Let $A$ and $B$ be two separable $C^\star$ algebras, and $\phi:A\to B$ a bounded linear map. A faithful state on a $C^\star$ algebra $A$ is a state such that for $a$ in $A$, $\phi(a^\star a)=0$ iff $a=0$. Every separable $C^\star$ algebra has a faithful state (here is an answer that discusses this). If $\phi$ is a faithful state on a $C^\star$ algebra $A$, then $\langle a,b \rangle _\phi=\phi(b^\star a)$ for all $a,b$ in $A$ yields an inner product on $A$, and thus induces a norm $\lVert a\rVert_\phi =\sqrt{\phi(a^\star a)}$. Of course, $A$ might not be Cauchy complete with respect to the norm $\lVert \cdot \rVert_\phi$, in which case we denote the completion of $A$ with respect to $\lVert \cdot \rVert_\phi$ as $L^2 _\phi (A)$. It is direct to see that when $A$ is finite-dimensional, it is in fact Cauchy complete with respect to $\lVert \cdot \rVert_\phi$, that is, $L^2 _\phi(A)$ and $A$ are isomorphic as linear spaces.
My first question is:
(1) For infinite dimensional separable $A$, does there always exist a faithful state $\phi$ such that $A$ is Cauchy complete with respect to $\lVert \cdot \rVert_\phi$? That is, $L^2 _\phi(A)$ and $A$ are isomorphic as linear spaces? If not, what are necessary and / or sufficient conditions that must be imposed on $A$ for this to hold true?
Now, let us consider finite dimensional $C^\star$ algebras $A,B$, with faithful states $\phi,\psi$. Let $\alpha:A\to B$ be a linear map (and hence bounded: $A,B$ being finite dimensional), we denote by the same symbol the corresponding linear map between $L^2 _\phi(A)$ and $L^2 _\psi(B)$ (which are isomorphic to $A,B$ respectively as linear spaces), and is also bounded (finite dimensional). Now, we denote by $\alpha^\star:L^2 _\psi(B)\to L^2 _\phi(A)$ the Hilbert space adjoint of $\alpha$, an denote by the same symbol the corresponding linear map between $B,A$.
My second question is:
(2) Given two separable infinite dimensional $C^\star$ algebras $A,B$, with separable states $\phi,\psi$, and a bounded linear map $\alpha:A\to B$, is it possible to define a bounded linear map $\alpha^\star:B\to A$, such that $\langle \alpha(a), b \rangle_\psi = \langle a, \alpha^\star(b) \rangle_\phi$ for all $a$ in $A$ and $b$ in $B$? (I am looking for an answer even when $A\subsetneq L^2 _\phi(A)$, $B\subsetneq L^2 _\psi(B)$ as linear spaces).
My attempts (unsuccessful):
For the first question, I tried out several (simple) examples of $C^\star$ algebras in infinite dimensions, like $C(\mathbb{T})$ (continuous functions on the unit circle, with the supremum norm, multiplication pointwise, and the involution given by complex conjugation), and $\ell^\infty$ (with involution given by the complex conjugation, multiplication entrywise), and a couple of similar examples. However, with the half a dozen faithful states I could cook up overall (for example, for $C(\mathbb{T})$, $f\mapsto\frac{1}{2\pi} \int_\mathbb{T} f d\mu$ with $\mu$ being the Lebesgue measure, and for $\ell^\infty$, $(x_n)_n \mapsto \sum ^{j=1} _{\infty} \frac{x_j}{2^j}$) and some others, all being different, in the sense of not being related by permutations, I failed to get a positive answer. But of course, this does not imply one way or the other given the question, other than perhaps saying that the answer might not be immediate. For the second question, I identified several problems: a bounded map $\alpha:A\to B$ might not even be bounded when we consider the $\lVert \cdot \rVert _\phi$, $\lVert \cdot \rVert _\psi$ norms on $A,B$ respectively, rather than the $C^\star$ norms: for example, considering $A=B\otimes_{min} B$, the multiplication map turned out to be unbounded in those norms, for the example of $\ell^\infty$ mentioned above. Even if we consider a case where it is bounded, it might not be Cauchy complete with respect to the norm of interest (for example, in the case of $C(\mathbb{T})$ mentioned above). The proof of the existence and uniqueness of the adjoint that I am familiar with, pivots around the Riesz Representation Theorem for bounded linear functionals on Hilbert spaces, which fails if either of the conditions of completeness or boundedness fail to hold (see here and here respectively, for answers that discuss these).
A little bit on my background, in case it relevant: I studied, for my Masters' thesis, basics of $C^\star$ algebras, positive and completely positive maps, dilations, and related topics. However, I am not (yet) familiar with the theory of unbounded linear operators.
Edit 1:
The motivation behind these questions are the following: any finite dimensional $C^\star$ algebra, as a linear space, can be "given" a Hilbert space structure, albeit with a different norm. Still, as everything is finite dimensional, all linear operators on such spaces are bounded. Moreover, for every linear map between two finite-dimensional $C^\star$ algebras, one can define "adjoint" maps, unique for each Hilbert space structure. These "adjoint" maps are sometimes useful in the study of quantum groups and graphs (I think). I was wondering how these carried on to infinite dimensions, if they did at all.
Edit 2:
Actually, even in finite dimensions, $\alpha^\star$ very much depends on $\phi$ and $\psi$, and not just $A$ and $B$, so it is perhaps not possible to have a very general idea of $\alpha^\star$ that is useful as well. However, I provide an example: Let $B=M_n$ ($n\times n$ complex matrices), $\psi(\cdot)=\frac{1}{n}Tr(\cdot)$, $A=B\otimes B$, $\phi=\psi\otimes\psi$ then, with $m:M_n\otimes M_n \to M_n$, $E_{i,j}\otimes E_{k,l} \mapsto \delta_{j,k} E_{i,l}$, where $E_{i,j}$ denotes the matrix with $1$ at the $i,j$-th entry and zero everywhere else. Of course, $\{E_{i,j}\}_{i,j=1}^n$ is a basis of $M_n$, and we write $\otimes_{min}$ as $\otimes$ as everything is nuclear. Then, $m^\star:M_n\otimes M_n \to M_n$, $E_{i,j}\mapsto \sum_{k=1}^n E_{i,k}\otimes E_{k,j}$. This is a "comultiplication", with respect to the trace.
Best Answer
The answer to your first question is no, and here is the reason.
Note that the identity mapping $$ i:A\to L^2_\phi(A) $$ is continuous. Thus, if $A=L^2_\phi(A)$, the open mapping theorem implies that $i$ is an isomorphism and hence $A$ is seen to be isomorphic$^{(1)}$ to a Hilbert space. This would in turn imply that every closed subspace of $A$ is also isomorphic to some Hilbert space.
This said, choose some self adjoint element $a$ in $A$ with infinite spectrum (these exist in any infinite dimensional $C^*$-algebra) and consider the closed $^*$-subalgebra $C^*(a)\subseteq A$ generated by $a$. It is well known that $C^*(a)$ is isometrically isomorphic to $C_0(\sigma (a)\setminus\{0\})$, so we conclude that $C_0(\sigma (a)\setminus\{0\})$ is isomorphic to some Hilbert space.
However it is a well known fact that $C_0(X)$ is not isomorphic to a Hilbert space for every infinite locally compact space $X$.
$^{(1)}$ Two normed spaces $E$ and $F$ are said to be isomorphic if there exists a map $T:E\to F$ which is linear, bijective, continuous, and whose inverse is continuous. If $T$ is moreover isometric, then we say that $E$ and $F$ are isometrically isomorphic.