Let $V$ be the space of real sequences $x_k$ so that $\sum_{k=1}^\infty x_k^2$ converges. Let $\langle x,y\rangle=\sum_{k=1}^\infty x_k y_k$

Prove that $\sum_{k=1}^\infty |x_k y_k|$ converges

My attempt: In order to prove that this sum converges I´m going to prove that this sum is a cauchy sequence. Let $\varepsilon>0$ then I need to prove that ther exist $N$ so that $\forall m,n>N$, $$|\sum_{k=1}^n|x_k y_k|-\sum_{k=1}^m |x_k y_k||<\varepsilon$$

Let $m>n$ (without lost of generality $n>m$) then $$|\sum_{k=1}^n|x_k y_k|-\sum_{k=1}^m |x_k y_k|| = ||x_{n+1}y_{n+1}| + |x_{n+2}y_{n+2}| + \cdots+|x_m y_m|| = |x_{n+1}y_{n+1}| +\cdots+|x_m y_m|$$

I want to give a bound to this sum, so that $\left(\sum_{k=n+1}^m x_k^2)\right) \left(\sum_{k=n+1}^m y_k^2\right)$ is in my bound

$|x_{n+1}y_{n+1}|$ is not always $\le |x_{n+1}y_{n+1}|^2$ but there exist $\lambda_{n+1} \in \mathbb R$ large enough so that $|x_{n+1}y_{n+1}|\le\lambda_{n+1} |x_{n+1}y_{n+1}|^2$

Hence there exist $\lambda_{n+2},\ldots,\lambda_{m} \in \mathbb R$ large enough so that: $$|x_{n+2}y_{n+2}|\le \lambda_{n+2}|x_{n+2}y_{n+2}|^2,\ldots,|x_m y_m|\le \lambda_m |x_m y_m|^2$$ and therefore: $$\sum_{k=n+1}^m |x_k y_k|\le \sum_{k=n+1}^m \lambda_k|x_k y_k|^2$$

Taking $\lambda=\max\{\lambda_{n+1},\ldots,\lambda_{m}\}$ $$\sum_{k=n+1}^{m}\lambda_{k}|x_{k}y_{k}|^{2}\le \sum_{k=n+1}^{m}\lambda|x_{k}y_{k}|^{2}$$

On ther other hand I see that the product $$\left(\sum_{k=n+1}^m x_k^2\right) \left(\sum_{k=n+1}^m y_k^2\right)=(x_{n+1}y_{n+1})^2 +\cdots+(x_{m}y_{m})^{2}+ \text{other terms}$$

then $$\sum_{k=n+1}^{m}\lambda|x_k y_k|^2 \le \sum_{k=n+1}^m \lambda(x_{k}y_{k})^2 + \lambda(\text{other terms}) =\lambda(\sum_{k=n+1}^m x_k^2)\left(\sum_{k=n+1}^m y_k^2\right)$$

therefore: $$ |\sum_{k=1}^n |x_k y_k|-\sum_{k=1}^m|x_k y_k||=\sum_{k=n+1}^m |x_k y_k|\le \lambda\left(\sum_{k=n+1}^m x_k^2\right) \left(\sum_{k=n+1}^m y_k^2\right)$$

By hypothesis I have that $\sum_{k=1}^{\infty}x_{k}^{2}$ and $\sum_{k=1}^{\infty}y_{k}^{2}$ converges, therefore the product $$\left(\sum_{k=1}^\infty x_k^2 \right) \left(\sum_{k=1}^\infty y_k^2\right)$$ also converges and $$\lambda\left(\sum_{k=1}^\infty x_k^2\right) \left(\sum_{k=1}^\infty y_k^2\right)$$ also converges

So taking $N=\max\{N_{1},N_{2}\}$, $$\lambda\left(\sum_{k=n+1}^m x_k^2\right) \left(\sum_{k=n+1}^m y_k^2\right) < \varepsilon$$ so that means that:

$$|\sum_{k=1}^n |x_k y_k|-\sum_{k=1}^m |x_k y_k||=\sum_{k=n+1}^m |x_k y_k| < \varepsilon$$

I would like you to tell me if this proof is correct. I would really appreciate your help :)

  • 4
    $\begingroup$ What prevents you from immediately concluding convergence from the Cauchy Schwarz inequality? $\endgroup$ – Alex R. Aug 26 '14 at 1:56
  • $\begingroup$ the absolute value in the series, I don´t see how I can get rid of it :( $\endgroup$ – user128422 Aug 26 '14 at 1:58
  • $\begingroup$ In fact I use the cauchy schwarz inequality to prove that $\sum_{k=1}^{\infty}x_{k}y_{k}$ converges $\endgroup$ – user128422 Aug 26 '14 at 2:00
  • $\begingroup$ The absolute value is irrelevant. You have $|x_iy_i|=|x_i||y_i|$ and then do Cauchy Schwarz on the latter two. $\endgroup$ – Alex R. Aug 26 '14 at 2:18
  • $\begingroup$ so $\sum_{k=n+1}^{m}|x_{k}y_{k}|\le ((\sum_{k=n+1}^{m}(x_{k})^{2}))^{1/2}((\sum_{k=n+1}^{m}(y_{k})^{2}))^{1/2})$? $\endgroup$ – user128422 Aug 26 '14 at 2:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.