I know that a functional is defined as a map $$f : M \mapsto \mathbb{R} \tag{1}$$ or $$f : M \mapsto \mathbb{C} \tag{2}$$ (or similar). Therefore, expressions like the following are functionals: $$ F[y] = \int_{a}^{b} \mathrm{d}x \sqrt{1-y'^{2}(x)} \tag{3}$$ $$ G[y, y'] = \int_{x_{1}}^{x_{2}} \mathrm{d}x \frac{y^{3}(x) + y'(x) +x^{2}}{\cos(y(x))} \tag{4}$$ $$ H[y] = \exp(y^{2}(x) \delta(x-x_{0})) \tag{5}$$ because they depend on the function ($y(x)$) and its derivatives only, but not on the independent variable. Thus by plugging an implicit map $y: z \mapsto f(z)$, one gets a real number (that is how I understand it, please correct me if I am wrong).
My question regards expressions which also involve the independent variable, like $$A[y(x)] = y(x) \sin(y(x)) \tag{6}$$ $$B[x(t)] = x(t) \int_{0}^{1} \mathrm{d}w \frac{e^{-w}}{w + t} x(w) \tag{7}$$ Can these be regarded as functionals? Are they some kind of generalised functionals?
The reason for my question is the fact that I am learning about functional derivatives. And while I have seen many examples of how to find the functional derivatives of $F, G, H$, I am not clear on how (or whether it is possible) to find the functional derivative of something like $A$ or $B$.
Thousands of people would disagree with that definition of a "functional" since the Feynman-Vernon influence functional does depend on the independent variable $t$ [see Eq. 9 of this 2012 paper or Eq. 8 of this 1995 paper]. These are just two of the many papers over a period of 6 decades that use this functional (an older paper from 1963, which has 3000+ citations on Google Scholar is this one).
In your comment, you said:
But no one said that an operator cannot be a functional.
Your $A$ is a composite function. It can also be written with round brackets: $A(y(x))$ to indicate that it's an ordinary function $A(x)$ in which the argument is an ordinary function $y(x)$. Knowing $x$ is enough to give you $A(y(x))$, whereas for $B[y(x)]$ we certainly need the square brackets to indicate that we need to know all of $y(x)$ (not just $y$ at a specific value of $x$) to obtain $B$.
Your $B$ is precisely the same as the following operator in Annie Cuyt's PhD thesis, except you changed $s$ to $w$ and the numerator was changed from $t$ to $\exp(w)$. It was differentiated in her thesis as you can see below:
She used the Frechet derivative, where $L$ is the derivative (see one page earlier in the same thesis):