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In the Wikipedia Page for the Fréchet Derivative they say that a function $f:V \to W$ is Fréchet differentiable at $x \in V$ If there exists a linear Operator $A: V \to W$ such that:

$$\lim_{||h|| \to 0} \frac{||f(x+h) -f(x) -A(h)||_W}{||h||_V} = 0$$

My question is: Is $\lim_{||h|| \to 0}$ the same as $\lim_{h \to 0_V}$? Because $||h|| = 0 \iff h = 0_V$.

Eduardo Magalhães
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    Technically it depends on the topology of $V$, but in the norm topology these are the same. – Ian Jul 24 '21 at 13:34
  • @Ian by definition of norm, only zero vector has norm $0$. If it were approaching another vector then the norm would differ from zero. Could you please elaborate a little? How does topology affect situation, is there any obvious example? – Salihcyilmaz Jul 24 '21 at 15:52
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    @Salihcyilmaz In other topologies you can converge to the zero vector without having the norm converge to zero. The classic example is the situation of the Riemann-Lebesgue lemma where rapidly oscillating sinusoids converge weakly in $L^2$ but not strongly in $L^2$ to 0. – Ian Jul 25 '21 at 00:23
  • Thanks for the comment @Ian, I will check it out – Salihcyilmaz Jul 25 '21 at 07:02

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