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Let $(X,d_{x})$ and $(Y,d_{y})$ be metric spaces. Define the function $f:X \to Y$ which satisfies
$d_{y}(f(x),f(y)) < d_{x}(x,y)$ for all $x,y \in X$.
Can we say that $f$ is a strict contraction?

I don't think that this is a strict contraction but I can't think of a proof/example, is there someone who has an idea?

Mee98
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  • Please define strict contraction. My guess is that it means that there is a $r<1$ such that $d_Y\bigl(f(x),f(y)\bigr)\leqslant rd_X(x,y)$, but I am not sure. – José Carlos Santos Jun 16 '17 at 09:37
  • Yes, that's the definition of strict contraction. English is not my native language so I don't know if you say it like that... Sorry – Mee98 Jun 16 '17 at 09:39

2 Answers2

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Hint: $f(x) = \frac{1}{2} x^2$ on $X=(0,1)$

Red shoes
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Just take $X=Y=\Bbb R$ with the same Euclidean metric $d$, let $f:X\to Y$ be $f(x)=\log(1+e^x)$. By Mean Value Theorem, there is some $c$ between $x,y$ such that $$d(f(x),f(y))=f'(c)d(x,y)\forall x,y\in X$$ As $0<f'(x)<1$, so $d(f(x),f(y))<d(x,y)$. But if $f$ is a strict contraction, by the Banach Fixed Point Theorem there would be some $a$ such that $f(a)=a$. But clearly $f$ has no fixed point.

Nick
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