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I came across a question in linear algebra, Prove that $W = \{ \langle a_n \rangle \in V : \sum_{n=1}^\infty a_i^2$ is finite } is a vector space, where V is vector space of sequences in R. Now, I tried to proceed as usual, but got stuck in proving that $\sum_{n=1}^\infty {ab}$ is finite. This question is not from a standard textbook, and hence I am unsure whether the statement is even true.

With some examples, it is easy to see that this statement is definitely true for series of the form $\sum \frac{1}{n^k}, k\in R$. However, whether this extends to all sequences, I am not sure. If your answer is in the affirmitive, please provide a proof. I did try to prove the statement, but couldn't (Cauchy series doesn't help). I think Cauchy's inequality might help, but I am unsure how to use it(It has been time since I studied it in functional analysis class while in undergraduate class).

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    $\begingroup$ Yes, Cauchy’s inequality will do it. $\endgroup$ Commented Jan 8, 2021 at 3:01
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    $\begingroup$ Please check that I've correctly made the edits you requested $\endgroup$ Commented Jan 8, 2021 at 3:04
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    $\begingroup$ Ah, Thanks @HallaSurvivor. $\endgroup$
    – IamThat
    Commented Jan 8, 2021 at 3:05
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    $\begingroup$ Try the Cauchy Bunyakovsky Schwarz Bieber inequality. $\endgroup$
    – copper.hat
    Commented Jan 8, 2021 at 3:07
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    $\begingroup$ And, in future, I recommend to stop using $*$ (*). In math, it usually denotes very special things (e.g. convolution), not multiplication. The latter is usually denoted by $\cdot$ (\cdot), $\times$ (\times), or just by absence of any symbol (which would be my choice in the present case). $\endgroup$
    – metamorphy
    Commented Jan 8, 2021 at 8:49

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Just remembered the inequality in its correct form (credit- Wikipedia). The statement itself is a one-line proof: $|\langle u,v\rangle|^2 \leq \langle u,u\rangle\cdot\langle v,v\rangle$ I was writing it as $\sum{a_ib_i \leq (\sum a_i)^{1/2}(\sum b_i)^{1/2}}$ Credit : Qiaochu Yuan and Copper.hat in the comments section.

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    $\begingroup$ The credit really goes to Qiaochu Yuan, I just wanted to make sure Justin got inappropriate credit. $\endgroup$
    – copper.hat
    Commented Jan 8, 2021 at 3:23
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    $\begingroup$ LOL the Bieber jokes just don't stop. $\endgroup$
    – IamThat
    Commented Jan 8, 2021 at 3:26

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