I had a couple of basic questions about the Frechet derivative.
1) The chain rule: If $F: X\rightarrow Y$ and $G: Y\rightarrow Z$, then $D(G\circ F)(x) = DG(F(x))DF(x)$. The left hand side has a composite map $D(G\circ F): X \rightarrow Z$ and is very clear to me but the right hand side is unclear. Should I be reading it as $DG(DF(x))$, where $DG$ is evaluated at $F(x)$ and $DF$ is evaluated at $x$?
Some others also present this with an additional $H$. That is, in the limit $H\rightarrow 0$, we have $D(G\circ F)[H]$ on the left hand side. The right hand side is written as $DG(F(X))DF(X)[DF(X)[H]]$ and I wasn't too sure how this worked.
2) If any function is linear, the Frechet derivative is the function itself. In this case, how does one express the right hand side of the chain rule when either $G$ is linear or $F$ is linear. Given my problems with the notation in the previous part, I'm not too sure how to rewrite the right hand side with $DG = G$ or $DF = F$.
Example
Here's an example to which I know the solution so it becomes even clearer. It is known that $D Tr(A\log X) = Tr(A X^{-1} dX)$ for constant $A$ but how do I use the chain rule to get this? My attempt is below
$$D Tr(A\log X) = D Tr(A\log X)D(A\log X) = DTr(A\log X) AX^{-1}dX = ??$$
Clearly, I don't know exactly how to use the rule correctly. Please let me know how to proceed. Thank you
1) It's not clear to you since what you wrote is incorrect. It would be correct however you wrote $D(G \circ F)(x): X \rightarrow Z$ (i.e, what you had but evaluated at some x in domain of $F$), or you could say $D(G \circ F):X \rightarrow \mathcal{L}(X;Z)$ , and the RHS is a composition, so should have a '$\circ$' in between, i.e $DG(F(x)) \circ DF(x)$. With this you should now be able to see what is going on with $H$, it's an element of $X$ and so we are mapping $H$ to $D(G \circ F)(x)(H) \in Z$
2) If $F: X \rightarrow Y$ is linear, then $DF(x): X \rightarrow Y$ is equal to F (that is, F=DF(x)), notice that derivative is being evaluated at a point in $X$ whilst the original function is not, this is crucial in understanding the Frechet derivative.