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Math Help - cauchy sequence

  1. #1
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    cauchy sequence

    I am considering the sequences (f_{n}) of continuous functions f_{n}:[-1,1]\to \mathbb{R} defined by:

    <br />
\begin{displaymath}<br />
f_{n}(x) = \left\{<br />
\begin{array}{lr}<br />
1 & : -1\leq x \leq 0\\<br />
1-nx & : 0 \leq x \leq \tfrac{1}{n}\\<br />
0 & : \tfrac{1}{n} \leq x \leq 1<br />
\end{array}<br />
\right.<br />
\end{displaymath} <br />

    How do I show that (f_{n}) is a cauchy-sequence in the normed vector space (C([-1,1],\mathbb{R}), \Vert \cdot \Vert_{1}) and how do I examine if the mentioned normed vector-space is a banach space (i know that every finite dimensional normed vector space is a banach space).

    (C([-1,1],\mathbb{R}), \Vert \cdot \Vert_{1}) is the normed vector space of continuous real-valued functions defined in the closed interval [-1,1].

    Thanks.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Have you found \left\|{f_m-f_n}\right\|_1 ?


    Fernando Revilla
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  3. #3
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    If m> n then 1/m< 1/n so

    f_n- f_m (x)= \left\{\begin{array}{cc}0 & -1< x< 0 \\ (m-n)x & 0<x< 1/m \\ 1- nx & 1/m< x< 1/n \\ 0 & 1/n< x< 1\end{array}

    What is the norm of that? Does it go to 0 as m and n go to infinity.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by surjective View Post
    I am considering the sequences (f_{n}) of continuous functions f_{n}:[-1,1]\to \mathbb{R} defined by:

    <br />
\begin{displaymath}<br />
f_{n}(x) = \left\{<br />
\begin{array}{lr}<br />
1 & : -1\leq x \leq 0\\<br />
1-nx & : 0 \leq x \leq \tfrac{1}{n}\\<br />
0 & : \tfrac{1}{n} \leq x \leq 1<br />
\end{array}<br />
\right.<br />
\end{displaymath} <br />

    How do I show that (f_{n}) is a cauchy-sequence in the normed vector space (C([-1,1],\mathbb{R}), \Vert \cdot \Vert_{1}) and how do I examine if the mentioned normed vector-space is a banach space (i know that every finite dimensional normed vector space is a banach space).

    (C([-1,1],\mathbb{R}), \Vert \cdot \Vert_{1}) is the normed vector space of continuous real-valued functions defined in the closed interval [-1,1].

    Thanks.
    Every finite dimensional vector space is a Banach space just by the "accident" that on finite dimensional vector spaces all norms are equivalent. For infinite dimensional vector spaces such as \mathcal{C}[-1,1] this isn't always the case. You need to show that the norm induces a complete metric space structure on your vector space. Does it here?
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  5. #5
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    cauchy sequence

    Stupid question: How did you get \Vert f_{n}-f_{m} \Vert_{1}?


    Since I have shown that:

    \Vert f \Vert _{1}=\int_{-1}^{1} \left| f(x) \right| dx

    the norm you are asking for becomes:

     \vert f_{n}-f_{m} \Vert_{1}= <br />
\int_{-1}^{1} \left| f_{n}-f_{m} \right| dx=<br />
 \int_{0}^{\frac{1}{m}} \left| (m-n)x   \right| dx + \int_{\frac{1}{m}}^{\frac{1}{n}} \left| 1-nx \right| dx =<br />
\frac{1}{2m} \left| \frac{m-n}{m} \right|\cdot\left( 1-\frac{1}{n}(m-n) \right)

    I would say that this expression goes to 0 as n and m tend towards infinity.

    Per definition i know that for a sequence to a be a cauchy-sequence it must hold that: n,m\geq n_{0} \Rightarrow \Vert f_{n}-f_{m} \Vert_{1} < \epsilon

    Could I conclude that since the expression above tends towards 0 (which is smaller than \epsilon) for n,m going towards infinity, that f_{n} is a cauchy-sequence?
    Last edited by surjective; January 13th 2011 at 12:06 PM.
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  6. #6
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    Hmmmm, not sure how to do that. To be honest I don't understand what it means for a norm to induce a complete metric structure. Elaboration would be appreciated.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by surjective View Post
    Hmmmm, not sure how to do that. To be honest I don't understand what it means for a norm to induce a complete metric structure. Elaboration would be appreciated.
    You need to show that every Cauchy sequence converges, or find a counterexample.
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  8. #8
    mus
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    I can come up with the following counter example: If n \to \infty then:

    f_{n} \to \begin{displaymath}<br />
 \left\{<br />
\begin{array}{lr}<br />
1 & : -1\leq x \leq 0\\<br />
- & : 0 \leq x \leq \tfrac{1}{n}\\<br />
0 & : \tfrac{1}{n} \leq x \leq 1<br />
\end{array}<br />
\right.<br />
\end{displaymath}

    The "-" means that the values is not defined.
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