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Take a Hilbert space $(\mathcal H,(\cdot,\cdot)_{\mathcal H})$ and two equivalent inner products $(\cdot,\cdot)_1$ and $(\cdot,\cdot)_2$ on $\mathcal H$, i.e. such that there are $a,b \in \mathbb R$ with $0 < a \leq b$ satisfying $$ a (f,f)_1 \leq (f,f)_2 \leq b(f,f)_1 \quad \forall f \in \mathcal H $$ Suppose that the metric induced by $(\cdot,\cdot)_i$ makes $(\mathcal H,(\cdot,\cdot)_i)$ a Hilbert space. In this way we get an equivalence relation $\sim$ on the set $X$ of the inner products on $\mathcal H$.

  • If $A : \mathcal H \to \mathcal H$ is, with respect to the inner product of $\mathcal H$, a positive, invertible self-adjoint operator then $$ \mathcal H \times \mathcal H \ni (f,g) \mapsto (Af,g) \in\mathbb C $$ is an inner product on $\mathcal H$ equivalent to $(\cdot,\cdot)_{\mathcal H}$.
  • If $H,K \in X$ are such that $H \sim K$, is it true that $H$ and $K$ are related by an operator $A : (\mathcal H,H) \to (\mathcal H,H)$ as in the previous point?
  • What can be said on $\tilde X := X/\sim$? How many inequivalent inner products on $\mathcal H$ are there? Is there a way to classify them?

For the second point, assuming that $(\mathcal H,H)$ and $(\mathcal H,K)$ are separable, I thought to take two complete orthonormal systems $\{a_i\}_{i \in I}$ on $(\mathcal H,H)$ and $\{b_i\}_{i \in I}$ on $(\mathcal H,K)$ and define $A := T^* T$, where $T : (\mathcal H,H) \to (\mathcal H,H)$ is defined by $$ f \mapsto Tf := \sum_{i \in I} K(f,b_i) a_i $$ In this way, using $H(a_i,a_j) = \delta_{ij} = K(b_i,b_j)$ , $$ \begin{align*} H(f,Ag) = H(Tf,Tg) &= \sum_{i,j,k \in I} K(f,b_i) H(a_i,a_j) K(b_j,g) \\ &= \sum_{i,j,k \in I} K(f,b_i) K(b_i,b_j) K(b_j,g) = K(f,g) \end{align*} $$ Is this reasoning valid? What can be said in general?

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The Riesz representation theorem defines an invertible mapping $\Delta: \mathcal{H}\to\mathcal{H}^*$. You can define $\Delta_i$ for each of your inner products. Then your $T = \Delta_2^{-1}\Delta_1$. –  Willie Wong Jul 21 '11 at 23:39
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In the separable case, all infinite dimensional separable Hilbert spaces are isometrically isomorphic to $\ell^2$, which should answer your other question. –  Willie Wong Jul 21 '11 at 23:47
    
@WillieWong: thanks for your interesting comments! –  Robert Jul 22 '11 at 20:29
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1 Answer 1

up vote 11 down vote accepted

The first observation to make is that, given a bounded sesquilinear form $B$ on $\mathcal{H}$, i.e., a form such that $|B(x,y)| \leq \|B\|\,\|x\|\,\|y\|$, there exists a unique bounded linear operator $T$ on $\mathcal{H}$ such that $B(x,y) = \langle x,Ty \rangle$ and $\|T\| = \|B\|$ by the Riesz representation theorem (see Pedersen, Analysis Now, Lemma 3.2.2, p.89 for details).

If $B$ is Hermitian, the identity $$\langle x, Ty \rangle = B(x,y) = \overline{B(y,x)} = \overline{\langle y, Tx \rangle} = \langle Tx, y\rangle = \langle x, T^{\ast}y\rangle$$ shows that $T = T^{\ast}$, so $T$ is self-adjoint.

If $B$ is positive definite then of course $T$ must be injective, but it need not be invertible (an injective self-adjoint operator has dense range, of course).

It is not hard to show that your equivalence relation on scalar products implies that the operator $T$ is bounded and bounded away from zero, and as it is self-adjoint, this means that $T$ is invertible, see Pedersen, Proposition 3.2.6, page 90.

Conversely, if $T$ is bounded, self-adjoint and invertible, then $T$ is bounded away from zero and hence the scalar products $\langle x, Ty \rangle$ and $\langle x,y\rangle$ are equivalent.

It is plain that for a scalar product $B$ the condition $$a \langle x,x \rangle \leq B(x,x) \leq b\langle x,x\rangle \quad \text{for all } x \in \mathcal H$$ implies that $\mathcal{H}$ is complete with respect to $B$, so we can sum up:

If a scalar product $B$ satisfies $a \|x\|^2 \leq B(x,x) \leq b \|x\|^2$ for all $x$ then $B(x,y) = \langle x, Ty \rangle$ for a unique bounded, invertible self-adjoint operator $T$. Conversely, for every bounded, invertible self-adjoint operator we get equivalent scalar products.

In particular, there is only one equivalence class of scalar products. It is clear that you can perform all this with bases as well, and, as Willie said, your reasoning is correct. However, these calculations seem more involved to me than the above considerations.


Added:

A closely related result that I should have mentioned is the Lax–Milgram theorem.

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Thanks for the reference and the clear explanation! I have to admit that the basis-free approach is more clear and elegant. –  Robert Jul 22 '11 at 20:30
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