Let $H$ denote a Hilbert space , and $T$ is a map from $A$ to $H$ . $A$ is isomorphic to $H$ with the map $T$ , can we show that $A$ is a Hilbert space ? Moreover , if $H$ is separable , can we prove $A$ is separable ?
My attempt:
$(1)$ $A$ defines an inner product $$(a,b)_A=(T(a),T(b))_H \,\,\,\,\,\,\,\text{$a,b \in A$}$$
$(2)$ If the operator $T$ and $T^{-1}$ are bounded , then we can easily see $A$ is complete . Indeed , if $\{f_n\}$ is a Cauchy sequence in $A$ , then $\{T(f_n)\} $ is a Cauchy sequence in $H$ . Since $T(f_n) \to T(f)$ and $T^{-1}$ is bounded , $f_n$ converges to $f$ .
Can we show that $T$ or $T^{-1}$ defined above are nesscessarily bounded ?
Since $A$ doesn't carry any metric structure and $T$ is not map in the category of (let's say) Banach-spaces, we can't say anything about the metric-structure of $A$ with the help of $T$ and $H$. Of course, $T$ induces a Hilbert-norm on $A$. But if $A$ is a Banach-space with some arbitary norm, this new induced Hilbert-norm can be different.
Example: Take a Hamel-Basis $(e_i)_{i \in I}$ of $L^2(0,1)$ and $(u_i)_{i \in I}$ of $L^1(0,1)$. One can show that the Hamel-dimension of an infinite-dimensional banach space is its cardinality, see the discussion here. I have this used already in the sense that in both cases the index set is the same! (Need: Axiom of choice.) This induces the map $$L^2(0,1) \rightarrow L^1(0,1) \quad \text{via} \quad e_i \mapsto u_i$$ and linear continuation. (In fact, we need the axiom of choice here, too.) This map is obviously one-to-one and surjective. Thus $L^2(0,1)$ and $L^1(0,1)$ are isomorphic as vector spaces. However $L^1(0,1)$ with the usual norm $\|f\|_{L^1} = \int |f| \, \mathrm{d} \lambda$ is not a Hilbert space! (And is not isometric isomorphic to a Hilbert space, because it isn't a dual space of some banach space. This is a consequence of the Krein–Milman theorem.)
One remark to (2): If $T$ is bounded and $A$ a Banach-space, then $T^{-1}$ is by the open map always continuous.