Let $\varphi(x)$ and $j(x)$ be two field configurations. Let $\Gamma[\varphi]$ be a functional of the field $\varphi$ defined by:
$$ \Gamma[\varphi] := \sup_j \ F[\varphi, j] = F[\varphi, j_\varphi] $$
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,
$$ \frac{\delta}{\delta j} F[\varphi,j] \Big\rvert_{j=j_\varphi} = 0 \,.$$
Now suppose that the structure of $F$ allows the following identity to hold:
$$ \frac{\delta}{\delta\varphi} \Gamma[\varphi] + j_\varphi = 0 \,.$$
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:
$$ \frac{\delta}{\delta\varphi} \Gamma[\varphi] + j = 0 \,,$$
whose solution is, of course, $j=j_\varphi$ for any given $\varphi$. Alternatively, however, we can fix $j$ and ask for the $\varphi =: \varphi_j$ that solves it.
We then have an equivalent identity:
$$ \frac{\delta}{\delta\varphi} \Gamma[\varphi]\Big|_{\varphi=\varphi_j} + j = 0 \,. $$
In such a scenario, would it be true to say that
$$ \Gamma[\varphi_j] = F[\varphi_j, j] $$
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 $$ j_{\varphi_j} = j \,.$$
What does that even mean?
Thank you very much for your time!
[Motivation: Effective Quantum Action]
I believe so. Here's a possible justification. To avoid confusion with our fixed $j$, I will denote $J_\varphi$ as the solution $k$ of $\frac{\delta}{\delta k}F[\varphi,k]=0$. Now, by definition of $J_{\varphi_j}$ and the structure identity of $F$, we have $$ \frac{\delta}{\delta\varphi}\Gamma[\varphi_j]+J_{\varphi_j}=0. $$ By definition of $\varphi_j$, we have $$ \frac{\delta}{\delta\varphi}\Gamma[\varphi]|_{\varphi=\varphi_j}+j=0. $$ But $\frac{\delta}{\delta\varphi}\Gamma[\varphi_j]\equiv \frac{\delta}{\delta\varphi}\Gamma[\varphi]|_{\varphi=\varphi_j}$ is the same thing in two different notations, so we get $J_{\varphi_j}=j$.