TITLE: Equivalence of 2 definitions of Differentiability
QUESTION [6 upvotes]: Let $X,Y$ be Banach spaces. I would like to prove the equivalence of the following definitions of differentiability. Let $f:X\to Y$ and $a\in X$

There is a map $\Delta : X \to L(X,Y)$ continuous at $a$, s.t.
$$f(x)=f(a)+\Delta(x)(x-a)$$
There is a map $D_af\in L(X,Y)$ s.t.

$$\lim_{x\to a}\frac{f(x)-f(a)-D_af(x-a)}{\|x-a\|_X}=0$$
The implication 1 => 2 is easy by picking $D_af=\Delta(a)$.
Im however stuck on the other direction, do I have to assume that $X,Y$ are finite dimensional?
1 => 2: Assume 1 holds, then
$$\begin{align*}\frac{f(x)-f(a)-\Delta(a)(x-a)}{\|x-a\|_X}&=\frac{\Delta(x)(x-a)-\Delta(a)(x-a)}{\|x-a\|_x}\\
&=[\Delta(x)-\Delta(a)]\left(\frac{x-a}{\|x-a\|_X}\right)\to0\end{align*}$$ since  $\Delta$ is continuous at a.
For 2=>1 I can do the 1-dimensional case. Then the map $D_af$ is just multiplication with the element $f'(a)$. If I let
$$R(x):= \frac{f(x)-f(a)-D_af(x-a)}{x-a}$$
And define the map $\Delta(x)$ to be multiplication with the element $(f'(a) + R(x))$ (for $x\neq a$ and $D_af$ else) everything works out fine.
So I thought the general finite dimensional case should work similar by defining $\Delta(x)$ to be the map given by the Jacobi Matrix of $D_af$ with every entry increased by $R(x):=\frac{\|f(x)-f(a)-D_af(x-a)\|_Y}{\|x-a\|_X}$. This surely gives continuity at $a$ but I don't see how the equality in 1 follows.
For the infinite dimensional case I have no idea how to proceed.

REPLY [1 votes]: Here is a generalization of your approach for the one-dimensional case to all cases, making heavy use of the axiom of choice.
For convenience, set $\eta(x):= f(x) - f(a) - D_a f(x-a)$.
Define $R \colon X \to L(X, Y)$ by choosing at each $x \in X$ an
$R(x) \in L(X, \operatorname{span}\{\eta(x)\}) \subset L(X,Y)$
such that
$R(x)(x-a) = \eta(x)$
and
$$\|\eta(x)\|_Y = \|R(x)\|_{L(X,Y)} \cdot \|x-a\|_X;$$
this is possible due to the Hahn–Banach theorem. Note that $R(a) = 0$. We take $\Delta := D_a f + R$, which satisfies the equality in 1.
By 2 (which is the usual definition of Fréchet differentiability), there is for each $\epsilon > 0$ a $\delta > 0$ such that $\|x-a\|_X < \delta$ implies
$$\|\Delta(x) - \Delta(a)\|_{L(X,Y)} = \|R(x)\|_{L(X,Y)} < \epsilon.$$
This is the continuity of $\Delta$ at $a$. (Note that there is zero guarantee of continuity anywhere else.)
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