I'm trying to prove the uniqueness of the derivative.
Let $(E,\|\cdot\|)$ and $(F,\|\cdot\|)$ be Banach spaces over the field $\mathbb{K}$, and $X$ an open subset of $E$. A function $f: X \rightarrow F$ is differentiable at $a \in X$ if there is an $A \in \mathcal{L}(E, F)$ such that $$f(x)=f\left(a\right)+A\left(x-a\right)+o\left(\left\|x-a\right\|\right) \quad\left(x \rightarrow a\right)$$ Prove that the map $A$ above is unique.
Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
My attempt:
Lemma: Assume $A,B \in L(E,F)$ and $(A-B)(x)=0$ for all $x \in E$ such that $\|x\|=1$. Then $A=B$.
Proof of the Lemma: Clearly, $(A-B)(0)= (A-B)(0x) = 0(A-B)(x)=0$ by the linearity of $A-B$. Suppose that $x \neq 0$, so $\|x\|>0$. We have $(A-B)(x) =(A-B)((x/\|x\|)\|x\|) = \|x\|(A-B)(x/\|x\|) = 0$ because $\|x/\|x\|\|=1$. Hence $(A-B)(x)=0$ and consequently $A(x)=B(x)$ for all $x \in X$. Thus $A=B$.
Let $B \in \mathcal{L}(E, F)$ such that $$f(x)=f\left(a\right)+B\left(x-a\right)+o\left(\left\|x-a\right\|\right) \quad \left(x \rightarrow a\right)$$
As such, $$A(x-a)- B(x-a) = o\left(\left\|x-a\right\|\right) \quad \left(x \rightarrow a\right)$$ or equivalently $$(A-B)(x-a) = o\left(\left\|x-a\right\|\right) \quad \left(x \rightarrow a\right)$$ and consequently $$\lim_{x \to a} \frac{(A-B)(x-a)}{\|x-a\|} = 0$$
Let $y \in X$ such that $\|y\|=1$. Define a sequence $(x_n)$ in $X$ by $x_n= a+y/n$. It follows that $\lim x_n =a$ and $y = (x_n-a)/\|x_n-a\|$ for all $n \in \mathbb N^*$. As such, $$(A-B)y =(A-B) \left ( \frac{x_n-a}{\|x_n-a\|} \right ) = \frac{(A-B)(x_n-a)}{\|x_n-a\|}$$
Take the limit $n \to \infty$, we have $$(A-B)y = \lim_{n \to \infty} \frac{(A-B)(x_n-a)}{\|x_n-a\|} = \lim_{x \to a} \frac{(A-B)(x-a)}{\|x-a\|} = 0$$
By our lemma, $A=B$.