Let $H$ be an infinite dimensional Hilbert space.

claim: $H\times H$ with the norm $\|(x,y)\|=\|x\|+\|y\|$ is an Hilbert space.

I can't find a counterexample..

  • $\begingroup$ What is the norm $\|\cdot\|$? $\endgroup$ – Ellya May 2 '14 at 6:58
  • $\begingroup$ this is not given $\endgroup$ – sky90 May 2 '14 at 6:59
  • $\begingroup$ Are you told if $(H,\|\cdot\|)$ is a banach space or anything? $\endgroup$ – Ellya May 2 '14 at 7:03

If you knw the norm itself I.e. you kbe w what $\|\cdot\|$ is, then you need to find $u,v\in H\times H$ such that:

$\|u-v\|^2+\|u+v\|^2\ne 2(\|u\|^2+\|v\|^2)$

I think there are two possibilities.

  1. Assume $(H,\|\cdot\|)$ is a Hilbert space, then the above holds in $H$

Then $\|(x,y)-(u,v)\|^2+\|(x,y)+(u,v)\|^2=(\|x-u\|+\|y-v\|)^2+(\|x+u\|+\|y+v\|)^2$


$\ne 2(\|x\|+\|y\|)^2+2(\|u\|+\|v\|)^2=2(\|(x,y)\|^2+\|(u,v)\|^2$

Thus $H\times H$ is not a Hilbert space.

2.Assume $(H,\|\cdot\|)$ is not a Hilbert space, then one or more of the Hilbert space axioms would fail for $H\times H$ I.e non completeness etc.

Thus $H\times H$ is not a Hilbert space.

  • $\begingroup$ Yes, sorry I meant $H\times H$ rather than just $H$ :) $\endgroup$ – Ellya May 2 '14 at 7:07
  • $\begingroup$ @sky90 I have put in an edit that I think solves the problem. $\endgroup$ – Ellya May 2 '14 at 7:28
  • $\begingroup$ Yes I followed your first strategy. Thank you :) $\endgroup$ – sky90 May 2 '14 at 7:28
  • $\begingroup$ sorry yesterday I did not notice it... but I think there is a mistake in the second equality in point 1... isn't it? $\endgroup$ – sky90 May 3 '14 at 5:58
  • $\begingroup$ Which bit sorry? $\endgroup$ – Ellya May 3 '14 at 6:02

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