Prove that $f:\mathbb{R}^2\to\mathbb{R}^2$ has a fixed point in a subset of $\mathbb{R}^2$ 
Let $f:\mathbb{R}^2\to\mathbb{R}^2$ by defined by $f(x_1, x_2) = \begin{pmatrix}
\frac{1}{3}x^2_2 + \frac{1}{8} \\\
\frac{1}{4}x^2_1 - \frac{1}{6}
\end{pmatrix}$ and let $D = \{ x \in \mathbb{R}^2 : \| x \|_{\infty}\ \leq 1 \}$. Prove that $f$ has a fixed point in $D$.

Obviously the Banach fixed point theorem doesn't apply here, because $f$ isn't lipschitz. I was trying to come up with a solution by calculating the intersections of the two quadratic component functions with $h(x) = x$ separately. I thought i could maybe find the fixed point by comparing the solutions but this didn't work either. I guess there's a simple solution.
 A: $\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$
You can use the Banach fixed point theorem to solve this problem. As you remarked, the function $f$ is not a contraction, but we are only interested in what happens in $D = [-1,1] \times [-1,1]$. So let us look at the restriction $f |_D$. It is easy to see that it maps $D$ to itself, and it is in fact a contraction. Indeed, for any $(x_1, x_2), (y_1,y_2) \in D$ we have:
\begin{align}
\norm{f(x_1,x_2)-f(y_1,y_2)}^2 &= 
\norm{ \begin{pmatrix} 1/3 (x_2^2-y_2^2) \\ 1/4(x_1^2-y_1^2) \end{pmatrix}}^2 \\
&=\frac{1}{9} (x_2^2-y_2^2)^2 +\frac{1}{16} (x_1^2-y_1^2)^2 \\
&=\frac{1}{9} (x_2+y_2)^2(x_2-y_2)^2 + \frac{1}{16} (x_1+y_1)^2(x_1-y_1)^2\\
&\leq \frac{4}{9} (x_2-y_2)^2 + \frac{1}{4} (x_1-y_1)^2 \qquad 
\text{since } x_1+y_1 \leq 2 \; \text{and} \; x_2+y_2 \leq 2 \\
&\leq \frac{4}{9} \left( (x_2-y_2)^2 + (x_1-y_1)^2 \right) \\
&=\frac{4}{9} \norm{ \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} - \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} }^2
\end{align}
Taking square roots, we see that $f|_D$ is a contraction. Now, since $D$ is a closed subset of $\mathbb{R}^2$ (which is complete), it is a complete metric space. By the Banach fixed point theorem, we conclude that $f$ has a unique fixed point in $D$.
A: Hint: try applying the Brouwer Fixed Point Theorem.
A: Here is a more elementary approach. The function $f$ has a fixed point at $(x_1, x_2)$ if and only if the following equations are satisfied:
\begin{align}
1/3 \; x_2 ^2 -x_1 +1/8 =0 \\
1/4 \; x_1 ^2-x_2-1/6 = 0 
\end{align}
Now you can solve the first equation for $x_2$ and insert this in the second equation to obtain a function of $x_1$ only. Use the continuity of this function and the fact that it changes signs to prove it has a root in some suitable interval. You will need to find a small enough interval to guarantee that the corresponding value for $x_2$ will be in $[-1,1]$. (I think $(1/8,1/4)$ does the trick.)
A: If you fancy calulating with differentials, it's fairly easy.
As previously remarked $f$ restricted to $D$ maps $D$ into $D$ (easy). The derivative equals
$$Df = \left( \begin{matrix} 0 & \frac23 x_2 \\ \frac12 x_1 & 0 \end{matrix} \right).$$ The norm (in $\ell^\infty$) is 2/3 on $D$, whence $f_{|D}$ is a contraction. $D$ is closed and connected so there is a unique fixed point.
