Let $X=\mathbb{Q}\cap [1,2]$, i.e $X$ is the set of rational number between 1 and 2 inclusive. We can consider $X$ to be a metric space by endowing it with the usual distance function, i.e for $x,y \in X$ we put $d(x,y)=|x-y|$. Now we define $f:X\rightarrow X$ by $f(x)=x-\dfrac{x^2-2}{2x}$. Ones should check that if$x\in X$ , then $f(x)\in X$ as well. This means checking that if $x$ is rational and in $[1,2]$ then $f(x)$ is also rational and in $[1,2]$. Prove that $f$ is a contraction mapping but $f$ does not has a fixed point. This is my homework exercise, but I am not good at mathematics, so please feel freely helping me. Thank you very much !
Contraction mapping does not hold in metric space
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real-analysis
metric-spaces
examples-counterexamples
fixed-point-theorems
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0You have three things to prove: (a) $f$ is a function from $x$ to $X$. (b) $f$ is a contraction. (c) $f$ has no fixed point. Can you prove any of these statements? What have you tried? – 2011-10-29
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0By definition of *contraction mapping*, you need to show that there exists $0 \leq k < 1$ such that, for all $x, y \in X$... (among other things) – 2011-10-29
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0Yes, I can prove that $f$ is a contraction.Infact, I can prove that $|f(x)-f(y)|\le \dfrac{3}{2}|x-y|$. But what about the proof for fixed point ? is it too obvious that we have to solve the equation $f(x)=x$? – 2011-10-29
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0@msnaber: look again at the definition of a contraction: what you have proven is not enough. Yes, a fixed point of $f$ is exactly a solution (in $X$) of $f(x)=x$. – 2011-10-29
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1Ah, I remember false, $\frac{3}{2}\ge 1$. but that's why I need you help in solving this problem. I may be need a more precise solution. sorry because I am not good at mathematic. – 2011-10-29