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Let's consider an application $T : \mathbb R^n \longrightarrow \mathbb R^n$ such as $T:x \longmapsto x+\phi(x)$ where $\phi: \mathbb R^n \longrightarrow \mathbb R^n$ is $\frac{1}{2}$-Lipschitz continuous.

I'm trying to show that T is a bijective application and that it is also an homeomorphism. I tried using the Banach fixed point theorem but i can't seem to find the right function.

I think not really understanding the standard proof keeps me from getting this one.

Alex M.
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C.Patrick
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  • To show that $T$ is a bijection, you need to show that for every $y\in \mathbb{R}^n$ there is exactly one $x\in \mathbb{R}^n$ with $T(x) = y$. To apply Banach's fixed point theorem, you need to rewrite $T(x) = y$ into a fixed point equation, so find an $S$ such that $T(x) = y \iff S(x) = x$. – Daniel Fischer Oct 31 '16 at 16:38
  • Yes, that's what i did. If T is bijective then y+x-T(x)=x so if we name thé left part g we have g(x)=x. Now we try to show that g is à contraction mapping which would mean that T is bijective. I tried several things but i can't seem to find thé right inequality for g – C.Patrick Oct 31 '16 at 16:42
  • Whether $T$ is bijective or not, we have $T(x) = y \iff y + x - T(x) = x$. A good start. Can you simplify $y + x - T(x)$? – Daniel Fischer Oct 31 '16 at 16:46
  • I get $g(x)=y-\phi(x)$ but with the distance i can't seem to get the right inequality. $d(g(z),g(x))=d(y-\phi(x),y-\phi(z))$ am i supposed to use the triangular inequality afterwards? – C.Patrick Oct 31 '16 at 17:32
  • Well, things depend on the used distance. If nothing else is explicitly specified, on $\mathbb{R}^n$ the default is the Euclidean distance. That comes from a norm, $d(a,b) = \lVert a - b\rVert$. This simplifies the expression for $d(g(z),g(x))$. – Daniel Fischer Oct 31 '16 at 18:11
  • Using the norm is really much easier in this case. I was worried that it might not be correct but since it's a banach space you're absolutely right. Then whe just have $|g(x)-g(z)|=|\phi(x)-\phi(z)|\leq \frac{1}{2}|x-z|$ – C.Patrick Nov 01 '16 at 08:12
  • Is T also an homeomorphism? To prove that is it correct to just prove that $T^{-1}$ is also a lipschitz function? – C.Patrick Nov 01 '16 at 08:19
  • Yes, and yes. Once you've proved that $T^{-1}$ is Lipschitz, it follows that it is continuous, hence $T$ is a continuous bijection with continuous inverse. – Daniel Fischer Nov 01 '16 at 08:50
  • Thank you very much for your help. I'm also having trouble with $T^{-1}$. Am i supposed to write $T^{-1}=Id+\phi^{-1}$ and do the same with the norm? Because that seems too simple to be right – C.Patrick Nov 01 '16 at 13:39
  • Did you forget parentheses, $T^{-1} = (\operatorname{id} + \phi)^{-1}$? That would be correct, but doesn't help (much). As written, it's wrong. First, $\phi^{-1}$ need not exist, second, even if it exists, the inverse of $\operatorname{id} + \phi$ is hardly ever $\operatorname{id} + \phi^{-1}$. To show that $T^{-1}$ is Lipschitz, find a lower bound for $\lVert T(x) - T(y)\rVert$ in terms of $\lVert x-y\rVert$. – Daniel Fischer Nov 01 '16 at 13:48
  • Yes sorry about the parentheses, i'm currently writing this from my phone and forgot them.$|T(x)-T(y)|=\Vert x-y+\phi(x)-\phi(y)\Vert \leq \Vert x-y \Vert+\Vert \phi(x)-\phi(y)\Vert \leq \frac{3}{2}\Vert x-y\Vert$ Now i've proved that T is Lipschitz but what does that tell me about T^{-1}? – C.Patrick Nov 01 '16 at 14:31
  • You need a lower bound, $\lVert T(x) - T(y)\rVert \geqslant C\lVert x-y\rVert$. – Daniel Fischer Nov 01 '16 at 14:41
  • $T^{-1}(y_1)=x_1$ and $T^{-1}(y_2)=x_2$. So $\Vert T^{-1}(y_1)-T^{-1}(y_2)\Vert = \Vert x_1-x_2\Vert \leq \Vert T(x_1)-T(x_2)\Vert + \Vert g(x_1)-g(x_2)\Vert \leq \Vert T(x_1)-T(x_2)\Vert +\frac{1}{2}\Vert x_1-x_2\Vert$. So $\Vert x_1-x_2\Vert \leq 2\Vert T(x_1)-T(x_2)\Vert$ and by taking the inverse we have $\Vert T^{-1}(y_1)-T^{-1}(y_2)\Vert \leq 2\Vert y_1-y_2\Vert$ – C.Patrick Nov 01 '16 at 14:49
  • Yup, that's it. – Daniel Fischer Nov 01 '16 at 14:52
  • Alright! Thank you very much for your time and patience i really appreciate It – C.Patrick Nov 01 '16 at 14:52

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