Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be a continuously differentiable function. Suppose that the Jacobian determinant $Df(0,0)$ is equal to zero. show that for $\epsilon >0$ there exist $M, \delta>0$ with the following property:
If $B_r$ is the closed disk of radius $r<\delta$ centered at $(0,0)$, then $f(B_r)$ is contained in a rectangle with side $Mr$ and $ \epsilon r$.
Here the $Df(0,0)$ is equal to zero, so I cannot apply the Inverse Function theorem here. Can anyone suggest me it's solution?
WLOG $f(0,0)=(0,0)$. Since $\det Df(0,0)=0$, we rotate both domain and codomain $\mathbb{R}^2$ independently so $Df(0,0)$ is $$ \begin{pmatrix}a&0\\0&0\end{pmatrix},\text{ some }a\geq 0. $$ Choose $\delta>0$ such that on $B_\delta$, $Df(x,y)$ deviates from $Df(0,0)$ by less than $\frac12\epsilon$ in every entry. Then, for all $(x,y)\in B_r$, $r<\delta$, we have $\lvert x\rvert+\lvert y\rvert\leq\sqrt{2}r<2r$ and \begin{align*} \lvert f_1(x,y)-ax\rvert& =\left\lvert\int_0^1 \left[D_{(x,y)}f_1(tx,ty)-D_{(x,y)}f_1(0,0)\right]\mathrm{d}t\right\rvert\\ &\leq\int_0^1\Big\lvert [Df_1(tx,ty)-Df_1(0,0)](x,y)\Big\rvert\,\mathrm{d}t\\ &\leq\int_0^1\Big[\underbrace{\lvert\partial_1f_1(tx,ty)-a\rvert}_{\leq\frac12\epsilon}\lvert x\rvert+\lvert\underbrace{\partial_2f_1(tx,ty)\rvert}_{\leq\frac12\epsilon}\lvert y\rvert\Big]\,\mathrm{d}t\\ &\leq\frac12\epsilon(\lvert x\rvert+\lvert y\rvert)< r\epsilon \end{align*} and similarly $\lvert f_2(x,y)\rvert<\epsilon r$. So $M=a+\epsilon$ would do.