I have recently read in (these German notes) that the functional derivative (a.k.a variational derivative) is the same as the Frechet derivative as long as we are over open sets. However, I can not find a proof for this. Could you please point me to some reference?


For the sake of completeness here are the definitions:

In my lecture we defined Frechet derivatives in the following way: Let $X$, $Y$ be Banach spaces and let $f: X \rightarrow Y$ be locally defined around $a \in X$. We say that $f$ is Frechet differentiable in $a$ if there is a mapping $Df_a \in \mathcal{L}(X,Y)$; i.e. $Df_a$ is bounded and linear, such that

$$f(a+h) = f(a) + Df_a(h) + o(\lvert\lvert h \rvert\rvert)$$

as $h \rightarrow 0$. We call $Df_a$ the Frechet derivative of $f$ at $a$ or the directional derivative of $f$ at $a$ in direction of the vector $v$.

In the notes linked above the author defines on page 60 (translation by me):

Let $B$ be an arbitrary Banach space and let $Z \subseteq B$. Let further be $V \subseteq B$ be a linear subspace such that $\varphi \pm \varepsilon \eta \in Z$ for all $\varphi \in Z, \eta \in V$ and sufficiently small $\varepsilon > 0$.

The variational derivative (I am pretty sure that another name is functional derivative) of a functional $I: Z \rightarrow \mathbb{R}$ in direction $\eta \in V$ at $\varphi \in Z$ is defined as

$$\partial_\eta I(\varphi) := \frac{d}{d\varepsilon} I(\varphi+\varepsilon \eta) \vert_{\varepsilon = 0} = lim_{\varepsilon \rightarrow 0} \frac{I(\varphi + \varepsilon \eta) - I(\varphi)}{\varepsilon}$$

Directly below the author claimes that for $B = V = \mathbb{R}^n$ and $Z$ an open subset of $\mathbb{R}^n$ directional derivatives and variational derivatives are the same.

On a side note, this question seems to be similar.

  • 1
    $\begingroup$ What is the "functional derivative" exactly. When I saw it in the Euler-Lagrange equations, it didn't seem to be a Frechet derivative. $\endgroup$
    – Mason
    Nov 17, 2022 at 2:48
  • $\begingroup$ I made an edit. On a side note, on page 61 and 62 the author uses functional derivatives to show the Euler Lagrange equation. $\endgroup$
    – 3nondatur
    Nov 17, 2022 at 17:59
  • 1
    $\begingroup$ Yes $\partial_{\eta}I(\phi)$ is the directional derivative of $I$ in the direction of $\eta$, known as the Gateaux derivative. If $I$ is Frechet differentiable, then by the chain rule $\partial_{\eta}I(\phi) = DI(\phi)\eta$. $\endgroup$
    – Mason
    Nov 17, 2022 at 21:18

1 Answer 1


I'm not sure how to exactly relate this to what you asked but essentially a functional derivative IS a directional derivative where the direction is a function.

We start by recalling the definition of a directional derivative. Let $x, v \in \mathbb{R}^n$, $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$

$$ D_{v}[f] = \lim_{h \rightarrow 0} \frac{f(x + vh) - f(x)}{h} = \frac{\partial F}{\partial x_0} v_0 + \frac{\partial F}{\partial x_1} v_1 + ... = \nabla F \cdot (v_0, v_1 ... v_n) $$

Now lets look at the functional derivative. Given an operator $O: \left( \mathbb{C} \rightarrow \mathbb{C} \right) \rightarrow \left( \mathbb{C} \rightarrow \mathbb{C} \right) $ and a function $\lambda$

$$ \delta_{\lambda}[O[f]] = \lim_{h \rightarrow 0} \frac{O[f+h\lambda]-O[f]}{h} = \frac{\partial O}{\partial \lambda} \lambda + \frac{\partial O}{\partial \lambda '} \lambda' + \frac{\partial O}{\partial \lambda ''} \lambda'' +... = \nabla O \cdot (\lambda, \lambda', \lambda'' , ...) $$

So here a subtle thing is happening, a function is basically an infinite vector given by its derivatives and the functional derivative is just a regular old directional derivative w.r.t to this "functional" direction.


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