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If we have a Hilbert space that has two equivalent norms (and inner products), are the Riesz maps (from Riesz representation theorem) associated with each inner product the same?

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up vote 1 down vote accepted

Note: of course, it fails already in $\mathbb{C}$. Just consider $(z,w)=z\overline{w}$ and $(z,w)'=2z\overline{w}$. And you see how to generalize this dilation idea. But I thought it could be more interesting to give the slightly more general viewpoint below.

If $(x,y)$ is one inner product and if $P$ is invertible in $B(H)$, then $$ (x,y)':=(Px,Py) $$ is an equivalent inner product. If $P$ is not an isometry for the latter, take $x$ such that $\|Px\|\neq \|x\|$. Then $$ (x,x)'=(Px,Px)=\|Px\|^2\neq \|x\|^2=(x,x). $$ In particular, the functionals $(x,\cdot )'$ and $(x,\cdot )$ are distinct.

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Hmm. But your norm on the left of the $\neq$ sign is with respect to $(\cdot,\cdot)'$ and the norm on the right of the sign is wrt. the original inner product. – aere Apr 29 '13 at 12:46
Consider $P$ constant > 0 . – aere Apr 29 '13 at 12:48
@aere No. The only thing that is wrt to the new inner product is $(x,x)'$. By definition, this is $(Px,Px)$ in the old inner product. – 1015 Apr 29 '13 at 12:48
Thanks. If P is non-zero constant, then it seems to be OK? – aere Apr 29 '13 at 12:52
@aere Yes $P$ scalar $P=cI$ with $c\neq 0$ is the example I gave at the beginning. I just added the rest because it is a standard way to build new inner products from old ones, so I thought you would find it interesting. Do you understand the argument, now? – 1015 Apr 29 '13 at 12:53

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