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from the definition of functional derivative it is possible to write the Taylor expansion of a functional: $$f[g(x)]=f[g_0(x)]+\int dx\frac{\delta f[g_0]}{\delta g(x)}\lambda(x)+\tfrac{1}{2!}\int dxdx'\frac{\delta^2f[g_0]}{\delta g(x')\delta g(x)}\lambda(x)\lambda(x')+\ldots$$ but is this applicable to operators as well as functionals? I have found this question where it's clear that the functional derivative operators exist,so my guess it that it can be used but with math you never know. and my second question is does it converge always or is there only a region of convergence?

MtTriolet
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Yes, you can write (finite) Taylor expansions for any regular enough map between two Banach spaces for example (see e.g. the first answer in this other question).

However, no, it does not always converge, not even in a small region. But this is already the case for Taylor expansions of function from $\mathbb{R}$ to itself. For $f : \mathbb{R} \to \mathbb{R}$, the Taylor series does not need to converge to $f$. The typical counter-example is $$ f := \begin{cases} 0 & \text{for } x \leq 0, \\ \exp(-1/x^2) & \text{for } x > 0. \end{cases} $$ You can check that all terms in the Taylor series of $f$ at $x = 0$ vanish. But $f$ is not identically null.

cs89
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  • is the condition of a 'nice enough' functional good for convergence or not even that? and how does it not converge, do the terms go to infinity? – MtTriolet May 23 '25 at 13:42
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    The typical condition for convergence is analyticity. I would advise you start by getting familiar with such convergence issues and results for the case of function from $\mathbb{R}$ to $\mathbb{R}$. The terms need not go to infinity. For the example I gave, all terms of the Taylor expansion are zero (since all derivatives of $f$ vanish at $0$). But $f$ is not zero so it cannot be the sum of its Taylor series (which is $0$). – cs89 May 23 '25 at 14:02