# Is the function $f(x, y) = (\cos(2x - y), \frac{1}{2} \cdot \ln(1 + x^2 + y^2))$ a contraction mapping?

Given the mapping

$$f(x, y) = (\cos(2x - y), \frac{1}{2} \cdot \ln(1 + x^2 + y^2))$$

and the region $$R = [-1, 1] \times [-1, 1]$$, I want to know if I can use contraction mapping iteration in order to find a fixed point.

I'm pretty sure the answer is NO because I don't think $$\ln(1 + x^2 + y^2)/2$$ is a contraction; however, I'm having a difficult time showing this since the function is multivariate.

I've seen a similar post here, but I'm still not able to make much progress.

Define $$g(x, y) = \ln(1 + x^2 + y^2)/2$$. To show $$g$$ is not a contraction, I need to show that there is no constant $$C \in (0, 1)$$ for which

$$|g(p_1) - g(p_2)| \leq C|p_1 - p_2|$$ for all points $$p_1, p_2 \in R$$. I am not so sure about how to prove this claim though, and I would really aprpeciate any help. I've tried many things (e.g. taking a limit), but I have not had any luck.

• If it's the multivariate nature of the function that's bothering you, note that $\ln(1 + x^2 + y^2)$ is a function of the norm of $\|(x, y)\|$. It's not hard to show that, given a univariate function $f : [0, \infty) \to \Bbb{R}$, that $f(\|(x, y)\|)$ is a contraction if and only if $f$ is a contraction. So, look at $\ln(1 + x^2)$ instead. – user744868 Feb 25 at 2:39
• What are we dividing by $2$: the logarithm or the argument of the logarithm? – bjorn93 Feb 25 at 2:58
• Let me make it more clear @bjorn93. I edited my post. – user728115 Feb 25 at 2:58
• Also, is it enough to show that both $\cos(2x - y)$ and $\frac{1}{2} \cdot \ln(1 + x^2 + y^2)$ are contractions to justify using the contraction mapping iteration procedure? – user728115 Feb 25 at 2:59

$$g$$ is a contraction; the mean value theorem in $$2$$ dimensions gives $$|g(q) - g(p)| \leq |\nabla g((1 - t)p + tq)| |q - p|$$ where $$p, q$$ play the role of $$p_1, p_2 \in \mathbb [-1, 1]^2$$. Here $$\nabla g: \mathbb [-1, 1]^2 \rightarrow \mathbb R^2$$ is the gradient: $$\nabla g(x, y) = \left(\frac{\partial g}{\partial x}(x, y), \frac{\partial g}{\partial y}(x, y)\right) = \frac{1}{2}\left(\frac{2x}{1 + x^2 + y^2}, \frac{2y}{1 + x^2 + y^2}\right).$$ The norm of the gradient is $$|\nabla g(x, y)| = \frac{1}{2}\sqrt{\frac{4x^2}{(1 + x^2 + y^2)^2} + \frac{4y^2}{(1 + x^2 + y^2)^2}} = \frac{\sqrt{x^2 + y^2}}{1 + x^2 + y^2}.$$ If you take $$z = x^2 + y^2$$, you should be able to find a bound for $$|\nabla g(x, y)| = |\nabla g(z)|$$.
• Thank you. And since $\cos(2x - y)$ is also a contraction, that means I can use fixed-point iteration? – user728115 Feb 25 at 2:46
• I didn't think about whether $\cos(2x-y)$ was a contraction, but that reasoning is wrong. If $g = \cos$ is a contraction and $h(x, y) = 2x - y$, then $f(x, y) = g(h(x, y))$. But then $|f(p_2) - f(p_1)| = |g(h(p_2)) - g(h(p_1))| \leq C|h(p_2) - h(p_1)|$ but there is no guarantee $h$ is also a contraction. In this case, take $p_1 = (x, y)$ and $p_2 = (z, w)$ and take $y = z = w = 0$. Then $|p_1 - p_2| = |x|$ and $|\cos(p_1) - \cos(p_2)| = |\cos(2x) - 1|$, and $|\cos(2x) - 1|/|x| \rightarrow 4/\pi$ as $x \rightarrow \pi/4$. – Riley Feb 25 at 3:22
• It's necessary that each component is a contraction, since $|f(p_2 ) - f(p_1)| = |(h(p_2) - h(p_1), g(p_2) - g(p_1))| \geq |h(p_2) - h(p_1)|$, so the fact $h(x, y) = \cos(2x - y)$ is not is enough to show $f$ is not a contraction either. I suspect that even when both components are contractions with constants $C_g, C_h$ it is not necessary (but it is sufficient) to have $C_g^2 + C_h^2 < 1$, since it's possible the extremal points (where the contraction constants are attained) of each component are far away, but I can't give a counterexample. – Riley Feb 25 at 3:29
• Sorry, $p_1, p_2$ are points in the square, yes. – Riley Feb 25 at 3:32
• Okay, here is a quick counterexample: take $f(x, y) = (\frac{3}{4}x, \frac{3}{4}y)$; then if $f(x, y) = (g(x, y), h(x, y))$ each $g, h$ is a contraction, since $|g(x, y) - g(z, w)| = \frac{3}{4}|x - z| \leq \frac{3}{4}\sqrt{(x - z)^2 + (y - w)^2} = \frac{3}{4}|(x, y) - (z, w)|$, so each contraction constant can be taken as $\frac{3}{4}$, but $2(\frac{3}{4})^2 = 18/16 > 1$. But $f$ is still a contraction. – Riley Feb 25 at 3:47