# Show that $f+\epsilon g\in{\rm Diff^1}(\Bbb R^m)$ for all $\epsilon\in(-\epsilon_0,\epsilon_0)$

This is an exercise on page 220 of Analysis II of Amann and Escher Here $\rm Diff^k$ means the set of diffeomorphisms where the $k$-th derivative is also an homeomorphism. My work below.

The exercise have a hint that Ive not used because I dont see the relation to the exercise. The hint says "consider the function ${\rm id_{\Bbb R^m}}+f^{-1}\circ(\epsilon g)$ and apply exercise 7".

We want to show for each case that exists some $\epsilon_0>0$ such that $f+\epsilon g\in{\rm Diff^1}(\Bbb R^m)$ for all $\epsilon\in(-\epsilon_0,\epsilon_0)$.

(a) We have that

1. $\exists \alpha>0: |f(x)-f(y)|\ge \alpha |x-y|,\, x,y\in\Bbb R^m$.
2. $\exists\beta>0:|g(x)-g(y)|\le\beta|x-y|,\, x,y\in\Bbb R^m$.
3. If $f+\epsilon g$ is not injective then exists $x,y\in\Bbb R^m$ such that $f(x)-f(y)=-\epsilon(g(x)-g(y))$.
4. Clearly $f+\epsilon g\in C^1(\Bbb R^m)$ for any $\epsilon\in\Bbb R$.

From the third statement we have that $|f(x)-f(y)|=|\epsilon|\,|g(x)-g(y)|$, and from the first two statements we have that $$\alpha|x-y|\le|f(x)-f(y)|=|\epsilon|\,|g(x)-g(y)|\le\beta|\epsilon|\,|x-y|\tag1$$ Hence if we take $\epsilon_0=\alpha/\beta$ then we get a contradiction in $(1)$ so the function is injective. And $$|f(x)-f(y)+\epsilon(g(x)-g(y))|\ge\big||f(x)-f(y)|-|\epsilon||g(x)-g(y)|\big|\\\ge\big|\alpha|x-y|-|\epsilon|\beta|x-y|\big|=\big|\alpha-|\epsilon|\beta\big||x-y|\tag2$$ so $(f+\epsilon g)^{-1}​$ is Lipschitz for each $\epsilon\in(-\alpha/\beta,\alpha/\beta)​$, so it is continuous. From it continuity we find that $f+\epsilon g​$ is surjective because the image of it inverse is $\Bbb R^m​$.

It remains to show that $\partial(f+\epsilon g)(x)\in\mathcal L{\rm is}(\Bbb R^m)$ for each $x\in\Bbb R^m$. From $(2)$ we have that $$|(f+\epsilon g)(x)-(f+\epsilon g)(y)|\ge K |x-y|,\quad\forall x,y\in\Bbb R^m,\, K:=|\alpha-\beta|\epsilon||>0$$ Thus, from the definition of directional derivative, it follow that $\partial (f+\epsilon g)(x)h\neq 0$ for all $x,h\in\Bbb R^m\setminus\{0\}$.

(b) Without lose of generality suppose that this bounded set, namely $X$, is open and convex. Then $\overline X$ is compact, so its easy to see that $g$ is bounded. Also $\partial g$ is zero outside of $\overline X$ so $\partial g$ is also bounded and, by the MVT, $g$ is Lipschitz.

By the same reasons $f^{-1}|_{f(X)}$ is also Lipschitz so we knows that $f+\epsilon g\in{\rm Diff^1}(X,Y)$ for some $Y\subset\Bbb R^m$. It remains to see that $f+\epsilon g$ is bijective in $\Bbb R^m$.

To show injectivity we need to show that $f(x)-f(y)\neq\epsilon g(y)$ for all $x\in\Bbb R^m\setminus X$ and all $y\in X$, what is equivalent to show that $f(X)=(f+\epsilon g)(X)$.

My questions:

1. Is the part a) correctly done?

2. Im stuck in the part b), Im unable to prove the injectivity of $f+\epsilon g$ in $\Bbb R^m$ knowing that this function is injective, separately, in the domains $\Bbb R^m\setminus X$ and $X$. I need some help here.

## UPDATE:

I think I finally solved it. Observe that $$(f+\epsilon g)(x)=({\rm id}_{\Bbb R^m}+\epsilon g\circ f^{-1})(f(x))\tag3$$ I set $h:=g\circ f^{-1}$. Then by the mean value theorem we knows that

$$|h(y)-h(x)|\le\sup_{t\in[0,1]}|\partial h(xt+y(1-t))||x-y|\tag4$$

and because $\partial h(f(x))=\partial g(x)[\partial f(x)]^{-1}$ is continuous and vanishes if $x\notin X$ then we can conclude that $\partial h$ is bounded, so $h$ is Lipschitz with some Lipschitz constant $K>0$. Thus using $(3)$ we can see that

$$\frac{|x'+\epsilon h(x')-y-\epsilon h(y')|}{|x'-y'|}\ge\left|1-|\epsilon|\frac{|h(x')-h(y')|}{|x'-y'|}\right|\ge |1-|\epsilon|K|\tag5$$

whenever $|\epsilon| K<1$, and where $x':=f(x)$ and $y':=f(y)$. This shows that for $\epsilon\in(-1/K,1/K)$ we have that

$$|({\rm id}_{\Bbb R^m}+\epsilon g\circ f^{-1})(x')-({\rm id}_{\Bbb R^m}+\epsilon g\circ f^{-1})(y')|\ge |1-|\epsilon|K||x'-y'|\tag6$$

From a previous result I know that a function $r\in C^1(\Bbb R^m,\Bbb R^m)$ such that $|r(x)-r(y)|\ge\alpha|x-y|$ for all $x,y\in\Bbb R^m$ and some $\alpha>0$ is a diffeomorphism in the whole $\Bbb R^m$, then ${\rm id}_{\Bbb R^m}+\epsilon g\circ f^{-1}$ is a diffeomorphism, and by $(3)$ we find that $f+\epsilon g$ it is also.$\Box$

Can someone confirm if the above is correct? Thank you.

• You can with the Radamacher theorem:en.wikipedia.org/wiki/Rademacher%27s_theorem . which asserts that g is a.e-diffenrentiable – Guy Fsone Nov 1 '17 at 18:06
• @GuyFsone but it is stated in the exercise that $g$ is continuously differentiable in $\Bbb R^m$. How this can help me to solve the injectivity of $f+\epsilon g$ in part b)? – Masacroso Nov 1 '17 at 18:12
• sorry what is Diff${ }^1$? – Guy Fsone Nov 1 '17 at 18:13
• dg(x) is bounded, then, $df(x) +\varepsilon dg(x)$ is invertible for small epsilon – Guy Fsone Nov 1 '17 at 18:21