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Let $H$ be a vector space equipped with two inner products $\langle \cdot,\cdot\rangle_1, \ \langle \cdot,\cdot\rangle_2$, s.th. $(H,\langle \cdot,\cdot\rangle_j)$ is a Hilbert space for $j=1,2$. Suppose that $R_1$ and $R_2$ are linear operators on $H$, which are bounded w.r.t. $\|\cdot\|_1$ and $\|\cdot\|_2$,respectively. Moreover, let us assume that $R_1$ is injective and for any $x,y\in H \ \langle R_1x,y\rangle_1=\langle R_2x,y\rangle_2$. Is it true that norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent ?

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  • $\begingroup$ I find it hard to develop an intuition for this. Operators aside, do you have an example of a vector space with two inequivalent Hilbert norms? $\endgroup$ Dec 28, 2015 at 16:46
  • $\begingroup$ No, I don't have such an example. $\endgroup$
    – mikis
    Dec 28, 2015 at 17:37
  • $\begingroup$ That's a question within the question, then. Do you mind if I post it separately? Or you can do so, as well. $\endgroup$ Dec 28, 2015 at 20:14
  • $\begingroup$ Ok. I did it : math.stackexchange.com/questions/1591995/… $\endgroup$
    – mikis
    Dec 28, 2015 at 20:28

1 Answer 1

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Yes.

Say $H_j=(H,<.>_j)$. Define $T:H_1\to H_2$ by $Tx=x$. We need only show that $T$ is bounded (then the open mapping theorem implies that $T^{-1}$ is also bounded, so the two norms are equivalent).

By the closed graph theorem we need only show this: If $||x_n-x||_1\to0$ and $||x_n-y||_2\to0$ then $x=y$. But for every $z$ we have $$\langle R_1x,z \rangle_1 = \lim\langle R_1x_n,z \rangle_1 = \lim \langle R_2x_n,z \rangle_2 = \langle R_2y,z \rangle_2 = \langle R_1y,z \rangle_1.$$

So $R_1x=R_1y$ and hence $x=y$, since $R_1$ is injective.

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