Functional Taylor series

Question

The functional Taylor series of the functinal $f$ about the function $g$ is defined as

$f[g+\epsilon\lambda(x)]=f[g_0]+\int dx \frac{\delta f[g_0]}{\delta g(x)}\lambda(x)+\frac{1}{2!}\int dx dx^\prime \frac{\delta^2f[g_0]}{\delta g(x^\prime)\delta g(x)}\lambda(x)\lambda(x') + ... $

(it is assumed that $\lambda(x)=0$ outside the interval in question)

but in what sense does this converge? Usually this means that the derivatives of both sides are the same.

Following the text here. How does the author derive

$D_{\lambda_1}D_{\lambda_2}(f(g))=\int dx dx^\prime \frac{\delta^2f[g_0]}{\delta g(x^\prime)\delta g(x)}\lambda(x)\lambda(x') $

I know how to compute derivatives by definition as described here, or by a way similar to the derivation of Euler-Lagrange equations. Where does the double integral come from in the equation above?

I have my own way (maybe it's wrong) to construct such integrals as limiting differentials

$$\Delta g=g+\epsilon\lambda-g=\epsilon\lambda$$ $$df=\sum_{i=0}^n\frac{\partial f}{\partial g_i}\Delta g=\sum_{i=0}^n\frac{\partial f}{\partial g(x_0+\epsilon i)}\epsilon\lambda(x_0+\epsilon i)$$ Taking the limit, treating each point as a separate variable, we get $$df=\int \frac{\partial f}{\partial g(x)}\lambda(x) dx$$ Similarly for second order. But I fail to deduce, how this implies convergence of the Taylor series (first equation). How, or which derivatives match comparing left side with the right side of the equality? I would greatly appreciate any clarification.

Please forgive any lapses, I am an amateur exploring his interest.

Answer 1

Looking at taylor series of functions,

$$T_f(x)=\sum_{i=0}^{\infty}\frac{d^i}{dx^i}f(x_0)\frac{(x-x_0)^i}{i!}$$ where the derivatives are explicitly evaluated at a point $x_0$, and are as such treated as constants, the derivative of $T_f$ with respect to $x$ is to be taken, and not $x_0$.

It is the same with functional analog of the taylor series, the term

$$\frac{\delta F}{\delta g(x)}$$ is to be treated as a constant, because the variation was evaluated at a particular function $g$. You are to take the derivative of your expression with respect to $\lambda(x)$, and not $g$, just as we took derivatives of $T_f(x)$ with respect to $x$ and not $x_0$.