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I need to show whether $f:X \rightarrow X$ is a contraction mapping where $$f(x,y)=(-y,x^3), X = \{(x,y) \in \mathbb R^2 : x+y\geq 1 \}$$

I know we have to show that $d(f(x),f(y)) \leq c.d(x,y)$ for some $0 \leq c < 1$ but this is my attempt at it: $$d((-y_1,x_1^3),(-y_2,x_2^3)) = \sqrt{(-y_1+y_2)^2+(x_1^3-x_2^3)^2}=\sqrt{(y_1-y_2)^2+[(x_1-x_2)(x_1^2+x_1x_2+x_2^2)]^2} \leq d((x_1,y_1),(x_2,y_2)).(x_1^2+x_1x_2+x_2^2)$$

which implies that $f$ is a contraction mapping if and only if $0 \leq x_1^2+x_1x_2+x_2^2 < 1$. So then would this mean that $f$ isn't actually a contraction since the above is not true $\forall (x,y) \in X$. Thank you in advance for the help.

RobPratt
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wantingtoimprove
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1 Answers1

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As said in my comment $f$ is not a well defined (So it is not even a function). And a contraction is a function F defined from a metric space $(M,d)$ to itself with a constant $0 \leq c <1$ such that $$d(F(x),F(y)) \leq cd(x,y)$$ for all $x,y \in M$. So if it is not a function there is no point in trying to show it is a contraction. Take a look at these links if you want more information about what it means to be well defined. What are well-defined functions? "Well defined" function - What does it mean?

PAM1499
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  • Thank you. I haven't encountered such a problem where I had to worry about a mapping being well-defined but your links have helped immensely. – wantingtoimprove Nov 02 '20 at 19:05
  • You are welcome ;) I don't know if this is an exercise or not but it is kind of weird if you were asked to show that something, which is not even a function, is a contraction map. Perhaps it was a mistake in defining set X. – PAM1499 Nov 02 '20 at 19:11
  • It was part of an exercise sheet given to me by my professor so it is quite possible that it is an error in defining the set or maybe he wanted to catch us out. It has definitely shown me not to take things for granted anymore though. – wantingtoimprove Nov 02 '20 at 22:34