# 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 related question: math.stackexchange.com/q/123007/8157 – Giuseppe Negro Sep 1 '12 at 8:15

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).

-
Why not just write $L(E^r,F)$ instead of $L^r(E,F)$? – Ben Millwood Sep 1 '12 at 9:28
Because it's the commonly used notation. – Ilies Zidane Sep 1 '12 at 9:38

Regarding Ilies Zidane's answer, it's 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.

-

Just as the first derivative is the linear term in the Taylor expansion: $$f(x+h)=f(x)+Df(x)h+\text{higher order},$$ 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 $$Q(h,k)=Q(h)(k),$$ and, defining $D^2f(x)$ to be the derivative of $Df(x)$ with this identification, turns out that Taylor's formula holds: $$f(x+h)=f(x)+Df(x)h+\frac{1}{2}D^2f(x)(h, h)+o\left(\lVert h \rVert^2\right).$$ This completely justifies our definition.

-
Thank you, that was very clear and precise. – Vishesh Sep 1 '12 at 9:06