Analysis II Basic issue applying Frechet chain rule

Question

Let $f$ be a $C^1$ function from $\mathbb{R^n}$ to itself. Let $T=f'(a)$ and $h(x)=T^{-1}( f(x+a)-f(a))$. Then $h'(x)=T^{-1}\circ f'(x+a)$ by the chain rule.

I can derive this result using little $o$ notation, but I clearly I've misunderstood something before because I can't see for the life of me how to get this from the chain rule.

We have $(g\circ f)'(c)=(Dg)[f(c)]\circ Df(c)$ Then in the example, set $g= T^{-1}$ and $f= f(x+a)$. We have $$\begin{aligned} Dg&= g=T^{-1}\\Df&=f'(x+a)\end{aligned}$$ so, ignoring the constant, $h'(x)=T^{-1}f(x+a)\circ f'(x+a)$

What went wrong? Thanks

Answer 1

You have $Dg=T^{-1}$ at every point of the domain of $g$. In particular, $Dg(f(x+a))=T^{-1}$.

Answer 2

It seems that you tried to obtained an expression for $h'(x)$ from the definition of derivative in $\mathbb{R}^n$. Here is one set of computations that you might have gone through already: \begin{align} h(x+k)&=T^{-1}\big(f(x+a+k)-f(x+k)\big)\\ &=T^{-1}\big(f(x+a)+f'(x+a)k+r(x+a;k)-f(x)-f'(x)-r(x;k)\big)\\ &=h(x)+\big(T^{-1} (f'(x+a)-f'(x))\big)k + R(x;k) \end{align} where

1. $\lim_{k\rightarrow0}\frac{|r(x+a;k)|}{|k|}=0=\lim_{k\rightarrow0}\frac{|r(x;k)|}{|k|}$,
2. $R(x;k)=T^{-1}(r(x+a;k)-r(x;k))$

($|\;|$ is the Euclidean norm in $\mathbb{R}^n$). Notice that $$|R(x;k)|\leq \|T^{-1}\|\big(|r(x+a;r)|+|r(x;k)|\big)$$

Obs: The space of matrices $\mathbb{M}_n(\mathbb{R})$ is normable. For example $\|A\|:=\sup_{|x|=1}|Ax|$ defines a norm on $\mathbb{M}_n(\mathbb{R})$ with the property that $|Ax|\leq \|A\||x|$ for any $x\in\mathbb{R}^n$.

Hence $\lim_{k\rightarrow0}\frac{|R(x;k)}{|k|}=0$, whence $h'(x)=T^{-1}\circ(f'(x+a)-f'(x))$ (as composition of linear operators) or $h'(x)=T^{-1}( f'(x+a)-f'(x))$ as product of matrices.