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I'm having problems showing this result:

Let $f:U\rightarrow\mathbb{R}^m$ be a Lipschitz function with $a\in U$ and $g:V\rightarrow\mathbb{R}^p$, $V$ open, with $f(U)\subset V$ and $b=f(a)$. If $g'(b)=0$ then $g\circ f:U\rightarrow \mathbb{R}^p$ is differentiable in $a$, with $(g\circ f)'(a)=0$.

I have tried to use the definition of $g$ being differentiable in $b=f(a)$ and using $f(v)$ as the increment $h$ so:

$g(f(a)-f(v))-g(f(a))=g'(f(a))\cdot f(v)+r(f(a),f(v))$

Since $g'(b)=g'(f(a))=0$, we have:

$g(f(a)-f(v))-g(f(a))=r(f(a),f(v))$

I have tried to do through the inverse way but couldn't do it either. Can someone help me?

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    You may start by using that $f$ is Lipschitz, to replace the first term in you last equation.2012-04-13
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    You need to show that $||f\circ g(a+h)-f\circ g(a)||=||h||R(h)$ where $R(h)\to 0$ as $h\to 0$. Now, you have $||f\circ g(a+h)-f\circ g(a)|| \le ||g(a+h)-g(a)||=||h||R(h)$ since $g'(a)=0$.2012-04-13
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    @AndresCaicedo, the Lipschitz property is only valid under absolute value.2012-04-13
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    @Ivan, I think you confused f with g when doing the composition.2012-04-13
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    @GustavoMarra Yes, so: Do you see what to do about that? Where are the absolute values going to come from?2012-04-13
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    @AndresCaicedo, I have no idea...2012-04-13

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