I'm currently trying to solve an exercise and I'm having a notational issue here. Maybe someone knows that kind of notation and could help me out.
If we want to compute $$\frac{\partial}{\partial t} Df(y(t)) \cdot f(y(t))\vert_{t = t_0}$$ for $f : \mathbf{R}^n \rightarrow \mathbf{R}^n$ sufficiently smooth, $y(t_0) = y_0$ and $y' = f(y)$, we should get the following: $$D^2f(y_0)(f(y_0), f(y_0)) + Df(y_0) Df(y_0) f(y_0)$$
But what is meant by $D^2f(y_0) (f(y_0), f(y_0))$? Isn't it more like $D^2f(y_0) \cdot f(y_0)$? Or does that mean the same?
Thanks for any help!
The second derivative is, by definition, a (continuous) bilinear map. Well, I'd rather say it can be identified in an isometric way with a bilinear map. Hence, if $f \colon X \to \mathbb{R}$, then $$ D^2 f(u_0) \colon X \times X \to \mathbb{R} $$ is a bilinear map on $X$, and its action is denoted by $D^2f(u_0)(h,k)$ of the generic couple $(h,k) \in X \times X$. Think of the case $X = \mathbb{R}^n$, when you have the standard identification of $D^2 f(u_0)$ with the hessian matrix. This matrix, call it $H(u_0)$, acts as $x^t Ay$ on $x$, $y \in \mathbb{R}^n$.