In Appendix A of Density Functional Theory An Advanced Course by Eberhard Engel, Reiner M. Dreizler functional derivative is introduced by studying the Taylor expansion of some functional $F$ around function $f$. $F$ is treated as a function of scalar $\epsilon$, expanded around $\epsilon = 0$:

\begin{equation} F[f + \epsilon\eta] = F[f] + \frac{dF[f + \epsilon\eta]}{d\epsilon} \Bigg \rvert_{\epsilon=0} \epsilon + \frac{1}{2}\frac{d^2F[f + \epsilon \eta]}{d\epsilon^2}\Bigg \rvert_{\epsilon=0} \epsilon^2 + \ldots \end{equation}

In the next step, the following equality is imposed, and if the integral on the right actually exists, the object $\frac{\delta F[f]}{\delta f(x_1)}$ is said to be the (first order) functional derivative of $F$.

\begin{equation} \frac{dF[f + \epsilon\eta]}{d\epsilon}\Bigg \rvert_{\epsilon=0} =: \int dx_1 \frac{\delta F[f]}{\delta f(x_1)} \eta(x_1) \end{equation}

One way to think about functional derivative, the text goes, is to understand it as an expansion of the concept of the total differential of a multivariate function to the case where the number of variables $x_i$ goes to infinity:

\begin{equation} df = \sum_i \frac{\partial{f}}{\partial{x_i}} dx_i \end{equation}

Following this reasoning, the second order functional derivative is defined analogous to how second order total differential is defined.

\begin{equation} \frac{d^2F[f + \epsilon \eta]}{d\epsilon^2}\Bigg \rvert_{\epsilon=0} =: \int dx_1 dx_2 \frac{\delta^2 F[f]}{\delta f(x_1) \delta f(x_2)} \eta(x_1) \eta(x_2) \end{equation}

Although this does intuitively make sense to me, I wonder if there is a more direct way of defining the second order functional derivative, possibly by somehow applying the definition of the first order one? Similarly as the second order total differential can be understood as a differential of the first order total differential:

\begin{equation} d^2f = d (df) = \sum_j \frac{\partial (df)}{\partial x_j} = \sum_{i,j} \frac{\partial^2 f}{\partial x_i \partial x_j} \end{equation}

My attempt:

Assuming that we can find some functional $G$ such that:

\begin{equation} G[f + \epsilon \eta] = \frac{d F[f + \epsilon \eta]}{d \epsilon} \end{equation}

Now, it is not clear why this should be the case in general (although for $\rvert_{\epsilon=0}$, this assumption seems to be fulfilled if the first order functional derivative exists), but assuming it is, we can write:

\begin{equation} \frac{d^2F[f + \epsilon \eta]}{d\epsilon^2}\Bigg \rvert_{\epsilon=0} = \frac{dG[f + \epsilon \eta]}{d\epsilon}\Bigg \rvert_{\epsilon=0} = \int dx_1 \frac{\delta G[f + \epsilon\eta]}{\delta f(x_1)} \eta(x_1)\Bigg \rvert_{\epsilon=0} \\= \int dx_1 \eta(x_1) \frac{\delta}{\delta f(x_1)} \int dx_2 \eta(x_2) \frac{\delta F[f]}{\delta f(x_2)} = \int dx_1 \int dx_2 \frac{\delta^2 F[f]}{\delta f(x_1)\delta f(x_2)} \eta(x_2) \eta(x_1) \end{equation}

Where in the third term, $\epsilon$ in re-introduced in but effectively only evaluated at $\epsilon = 0$. Although the resulting term does correspond to the definition of the second order functional derivative, the reasoning behind the steps does not seem very clear and formal.

Is there is a better way to think about this?

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    $\begingroup$ This question reminds me of math.stackexchange.com/questions/123007/… (actually, this answer might be a bit more useful to you) $\endgroup$ Commented Mar 28, 2021 at 14:39
  • $\begingroup$ Thanks, that is definitely instructive for a more general understanding of derivatives. Just to clarify, using notation from the linked question, for $ Df: E \to L(E,F)$ besides being unique, there is no other restriction for this mapping? Also, would it be correct to say that in most cases, when dealing with well behaved functions, $Df$ would map different $e \in E$ to the same $l \in(E,F)$? $\endgroup$
    – cHewap
    Commented Mar 28, 2021 at 18:36
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    $\begingroup$ I don't think I'll have any time to answer your questions, unfortunately. I recommend Part 4 of Serge Lang's Undergraduate Analysis. That's where I learned all of this and I think it is a great book. $\endgroup$ Commented Mar 30, 2021 at 10:33
  • $\begingroup$ Great! Thank you for the reference $\endgroup$
    – cHewap
    Commented Mar 30, 2021 at 20:53

1 Answer 1


That is the "physicist" version which assumes the measure we get from Riesz representation $C_0(\mathbb{R}^n)'\cong M(\mathbb{R}^n)$ is absolutely continuous wrt Lebesgue so take the Radon-Nikodym derivative of $DF\rvert_f$ with respect to Lebesgue measure $\mathcal{L}$ and get an element of $L^1(\mathcal{L}^n)$ $$ \frac{\delta F[f]}{\delta f(x)}:=\frac{\mathrm{d}(DF\rvert_{f})}{\mathrm{d}\mathcal{L}}(x)\quad\mathcal{L}\text{-a.e. }x $$(and if we don't have $DF\rvert_f\ll\mathcal{L}$, fudge it like the Dirac delta "function").

Then again for nice enough $F$, for almost every $x_1\in\mathbb{R}^n$, the map $$ x_1\mapsto\frac{\delta F[f]}{\delta f(x_1)} $$ is in $C_0(\mathbb{R}^n)$, so we can take its functional derivative $$ \frac{\delta}{\delta f(x_2)}\frac{\delta F[f]}{\delta f(x_1)} $$ and call it the second functional derivative of $F$.

  • $\begingroup$ So this is as formal as it gets in the physicist version? What would then be a more mathematical version? $\endgroup$
    – cHewap
    Commented Mar 28, 2021 at 12:40
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    $\begingroup$ The mathematical version is just the second (Fréchet) derivative of the map $F\colon C_0(\mathbb{R}^n)\to\mathbb{R}$. It is exactly the same treatment you get as maps between finite-dimensional vector spaces $f\colon E\to F$ (i.e., a linear-map valued $E\to L(E\otimes E; F)$) but you don't get it as a function on $\mathbb{R}^n\times\mathbb{R}^n$. $\endgroup$ Commented Mar 28, 2021 at 13:58

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