# product of Hilbert spaces

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..

• What is the norm $\|\cdot\|$? – Ellya May 2 '14 at 6:58
• this is not given – sky90 May 2 '14 at 6:59
• Are you told if $(H,\|\cdot\|)$ is a banach space or anything? – 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 $\|(x,y)-(u,v)\|^2+\|(x,y)+(u,v)\|^2=(\|x-u\|+\|y-v\|)^2+(\|x+u\|+\|y+v\|)^2$

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

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

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.

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