Fréchet -Differentiability on Banach Algebras I have a problem with the following statement, hoping someone can help me out. 
Supose $A$ is a Banach algebra and $Inv(A)$ is the set of all invertible elements of $A$. Let H be the operator defined as 
$$ H(a)= a^{-1} \quad H: Inv(A) \to Inv(A) $$
the statement is the following. 
 The operator is Fréchet differentiable and its derivative is:
$$ H'(i, a) = -a  \quad  \forall a \in A $$
The thing is that if we compute the Gataux derivative in the point $i$ , meaning, the neutral (or identity) element in the direction of $a$  we have: 
$$ H'(i; a) = \frac{1}{h} \Big\{ H(a+hi)-H(a) \Big\}= a^{-1} i a^{-1}  $$
so the question is: is there any result in Banach Algebras that coud lead to have $-a$ indeed. Baucause for example if we take to be $\mathbb{R}=A$ then (I think)
$$ H'(i; a)= -a^{-2}$$
One thing that I'd have to let you know is that I know nothing about Banach Algebras besides the basics,but the result I am dealing with uses it. 
Thanks so much. 
 A: I think part of your confusion arises from using similar notation for the two notions of differentiability, which is why you might be confused about where the $a$ goes and where the $i$ goes (you need to be very careful about the placement of $a$ and $i$).

The map $H: \text{Inv}(A) \to \text{Inv}(A)$ defined by $H(a) = a^{-1}$ is Frechet differentiable (in fact $C^{\infty}$) at every point of its domain, and for every $\xi\in \text{Inv}(A)$ and all $a \in A$, the Frechet derivative of $H$ at the point $\xi$ (which I denote by $dH_{\xi}$) applied to the vector $a$ is given by the formula
\begin{align}
dH_{\xi}(a) = -\xi^{-1} a \xi^{-1} \tag{$*$}
\end{align}
In particular, if we choose to differentiate at the identity $i \in \text{Inv}(A)$, then the RHS can be simplified to give:
\begin{align}
dH_{i}(a) = -i^{-1} a i^{-1} = -a
\end{align}
In other words, $dH_i(\cdot) = -\text{id}_{A}(\cdot)$.
Equation $(*)$ does indeed agree with intuitive expectations, because if we choose $A= \Bbb{R}$, then for any $\xi \in \text{Inv}(\Bbb{R}) = \Bbb{R} \setminus\{0\}$, and any $a\in \Bbb{R}$, we have that
\begin{align}
dH_{\xi}(a) &= -\xi^{-1} a \xi^{-1} \\
&= \left( -\dfrac{1}{\xi^2} \right)\cdot a \\
&= H'(\xi) \cdot a
\end{align}
(Here I use the notation $H'(\xi)$ in the same way as in single variable calculus) The result $dH_\xi(a) = H'(\xi) \cdot a$ should be perfectly reasonable, because all it says is that $dH_{\xi}$ is the linear transformation which just multiplies its input $a$ by the slope $H'(\xi) = -\dfrac{1}{\xi^2}$.

A "quick way" of deriving this (and using sloppy notation) is to realise that $\xi \xi^{-1} = i$. So, applying the "product rule" and assuming that inversion is already a differentiable map, we get that
\begin{align}
d\xi \cdot \xi^{-1} + \xi \cdot d(\xi^{-1}) = d(i) = 0
\end{align}
Hence,
\begin{align}
d(\xi^{-1}) = - \xi^{-1} \cdot d\xi \cdot \xi^{-1}
\end{align}
Here, we interpret $d\xi$ to mean the linear map $a \mapsto a$ from $A$ into $A$. Now, this should look a lot like $(*)$. So if you ever forget where the inverse $^{-1}$ goes or whatever, you can always "rederive" it this way.
The argument can be made more rigorous if we explicitly define all the maps, their domains, target spaces and indicate where the derivative is being taken and where we are evaluating everything etc. However, it should be noted that the argument above already assumes that the inversion map $H$ is differentiable. So, once you prove $H$ is differentiable, the above argument immediately gives $(*)$.
However, to prove $H$ is differentiable, we have to use a direct argument (if you want a proof, let me know, I'll try to sketch it).

Addition after OP's request of outline of proof:
Very roughly speaking, the proof relies on the geometric series formula $1+x + x^2 + \dots = \dfrac{1}{1-x}$ or equivalently
\begin{align}
(1 +z)^{-1} = \sum_{n=0}^{\infty} (-z)^n
\end{align} which you're hopefully familiar with for real (even complex) numbers with $|x|<1$ (to get the second equation, I put $z=-x$ into the first). In what follows I'll be very sloppy with all the technical details; for this I refer you to Henri Cartan's amazing book on differential calculus, in particular Theorems $1.72,1.73,2.44$.
So, to prove $H$ is indeed differentiable (it's actually $C^{\infty}$, and I think even analytic), pick any point $a$. Then, 
\begin{align}
H(a+h) &= (a+h)^{-1} \\
&= \bigg( a(i + a^{-1}h)\bigg)^{-1} \\
&= (i + a^{-1}h)^{-1} \cdot a^{-1} \\
&= \bigg(\sum_{n=0}^{\infty} (-a^{-1}h)^n \bigg) \cdot a^{-1} \\
&= \bigg( i + -a^{-1}h + \phi(h) \bigg) \cdot a^{-1} \\
&= a^{-1} - a^{-1}h a^{-1} + \phi(h) \cdot a^{-1} \\
&= H(a) - a^{-1}h a^{-1} + \phi(h) \cdot a^{-1}
\end{align}
where $\phi(h)$ is the remainder of the series excluding the $n=0,1$ terms. In this form, we can already see that $H(a+h)$ equals $H(a)$ plus a term linear in $h$ (which will turn out to be the derivative $dH_a(h)$) plus a remainder term depending on $h$. The technical analysis comes in when trying to show that the remainder term converges to $0$ "fast enough" i.e $\lim \limits_{h \to 0} \dfrac{\phi(h)\cdot a^{-1}}{\lVert h\rVert} = 0$.
This is the rough idea of the proof, but this is no where near complete, because especially when we're dealing with such series, we have to be extremely cautious of convergence issues etc. Also, strictly speaking, before even speaking of $H$ being differentiable, we need to ensure its domain $\text{inv}(A)$ is an open subset of $A$ (proven in Cartan). There are several other technical issues which need to be dealt with so that we can properly justify each equal sign above. For this, take a look at the book by Cartan.
