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Let $f,g:U\to\mathbb{R}^n$ be differentiable at point $a\in U$, where $U\subset\mathbb{R}^m$ is an open set. Suppose $f(a)=g(a)$. Prove that $$\lim_{v\to0}\frac{f(a+v)-g(a+v)}{|v|}=0\;\;\;\;[\#]$$ if, and only if, $$f'(a)=g'(a)\;\;\;\;[*]$$

I've proved that $[*]\Rightarrow[\#]$ and I need help for the converse.

Definition: Suppose $U$ is an open set in $\mathbb{R}^m$, $f$ maps $U$ into $\mathbb{R}^n$, and $a\in U$. If there exists a linear transformation $T:\mathbb{R}^m\to\mathbb{R}^n$ such that $$f(a+v)-f(a)=Ta+r(h), \;\text{where}\; \lim_{v\to 0}\frac{r(v)}{|v|},$$ then we say that $f$ is differentiable at $a$, and we write $T=f'(a)$.

Thanks.

Pedro
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1 Answers1

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If $f$ and $g$ are differentiable at $a$, with $f(a)=g(a)$, then

$$f(a+v)-f(a)=\nabla f(a)\cdot v+O(\|v\|^2), $$

$$g(a+v)-g(a)=\nabla g(a)\cdot v+O(\|v\|^2), $$

and so

$$f(a+v)-g(a+v)=\left(\nabla f(a)-\nabla g(a)\right)\cdot v+O(\|v\|^2), $$

for all $v\in\mathbb R^m$ s.t. $a+v\in U$ (the for all is the important missing part in the OP). But then

$$0=\lim_{v\rightarrow 0}\frac{f(a+v)-g(a+v)}{\|v\|}=\lim_{v\rightarrow 0} \frac{\left(\nabla f(a)-\nabla g(a)\right)\cdot v+O(\|v\|^2) }{\|v\|}, ~~(*) $$ which implies

$$q(a):=\nabla f(a)-\nabla g(a)=0 $$

by arbitrariness of $v$. In fact, you can choose $v=te_i$ in $(*)$, for all $i=1,\dots m $ s.t. $a+te_i\in U$ and prove that all components $q_i(a)$ of the vector $q(a)$ are equal to zero.

Avitus
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    So, can we say that $0=\lim_{te_i\rightarrow 0} \frac{\left(\nabla f(a)-\nabla g(a)\right)\cdot te_i+O(|v|^2) }{|te_i|}=\lim_{te_i\rightarrow 0} \frac{\left(\nabla f(a)-\nabla g(a)\right)\cdot e_i }{|e_i|}=\left(\nabla f(a)-\nabla g(a)\right)\cdot e_i $? – Pedro Aug 15 '13 at 18:51
  • @Perguntador: with $t>0$ small enough (i.e. such that $a+te_i\in U$) you have $0=\lim_{t\rightarrow 0}\frac{t(\nabla f(a)-\nabla g(a))_i+O(t^2)}{t}=(\nabla f(a)-\nabla g(a))_i$. Do it for all $i$'s and you get the answer. Btw, in my notation, $e_i=(0,...,0,1,0,...,0)$. – Avitus Aug 15 '13 at 19:08