Can anyone fill in the dots in this proof?
Let $D := [0,\frac{1}{2}]^2$. Show there is exactly one $(x,y)=(x^*,y^*)\in D$ such that \begin{align*} x &= \frac{x^3}{2} + y^4 + \frac{1}{4} \,, \\ y &= x^4 + \frac{y^3}{2}+\frac{1}{5} \,. \end{align*}
Obviously I need to use Banach's fixed point theorem, so I need to show $F$ (as defined below) is contracting.
My try (the $q \in \mathbb R$, $q<1$, isn't filled in yet of course)
Let $F(x,y) = \begin{pmatrix} \frac{x^3}{2} + y^4 + \frac{1}{4} \\ x^4 + \frac{y^3}{2}+\frac{1}{5}\end{pmatrix}$, $x=(x_1,x_2),y=(y_1,y_2)$. Then \begin{align*} |F (x_1,x_2)-F(y_1,y_2)|^2 &= |F_1 (x_1,x_2)-F_1(y_1,y_2)|^2 - |F_2 (x_1,x_2)-F_2(y_1,y_2)|^2 \\ &= \left| \frac{x_1^3}{2} - \frac{y_1^3}{2} + x_2^4 - y_2^4 \right|^2 - \left| \frac{x_2^3}{2} - \frac{y_2^3}{2} + x_1^4 - y_1^4 \right|^2 \\ &\ \ \vdots \\ &\le q^2 |x_1-y_1|^2 + q^2 |x_2 - y_2|^2 \\ &= q^2 |x-y|^2 \,. \end{align*} So now we have $F(x)-F(y)|\le q |x-y|$ and thus $F\colon D \to D$ contracting, $D \subset \mathbb R^2$ closed, so by Banach's fixed point theorem there exists exactly one $(x^*,y^*)\in D$ as described. $\qquad \qquad \qquad \Box$
Hint: If $f$ is a continuously-differentiable real-valued function of two variables and $$ \gamma(t) = \bigl(x_{0} + t(x - x_{0}), y_{0} + t(y - y_{0})\bigr) $$ is the constant-speed parametrization of the segment from $(x_{0}, y_{0})$ to $(x, y)$ over $[0, 1]$, then $$ f(x, y) - f(x_{0}, y_{0}) = \int_{0}^{1} \nabla f\bigl(\gamma(t)\bigr) \cdot \gamma'(t)\, dt. $$ The triangle inequality for integrals and the Cauchy-Schwarz inequality bound the absolute value in terms of the maximum of $\|\nabla f\|$ and the distance from $(x_{0}, y_{0})$ to $(x, y)$.