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We define $\ell_p=\{(x_n)_{n\in{\mathbb{N}}}\in\mathbb{C}^\infty:\sum_n{|x_n|^p}<\infty\}$. With the usual usual norm $||.||_p$ this becomes a Bancach space. Also we have the usual inner product : $\langle x,y\rangle=\sum_i x_i \bar{y_i}$.

Now I have two questions.

Can we can define other norms and inner products on the set $\ell_p$? Obviously any constant multiple of the usual norm is also a norm. But can we have a norm that is not equivalent to the usual $p$-norm (i.e induces a different topology) on the set $\ell_p$?

I also know that $\ell_p$ space is a Hilbert space if and only if $p=2$. But this is true with the usual norm and inner product on the set $\ell_2$. What happens if I change them (assuming that answer to my first question is affirmative)?

Any help is appreciated. Thanks!

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  • $\begingroup$ @John, can you please explain why? Is it because there is an infinite sum involved? $\endgroup$ – ChesterX Oct 29 '14 at 7:57
  • $\begingroup$ That's an interesting question. $\endgroup$ – Rudy the Reindeer Oct 29 '14 at 8:12
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    $\begingroup$ Using this answer here you know that if $p \le q$ you can always endow $\ell^p$ with the $q$ norm. $\endgroup$ – Rudy the Reindeer Oct 29 '14 at 8:15
  • $\begingroup$ Once you equip the set $\ell_p$ with a norm we can define an inner product via the polarization identity only if the norm satisfies the parallelogram law. $\endgroup$ – Mustafa Said Oct 29 '14 at 8:43
  • $\begingroup$ @MattN., so if I understand it correctly, for $1\leq{}p\leq{}2$, $\ell_p$ can have the $||.||_2$-norm. Does this imply it is also Hilbert? $\endgroup$ – ChesterX Oct 29 '14 at 11:03
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There are many norms. This is but an example. Let $W=\{w_i\}_{i=1}^\infty$, $w_i>0$, be a weight-sequence. Then $$ \|x\|_{p,W}=\sum_{í=1}^\infty|x_i|^pw_i $$ defines a norm.

If $W$ is bounded, then it is a norm on $\ell^p$. If moreover $\inf_{i} w_i>0$, then it is equivalent to the $\ell^p$ norm. But if $\inf_{i}w_i=0$, the norms are not equivalent.

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