# alternative for Schwarz inequality

I wish to show that $$\sum_{n=1}^{\infty}|\langle e_n, x\rangle||\langle e_n,y \rangle| \leq \|x\|\|y\|$$

where $$\{e_n\}$$ is an orthonormal system. But my professor wants me to prove it without using Cauchy-Schwarz (or Hölder) inequality. His sugestion is to use the obvious inequality $$(a-b)^2 \geq 0$$ for the sequences $$a_n = |\langle e_n, x\rangle|$$ and $$b_n = |\langle e_n, y\rangle|$$. It follows that, in general, $$\sum_{n=1}^{\infty}\{(a_n - b_n)^2\} \geq 0 \implies \sum_{n=1}^{\infty}\{a_n^2 + b_n^2 - 2a_nb_n\}\geq 0$$ is this right? if so, by Bessel's inequality (i'm allowed to use it lol), $$\sum a_n^2 = \sum|\langle e_n, x\rangle|^2 \leq \|x\|$$ and $$\sum b_n^2 =\sum |\langle e_n, y\rangle|^2 \leq \|y\|^2$$. With this, I get to the inequality: $$2\sum_{n=1}^{\infty}|\langle e_n, x\rangle||\langle e_n,y \rangle| \leq \|x\|^2 + \|y\|^2$$ and... this is not quite what I want. Did i do something wrong or i'm forgetting something? how can I proceeed or restart it? any help will be very appreciated! thanks in advance.

• Just to make sure, $e_n$ vectors form an orthonormal basis? May 12 '21 at 17:44
• yea, i forgot to say that May 12 '21 at 17:45

Here is an idea I have using Cauchy product (I don't know if you are allowed to use it) :

Because the series are absolutely convergent then with Cauchy product :

$$\left(\sum_{n=1}^{\infty} a_n b_n \right)^2 = \sum_{n=1}^{\infty}\sum_{k=1}^n (a_k b_{k}) (a_{n-k}b_{n-k})$$

Moreover with what you have done using Bessel's inequality:

$$||x||^2||y||^2 \geq \left(\sum_{n=1}^{\infty} a_n^2 \right)\left(\sum_{n=1}^{\infty} b_n^2 \right)= \sum_{n=1}^{\infty}\sum_{k=1}^n a_k^2b_{n-k}^2$$ and using your professor's suggestion, for all $$n \geq k \geq 1$$ we have:

$$(a_kb_{n-k})(a_{n-k} b_{k}) \leq \dfrac{1}{2}(a_k^2 b_{n-k}^2+a_{n-k}^2b_k^2)$$

Therefore :

$$\left(\sum_{n=1}^{\infty} a_n b_n \right)^2 = \sum_{n=1}^{\infty}\sum_{k=1}^n a_k b_{n-k} a_{n-k}b_{k} \leq \sum_{n=1}^{\infty}\sum_{k=1}^n \dfrac{1}{2}(a_k^2 b_{n-k}^2+a_{n-k}^2b_k^2)$$

Furthermore :

$$\sum_{n=1}^{\infty}\sum_{k=1}^n \dfrac{1}{2}(a_k^2 b_{n-k}^2+a_{n-k}^2b_k^2) = \sum_{n=1}^{\infty} \dfrac{1}{2} \left(\sum_{k=1}^n a_k^2 b_{n-k}^2 + \sum_{k=1}^n a_{n-k}^2 b_{k}^2 \right) = \sum_{n=1}^{\infty}\sum_{k=1}^n a_k^2b_{n-k}^2$$

Hence: $$\left(\sum_{n=1}^{\infty} a_n b_n \right)^2\leq||x||^2||y||^2$$

Finally:

$$\sum_{n=1}^{\infty} |\langle e_n|x\rangle ||\langle e_n|y \rangle|\leq ||x|| ||y||$$

• Unfortunatelly i didn't see anything related to this Cauchy product in my classes, but I still tried following this way and could avoid the Cauchy product. This article: uni-miskolc.hu/~matsefi/Octogon/volumes/volume1/article1_19.pdf shows a proof of Cauchy-Schwarz inequality using the fact that $(a-b)^2 \geq 0$. In the end he just wanted me to show every single step on the calculation. Thanks to you I could do it! May 12 '21 at 21:58
• @PedroItalo Glad I could help you! Well Cauchy product is convenient for infinite series that converge absolutely. In the article (that is by the way very interesting), they only prove the result for finite sequences, but you can extend it to your case.
– Axel
May 12 '21 at 22:34