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Math Help - Normed sequence space

  1. #1
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    Normed sequence space

    Let \ell^1(\mathbb{N},\mathbb{R}) denote the space of sequences x(j) such that \sum_{j=1}^{\infty}|x(j)|<\infty, equipped with the norm \|x\|_1=\sum_{j=1}^{\infty}|x(j)|. Let \|x\|_2 denote the usual norm (\sum_{j=1}^{\infty}|x(j)|^2)^\frac{1}{2} on \ell^2(\mathbb{N},\mathbb{R}).
    Show that \|x\|_2\leq\|x\|_1, and hence deduce that \ell^1(\mathbb{N},\mathbb{R})\subseteq\ell^2(\math  bb{N},\mathbb{R}).

    Any help would be appreciated. Thanks in advance
    Last edited by ejgmath; April 13th 2010 at 07:18 AM.
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  2. #2
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    Quote Originally Posted by ejgmath View Post
    Let \ell^1(\mathbb{N},\mathbb{R}) denote the space of sequences x(j) such that \sum_{j=1}^{\infty}|x(j)|<\infty, equipped with the norm \|x\|_1=\sum_{j=1}^{\infty}|x_j|. Let \|x\|_2 denote the usual norm (\sum_{j=1}^{\infty}|x(j)|)^\frac{1}{2} on \ell^2(\mathbb{N},\mathbb{R}).
    Show that \|x\|_2\leq\|x\|_1, and hence deduce that \ell^1(\mathbb{N},\mathbb{R})\subseteq\ell^2(\math  bb{N},\mathbb{R}).

    Any help would be appreciated. Thanks in advance
    Use the fact that since the series converges the general term |x_j| \rightarrow 0 and that for |t|<1 it is valid |t|^2<|t|
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  3. #3
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    Quote Originally Posted by Jose27 View Post
    Use the fact that since the series converges the general term |x_j| \rightarrow 0 and that for |t|<1 it is valid |t|^2<|t|
    This will prove the inclusion, but not the comparison of norms, I think.

    For that one, you can write \left(\sum_{j=1}^n |x_j|\right)^2\geq \sum_{j=1}^n |x_j|^2 for all n, which comes from the expansion of the left-hand side (all terms of the expansion are nonnegative, and the right-hand side consists in a few of those). And let n\to\infty.
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  4. #4
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    Hi, thanks for the response, I am really sorry but I made a small mistake in the original post. I missed the squared out of the 2-norm, does this change much. Also, how do you deduce the second part of the question? I have included the revised version below. Thanks

    Quote Originally Posted by ejgmath View Post
    Let \ell^1(\mathbb{N},\mathbb{R}) denote the space of sequences x(j) such that \sum_{j=1}^{\infty}|x(j)|<\infty, equipped with the norm \|x\|_1=\sum_{j=1}^{\infty}|x(j)|. Let \|x\|_2 denote the usual norm (\sum_{j=1}^{\infty}|x(j)|^2)^\frac{1}{2} on \ell^2(\mathbb{N},\mathbb{R}).
    Show that \|x\|_2\leq\|x\|_1, and hence deduce that \ell^1(\mathbb{N},\mathbb{R})\subseteq\ell^2(\math  bb{N},\mathbb{R}).

    Any help would be appreciated. Thanks in advance
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  5. #5
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    Quote Originally Posted by ejgmath View Post
    Hi, thanks for the response, I am really sorry but I made a small mistake in the original post. I missed the squared out of the 2-norm, does this change much.
    Nothing; this was an obvious mistake we took into account...

    Also, how do you deduce the second part of the question?
    I think you can find it yourself. What is \ell^1(\mathbb{N},\mathbb{R}) by definition?...

    (Also, Jose27 suggested a direct proof, without the norm inequality)
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  6. #6
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    Could someone help explain the logic here?

    I don't understand how \ell^1(\mathbb{N},\mathbb{R})\subseteq\ell^2(\math  bb{N},\mathbb{R}) follows from \|x\|_2\leq\|x\|_1.

    If the vectors in \ell^1 have a greater length than the ones in \ell^2 how can \ell^1 be a subset of \ell^2?

    Many thanks to anyone who can help
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  7. #7
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    Just think that x\in l_1 means that \|x\|_1 is finite, and the same for l_2 with \|\cdot\|_2. You can look your geometric argument by the point of view that there are vectors with "infinity length" with the measure \|\cdot\|_1, but with finite length with the smaller masure \|\cdot\|_2, as for example x=(1/n)_n. The contrary is not possible because this inequality
    Last edited by Enrique2; April 14th 2010 at 05:54 AM.
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