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I found the following question on the internet, and I think it would be useful to solve it as part as studying to my midterm:

Let $f:\mathbb{R}^n \to \mathbb{R}^n $ having continuous partial derivatives up to first order , such that: $ \| f(x)- f(y) \| > \frac{1}{10} \|x-y\| $ for every $x,y\in \mathbb{R}^n $ such that $x\neq y$.

  1. Prove that $J(f) \neq 0 $

  2. Prove that $f$ sends $\mathbb{R}^n $ onto $\mathbb{R}^n $ .

Thanks in advance

copper.hat
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homogenity
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2 Answers2

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Part 1:

Since $f$ is differentiable at $x$, for all $\epsilon>0$ there is some $\delta>0$ such that for all $\|h\| < \delta$, we have $\|f(x+h)-f(x) -DF(x)h\| \le \epsilon \|h\|$. Choose $\epsilon = \frac{1}{20}$. Then \begin{eqnarray} \|Df(x)h\| &=& \|DF(x)h - (f(x+h)-f(x))+(f(x+h)-f(x))\| \\ &\ge& \|f(x+h)-f(x)\| - \|DF(x)h - (f(x+h)-f(x))\| \\ &\ge& \frac{1}{20} \|h\| \end{eqnarray} And so $DF(x)$ is invertible and $J(f(x)) \neq 0$.

Part 2:

Let $Y= f(\mathbb{R}^n)$.

Since $DF(x)$ is invertible, the implicit function theorem shows that $f$ has a local inverse, and hence $f$ maps open sets into open sets. Hence $Y$ is open.

Suppose $y_n \in Y$, and $y_n \to y$. Since $y_n \in Y$, there is some $x_n$ such that $y_n = f(x_n)$. Since $\|y_n-y_m \| = \|f(x_n)-f(x_m)\| \ge \frac{1}{10} \|x_n-x_m\|$, we see that $x_n$ is Cauchy, and hence $x_n \to x$ for some $x$. Since $f$ is continuous, we have $f(x) = y$, and so $y \in Y$. This shows that $Y$ is closed.

Since $\mathbb{R}^n$ is connected, and $Y \ne \emptyset$, we have $Y = \mathbb{R}^n$.

copper.hat
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  • Hi copper.hat, Thanks for your answer . Will you please explain to me why the fact that $||Df(x)h||\geq \frac{1}{20} ||h|| $ implies that $Df(x)$ is invertible ? Thanks ! – homogenity Nov 10 '13 at 08:09
  • Sure; the inequality shows that if $h \neq 0$, then $Df(x)h \neq 0$. That is, $\ker DF(x) = {0}$. Since $DF(x)$ is square, it follows that it is invertible. – copper.hat Nov 10 '13 at 08:43
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Since $f: \mathbb{R}^n \to \mathbb{R}^n$ has continuous partial derivatives, $f$ is differentiable. So we have $$ \lim_{h \to 0} \frac{\|f(x + h) - f(x) - Df(x) \cdot h\|}{\|h\|} = 0. $$ Triangle inequality gives $$ \|f(x + h) - f(x)\| - \|Df(x) \cdot h\| \leq \|f(x + h) - f(x) - Df(x) \cdot h\|. $$ Hence $$ \lim_{h \to 0} \frac{\|f(x + h) - f(x)\|}{\|h\|} - \lim_{h \to 0} \frac{\|Df(x) \cdot h\|}{\|h\|} \leq 0 \\ \implies \lim_{h \to 0} \frac{1}{\|h\|}\frac{1}{10}(\|(x + h) - x\|) - \lim_{h \to 0} \frac{\|Df(x) \cdot h\|}{\|h\|} < 0 \\ \implies \frac{1}{10} < \lim_{h \to 0} \frac{\|Df(x) \cdot h\|}{\|h\|} \leq \lim_{h \to 0} \frac{\|Df(x)\| \cdot \|h\|}{\|h\|} \\ \implies \|Df(x)\| > \frac{1}{10} $$ where $\|Df(x)\|$ is the operator norm. So $Df(x) \neq 0$.

Pratyush Sarkar
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  • Hi Pratyush, does $||Df(x)|| \neq 0 $ where $$||$$ is the operator norm imply that $J(f) \neq 0 $ ? – homogenity Nov 10 '13 at 08:11
  • @homogenity You're right, it doesn't. Initially I thought you meant $Df$ by the notation $J(f)$, but now I realize you meant the determinant. You can just follow copper.hat's answer. My answer is essentially the same ... from the second last step you make the $\epsilon$-$\delta$ argument (with $\epsilon = \frac{1}{20}$) to get the same result as copper.hat. – Pratyush Sarkar Nov 10 '13 at 15:52