Fréchet derivative, redundant norm in the definition?

Question

The Fréchet derivative of a map $f:V \to W$ between Banach space at $x \in V$ is defined as the linear map $A:V \to W$ such that $$ \lim_{h\to 0} \frac{\| f(x-h) - f(x) - Ah \|_W}{\| h \|_V} = 0 \in \mathbb R $$ If the Quotient goes to zero then and the denominator goes to zero then numerator must as well, and so $$ \lim_{h\to 0} \| f(x-h) - f(x) - Ah \|_W = 0 \implies \\ \| \lim_{h\to 0} (f(x-h) - f(x) - Ah) \|_W = 0 \iff\lim_{h\to 0} (f(x-h) - f(x) - Ah) = 0 $$ I think that the norm in the numerator is redundant so the definition could be written as $$ \lim_{h\to 0} \frac{f(x-h) - f(x) - Ah}{\| h \|_V} = 0 \in W $$ Is there any counter example why this may not be equivalent ?

Answer 1

Well, not really, because the norm specifies the topology in which the limit is taken. Yes, of course, the only choice of topologoy we have at our disposal is the one that is induced by $\lVert \cdot \rVert_W$. But then, what does the following mean? $$ \lim_{h \rightarrow 0} \frac{f(x-h) - f(x) -Ah}{\lVert h \rVert_V} =0 $$ This is defined as $$ \lim_{h \rightarrow 0} \left \lVert \frac{f(x-h) - f(x) -Ah}{\lVert h \rVert_V} \right \rVert =0 \iff \lim_{h \rightarrow 0} \frac{\lVert f(x-h) - f(x) -Ah \rVert_W}{\lVert h \rVert_V} =0. $$