# Functional derivative same as Frechet derivative over open sets?

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?

Edit:

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.

• What is the "functional derivative" exactly. When I saw it in the Euler-Lagrange equations, it didn't seem to be a Frechet derivative. Nov 17, 2022 at 2:48
• I made an edit. On a side note, on page 61 and 62 the author uses functional derivatives to show the Euler Lagrange equation. Nov 17, 2022 at 17:59
• 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 $\partial_{\eta}I(\phi) = DI(\phi)\eta$. Nov 17, 2022 at 21:18

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