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In a proof to show that $D(L\circ f)_a = L \circ Df_a$, where $f: U \subset E \rightarrow F$ and $L \in L_c(F,G)$.

They take the following limit to show that the frechet derivative exists,

$$\lim_{h\rightarrow 0} \frac{(F\circ f)(a+h) - (L \circ f)(a) - (L\circ Df_a)(h)}{\lVert h \rVert}$$

My question: normally, in the definition of the frechet derivative the numerator is contained within a norm. Here, this is not the case. My guess is that we are allowed to do this, since the norm is continuous, and $\lVert x \rVert = 0$ iff $x = 0$. But aren't we always allowed to do this? Furthermore, if we are always allowed to do this, then why does the frechet derivative include a norm in its numerator?

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up vote 1 down vote accepted

The limit above being $0 \in G$ is defined as the limit of the norm being $0 \in \mathbb{R}$.

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Well, in any normed space, $x_n \to 0$ if and only if $\|x_n\| \to 0$ as $n \to +\infty$. Why? Easy: the first statement means

for every $\epsilon >0$ there exists $N \in \mathbb{N}$ such that $n>N$ implies $|x_n|<\epsilon$

which is exactly the second statement, since $\||x_n|\| = \|x_n\|$.

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