Showing that if the operator norm of $Dg$ is $< 1$, then $g$ has a fixed-point

Question

I'm trying to solve this following question:

Let $g:\mathbb{R^{n}\to R^{n}}$ be a differentiable function. Show that if exists $0\le r<1$ s.t. $\forall a\in \mathbb{R^{n}} \quad \vert\vert{(Dg)_a}\vert\vert_{op}\le r$, when $\vert\vert{(Dg)_a}\vert\vert_{op} =\max \left\{\vert\vert(Dg)_a \vert\vert_2: \vert\vert x\vert\vert _2=1\right\} $, then $g$ has a fixed-point in $\mathbb{R^{n}}.$

I wanted to show that $g$ is a contraction mapping, and since $\mathbb{R^{n}}$ is a complete metric space - g admits an unique fixed-point (Banach fixed-point theorem), but I couldn't prove that $g$ is indeed a contraction mapping.

Any hint would be appreciated. Thank you!

Answer 1

Surely, in your problem statement, you meant $0\leq r<1$.

Apart from that typo, use the mean-value inequality: $\|g(x)-g(y)\|\leq \bigg(\sup\limits_{t\in [0,1]}\|Dg_{x+t(y-x)}\|_{\text{op}}\bigg)\cdot \|x-y\|$. For the proof of this theorem, see Loomis and Sternberg's Advanced Calculus, section 3.7, theorems 7.3 and 7.4.

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Remark.

Anytime you have to relate information about derivatives to information about the function, your first thought should be to somehow invoke the mean-value theorem (in single-variable calculus) or the mean-value inequality (in multivariable calculus).