Bogoliubov Transformation Let $\mathcal{A}_{CAR}(\mathcal{H})$ be a CAR algebra over a Hilbert space $\mathcal{H}$.
Consider a linear $S$ and an antilinear $T$ both bounded operators acting on $\mathcal{H}$ satisfying:
$$S^*S+T^*T=1=SS^*+TT^*\\
S^*T+T^*S=0=ST^*+TS^*$$
Define a Bogoliubov transformation by:
$$\gamma[a(f)]:=a(Sf)+a^*(Tf)\text{ and }\gamma[1]:=1$$
extended on the span of $a(f)$, i.e.:
$$\gamma[\lambda a(f)]:=\lambda\gamma[a(f)]\\
\gamma[a(f)+a(g)]:=\gamma[a(f)]+\gamma[a(g)]\\
\gamma[a(f)a(g)]:=\gamma[a(f)]\gamma[a(g)]\\
\gamma[a(f)^*]:=\gamma[a(f)]^*$$
Similarly define its backtransformation by:
$$\delta[a(f)]:=a(U^*f)+a(V^*f)^*\text{ and }\delta[1]:=1$$
Question 1: Why are the Bogoliubov transformation and its backtransformation well defined?
My problem is that the $a(f)$ are not independent to each other:
$$a(\lambda f)=\overline{\lambda}a(f)\text{ and }a(f+g)=a(f)+a(g)\\
\{a(f),a(g)\}=0\text{ and }\{a(f),a(g)^*\}=\langle f,g\rangle 1$$
Question 2: Why are the Bogoliubov transformation and its backtransformation inverse to each other?
Question 3: Why are the Bogoliubov transformation and its backtransformation continuous?
 A: Although this is an old question, I would still like to add that the Bogoliubov transform is actually induced by a unitary transform on the fermion Fock space $\mathscr{F}$, i.e., for every $S,T$ which satisfy $\operatorname{tr}{(TT^*)}<\infty$, there exists a unitary operator $W:\mathscr{F} \rightarrow \mathscr{F}$ such that
\begin{equation}
W a^*(f)W^* =a^* (Sf)+a(Tf)
\end{equation}
Your questions then follow from properties of the unitary transform.
Just a comment.
I personally think that this is very important to understanding the Bogoliubov transform, but unfortunately, I have almost never seen it mentioned in textbooks or papers on the subject. The only reference I know that proves this rigorously is "Bach, V., Lieb, E.H. & Solovej, J.P. Generalized Hartree-Fock theory and the Hubbard model. J Stat Phys 76, 3–89 (1994)." (see Theorem 2.2) (though I admit that there are a few steps in the proof that I have not checked carefully)
A: The Bogoliubov transformation is well defined since it respects the antilinearity:
$$a(U(\lambda f))+a(V(\lambda f))^*=\gamma[a(\lambda f)]=\gamma[\overline{\lambda}a(f)]=\overline{\lambda}\gamma[a(f)]
\\=\overline{\lambda}a(Uf)+\overline{\lambda}a(Vf)^*=a(U(\lambda f))+a(V(\lambda f))^*\\
a(U(f+g))+a(V(f+g))^*=\gamma[a(f+g)]=\gamma[a(f)+a(g)]=\gamma[a(f)]+\gamma[a(g)]
\\=a(Uf)+a(Ug)+a(Vf)^*+a(Vg)^*=a(U(f+g))+a(V(f+g))^*
$$
and the anticommutation relations:
\begin{align}
0&=\gamma[0]=\gamma[\{a(f),a(g)\}]=\{\gamma[a(f)],\gamma[a(g)]\}\\
&=\{a(Uf),a(Ug)\}+\{a(Uf),a(Vg)^*\}+\{a(Vf)^*,a(Ug)\}+\{a(Vf)^*,a(Vg)^*\}\\
&=0+\langle Uf,Vg\rangle 1+\langle Ug,Vf\rangle 1+0=\langle f,(U^*V+V^*U)g\rangle 1=0\\
\langle f,g\rangle 1&=\gamma[\langle f,g\rangle 1]=\gamma[\{a(f),a(g)^*\}]=\{\gamma[a(f)],\gamma[a(g)]^*\}\\
&=\{a(Uf),a(Ug)^*\}+\{a(Uf),a(Vg)\}+\{a(Vf)^*,a(Ug)^*\}+\{a(Vf)^*,a(Vg)\}\\
&=\langle Uf,Ug\rangle 1+0+0+\langle Vg,Vf\rangle 1=\langle f,(U^*U+V^*V)g\rangle 1=\langle f,g\rangle 1
\end{align}
The Bogoliubov backtransformation is well defined too since it also respects the antilinearity:
$$a(U^*(\lambda f))+a(V(\lambda f))^*=\delta[a(\lambda f)]=\delta[\overline{\lambda}a(f)]=\overline{\lambda}\delta[a(f)]
\\=\overline{\lambda}a(U^*f)+\overline{\lambda}a(V^*f)^*=a(U^*(\lambda f))+a(V^*(\lambda f))^*\\
a(U^*(f+g))+a(V^*(f+g))^*=\delta[a(f+g)]=\delta[a(f)+a(g)]=\delta[a(f)]+\delta[a(g)]
\\=a(U^*f)+a(U^*g)+a(V^*f)^*+a(V^*g)^*=a(U^*(f+g))+a(V^*(f+g))^*
$$
and the anticommutation relations:
\begin{align}
0&=\delta[0]=\delta[\{a(f),a(g)\}]=\{\delta[a(f)],\delta[a(g)]\}\\
&=\{a(U^*f),a(U^*g)\}+\{a(U^*f),a(V^*g)^*\}+\{a(V^*f)^*,a(U^*g)\}+\{a(V^*f)^*,a(V^*g)^*\}\\
&=0+\langle U^*f,V^*g\rangle 1+\langle U^*g,V^*f\rangle 1+0=\langle f,(UV^*+VU^*)g\rangle 1=0\\
\langle f,g\rangle 1&=\delta[\langle f,g\rangle 1]=\delta[\{a(f),a(g)^*\}]=\{\delta[a(f)],\delta[a(g)]^*\}\\
&=\{a(U^*f),a(U^*g)^*\}+\{a(U^*f),a(V^*g)\}+\{a(V^*f)^*,a(U^*g)^*\}+\{a(V^*f)^*,a(V^*g)\}\\
&=\langle U^*f,U^*g\rangle 1+0+0+\langle V^*g,V^*f\rangle 1=\langle f,(UU^*+VV^*)g\rangle 1=\langle f,g\rangle 1
\end{align}
They're inverse to each other:
\begin{align}
\delta[\gamma[a(f)]]&=\delta[a(Uf)+a(Vf)^*]=\delta[a(Uf)]+\delta[a(Vf)]^*\\
&=a(U^*Uf)+a(V^*Uf)^*+a(U^*Vf)^*+a(V^*Vf)\\
&=a((U^*U+V^*V)f)+a((U^*V+V^*U)f)^*=a(f)\\
\gamma[\delta[a(f)]]&=\gamma[a(U^*f)+a(V^*f)^*]=\gamma[a(U^*f)]+\gamma[a(V^*f)]^*\\
&=a(UU^*f)+a(VU^*f)^*+a(UV^*f)^*+a(VV^*f)\\
&=a((UU^*+VV^*)f)+a((UV^*+VU^*)f)^*=a(f)
\end{align}
Moreover both are continuous:
$$\|\gamma[A]\|^2=\|\gamma[A^*A]\|=\sup\sigma(\gamma[A^*A])\leq\sup\sigma(A^*A)=\|A^*A\|=\|A\|^2\\
\|\delta[B]\|^2=\|\delta[B^*B]\|=\sup\sigma(\delta[B^*B])\leq\sup\sigma(B^*B)=\|B^*B\|=\|B\|^2$$
(Besides, they are therefore even isometries: $\|\gamma[A]\|=\|A\|\text{ and }\|\delta[B]\|=\|B\|$)
Finally, by the uniform extension principle it follows:
$$\bar{\delta}[\bar{\gamma}[A]]=\bar{\delta}[\bar{\gamma}[\lim_n A_n]]=\bar{\delta}[\lim_n\gamma[A_n]]=\lim_n\delta[\gamma[A_n]]=\lim_n A_n=A\\
\bar{\gamma}[\bar{\delta}[A]]=\bar{\gamma}[\bar{\delta}[\lim_n A_n]]=\bar{\gamma}[\lim_n\delta[A_n]]=\lim_n\gamma[\delta[A_n]]=\lim_n A_n=A$$
