Higher order Derivatives. Now we know that given two Banach spaces $E$ and $F$ and a function $\ f:E \to F $ , the derivative $ Df(x) $ is a linear map from $E$ to $F$ at some point $ x $ in $E$. Briefly $ \ Df: E \to L(E,F) $  where $ L(E,F) $ is space of all linear mappings from $E$ to $F$. Now my question is how would a second order derivative look like. Since the derivative map of a linear map is linear map itself does this mean $ \ D^2(f(x))=D(f(x)) $ which doesn't seem likely to me. Or is it something else? Also, is the following right?
$$D^2(f): E \to L(L(E,F),F) $$
 A: Regarding Ilies Zidane's comment, it is important to point out that it isn't just matter of widely used notation, but actually $\mathcal{L}(E^r,F)$ and $\mathcal{L}^r(E,F)$ are just not the same thing: the former is the space of all linear continuous maps from $E^r$ to $F$, while the latter is the one of all $r$-linear continuous maps from $E$ to $F$. So, for instance, if $E = F = \mathbf{R}$, then $\mathcal{L}(\mathbf{R}^r,\mathbf{R}) = {(\mathbf{R}^r)}^*$ is an $r$-dimensional vector space, while $\mathcal{L}^r(\mathbf{R},\mathbf{R})$ is just a $1$-dimensional one!
As for $\mathcal{L}^r(E,F)$ and $\mathcal{L}(E,\mathcal{L}^{r-1}(E,F))$, they aren't equal to each other either, but isomorphic as Banach spaces.
A: Just as the first derivative is the linear term in the Taylor expansion:
\begin{equation}
f(x+h)=f(x)+Df(x)h+\text{higher order}, 
\end{equation}
the second derivative should be the quadratic term in the Taylor expansion:
\begin{align}
D^2f(x)\ \text{should be}\ Q\colon E \times E \to F, \text{where}& f(x+h)=f(x)+Df(x)h+\frac{1}{2}Q(h, h)+\ldots
\end{align}
Here $Q$ is linear in each variable, that is, $Q$ is a bilinear operator. 
Since $Df$ is a mapping from $E$ to the Banach space $L(E;F)$ it can be differentiated yielding a mapping from $E$ to $L(E; L(E;F))$, as you point out in the first post. But, as Ilies notes, this last space is exactly the space of bilinear operators, up to the natural identification
\begin{equation}
Q(h,k)=Q(h)(k),
\end{equation}
and, defining $D^2f(x)$ to be the derivative of $Df(x)$ with this identification, turns out that Taylor's formula holds:
\begin{equation}
f(x+h)=f(x)+Df(x)h+\frac{1}{2}D^2f(x)(h, h)+o\left(\lVert h \rVert^2\right).
\end{equation}
This completely justifies our definition.
A: Let $U$ be an open set of $E$ and $f : U \longrightarrow F$. The differential of $f$ is a map:
$$ D: U \longrightarrow L(E,F), \, \, \ u\mapsto Df(u) $$
And we can define,
$$ D^r = D(D^{r-1})(f) : U \longrightarrow L(E,L^{r-1}(E,F)) $$
Since $$ L(E,L^{r-1}(E,F)) \simeq L^r(E,F) $$ we usually use
$$D^r : U \longrightarrow L^r(E,F) $$
Where $L^r(E,F) = L(E \times \cdots \times E,F)$ ($r$-times).
