If $f$ is Lipschitz and g is differentiable, then $g\circ f$ is differentiable where $g'(b)=0$.

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. –  Andres Caicedo Apr 13 '12 at 18:34
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 $g'(a)=0$. –  Ivan Apr 13 '12 at 18:36
@AndresCaicedo, the Lipschitz property is only valid under absolute value. –  Marra Apr 13 '12 at 19:28
@Ivan, I think you confused f with g when doing the composition. –  Marra Apr 13 '12 at 19:28
@GustavoMarra Yes, so: Do you see what to do about that? Where are the absolute values going to come from? –  Andres Caicedo Apr 13 '12 at 20:19

The definition of differentiability of $g$ is more commonly written $$g(y)=g(x)+g'(x)\cdot (y-x)+o(|y-x|)\qquad\text{as } y\to x.$$ Substitute in $x=f(a)$ and $w=f(v)$ and let $v\to w$.
@GustavoMarra: You need it in order to see that $o(|f(v)-f(a)|)=o(|v-a|)$. –  Harald Hanche-Olsen Apr 14 '12 at 7:21