# Thread: Unique Fixed Point for B-Contraction

1. ## Unique Fixed Point for B-Contraction

Having trouble with this one...

Question: (S,rho) a metric space. The function L: S --> S is a contraction with modulus A. L(S) belongs to K, a compact space with at least one element. I need to show that there's one fixed point in S.

Any ideas?

Having trouble with this one...

Question: (S,rho) a metric space. The function L: S --> S is a contraction with modulus A. L(S) belongs to K, a compact space with at least one element. I need to show that there's one fixed point in S.

Any ideas?
Hint:

Spoiler:

Consider the sequence $\displaystyle \left\{f^n(k)\right\}_{n\in\mathbb{N}}$ where $\displaystyle k\in K$.

Uniqueness also clearly follows since if one assumes that $\displaystyle x\ne y$ but $\displaystyle f(x)=x,f(y)=y$ then $\displaystyle d(f(x),f(y))=d(x,y)\leqslant A d(x,y)<d(x,y)$

3. Hm, let me see...

If f(x0)=x1, f(x1)=x2 and so on...

d(x_n,x_n-1) <= A^n d(x1,x0)

For n high enough, we should have A^n-->0, so:

d(x_n,x_n-1) = 0
So x_n=x_n-1, so f^n(x0)=f^(n-1)(x0)

This implies that f(x_n-1)=x_n-1, so lim x_n-1 = v is a fixed point? Can I do that?

Hm, let me see...

If f(x0)=x1, f(x1)=x2 and so on...

d(x_n,x_n-1) <= A^n d(x1,x0)

For n high enough, we should have A^n-->0, so:

d(x_n,x_n-1) = 0
So x_n=x_n-1, so f^n(x0)=f^(n-1)(x0)
My bad!!! I was thinking of a different argument. Try thinking about the mapping which maps each point of $\displaystyle K$ to it's distance between itself and it's image under $\displaystyle f$

Spoiler:
Note that since $\displaystyle f:K\to K$ is continuous so is the mapping $\displaystyle \text{id}\oplus f:K\to K\times K:k\mapsto (k,f(k))$ (to see this look at the projections). But, it is an elementary fact that the mapping

$\displaystyle d:K\times K\to \mathbb{R}k_1,k_2)\mapsto d_K(k_1,k_2)$ is continuous. It then follows that $\displaystyle d\circ\left(\text{id}\oplus f\right):K\to\mathbb{R}:k\overset{\text{id}\oplus f}{\longmapsto}(k,f(k))\overset{d}{\longmapsto} d(k,f(k))$ is continuous. In particular, by the compactness of $\displaystyle K$ and

the extreme value theorem the mapping $\displaystyle d\circ\left(\text{id}\oplus f\right)$ assumes a minimum at some point $\displaystyle k_0\in K$. Now, assume that $\displaystyle d\circ\left(\text{id}\oplus f\right)(k_0)=d(k_0,f(k_0))\ne 0$. Then,

$\displaystyle d(f(f(k_0)),f(k_0))\leqslant A\text{ }d(k_0,f(k_0))<d(k_0,f(k_0))$ which contradicts the minimality of $\displaystyle k_0$