Functional derivative of the integral of a Taylor expansion?

Question

I am asking this question coming from a background in continuum modelling. This is a simplified example which should illustrate the problem I am facing.

Consider the functional \begin{equation} F \left[ f \right] = \int_{ 0 }^{ L } f \left( x \right) \, dx, \end{equation} then the functional derivative is just \begin{equation} \frac{ \delta F }{ \delta f } = \frac{ \partial f }{ \partial f } = 1. \end{equation}

Alternatively, \begin{equation} \left< f \right> = \frac{ 1 }{ L } \int_{ 0 }^{ L } f \left( x \right) \, dx. \end{equation} So it is also valid to write \begin{equation} F = \int_{ 0 }^{ L } \frac{ 1 }{ L } \left[ \int_{ 0 }^{ L } f \left( x \right) \, dx \right] \, dx^{ \prime } \end{equation} $ f \left( x \right) $ may be expressed as a Taylor expansion from some point $ x^{ \prime } $, i.e., \begin{equation} f \left( x \right) = f \left( x^{ \prime } \right) + \left( x - x^{ \prime } \right) f^{ \prime } \left( x \right) + \frac{ 1 }{ 2 } \left( x - x^{ \prime } \right)^{ 2 } f^{ \prime \prime } \left( x^{ \prime } \right) + ... \end{equation} If the expansion is infinite, then it shouldn't matter from which point $ x^{ \prime } $ the perturbation occurs. In this sense, $ x^{ \prime } $ could be continuously varied. Substituting this into $ F $ gives \begin{equation} F \left[ f \right] = \frac{ 1 }{ L } \int_{ 0 }^{ L } \left[ \int_{ 0 }^{ L } \left[ f \left( x^{ \prime } \right) + \left( x - x^{ \prime } \right) f^{ \prime } \left( x^{ \prime } \right) + ... \right] \, dx \right] \, dx^{ \prime }. \end{equation} However taking the functional derivative yields \begin{equation} \frac{ \delta F }{ \delta f } = \frac{ 1 }{ L } \int_{ 0 }^{ L } \left[ \frac{ \partial f }{ \partial f } - \frac{ \partial \left( x - x^{ \prime } \right) }{ \partial x^{ \prime } } \frac{ \partial f^{ \prime } }{ \partial f^{ \prime } } + ... \right] \, dx = \frac{ 1 }{ L } \int_{ 0 }^{ L } \left[ 1 + 1 + 1 +... \right] \, dx = \infty. \end{equation} This divergence is clearly incorrect. However, conceptually I don't see what my mistake is, although my suspicion is that it is something to do with treating $ x^{ \prime } $ as a variable. Is there a way to make this work?

Answer 1

$$\frac{ \delta F }{ \delta f } = \frac{ \partial f }{ \partial f } = 1.$$

This is completely incorrect and is the source of the erroneous final calculation.

Let $\mathcal{V} = L^1_{loc}(\mathbb{R}\to\mathbb{R})$, that is, the space of all functions $\mathbb{R}\to\mathbb{R}$ that are integrable over compact intervals $[a,b]$. Now if we define $F(f) = \int_0^L f(x)~\mathrm{d}x$, we have $F:\mathcal{V}\to\mathbb{R}$ and $f\in\mathcal{V}$, so $f:\mathbb{R}\to\mathbb{R}$. Recall that the derivative of a function at a point is a linear map between its domain and range. Therefore, $\frac{ \delta F }{ \delta f }$ and $\frac{ \delta f }{ \delta f }$ cannot be equal since they are linear maps with different ranges (codomains). However, we can compute these derivatives.

Notice that $F:\mathcal{V}\to\mathbb{R}$ is a linear map already, therefore its derivative is just itself: $$\frac{ \delta F }{ \delta f }[g] = \int_0^L g(x)~\mathrm{d}x.$$ To compute $\frac{ \delta f }{ \delta f }$, we need to understand what we are actually differentiating. The most sensible way to interpret this is as computing the derivative of the identity map $\mathrm{id}:\mathcal{V}\to\mathcal{V}$ that satisfies $\mathrm{id}(f) = f \ \forall f\in\mathcal{V}$. This map is also linear, so its derivative is just itself as well. Notice that the identity map on $\mathcal{V}$ is not equal to $1$, but it can be represented as a pointwise multiplication by the constant function $1\in\mathcal{V}$.

Now to compute the derivative of your final map, we notice that $$G(f) = \frac{1}{L}\int_0^L\int_0^L f(x)~\mathrm{d}x~\mathrm{d}x'.$$ This map is linear in $f$ and hence its derivative is itself just as in the previous examples. To see this from your Taylor expansion work, consider the map $T:\mathcal{V}\to\mathcal{V}$ that generates a Taylor expansion about $x'$: $$[Tf](x) = f(x') + (x-x')f'(x') + \frac{(x-x')^2}{2}f''(x') + \dots.$$ We know that for analytic $f$, this map is the identity, but your work seems to disagree.

First notice that technically $T$ is only defined over analytic functions, so lets just restrict ourselves to those. The key thing to notice is that multiplication by fixed functions of $x$ (e.g. $x-x'$, $(x-x')^2$, etc.) is a linear operation from $\mathcal{V}$ to itself. Similarly, the map $f\mapsto \frac{d^n}{dx^n}f$ is a linear map $\mathcal{V}\to\mathcal{V}$ for all $n$. The final operation we need is pointwise evaluation. Indeed, the map $f\mapsto f(y)$ is a linear map from $\mathcal{V}\to\mathbb{R}$. Let us denote these operations by the following: $$ \begin{aligned} P_n &: f\mapsto \frac{(x-x')^n}{n!}f \in\mathcal{V} \\ D_n &: f \mapsto \frac{d^n f}{dx^n}\in\mathcal{V} \\ E &: f \mapsto f(x') \in\mathcal{V}. \end{aligned} $$ We can then write $T = \sum_{k=0}^\infty P_nED_n$. This operator computes the $n$th derivative, evaluates it at $x'$, multiplies it by the correct polynomial, then sums them all up. From this we can see that all these operations are linear, so $T$ is linear and therefore we must have $\frac{\delta T}{\delta f} = T$. However, if $f$ is analytic, then $f$ is equal to its Taylor series everywhere. In that case, we have $\frac{\delta T}{\delta f}(f) = T(f) = f$. Therefore, $T = \mathrm{id}$ and $\frac{\delta T}{\delta f} = \mathrm{id}$.

As a general tip when computing these functional derivatives, it really helps to actually write down what you are actually differentiating and what its domain and range are. This along with computing derivatives using variational methods like a Frechet derivative are much harder to mess up that memorizing various interactions between all sorts of operators, densities, and various notions of functional derivatives as it seems happened in your Taylor series computation.