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Thread: Operator well-defined

  1. #1
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    Operator well-defined

    Hello,

    I just can't see why the following operator is well-defined: Let $\displaystyle x \in \ell^{2}(\mathbb{N}), a \in \ell^{1}(\mathbb{N}) $. Define the operator $\displaystyle A$ so that $\displaystyle Ax = (\sum_{k=1}^{\infty}a_{k+n-1}x_{k})_{n \in \mathbb{N}}$. Also, assume that $\displaystyle a_n$ is monotonically decreasing. Can anybody show me why $\displaystyle A: \ell^{2}(\mathbb{N}) \rightarrow \ell^{2}(\mathbb{N})$ is supposed to be well-defined?

    Thanks for your help,
    Best regards,
    Stiwan
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  2. #2
    Super Member Rebesques's Avatar
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    Re: Operator well-defined

    So, we are to prove that

    $\displaystyle Ax\in \ell^2(\mathbb{N}), \forall x\in \ell^2(\mathbb{N}).$


    This will be true if we can establish that

    $\displaystyle ||Ax||^2=\sum_n|(Ax)_n|^2<\infty \ \ (1) $


    Based on the facts that $\displaystyle a_n\in\ell^1$ and that $\displaystyle a_n$ is decreasing, we can use the Cauchy-Schwartz inequality
    to obtain that, eventually
    $\displaystyle ||Ax||^2=\sum_n\sum_k|a_{k-n+1}x_k|^2\leq (\sum_k|x_k|)^2\sum_n(\sum_k|a_{k-n+1}|) \ \ (2) $

    Now, we can use elementary calculus to show that, for large $\displaystyle n\in \mathbb{N}$, we can have
    $\displaystyle \sum_l|a_{k-n+1}|\leq n^{-2}$

    and so, (2) implies that (1) is true.
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  3. #3
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    Re: Operator well-defined

    I have come up with an answer myself in the meantime, and I don't think that yours is entirely correct.

    Quote Originally Posted by Rebesques View Post

    $\displaystyle ||Ax||^2=\sum_n\sum_k|a_{k-n+1}x_k|^2 $
    I don't think this is correct, in my opinion it should look like this:

    $\displaystyle \|Ax\|_{\ell^{2}}^{2} = \sum_{n=1}^{\infty} \left|\sum_{k=1}^{\infty} a_{k+n-1}x_k \right|^2 \leq \sum_{n=1}^{\infty} \left( \sum_{k=1}^{\infty} |a_{k+n-1}x_k | \right)^2 \leq (\text{Cauchy-Schwarz}) \sum_{n=1}^{\infty} \left(\sum_{k=1}^{\infty}|a_{k+n-1}|^2 \cdot \sum_{k=1}^{\infty}|x_k|^2 \right) = \|x\|_{\ell^{2}}^2 \cdot \sum_{n=1}^{\infty} \sum_{k=1}^{\infty}|a_{k+n-1}|^2 $

    Now, because of the monotonicity of $\displaystyle (a_n)_{n\in\mathbb{N}}$, we can establish that $\displaystyle |a_{n+k-1}| \leq |a_{k}| \; \forall n,k \in \mathbb{N} $ and in the same way $\displaystyle |a_{n+k-1}| \leq |a_{n}| \; \forall n,k \in \mathbb{N} $. Using this, we get

    $\displaystyle \|x\|_{\ell^2}^2 \cdot \sum_{n=1}^{\infty} \sum_{k=1}^{\infty}|a_{k+n-1}|^2 \leq \|x\|_{\ell^2}^2 \cdot \sum_{n=1}^{\infty}\sum_{k=1}^{\infty} |a_n|\cdot |a_k| = \|x\|_{\ell^{2}}^{2} \cdot \|a\|_{\ell^{1}}^{2} $.

    Thank you for your help,
    Best Regards,
    Stiwan
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  4. #4
    Super Member Rebesques's Avatar
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    Re: Operator well-defined

    Υes, you are right about the square on $\displaystyle |a_{k-n+1}x_k|$, it was a lapse. And your solution is elegant.

    But my gravest errors are these:

    1) In the last series. I have written
    $\displaystyle \sum_k a_{k-n+1}\leq n^{-2}$

    whereas, it should be something more in the like of

    $\displaystyle \sum_{k\geq k_n} |a_{k-n+1}|\leq n^{-2}$.


    2) The indices of the series are $\displaystyle a_{n+k-1}$ and not my $\displaystyle a_{k-n+1}$, for which one cannot invoke monotonicity save
    for the eventual $\displaystyle |a_{k-n+1}|^2\leq |a_{n-k+1}|$.
    So actually, i was dealing with a different exercise.... time for a myopia check.
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