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The lemma rests on two items: the definition of a function of n variables differentiable in a point "a" and the Mean Value Theorem for functions of n variables.

I. A function [tex] f:U\rightarrow{R}, [/tex] defined in an open set [tex] U \subset R^n,[/tex] is said to be differentiable in a point [tex] (a_1,...,a_n) \in U[/tex] when it fulfills these conditions:

1. There exist the partial derivatives [tex] \frac{\partial}{\partial x_1}f(a_1,...,a_n),..., \frac{\partial}{\partial x_n}f(a_1,...,a_n)[/tex].

2. For every [tex] v = (v_1,...,v_n) [/tex] such that [tex] a + v \in U [/tex] we got

[tex] f(a+v) - f(a) = \sum_{i=1}^{n} v_i \frac{\partial}{\partial x_i}f(a) + r(v), [/tex] where [tex] \lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert}=0 [/tex].

II. The Mean Value Theorem.

Let the function [tex] f:U\rightarrow{R} [/tex] be differentiable in the open set [tex] U \subset R^n,[/tex] and the line [tex] [a, a+v] \subset U[/tex]; then we can find a [tex] \theta \in (0,1) [/tex] such that

[tex] f(a+v) - f(a) = \sum_{i=1}^{n} v_i \frac{\partial}{\partial x_i}f(a+ \theta v) [/tex].

Now I state the

Lemma.- Let be the function [tex]r:B\rightarrow{R}[/tex] of class [tex]C^2[/tex] in the open ball [tex]B \subset R^n[/tex] of center [tex] (0,...,0).[/tex] If for every [tex] i = 1,..., n [/tex] we got [tex]r(0,...,0) = \frac{\partial}{\partial x_i}r(0,...,0) = \frac{\partial^2}{\partial x_j \partial x_i}r(0,...,0) = 0, [/tex] then [tex] \lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert^2}=0 [/tex].

And here I copy literally the proof of the author:

"Proof.-

1. "Being [tex]r:B\rightarrow{R}[/tex] a function of class [tex]C^1[/tex] (therefore differentiable) that gets null in the point [tex] (0,...,0)[/tex] (and the same for its derivatives [tex] \frac{\partial}{\partial x_i}r [/tex]), it follows from the definition of differentiable function that [tex] \lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert}=0 [/tex]". (My note: OK, this is fine).

2. "By the Mean Value Theorem, for each [tex] v = (v_1,..., v_n) \in B[/tex] exists [tex] \theta \in (0,1) [/tex] such that [tex] r(v) = \sum_{i=1}^{n} v_i \frac{\partial}{\partial x_i}r(\theta v). [/tex] Therefore [tex] \frac{r(v)}{\Vert{v}\Vert^2}= \sum_{i=1}^{n} {\frac {1}{\Vert{v}\Vert}v_i \frac{\partial}{\partial x_i}r(\theta v) [/tex]." (OK, this is fine also).

3. "Every partial derivative [tex] \frac{\partial}{\partial x_i}r [/tex]

and its derivatives [tex] \frac{\partial^2}{\partial x_j \partial x_i}r, [/tex] gets null in the point [tex] (0,...,0) [/tex]. Hence, from our initial observation (I suppose he refers to paragraph 1? ) it follows that (I do not understand this) [tex] \lim_{\Vert{v}\Vert\rightarrow 0} {\frac {1}{\Vert{v}\Vert}}{\frac{\partial}{\partial x_i}r(\theta v)} = 0 [/tex] for all [tex] i = 1,...,n.[/tex]"

4. "Furthermore, each quocient [tex] \frac {v_i}{\Vert{v}\Vert} [/tex]has absolute value [tex]\leqq 1[/tex]. Therefore [tex] \lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert^2}=0 [/tex]".

End of proof.

As I've said, the results of paragraphs 1 and 2 are OK. My trouble is the inference of paragraph 3. I know that each partial derivative [tex] \frac{\partial}{\partial x_i}r [/tex] is on its own right a function differentiable in [tex] (0,...,0) [/tex], so applying the definition we've seen before the lemma we got for every [tex] i = 1,...,n [/tex] that [tex]\lim_{\Vert{v}\Vert\rightarrow 0} {\frac {1}{\Vert{v}\Vert} \frac{\partial}{\partial x_i}r(\theta v) = 0 [/tex]. But I don't catch up how this fact leads to the result of paragraph 3. Or maybe he gets that result in another way which escapes me.

Can I ask for your assistance?

P Castilla.