I don't quite understand the following functional derivative computation when I read a variational inference literature, can someone explain? $$L[q] = E_{q(Y)}[f(Y)]$$ $$\frac{\delta L[q]}{\delta q} = f(Y)$$ Here, $q(Y)$ is a probability density function. The original deduction is here, the relevant steps are equation (14) (17) (under assumption (11)).
I don't know why $q \rightarrow 0 \Rightarrow \int f(Y)q(Y)dY \rightarrow f(Y)$, it seems $q$ acts as a Dirac delta function in the limit case. Am I right? If so, how to argue? Any insights are welcome.
The paper you linked makes several abuses of notation. Here, since the considered functional doesn't contain any derivative of $q$, the functional differential is given by $$ \delta L[q] = \delta\mathbb{E}_q[f(Y)] = \delta\int q(y)f(y) \,\mathrm{d}y = \int \frac{\partial}{\partial q}\left(q(y)f(y)\right)\delta q(y) \,\mathrm{d}y = \int f(y)\delta q(y) \,\mathrm{d}y, $$ hence $$ \frac{\delta L[q]}{\delta q(x)} = \int f(y)\frac{\delta q(y)}{\delta q(x)} \,\mathrm{d}y = \int f(y)\delta(y-x) \,\mathrm{d}y = f(x). $$ The relation $\frac{\delta q(y)}{\delta q(x)} = \delta(y-x)$ is the functional analog of the more classical $\frac{\mathrm{d}x_j}{\mathrm{d}x_i} = \delta_{ij}$.