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..
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..
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.
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.