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Let $H$ be Hilbert and let it have two innner products $(\cdot,\cdot)_1$ and $(\cdot,\cdot)_2$.

If the norms $|\cdot|_1$ and $|\cdot|_2$ are equivalent, does this ever imply: there exist constants $C_i$ such that $$C_1(a,b)_2 \leq (a,b)_1 \leq C_2(a,b)_2$$ (i.e. the inner products are equivalent in some sense)?

under what conditions on $H$ does this happen?

workl
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    Presumably you intend that the constants are (strictly) positive? (Without requiring strict positivity, the polarization identity would always give comparisons of the inner product from comparisons of the norms...) – paul garrett Jun 27 '13 at 19:54
  • @paulgarrett I see. Yes positive is what i was hoping for.. – workl Jun 27 '13 at 21:16

1 Answers1

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You have such constants that

$$C_1\langle a\mid b\rangle_2 \leqslant \langle a \mid b\rangle_1 \leqslant C_2\langle a\mid b\rangle_2$$

for all $a,\, b \in H$ if and only if the two norms are constant multiples of each other.

Firstly, such an inequality can only hold if $\langle\mid\rangle_2$-orthogonality implies $\langle\mid\rangle_1$-orthogonality.

Assume that to be the case.

Then let $(e_\nu)$ be a $\langle\mid\rangle_2$-ONB.

For $\nu \neq \mu$ (if $\dim H = 1$, there is nothing to show), we have $\langle e_\nu + e_\mu \mid e_\nu - e_\mu\rangle_2 = \lVert e_\nu\rVert_2^2 - \lVert e_\mu\rVert_2^2 = 0$, and therefore

$$\langle e_\nu + e_\mu \mid e_\nu - e_\mu\rangle_1 = \lVert e_\nu\rVert_1^2 - \lVert e_\mu\rVert_1^2 = 0,$$

i.e. $\lVert e_\nu \rVert_1$ is constant, and thus, up to scaling by a factor independent of $\nu$, $(e_\nu)$ is also an ONB for $\langle\mid\rangle_1$.

Daniel Fischer
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