# How to find that derivative of this function is invertible

The following question was part of my real analysis assignment and I having a hard time solving it.

Let $$f : \mathbb{R}^2 \to \mathbb{R}^2$$ be given by the formula $$f(x,y)=( 3x+2y+y^2 +|xy| , 2x+3y +x^2+|xy|)$$.

I have proved it to be differentiable at $$(0,0)$$ but I am unable to think whether Df(0,0) is invertible or not?

Jacobian at $$(0,0)$$ is $$5$$ so its invertible by inverse function theorem but the answer is it's not invertible.

what I am doing wrong ?

• How did you show that it is differentiable at $(0,0)$? – Michael Burr Nov 8 '20 at 14:14
• Can you give us the exact statement of the problem? – zhw. Nov 8 '20 at 22:26
• I think you are nervous for some specific reason... rest assure someone will help but can you rewrite your question so that it can be clearer? – user844292 Nov 14 '20 at 7:34
• @T.H.Shehadi Ya man , actually there were very less time remaining for exam . So, I asked in hurry. – Avenger Nov 14 '20 at 7:53
• @AlexRavsky Ok, thank you very much for your effort. So, I think question might be wrong ! I will update to you if I get anything concrete. – Avenger Nov 17 '20 at 7:13

Let $$f_1(x,y)\equiv (f_1(x),f_2(y))$$ for each $$(x,y)\in\Bbb R^2$$. Then the derivative $$f’(x,y)$$ of the function $$f(x,y)$$ is a matrix $$\begin{pmatrix}\frac{\partial{f_1}}{\partial{x}} & \frac{\partial{f_1}}{\partial{y}}\\ \frac{\partial{f_2}}{\partial{x}} & \frac{\partial{f_2}}{\partial{y}}\end{pmatrix}$$, and $$f’(0,0)=\begin{pmatrix} 3 & 2 \\ 2 & 3\end{pmatrix}$$ is invertible.

But the conditions of the inverse function theorem does not applicable to $$f$$ at $$(0,0)$$, because there are no neighborhood of $$(0,0)$$ where $$f$$ is continuously differentiable.

Nevertheless, the function $$f$$ is invertible on some open neighborhood of $$(0,0)$$. Let us show this.

Consider auxiliary functions $$f_+$$ and $$f_-$$ from $$\mathbb{R}^2$$ to $$\mathbb{R}^2$$ such that for each $$(x,y)\in\Bbb R^2$$ we have
$$f_+(x,y)=(3x+2y+y^2 +xy , 2x+3y +x^2+xy),$$ $$f_-(x,y)=(3x+2y+y^2 -xy , 2x+3y +x^2-xy).$$

Each of these two functions is continuously differentiable and has non-zero Jacobian at $$(0,0)$$. So, by the inverse function theorem, there exists $$0 such that these functions are invertible on a closed square $$S=\{(x,y)\in\Bbb R^2:|x|+|y|\le r\}$$. In particular, the restictions of these functions onto $$S$$ are injective.

Lemma. The restriction $$f_S$$ of $$f$$ onto $$S$$ is injective.

Therefore, by the invariance of domain, the restriction of $$f_S$$ (and so of $$f$$) onto the interior of $$S$$ is a homeomorphic embedding, and so invertible.

Proof of Lemma. The boundary $$B$$ of $$S$$ consists of four segments. We claim that $$f|B$$ is injective. For this suppose $$(x,y)\in B$$ and consider the following cases.

I)) $$x>0$$, $$y>0$$, $$y=-x+r$$. Then $$f(x,y)=(x-xr+r^2+2r , xr-x+3r)$$. Thus $$f_1(x,y)+f_2(x,y)=r^2+5r$$.

II)) $$x<0$$, $$y>0$$, $$y=x+r$$. Then $$f(x,y)=(5x+xr+r^2+2r, 5x-xr +3r)$$. Thus $$f_1(x,y)+f_2(x,y)=10x+r^2+5r$$.

III)) $$x<0$$, $$y<0$$, $$y=-x-r$$. Then $$f(x,y)= (x+xr+r^2-2r , -x-xr-3r)$$. Thus $$f_1(x,y)+f_2(x,y)=r^2-5r$$.

IV)) $$x>0$$, $$y<0$$, $$y=x-r$$. Then $$f(x,y)=(5x-rx+r^2-2r, 5x+xr-3r)$$. Thus $$f_1(x,y)+f_2(x,y)=10x+r^2-5r$$.

Using the above equations for $$f_1(x,y)+f_2(x,y)$$ and a condition $$0<|x| it is easy to check that $$f|B$$ is injective.

Also it is easy to check that $$f(B)$$ is a quadrilateral with vertices $$f(0,r)=r(2+r,3)$$, $$f(r,0)=r(3,2+r)$$, $$f(0,-r)=r(-2+r,-3)$$, and $$f(-r,0)=r(-3,-2+r)$$, which is symmetric with respect to a line $$x=y$$ and a point $$f(0,0)=(0,0)$$ is contained in the bounded component of $$\Bbb R^2\setminus f(B)$$.

Let $$G$$ be a geometric graph with vertices $$(0,0)$$, $$(0,r)$$, $$(r,0)$$, $$(0,-r)$$, and (eight) edges connecting them induced from $$B$$ and diagonals of $$S$$. It is easy to check that $$G$$ is three-connected.

Let $$L$$ be any of latin numbers from $$I$$ to $$IV$$ and $$f_L$$ be the restriction of $$f$$ onto $$L$$-th (closed) quadrant $$Q_L$$ of $$\Bbb R^2$$. Then $$f_L$$ coincides with the restriction onto $$L$$ of $$f_+$$ (if $$L=I$$ or $$L=III$$) or of $$f_-$$ (if $$L=II$$ or $$L=IV$$).

Let $$B^*$$ be the “support” of $$G$$, that is a union of $$B$$ with the diagonals. We claim that $$f|G$$ is injective. Indeed, let $$z$$ and $$z’$$ be distinct points of $$B^*$$. If $$z, z’\in B$$ then $$f(z)\ne f(z’)$$, because $$f|B$$ is injective. If $$z\in B^*\setminus B$$ then $$z$$ belongs to one of diagonals of $$S$$. Then $$z\in (Q_I\cup Q_{III})\cap (Q_{II}\cup Q_{IV})$$, so $$f(z)=f_+(z)=f_-(z)$$. If $$z’\in Q_I\cup Q_{III}$$ then $$f(z’)=f_+(z’)\ne f_+(z)=f(z)$$. If $$z’\in Q_{II}\cup Q_{IV}$$ then $$f(z’)=f_-(z’)\ne f_-(z)=f(z)$$.

Therefore $$f$$ induces a plane embedding $$\hat f$$ of $$G$$. By Whitney's theorem, all plane embeddings of a three-connected planar graph are equivalent, that is, obtainable from one another by a plane homeomorphism up to the choice of outer face. Since $$f(0,0)=(0,0)$$ is contained in the bounded component of $$\Bbb R^2\setminus f(B)$$, both $$G$$ and $$\hat f(G)$$ have as the outer face the unique face with four vertices. Therefore there exists a homeomorphism $$h$$ of $$\Bbb R^2$$ such that $$hf|B^*$$ is an identity map.

Let $$L$$ be any of Latin numbers from $$I$$ to $$IV$$. Since a triangle $$\Delta_L=S\cap Q_L$$ is compact and the function $$hf_L$$ is continuous and injective, $$hf_L|\Delta_L$$ is a homeomorphic embedding. Let $$B_L$$ be the boundary of $$\Delta_L$$. Since a boundary of a compact subset $$K$$ of $$\Bbb R^2$$ consists of points $$x$$ of $$K$$ having no neighborhood in $$K$$ homeomorphic to an open disk, $$hf_L(B_L)=B_L$$ is the boundary of $$hf_L(\Delta_L)$$. The set $$\Bbb R^2\setminus B_L$$ consists of two connected components $$\Delta_L\setminus B_L$$ and $$\Bbb R^2\setminus\Delta L$$, so a connected set $$C=hf_L(\Delta_L\setminus B_L)$$ is contained in one of them. It is easy to construct a continuous map $$r$$ from a set $$(\Bbb R^2\setminus\Delta L) \cup B_L$$ onto its boundary $$B_L$$ such that $$r|B_L$$ is the identity map, that is, $$r$$ is a retraction. If $$C\subset \Bbb R^2\setminus\Delta_L$$ then $$r|hf_L(\Delta_L)$$ is a retraction of the set $$hf_L(\Delta_L)$$ onto its boundary $$B_L$$, which contradicts the drum theorem, since $$hf_L(\Delta_L)$$ is homeomorphic to a closed disk. Therefore $$C=hf_L(\Delta_L\setminus B_L)\subset \Delta_L\setminus B_L$$. Since this holds for each $$L$$, we have that the map $$hf_S$$ is injective, and so is $$f_S$$. $$\square$$