The proposition that I'm trying to prove is:

Let $f:U\rightarrow \mathbb R^n$ be Lipschitz defined on $U\subset \mathbb R^m$ an open set and $a\in U$. Let $g:V\rightarrow \mathbb R^p$ differentiable in the open set $V\subset \mathbb R^m$ 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 $a$, and $(g\circ f(a))'=0$.

I've tried to prove by the definition, i.e., that $g\circ f(a+v)=g\circ f(a)+(g\circ f(a))'v+r(v)$, with $\lim_{v\rightarrow0}\frac{r(v)}{|v|}=0$.

So what I got is $\frac{r(v)}{|v|}=\frac{g( f(a+v))-g( f(a))}{|v|}$, I don't know how to solve this limit, I've tried to use the Lipschitz condition to change $|v|$ to $|f(a+v)-f(a)|$, and get $k\frac{g( f(a+v))-g( f(a))}{|f(a+v)-f(a)|}$, $k$ the Lipschitz constant, but I can't go further. Any tips?

Thanks in advance.

  • 1
    $\begingroup$ Take a look here $\endgroup$ – Masacroso Mar 25 '17 at 13:13

If $f(a+v)=f(a)$ then $r(v)=0$. If $f(a+v)\ne f(a)$, then $$\left\vert\frac{r(v)}{|v|}\right\vert=\left\vert\frac{g(f(a+v)-g(f(a))}{|f(a+v)-f(a)|}\right\vert\frac{|f(a+v)-f(a)|}{|v|}\le k \frac{\vert g(f(a+v)-g(f(a))\vert}{|f(a+v)-f(a)|}.$$ Using the fact that $g'(b)=0$ you have that $\frac{\vert g(y)-g(b)\vert}{|y-b|}\le\varepsilon$ whenever $|y-b|\le \delta$. Now, since $f$ is continuous, $f(a+v)\to f(a)=b$ when $v\to 0$ and so you can find $\eta>0$ such that if $|v|\le\eta$, then $|f(a+v)-b|\le \delta$. In turn, $$\frac{\vert g(f(a+v)-g(f(a))\vert}{|f(a+v)-f(a)|}\le \varepsilon.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.