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Let $U\subset \mathbb R^m$ and $V\subset \mathbb R ^n$ be an open sets and $f:U\to \mathbb R^n$ a Lipschitz map and $g:V\to \mathbb R^p$ a differentiable function. Suppose $f(U)\subset V$ and $b=f(a)$.

I want to prove that If $g'(b)=0$, then $g\circ f:U\to \mathbb R^p$ is differentiable at the point $a$ with $(g\circ f)'(a)=0$.

If $f$ were differentiable, this question would be just a straightforward application of the chain rule. Instead of this $f$ is just Lipschitz, i.e., there is a constant $c\in \mathbb R$ such that for every $x,y\in U$

$$|f(x)-f(y)|\le c|x-y|$$

In another words, for every $x\in U$ and a sufficiently small $u\in \mathbb R^m$ we have:

$$|f(x+u)-f(x)|\le c|u|$$

We also know that $\frac{g(b+v)-g(b)}{|v|}\to 0$, as $v\to 0$ and we want to prove that $\lim_{v\to 0}\frac{g(f(a+v))-g(f(a))}{|v|}=0$.

My attempt was trying to use triangle inequality and the squeeze theorem but I didn't see any relation between the absolute value of these expressions above.

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By the differentiability of $g$ at $b$, there exists a function $r:\mathbb{R}^n \rightarrow \mathbb{R}^p$ such that $g(b+v)=g(b)+g'(b)v + r(v)$ with $\frac{|r(v)|}{|v|} \rightarrow 0$ as $v \rightarrow 0$. So, $\frac{|g(f(a+v)) - g(f(a))|}{|v|} = \frac{|r(f(a+v) - f(a))|}{|v|} = \frac{|r(f(a+v) - f(a))|}{|f(a+v) - f(a)|} \frac{|f(a+v) - f(a)|}{|v|} \le \frac{|r(f(a+v) - f(a))|}{|f(a+v) - f(a)|} c$ and this last expression goes to $0$ as $v \rightarrow 0$.

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    Thank you for your answer. Why does this $\sup$ exist? – 2017-01-07
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    In another words, why is $g$ $C^1$ – 2017-01-07
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    I think you can't assume that, that's why I edited the post with a new approach and deleted the first one. – 2017-01-07