How to show map is non-singular
Nonsingularity of $Df(x)$ follows from the general theory of matrices. If a matrix $A$ satisfies $\|A\|<1$, then $I-A$ is invertible. Its inverse is give by Neumann series $$(I-A)^{-1}=\sum_{n=0}^\infty A^n.$$
Related videos on Youtube
user118224
Updated on February 10, 2020Comments
-
user118224 over 3 years
Let $f:\;\mathbb{R}^n\to\mathbb{R}^n$ be differentiable. Suppose that for all $x\in\mathbb{R}^n:$ $$\lVert \mathrm{D}f(x)-\mathrm{I}\rVert\leq \frac{1}{2}$$ where $\lVert\cdot\rVert$ is the operator norm. I need to show $f$ must be a diffeomorphism.
By using the contraction mapping theorem I have shown that $f$ is surjective, and also I have shown that $\lVert f(x)-f(y)\rVert\geq\frac{1}{2}\lVert x -y\rVert$ so $f$ must be injective.
I'd like to use the inverse function theorem, but to do that I need to show that $\mathrm{D}f$ is non-singular. And this is where I'm stuck. Help?
-
reuns over 7 yearswhat is the definition of your $D f(x) -I$ operator ? how does it act (on functions) ? and what is the norm you are using / your Banach space to define the operator norm ?
-