# Complete a proof that $F(x,y)$ is contracting.

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. $\qquad \qquad \qquad \Box$

• Anybody any idea why this questions seems to attract low attention? Any possible improvement in the question? Commented Jun 25, 2016 at 20:02
• Speaking for myself...I felt the hint given was adequate in light of your awareness of the Banach fixed-point theorem (which suggests this is an upper-level or beginning grad-level analysis course). If that's not the case, it would help to know exactly where you're stuck applying the hint; particularly, have you calculated $\nabla f$ and calculated its maximum magnitude in $D$? It might also help to know what's your background in multivariable calculus. Thanks. :) Commented Jun 26, 2016 at 14:45
• @AndrewD.Hwang Of course, sorry I thought my comment on your answer provided enough detail where I got stuck. I know it is hard to guess what the level is of the OP, so that's why it was good that you started with hints as you did and expanded when I mentioned I still don't understand. I can calculate the Jacobian (matrix) $\nabla F(x,y)$ but I don't see how you can value a matrix against some maximum magnitude (?) as you say... background: first year bachelor maths study at a Dutch university. :) Probably the teacher intended a different solution but I'm always eager to learn more methods! Commented Jun 26, 2016 at 15:14
• Actually, I'm thinking of working component by component: Find the maximum magnitude of $\nabla F_{1}$ (call it $M_{1}$, say), and use the hint to show$$|F_{1}(x, y) - F_{1}(x_{0}, y_{0})| \leq M_{1}\|(x - x_{0}, y - y_{0})\|.$$Then do the same for $F_{2}$. That should suffice to fill in your $\vdots$. Commented Jun 26, 2016 at 15:37
• @AndrewD.Hwang Aaah of course, just per component. Somehow I had misinterpreted/overlook that important detail, now using the already mentioned other question I managed! Thanks a lot! Commented Jun 26, 2016 at 16:28

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 $(x, y)$.
• Do you suggest to write $|F(x)-F(y)|$ in the form of something with a $\gamma$ and an integral? Then I see I can use the triangle inequality, but I don't see how I can then use the Cauchy-Schwarz inequality to get what I want, especially because the $\gamma$ is still in there. How can you take a maximum of that expression? Commented Jun 25, 2016 at 14:19
• Cauchy-Schwarz gives $|\nabla f \cdot \gamma'| \leq \|\nabla f\|\, \|\gamma'\|$, and the integral of $\|\gamma'\|$ is the distance between the endpoints. :) Commented Jun 25, 2016 at 14:47
• Ah, thanks! I'm not really familiar with parametrizations so I didn't know that, but anyway, now I have $|x-y|\cdot \int |\nabla F(\gamma(t))|$, but how do I show that the integral is smaller than 1? Because I have no idea what to do with that Jacobian in there. Commented Jun 25, 2016 at 15:12
• I feel I'm getting somewhat close to where this answer: math.stackexchange.com/a/1837605/293580 starts, but I can't quite connect yet where $d_{(x,y)} F \cdot (h,k)$ comes from in that answer. Commented Jun 25, 2016 at 15:19