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Let $U$ be open and convex, $f:U \rightarrow \Bbb R^n$ continuously differentiable, where $\|Df(x)-Id\| < 1$, for all $x \in U$.

Then, $f$ is injective on $U$ (thus, $f:U \rightarrow f(U)$ is globally invertible.)

Attempt at a solution:

Let $g(x) = f(x) - x$ and choose $a,b \in U$ where $a \neq b$.

\begin{align} ||g(b)-g(a)|| &=||f(b)-b- \left(f(a)-a) \right)|| \\ &= \left|\left|\int_{0}^{1} \left[\frac {d}{dt}\left(\left(1-t\right)a + tb-f\left(\left(1-t\right)a+tb\right)\right) dt \right]\right|\right| \\ &\le\left|\left|\int_{0}^{1} \left[\left(b-a\right)-Df\left(\left(1-t\right)a + tb\right)\left(b-a\right) dt \right]\right|\right| \\ &\le\left|\left|b-a\right|\right|\int_{0}^{1} \left|\left|Id-Df\left(\left(1-t\right)a + tb\right)\right|\right| dt \lt\left|\left|b-a\right|\right|\int_{0}^{1}dt \end{align}

Thus, $$||g(b)-g(a)||\lt\left|\left|b-a\right|\right|$$

It is at this point I do not come any further.

Yiorgos S. Smyrlis
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Swiftprotector
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1 Answers1

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By the reverse triangle inequality, if $a \neq b$ then $$ \|f(b) - f(a)\| = \|b - a + g(b) - g(a)\| \geq \bigl|\|b - a\| - \|g(b) - g(a)\|\bigr| > 0. $$

Andrew D. Hwang
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