For a linear transformation $A:\mathbb{R}^n\to\mathbb{R}^n$, I'd like to know what it's second derivative is. When I write it out in terms of the Hessian matrix in two dimensions, I get $(D^2A)_{x_0}$ is zero, where $(D^2A)_{x_0}$ means $D^2A$ evaluated at $x_0$.
I know that for any $x_0\in\mathbb{R}^n$, $(DA)_{x_0}=A_{x_0}=A$. To find $(D^2A)_{x_0}$, I need the bilinear map $D(DA)_{x_0}$ to be such that $(DA)_{x_0+h}-(DA)_{x_0}-D^2(DA)_{h,h}$ to go to zero fast. But since $(DA)_{x_0+h}=(DA)_{x_0}=A$, it should be that $D^2(DA)_{h,h}=0$ for all $h$. Is this correct?
Also, it may just be a notational thing, but the second derivative of $A$ at a point $x$, denoted $(D^2A)_{x_0}$, is not the same as the derivative of the derivative of $A$ evaluated at $x$, denoted $D((DA)_{x})$, because $D((DA)_x)=D(A)?$
Thank you
Let $f: X\to Y$ be a linear bounded map between Banach spaces, written say $f\in L(X,Y)$. The derivative of $f$ is a map $ df: X\to L(X,Y)$, and it is known that the derivative is constant, taking the value $df(x) = f\in L(X,Y)$ for every $x$.
It is also known that $L(X,Y)$ is then a Banach space, and constant maps between Banach spaces have 0 derivative. Hence, the map $d^2 f := d(df): X \to L(X, L(X,Y) )$ is identically 0.