Calculus in Banach space doubts about the notation $Df(\cdot)\cdot w.$

Question

I was self-studying the book of Abraham "Manifolds, Tensor Analysis and Applications." I was reviewing the part of calculus in Banach spaces when I came across to this notation at pag $81:$ $Df(\cdot)\cdot w.$
The application $f$ is twice differentiable, real valued and defined on an open set of $\mathbb{R}^{2}.$ I do not understand the following statement $$ \begin{gather*} Df(\cdot)\cdot w = T_{w}: \mathcal{L}(\mathbb{R}^{2},\mathbb{R}) \to \mathbb{R}\\ A \to A\cdot w. \end{gather*} $$ What is exactly $Df(\cdot)\cdot w$ ?
Why is it defined on the space $\mathcal{L}(\mathbb{R}^{2},\,\mathbb{R})$ ? From the abstract definition of differential I could only deduce that $D^{2}f(u)\cdot w \in \mathcal{L}(\mathbb{R}^{2},\,\mathbb{R}).$ Where did I go wrong?

Answer 1

In the book it is written like this: $$\text{Applying the chain rule to}\ Df(\cdot)\cdot w = T_{w}: A\in\mathcal{L}(\mathbb{R}^{2},\mathbb{R}) \to A\cdot w\in\mathbb{R}$$ The key is the chain rule here.

Note that $Df(\cdot)\cdot w$ is the composition of the functions $T_w: A\to A\cdot w\in\mathcal{L}(\mathbb{R}^{2},\mathbb{R})\to\mathbb R$ and $g: u\to Df(u)\in \mathbb R^2\to\mathcal{L}(\mathbb{R}^{2},\mathbb{R})$. So $Df(\cdot)\cdot w=T_w(Df(\cdot))=T_w(g(\cdot))$. And the chain rule is applied on this.

Answer 2

In my opinion the exposition in the book is confusing.

To prove this, note that by definition $$D^2f(u)\cdot (v,w) = D(Df(\cdot)\cdot w)(u)(v) \tag{1}$$ Applying the chain rule to $Df(\cdot)\cdot w = T_w : A \in L(\mathbb R^2,F) \mapsto A \cdot w \in F$, the above equals $$D(D(Df(\cdot)\cdot w)(u)(v) \\ = \ldots$$

1. By definition we have nothing else than $$D^2f(u)\cdot (v,w) = D^2f(u)(v,w) = D(Df)(u)(v)(w) . \tag{2}$$ Noting that equation $(1)$ holds by definition is a gross exaggeration. In fact we have to apply the chain rule to prove that "the above" (i.e. the expression $D(Df)(u)(v)(w)$) equals $D(D(Df(\cdot)\cdot w)(u)(v)$.

2. Writing $Df(\cdot)\cdot w = T_w : A \in L(\mathbb R^2,F) : A \in L(\mathbb R^2,F) \mapsto A \cdot w \in F$, is nonsense. As Koncopd explained in his answer, $T_w$ is the evaluation map $T_w : L(\mathbb R^2,F) \to F, T_w(A) = A\cdot w = A(w)$. Note that $T_w$ is linear. We do not have $Df(\cdot)\cdot w = T_w$. Okay, perhaps it was an erroneous omission and the intention was to write $Df(\cdot)\cdot w = T_w \circ Df, T_w : \ldots$.

Let us now apply the chain rule to $T_w \circ Df$. Since $T_w$ is linear, we have $DT_w(A) = T_w$ for all $A \in L(\mathbb R^2,F)$. This gives $$D(Df(\cdot)\cdot w)(u) = D(T_w \circ Df)(u) = DT_w(Df(u)) \circ D(Df)(u) = T_w \circ D(Df)(u) $$ and therefore $$D(Df(\cdot)\cdot w)(u)(v) = (T_w \circ D(Df)(u))(v) =T_w(D(Df)(u)(v)) = D(Df)(u)(v)(w) .$$ Using $(2)$ we obtain equation $(1)$.