2

I have problems with this demostration can anybody help me please? If $X$ is a compact space and $A:X\rightarrow X$ is such that $\rho(Ax,Ay)<\rho(x,y)$, $x\neq y, \Rightarrow A $ have only one fix point on X.

Martin Sleziak
  • 53,687
Aris Soria
  • 169

3 Answers3

1

Um, if x and y are two unequal fixed points then $\rho(Ax,Ay)= \rho(x,y)$.

So there can't be more than one fixed point.

fleablood
  • 124,253
0

Assuming $X\ne \emptyset.$

Let $f(x)=\rho (x,Ax).$ We have $$f(y)=\rho (y,Ay)\leq \rho (y,x) +\rho (x,Ax)+\rho (Ax,Ay)\leq$$ $$\leq \rho (y,x)+\rho (x,Ax)+\rho (x,y)=$$ $$=2\rho(x,y)+f(x).$$ $$\text {So }\quad f(y)-f(x)\leq 2\rho (x,y).$$ Interchanging $x$ and $y$ we have $$f(x)-f(y)\leq 2\rho (y,x).$$ From the last 2 inequlities above, we have $|f(x)-f(y)|\leq 2\rho (x,y).$

Therefore $f$ is continuous from $X$ to $\Bbb R.$

Now $X$ is compact and not empty and $f$ is continuous, so $f(X)$ is a non-empty compact subset of $\Bbb R,$ so $f(X)$ has a minimum value .

If $f(x)>0$ then $f(x)\ne\min f(X).$ Because with $y=Ax,$ we have $x\ne y$ (because $\rho (x,y)=\rho (x, Ax)=f(x)\ne 0$), so $$f(x)=\rho(x,Ax)=\rho (x,y)>\rho (Ax,Ay)=\rho (y,Ay)=f(y).$$

Therefore $\min f(X)=0.$ So there exists $x$ with $f(x)=0.$

Now $f(x)=0\iff \rho (x,Ax)=0 \iff x=Ax.$ But if $f(x)=f(y)=0$ with $x\ne y$ then $\rho (x,y)>\rho (Ax,Ay)=\rho (x,y),$ a paradox. So there is exactly one $x$ such that $x=Ax.$

DanielWainfleet
  • 57,985
-1

Under the lack of the compactness argument you can not assure that the infimum of the function $f(x)=\rho(Ax,x)$ equals zero so that the solution of the equation $Ax=x$ is unique.

In particular, on the metric space endowed by the interval $X=(0,1)$ and the Euclidean metric $\rho(x,y)=|x-y|$, the family of linear functions $Ax=ax$ $(0\leq a<1)$ has no fixed points.

Recall that the compactness argument is not a necessary condition to assure that a function has a unique fixed point: on the metric space $X=(-\frac{1}{2},\frac{1}{2})$, the quadratic function $Ax=x^2$ has a unique fixed point ($x=0$) and at the same time satisfies the condition $|Ax-Ay|<|x-y|$.

Nelson Faustino
  • 411
  • 2
  • 8
  • 1
    Lots of information, but no answer to the question. – Andreas Blass Jul 09 '18 at 00:27
  • The main goal here is to prove the existence of the fixed point (the uniqueness is rather obvious as explained on some previous comments).
    Here, we recall that the compactness of $X$ ensures the existence of a subsequence $(x_{n_k})k$ of $(A^nx_0)_n$ [$x_n=A^nx_0$] that converges to $x\in X$, that is $$ \lim{k\rightarrow \infty}d(x,x_{n_k})=0.$$

    Moreover, using the fact that $x\mapsto \rho(Ax,x)$ is uniformly continuous, one realize that $$ \rho(Ax,x)=\lim_{k\rightarrow \infty} \rho(Ax_{n_k},x_{n_k})=\lim_{k\rightarrow \infty} \rho(x_{n_k+1},x_{n_k})=0.$$

    Thus $Ax=x$, as desired.

     – Nelson Faustino Jul 09 '18 at 02:14
  • I agree that the uniqueness is obvious, but uniqueness, not existence, was the question. – Andreas Blass Jul 09 '18 at 21:19
  • Dear @AndreasBlass. Many thanks for your comments. At this stage I would like to argue with the following: When you employ the standard proofs of fixed point theory (e.g. proof of Banach Fixed Point Theorem), you need to check existence and uniqueness. I've tried to expand the discussion, first with some basic examples and afterwards with a detailed proof of existence. Why prove uniqueness nor existence? Do anyone clarifies this aspect before? – Nelson Faustino Jul 09 '18 at 22:16