Show that the function $g(x)= e^{-\|f(x)\|^2}$ is Lipchitz st $f$ it is?

Question

Let $F$ be an Euclidean space with norm || associated to the scalar product. $f: \mathbb{R} \rightarrow F$ Differentiable and Lipchitzian with $\lim _{x \rightarrow+\infty}\|f(x)\|=\lim _{x \rightarrow-\infty}\|f (x)\|=+\infty$. I want to show that the function $$g: \mathbb{R} \rightarrow \mathbb{R} ; x \mapsto e^{-\|f(x)\|^2}$$ is Lipchitz. I know that I should use the mean value inequality but in vain ! I have showed that : $|g'(x)|\leqslant 2M \langle f'(x),f(x)\rangle$ with $M=exp(-\inf_{x\in \mathbb R} \|f(x)\|^2)$ ! Any help please !

Answer 1

Write $g(x)=h(\|f(x)\|)$ where $h(x)=e^{-x^2}$. It's not hard to show that $|h'(x)|\leq \sqrt{\frac{2}{e}}=a$ for all $x$, making $h$ a Lipchitz function.

Now take $x,y\in \mathbb{R}$ arbitrarily. Then $$\begin{eqnarray*}|g(x)-g(y)| &\leq& a\Big|\|f(x)\|-\|f(y)\|\Big| \\ &\leq & a\|f(x)-f(y)\| \\&\leq& ak|x-y| \end{eqnarray*}$$ where $k$ is the Lipchitz constant for $f$.