Higher order functional derivatives

Question

Let $E, F$ be Banach spaces. A continuous bilinear functional ${\langle \cdot\,, \cdot \rangle }: E \times F \to \mathbb{R}$ is called $E$-non-degenerate if $\langle x,y\rangle = 0$ for all $y \in F$ implies $x=0$ (Similarly for $F$-non-degenerate). Equivalently, the two maps of $E$ to $F^{*}$ and $F$ to $E^{*}$ defined by $x \mapsto \langle x, \cdot \rangle$ and $y \mapsto \langle \cdot\,, y\rangle$, respectivelly, are one-to-one. If they are isomorphisms (*), $\langle \cdot\,, \cdot \rangle$ is called $E$ or $F$-strongly non-degenerate. We say that $E$ and $F$ are in duality if there is a non-degenerate bilinear functional $\langle \cdot\,, \cdot \rangle: E\times F \to \mathbb{R}$, also called a pairing of $E$ with $F$. If the functional is strongly non-degenerate, we say the duality is strong.

Consider the following definition.

Definition: Let $E$ and $F$ be normed spaces and $\langle \cdot, \cdot \rangle$ a $E$-non-degenerate pairing. Let $f: F \to \mathbb{R}$ be Fréchet differentiable at the point $\alpha \in F$ (denote this derivative as $Df(\alpha)$). The functional derivative $\delta f/\delta \alpha$ of $f$ with respect to $\alpha$ is the unique element in $E$, if it exists, such that: \begin{eqnarray} Df(\alpha)(\gamma) = \left\langle \frac{\delta f}{\delta \alpha}, \gamma\right\rangle\quad\forall\gamma \in F. \tag{1}\label{1} \end{eqnarray}

Now, I'd like to know how to define higher order derivatives of functional derivatives. In other words, suppose the Fréchet derivative of $f$ at $\alpha$, $Df(\alpha)$ is Fréchet differentiable at $\beta\in F$, is it possible to define $\dfrac{\delta^{2}f}{\delta \beta\delta\alpha}$?

Answer 1

Now, I'd like to know how to define higher order derivatives of functional derivatives. In other words, suppose the Fréchet derivative of $f$ at $\alpha$, $Df(\alpha)$ is Fréchet differentiable at $\beta\in F$, is it possible to define $$ \dfrac{\delta^{2}f}{\delta \beta\delta\alpha}? $$

Yes, it is possible to define higher order Fréchet derivatives directly as bilinear functionals in $\mathscr{L}_2(F,\Bbb R)$ (the space of bounded bilinear functionals from $F$ to $\Bbb R$). This is shown by Ambrosetti and Prodi ([1], pp. 23-29), for example, and while leaving to them the details, we can simply say that $f:F\to\Bbb R$ is twice Fréchet differentiable at $\alpha\in F$ iff

• $F$ is one time Fréchet differentiable and
• $Df[\alpha+\beta](\gamma)-Df[\alpha](\gamma)= \mathfrak{B}[\alpha](\beta,\gamma) + o(\beta)$ where $o(\beta)/\Vert\beta\Vert_F\to 0$ as $\Vert\beta\Vert_F\to 0$ and $\mathfrak{B}\in\mathscr{L}\big(F,\mathscr{L}(F,\Bbb R)\big)\simeq\mathscr{L}_2(F,\Bbb R)$ (again, the proof of this fact is found in [1] §1.3, p. 23, where it is precisely show that the isomorphism between these two spaces is actually an isometry).

This can bee seen also by using the definition of Fréchet derivative you give, but it seems to me that, when dealing with functional derivatives of order greater than one, it overshadows the intrinsic clarity and familiarity of the concept. Precisely, by using the standard definition of Fréchet differentiablity (as given for example in [1] §1.1, p. 9, definition 1.1) and \eqref{1} we get $$ \begin{split} Df[\alpha+\varepsilon\beta](\gamma)-Df[\alpha](\gamma) &= \left\langle \frac{\delta f}{\delta \alpha}(\alpha+\beta),\gamma\right\rangle -\left\langle \frac{\delta f}{\delta \alpha} (\alpha),\gamma\right\rangle\\ & = \left\langle\frac{\delta f}{\delta \alpha}(\alpha+\beta)- \frac{\delta f}{\delta \alpha} (\alpha),\gamma\right\rangle\\ &=\left\langle\mathfrak{L}[\alpha](\beta)+ o(\beta),\gamma\right\rangle \end{split} $$ where $\mathfrak{L}\in\mathscr{L}(F,E)$ is a linear bounded operator.

Bibliography

[1] Ambrosetti, Antonio; Prodi, Giovanni, A primer of nonlinear analysis, Cambridge Studies in Advanced Mathematics, 34. Cambridge: Cambridge University Press, pp. viii+171 (1993), ISBN: 0-521-37390-5, MR1225101, ZBL0781.47046.