The problem:
I am attempting to show
$$f(x,y)=\left(\frac{\cos x+\cos y-1}{2}, \cos x-\cos y\right)$$
has a fixed point using the mean value inequality, which states that for any $a \in \mathbb{R}^n$ and $f: B_r(a) \to \mathbb{R}^m$ differentiable on $B_r(a)$ with $\|Df(x)\| \leq M$ for all $x \in B_r(a)$, we have $|f(b_1) - f(b_2)\| \leq M\|b_1 - b_2\|$ for any $b_1, b_2 \in B_r(a)$.
If I can find such an $M$ on some ball, and show $M < 1$, then I am done by the contraction mapping theorem.
Here's what I've done:
Note that $|\frac{\cos x+\cos y-1}{2}| \leq \frac{3}{2}$ and $|\cos x-\cos y| \leq 2$ by applying the triangle inequality, so the fixed point should be on $B_{r}(a)$ with $r=2$ & $a=(0,0)$.
We can calculate that
$$Df(x,y)=\pmatrix{ -\frac{\sin x}{2} & -\sin x \\\ -\frac{\sin y}{2} & \sin y }$$
Now, I'm a bit stuck on the interpretation of the mean value inequality, specifically how to find and get $\|Df(x)\|$. If I'm interpreting things correctly, I believe this should be using the operator norm, which I'm not confident in using.
If so, I know that for $A \in L(\mathbb{R}^n, \mathbb{R}^m)$
$$|A|=\sup_{0 \neq x' \in \mathbb{R}^n} \frac{||Ax'||}{||x'||}$$
I've tried letting $x' = (u,v) \in B_{2}(0)$ be a vector on the ball and calculate
$$\sup_{0 \neq x' \in B_{2}(0)} \frac{\sqrt{((\sin^2 x + \sin^2 y) (\frac{1}{2}u+v)^2)}}{\sqrt{u^2+v^2}}$$
(the numerator is $|Df(x,y)x'|$) as an upper bound for $M$.
But I can't seem to get this bound lower than $1$ in order to use the contraction mapping theorem, I'm getting things above $1$.
I'm not sure where the problem is, perhaps I'm misunderstanding the theory. Is this calculation wrong? I'm thinking, perhaps, $x'$ should be $(x,y)$, i.e. the same variables used in $f$, but I'm not sure if this is correct.
My Method is like this :
When we state that there is a Stationary Point or fixed Point , what we imply is that there is a Solution to :
$(x,y) \equiv f(x,y) \equiv \left(\frac{\cos x+\cos y-1}{2}, \cos x-\cos y\right) \tag{1}$
We do not have to find the Solution(s) to (1) , we just have to show the Existence of a Solution.
OK , Here (1) gives us :
$2x = \cos x+\cos y-1 \tag{2}$
$y = \cos x-\cos y \tag{3}$
In (2) , We can see that $x$ is bounded between $ \pm 3 $ , while $y$ is unbounded.
Hence there is a Continuous curve from high above to deep down , somewhere near the y-axis.
In (3) , We can see that $y$ is bounded between $ \pm 3 $ , while $x$ is unbounded.
Hence there is a Continuous curve from far left to far right , somewhere near the x-axis.
[[ I have selected very loose Symmetric Bounds , to make it Simple to visualize : tighter un-Symmetric Bounds are Possible ]]
Due to Continuity , (2) & (3) must Naturally Intersect somewhere near the Origin.
That is the Stationary Point or fixed Point for $f(x,y)$
Diagrammatic View :
We see that the Curves Intersect [[ Image : curtesy of Wolfram Online tool ]] somewhere near the Origin.
Numerical Value : $(X,Y)=(+0.448432,−0.091647)$ : Close to the Origin.