If $a,b \in \ell^2 $ then $a\bar b \in \ell^1$

($\ell^1$ and $\ell^2$ on $\mathbb C$ and $\bar b$ is complex conjugate of $b$)

What did I write :

Since $a,b \in \ell^2 $, $a=(a_n)$ and $b=(b_n)$ such that $\sum_{n=1}^{\infty}|a_n|^2 \lt \infty$ and $\sum_{n=1}^{\infty}|b_n|^2 \lt \infty$

But I cannot realize that how can I show $\sum_{n=1}^{\infty}|a_n\bar {b_n}| \lt \infty$ from above?

Thanks in advance


Hint: apply the Cauchy-Schwartz inequality.

  • $\begingroup$ Thank you. Also I wonder how can I show it same thing for Lebesgue spaces. Is Cauchy-Schwarz inequality still valid for integrals? $\endgroup$
    – user519955
    Oct 12 '18 at 20:56
  • $\begingroup$ @user519955 Yes it is $\endgroup$
    – JavaMan
    Oct 13 '18 at 2:45
  • $\begingroup$ thanks a lot :) $\endgroup$
    – user519955
    Oct 13 '18 at 9:36

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.