Non-orthogonal extension of an inner product

Question

Let $W$ be a real linear vector space of possibly infinite dimension and let $<\cdot, \cdot>_W$ be an inner product on $W$. Further let $V$ be a superspace of $W$, such that holds $V = W \bigoplus span\{\psi\}$.

I am calling an inner product $<\cdot, \cdot>_V$ an extension of $<\cdot, \cdot>_W$ orthogonal to $\psi$, if it satisfies

1. $<\phi, \rho>_V = <\phi, \rho>_W \quad \forall\,\phi, \rho \in W$
2. $<\phi, \psi>_V = 0 \quad \forall\,\phi \in W$.

Any element $\phi \in V$ has a unique decomposition $\phi = \phi_w + a\cdot\psi$ with $\phi_w \in W$ and $a\in\mathbb{R}$. Hence the mapping $$\alpha_\psi: V \to \mathbb{R}, \; (\phi_w+ a\cdot\psi) \mapsto a$$ is well defined, linear and suffices $\alpha_\psi(\phi)=0$ for all $\phi \in W$.

We can decompose any extension $<\cdot, \cdot>_V$ of $<\cdot, \cdot>_W$ orthogonal to $\psi$ as follows $$\begin{align} <\phi, \rho>_V &= <\phi_w, \rho_w>_W + \alpha_\psi(\phi)<\rho,\psi>_V+\alpha_\psi(\rho)<\phi,\psi>_V + \alpha_\psi(\phi)\alpha_\psi(\rho)<\psi,\psi>_V\\ &= <\phi_w, \rho_w>_W + \alpha_\psi(\phi)\alpha_\psi(\rho)<\psi, \psi>_V \end{align}.$$ In other words: Any extension $<\cdot, \cdot>_V$ of $<\cdot, \cdot>_W$ orthogonal to $\psi$ is of the form $$<\phi, \rho>_V = <\phi_w, \rho_w>_W + \lambda\cdot \alpha_\psi(\phi)\alpha_\psi(\rho),$$ for some $\lambda > 0$.

My question: Can we say anything about the existence and uniqueness of extensions of the inner product $<\cdot, \cdot>_W$ to $V$ that are not orthogonal to $\psi$ (i.e. suffice 1. but not 2.)?

In finite dimension uniqueness certainly does not hold (but can we say anything about the cardinality of extensions w.r.t. the dimension?) In infinitely dimensions does existence holds? And is uniqueness possible?

Answer 1

Yes, you can, and it is very easy: suppose $A$ is a linear operator on $V$ that is bounded away from $0$ and $\infty$, i. e. there exist constants $0 < c < C$ such that \begin{align*} c \, \mathbb{1} \leq A \leq C \, \mathbb{1} \end{align*} holds. Note that in general $A$ mixes the subspaces $W$ and $\mathrm{span}\{ \psi \}$!

Now define a second scalar product on $V$ which is weighted by $A$, namely \begin{align*} \langle \varphi \, , \, \eta \rangle_A := \bigl \langle \varphi \, , \, A \eta \bigr \rangle_V . \end{align*} You can check that $\langle \, \cdot \, , \, \cdot \, \rangle_A$ satisfies all the axioms of a scalar product, and that the two norms induced by the scalar products $\langle \, \cdot \, , \, \cdot \, \rangle_A$ and $\langle \, \cdot \, , \, \cdot \, \rangle_V$ are equivalent. Thus, you obtain a second Hilbert space $V_A$ that agrees with $V$ as Banach spaces. That means that $A$ is still a bounded operator on $V_A$ that is bounded away from $0$ and $\infty$ (with different constants).

Then unless $A$ is block-diagonal with respect to the direct sum decomposition $V = W \oplus \mathrm{span} \{ \psi \} = V_A$ (the latter equality is the equality as Banach spaces!), the two subspaces $V_A$ (i. e. $V$ equipped with the alternative scalar product) are no longer orthogonal to each other. These arguments also work if you tack on an $N$- or infinite-dimensional subspace rather than a $1$-dimensional one.