attempt at a solution

    Let us take a Cauchy sequence [itex] \{f^n\}_{n=1}^{\infty} \in W[/itex], because
    [itex](C^1_{[0,1]},\|\cdot \|)[/itex] is a Hilbert space then the sequence [itex] \{f^n\}_{n=1}^{\infty} [/itex] converges to [itex] f\in C^1_{[0,1]} [/itex].
    Therefore it only remains to be shown that at the limit [itex] f(1)=0 [/itex].

    At this point I am stuck. I can see that the [itex] cosh [/itex] function is orthogonal
    to the set [itex] W [/itex] and I also tried to use the above "relevant equations" but
    I couldn't see what would be a possible proof.

    Any advice?
    Last edited: Oct 20, 2011
  2. jcsd
  3. Oct 20, 2011 #2
    First of all, I very much doubt that your space is a Hilbert space. It has an inner-product, but are you sure it is complete. Did you prove it??

    Anyway, you don't really need it for the proof. For the proof you must take a convergent sequence [itex](f_n)_n[/itex] in W. So you know that [itex]f_n(1)=0[/itex] for all n. You must prove that the limit f also has f(1)=0.

    You must not prove that [itex](f_n)_n[/itex] is Cauchy and you must not prove that it converges. You know convergence from the hypothesis.
  4. Oct 20, 2011 #3
    You are right, I was wrong about that.

    Ok, that was my thought from the very beginning. But if I need just the [itex]f(1)=0[/itex], then how do I know that the limit [itex] f [/itex] is also a continuous differentiable function? Don't I need that as well?
  5. Oct 20, 2011 #4
    No, you have that.

    So what you have is that a sequence [itex](f_n)_n[/itex] in W converges to a function [itex]f\in C^1_{[0,1]}[/itex]. You must show f to be in W.
  6. Oct 20, 2011 #5
    Oh yes, I am sorry, all that time I didn't pay any attention to the definition of W.
  7. Oct 20, 2011 #6
    I think that the solution is:

    [itex] |\left\langle f^n-f,cosh \right\rangle| \le \|f^n-f\|\|cosh\| [/itex]

    But [itex] \forall \epsilon \ \ \exists n_o(\epsilon) [/itex] such that for any
    [itex] n \ge n_0(\epsilon) [/itex] we have that:

    [itex] |\left\langle f^n-f,cosh \right\rangle| \le \epsilon \|cosh\| [/itex]

    Therefore [itex] \left\langle f^n-f,cosh \right\rangle \to 0 \Leftrightarrow [/itex] [itex] \left\langle f^n,cosh \right\rangle - \left\langle f,cosh \right\rangle \to 0 \Leftrightarrow [/itex]

    [itex] f(1)sinh(1) \to 0 \Leftrightarrow f(1)\to 0 [/itex]
    Last edited: Oct 20, 2011
  8. Oct 20, 2011 #7
    Looks ok. However in the last line, you don't want =0 but rather [itex]\rightarrow 0[/itex].

    So you got

    [tex]<f^n-f,cosh>\rightarrow 0[/tex]

    for example.
  9. Oct 20, 2011 #8
    Yes you are right again, thanks!
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