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I have to prove that a function going from $\Bbb R^n$ to $\Bbb R^n$ is a bijection. I fail at showing that it's a surjection. The properties of the function are:

  1. $\langle \mathbf{Df}(x)v,v\rangle \ge c||v||^2$ at any point $x$ and any $v, c>0$.
  2. $f$ is continuously differentiable at any point.

Thank for all your answers.

GNUSupporter 8964民主女神 地下教會
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1 Answers1

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By property 1, $Df$ never vanishes, so by the inverse function theorem, $f$ i a local diffeomorphism. (Because of your differentiability assumptions, I'm going to use "differentiable" to mean a continuously differentiable function, and diffeomorphism to mean a differentiable function with a differentiable inverse.)

You specifically asked about surjectivity, so I'll show that. Let $U$ be the image of $f$. It is open since $f$ is a local diffeomorphism and such functions are open maps. Since $\Bbb R^n$ is connected, it therefore suffices to show that $U$ is also closed.

Let $y$ be a limit point of $U$ (in the sense of point-set topology). Then there exists a differentiable path $\gamma : (0, 1) \to U$ such that $\lim_{t\to 1} \gamma (t) = y$. We may choose it in such a way that $\gamma$ has finite length. Now we can use the fact that $f$ is a local diffeomorphism to lift $\gamma$ to a differentiable path $\tilde{\gamma} : (0,1) \to \Bbb R^n$ such that $f\circ \tilde{\gamma} = \gamma$. (The idea here is to take each point $\gamma(t)$, find neighborhoods around $\gamma(t)$ and one of its preimages on which $f$ is a diffeomorphism, and then lift that local part of $\gamma$ using the diffeomorphism. Repeat for all points $\gamma(t)$ and show that you can choose the lifts in a compatible way.)

Reparametrize $\tilde{\gamma}$ so that it has unit speed and reparametrize $\gamma$ accordingly. Then by property 1, since $\gamma'(t) = Df(\tilde{\gamma}(t))\cdot \tilde{\gamma}'(t)$, we have that $||\gamma'(t)|| \geq c > 0$. Since $\gamma$ has finite length, it follows that the parameter $t$ is defined on a bounded interval. Thus $\tilde{\gamma}$ has finite length too, so in particular it is bounded and the limit $x_0 = \lim_{t \to 1} \tilde{\gamma}(t)$ exists. Finally, by continuity, we get that $f(x_0) = y$.

This proves that $U$ contains all of its limit points, i.e. $U$ is closed, so we're done.

Alex G.
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