0
$\begingroup$

I don't understand how to prove this, or where to start. Thanks

$\endgroup$

3 Answers 3

6
$\begingroup$
By definition, $$ f:\mathbb{R}^n\mapsto\mathbb{R}^m\mbox{ is differentiable at }a\mbox{ if there exists a linear transformation, }\lambda:\mathbb{R}^n\mapsto\mathbb{R}^m,\\ \ \\ \mbox{ such that }\,\,\, \lim\limits_{h\to0}\dfrac{\left\|f(a+h)-f(a)-\lambda h\right\|_{m}}{\left\|h\right\|_{n}}{}={}0\,. $$

See that the assertion follows, because

$$ \begin{eqnarray*} \lim\limits_{h\to0}\dfrac{\left\|f(a+h)-f(a)-\lambda h\right\|_{m}}{\left\|h\right\|_{n}}{}={}0&{}\iff{}&\left\|\lim\limits_{h\to0}\dfrac{1}{\left\|h\right\|_{n}}\left(f(a+h)-f(a)-\lambda h\right)\right\|_{m}{}={}0\newline &&\newline &{}\iff{}&\left\|\lim\limits_{h\to0}\dfrac{1}{\left\|h\right\|_{n}}\left(g_1,\ldots,g_m\right)\right\|_{m}{}={}0\newline &&\newline &{}\iff{}&\left\|\left(\lim\limits_{h\to0}\dfrac{1}{\left\|h\right\|_{n}}g_1,\ldots,\,\lim\limits_{h\to0}\dfrac{1}{\left\|h\right\|_{n}}g_m\right)\right\|_{m}{}={}0\newline &&\newline &{}\iff{}&\lim\limits_{h\to0}\dfrac{1}{\left\|h\right\|_{n}}g_i{}={}0,\mbox{ for each }i\newline &&\newline &{}\iff{}& \mbox{each }f_i\mbox{ is differentiable at }a\,.\newline &&\newline \end{eqnarray*} $$

Above, for notational convenience, I used $g_i$s which are defined as

$$ g_i{}:={}f_i(a+h)-f_i(a)-(\lambda h)_i\,. $$

$\endgroup$
4
$\begingroup$

This has nothing to do with differentiation per se. It is a simple consequence of the following basic fact about convergence in ${\mathbb R}^m$: $${\bf y}\to{\bf 0}\qquad\Longleftrightarrow\qquad y_i\to0\quad (1\leq i\leq m)\ .$$ It follows that $$\lim_{{\bf X}\to{\bf 0}}{{\bf f}({\bf p}+{\bf X})-{\bf f}({\bf p})-A.{\bf X}\over|{\bf X}|}={\bf 0}$$ iff we have $$\lim_{{\bf X}\to{\bf 0}}{f_i({\bf p}+{\bf X})-f_i({\bf p})-\sum_{k=1}^n a_{ik}X_k\over|{\bf X}|}=0\qquad(1\leq i\leq m)\ .$$

$\endgroup$
2
  • $\begingroup$ Can you elaborate how this can be used to prove the result? $\endgroup$
    – mdcq
    May 7, 2018 at 15:15
  • 1
    $\begingroup$ @philmcole: See my edit. $\endgroup$ May 7, 2018 at 17:41
1
$\begingroup$

General idea: Let $\varphi_1,\ldots,\varphi_m:\mathbb{R}^n\to\mathbb{R}$ be linear functionals. Then one can deduce a linear transformation $\mathbb{R}^n\to\mathbb{R}^m$ by $$v\mapsto(\varphi_1(v),\ldots,\varphi_m(v)).$$In your case, if each component is differentiable, then the differential of each component is a linear functional. The above construction yields the differential of $f$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .