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This came up as I was thinking about how to solve an assignment problem I've been given. I'm not posting the assignment problem here as I want to think about it myself, but one approach I am currently working on is to try to show that $$f:\mathbb{R}\to\mathbb{R},\quad f(x) = 1-\sin(x)$$ is a contraction mapping.

At the moment I have a feeling it isn't. For example take $x_1 < 0 < x_2$, then $$ \left|\frac{(1-\sin x_1)-(1-\sin x_2)}{x_1-x_2}\right| = \left|\frac{\sin(x_1) - \sin(x_2)}{x_1-x_2}\right| \to 1$$ as $x_1 \to 0^-, x_2 \to 0^+$.

Hence there does not exist a non-negative $k<1 \in \mathbb{R}$ such that $|\sin(x_1)-\sin(x_2)| \leq k|x_1-x_2|$ for all $x_1,x_2 \in \mathbb{R}$. (Just take $x_1,x_2$ arbitrarily close to $0$.) So $f$ is not a contraction mapping.

That was the idea. Thinking just a little bit further, we see that this occurs because $|f'(x)| = 1$ at $x=0$. So perhaps we have a necessary and sufficient condition for a function on the reals to be a contraction mapping, namely that $\forall x\in\mathbb{R},|f'(x)|\leq k<1$ for some non-negative real $k$. Is this the case? Or have I missed some subtle (or not-so-subtle) point?

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    from where to where. Note that a mapping $T$ is a contraction if $d(T(x),T(y))< \alpha \cdot d(x,y)$ for some $\alpha \in (0,1)$.2011-08-31
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    Sorry, I accidentally posted before I finished typing the question up. Will update shortly.2011-08-31
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    I believe this to be one of my more rambling posts; I apologize.2011-08-31
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    No need to apologize and you're exactly right. For the last part, you can simply apply the [mean value theorem](http://en.wikipedia.org/wiki/Mean_value_theorem) in one direction and estimate the absolute value of the derivative in the other direction. Of course, you need to assume the function to be differentiable everywhere. Think e.g. of $x \mapsto |x|/2$ which certainly is a contraction but isn't differentiable at zero.2011-08-31
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    There is a clear relation between the contraction rate $k$ of the mapping $f:A\to A$ for $A\subseteq \mathbb R$ and the best Lipschitz constant for the function $f$ on $A$. The last constant can be found through the absolute value of $f'$ if $f\in C^1(A)$ as Theo has already written.2011-08-31
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    Maybe you're interested in reading my answer [here](http://math.stackexchange.com/questions/69457) about a related question.2011-10-03

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That's the case if $f$ is differentiable on all of $\mathbb{R}$, then by Lagrange's theorem (mean value theorem) $$|f(x)-f(y)|=|f'(c_{x,y})||x-y|$$ for some $c_{x,y}$ between $x$ and $y$, for all $x,y\in\mathbb{R}$. From here, sufficiency is clear since if $|f'(x)|\leq K$ then $$|f(x)-f(y)|=|f'(c_{x,y})||x-y|\leq K|x-y|$$ Now, for necessity we only need to use the definition of the derivative: if $f$ is contracting with constant $K<1$, given $x\in \mathbb{R}$ then $$|f'(x)|=\lim_{h\to 0} \left|\frac{f(x+h)-f(x)}{h}\right|\leq\lim_{h\to 0} \frac{K\cdot|x+h-x|}{|h|}=K$$ using that the absolute value is continuous and that limits preserve inequalities. q.e.d.

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    I made some minor corrections. Note that in English Lagrange's theorem is usually called the mean value theorem.2011-08-31
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    Thank very much Iasafro and @Theo. :)2011-09-01
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    sorry, you haven't answered if $1-sin x$ is a contraction mapping.2015-08-20
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    @Yola I give a proof that a differentiable function $f:\mathbb{R}\rigtharrow\mathbb{R}$ is a contraction iff for some $K\in (0,1)$, for all $x\in \mathbb{R}$, |f'(x)|\leq K$, as conjectured and asked by the OP. From there it follow that $1-\sin(x)$ isn't a contraction immeditely since $-\cos(x)$ does not satisfy the later.2015-08-20