Concerning a functional of a functional of the former - classical fields in Quantum Action Let $\varphi(x)$ and $j(x)$ be two field configurations. Let $\Gamma[\varphi]$ be a functional of the field $\varphi$ defined by:
$$ \Gamma[\varphi] := \inf_j \ F[\varphi, j] = F[\varphi, j_\varphi] \tag{1}$$
where $F[\varphi, j]$ is a functional of both $\varphi$ and $j$, and for any fixed $\varphi$, the unique field configuration that extremizes $F$ is $j_\varphi$, which is to say,
$$ \frac{\delta}{\delta j} F[\varphi,j] \Big\rvert_{j=j_\varphi} = 0 \,. \tag{2}$$
Now suppose that the structure of $F$ allows the following identity to hold:
$$ \frac{\delta}{\delta\varphi} \Gamma[\varphi] + j_\varphi = 0 \,.\tag{3}$$
Please keep in mind that the above equality is an identity, but it can be easily promoted to the status of an equation as follows:
$$ \frac{\delta}{\delta\varphi} \Gamma[\varphi] + j = 0 \,,\tag{4}$$
whose solution is, of course, $$j=j_\varphi\tag{5}$$ for any given $\varphi$. Alternatively, however, we can fix $j$ and ask for the $$\varphi =: \varphi_j\tag{6}$$ that solves it.
We then have an equivalent identity:
$$  \frac{\delta}{\delta\varphi} \Gamma[\varphi]\Big|_{\varphi=\varphi_j} + j = 0 \,.\tag{7} $$
In such a scenario, would it be true to say that
$$ \Gamma[\varphi_j] = F[\varphi_j, j]\tag{8} $$
which allows us to write a functional of $j$ instead of $\varphi$? If yes, can you please provide a justification for the above claim?
Note that it is equivalent to claiming that $$ j_{\varphi_j} = j \,.\tag{9}$$
What does that even mean?
 A: We believe that OP's eqs. (1)-(3) essentially fixes the form of OP's function $F$ to be the underlying function 
$$F[\phi_{\rm cl},J]~=~W_c[J]-J_i \phi^i_{\rm cl} \tag{*}$$
of the QFT Legendre transformation $J_i\leftrightarrow \phi^i_{\rm cl}$ between the generating functional $W_c[J]$ for connected diagrams and the effective action 
$$\Gamma[\phi_{\rm cl}]~=~\inf_J F[\phi_{\rm cl},J] .$$
In this answer we will assume$^1$ that eq. (*) holds. Here $J_i$ are sources and $\phi^i_{\rm cl}$ are classical fields (hence the subscript "${\rm cl}$"). To make it look more familiar to the common physicists, we can recast the Legendre transformation in the language of classical mechanics via the dictionary
$$ \begin{array}{cccc}  v^i& L(v) & p_i & H(p) & h(v,p) \cr\cr\updownarrow &\updownarrow &\updownarrow &\updownarrow &\updownarrow  \cr\cr  J_i & W_c[J] &  \phi^i_{\rm cl} & -\Gamma[\phi_{\rm cl}] & -F[J,\phi_{\rm cl}] \end{array} $$
Using the notation & definitions of my Phys.SE answers here & here, 
$$h(v,p)~:=~p_j v^j -L(v), \qquad g_j(v)~:=~\frac{\partial L(v)}{\partial v^j}, \qquad f~:=~g^{-1} ,$$
then OP's formally correct equations (1)-(9) transcribe as
$$ H(p) ~:= \sup_v h(v,p)~=~h(f(p),p), \tag{1} $$
$$\left.  \frac{\partial h(v,p)}{\partial v^i}\right|_{v^i=f^i(p)}~=~0,\tag{2} $$
$$f^i(p)~=~\frac{\partial H(p)}{\partial p_i},\tag{3} $$
$$v^i~\approx~\frac{\partial H(p)}{\partial p_i},\tag{4} $$
$$ v^i~\approx~f^i(p),\tag{5} $$
$$ p_i~\approx~g_i(v),\tag{6} $$
$$v^i~=~\frac{\partial H(p)}{\partial p_i}\circ g(v),\tag{7} $$
$$H\circ g(v)~=~h(v,g(v)),\tag{8} $$
$$ f^i\circ g(v)~=~v^i, \tag{9}$$
respectively. All eqs. (2)-(9) can formally be derived. For instance, eq. (3) is derived in Section V of my Phys.SE answer here.
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$^1$ OP apparently wants to discuss a generalization of the Legendre transform. We have currently nothing to add to that interesting discussion.
