find the number of fixed points in the $A$

Question

[NBHM_PhD Screening Test_2005, Topology]

Let $f:[0,1] \rightarrow [0,1]$ such that $|f(x)-f(y)|\le\frac{1}{2}|x-y|$ $\forall x,y \in [0,1]$. Let

$A=${$x\in [0,1]:f(x)=x$}. How many number of elemets are there in $A$?

If $f:[0,1]\rightarrow[0,1]$ is uniformly continuous. Choosing $\delta(\epsilon)=2\epsilon$.Then, it must be having atleast one fixed point by Fixed-point Theorem. Can you please help me, where did I go wrong?

Answer 1

If $c$ is the fixed point, then setting $x=y=c$ in the given relation gives $0\lt 0$ which is clearly a contradiction. Hence, no fixed points exist.

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If the inequality is not strict, i.e., $|f(x)-f(y)|\leq\frac 12|x-y|$, then, by definition, $f$ is a contraction mapping on $[0,1]$ and by the Banach fixed point theorem, there is exactly $1$ fixed point $c$ of $f$ (which can be computed using fixed point iteration starting with an arbitrary seed).

Answer 2

Your condition is $|f(x)-f(y)|<\frac{1}{2}|x-y|$ and not $|f(x)-f(y)| \le\frac{1}{2}|x-y|$ !!

If $x_0$ is a fixed point, let $x=y=x_0$. Then you get $0<0$, which is absurd !