Day: March 15, 2023

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202303150916 Exercise 4.6.4

Let f:\mathbb{R}^2\to\mathbb{R}^2 be defined by

f(x,y)=(x^3+y^3,x^3-y^3) for (x,y)\in\mathbb{R}^2.

Prove that the Jacobian matrix J_{f,(0,0)} is the zero matrix. Show that nevertheless f is globally invertible on \mathbb{R}^2. [Hint: prove that f is \textrm{1--1} on \mathbb{R}^2.] Show also that the inverse function is not differentiable at f(0,0)=(0,0).

Extracted from P. R. Baxandall. (1986). Vector Calculus.

Roughwork.

\begin{aligned} f\bigg(\begin{bmatrix}x\\y\end{bmatrix}\bigg) & = \begin{bmatrix} x^3+y^3 \\ x^3-y^3 \end{bmatrix} \\ J_{f,(x,y)} & = \begin{bmatrix} \partial_x (x^3+y^3) & \partial_y(x^3+y^3) \\ \partial_x(x^3-y^3) & \partial_y(x^3-y^3) \\ \end{bmatrix} \\ & = \begin{bmatrix} 3x^2 & 3y^2 \\ 3x^2 & -3y^2 \\ \end{bmatrix} \\ J_{f,(0,0)} & = \begin{bmatrix} 3(0)^2 & 3(0)^2 \\ 3(0)^2 & -3(0)^2 \end{bmatrix} \\ & = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix} \\ \end{aligned}

The function f:\mathbb{R}^n\to\mathbb{R}^n is locally invertible at \mathbf{x}_0 if there is an \epsilon >0 and a function g:B_{\epsilon}(f(\mathbf{x}_0))\to\mathbb{R}^n such that

\begin{cases} f\circ g(\mathbf{y}) \equiv \mathbf{y}\quad\textrm{for }\mathbf{y}\in B_{\epsilon}(f(\mathbf{x}_0)) \\ g\circ f(\mathbf{x}) \equiv \mathbf{x}\quad \textrm{for }\mathbf{x}\in B_{\epsilon}(\mathbf{x}_0) \\ \end{cases}

Luca Rigotti. (2015). University of Pittsburgh, ECON2001 Lecture 12

I should have felt no hesitation in applying the inverse function theorem but for some difficulties when interpreting

\begin{aligned} &\quad\enspace\textrm{ locally invertible everywhere} \\ & \nLeftrightarrow \textrm{ globally invertible} \\ \end{aligned}

(to be continued)