# For which values of $a$ is $f(x) = \cos(ax)$ a contraction mapping?

I am studying contraction mapping and got stuck on this question:

Consider the following function:

$$f(x)=\cos(ax),\ a,x \in \mathbb{R}.$$

Find values of $a$ for which $f$ is a contraction mapping.

Any help with this question is hugely appreciated.

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Recall that the absolute value of the derivative can be useful in such problems. –  André Nicolas Apr 28 '13 at 16:06

$f$ is a contracting map if $\exists \alpha \in (0,1): \forall x,y \in \mathbb{R}, |f(x)-f(y)|< \alpha|x-y|.$

Let's evaluate the following expression for any x,y: $|f(x)-f(y)|=|\cos(ax)-\cos(ay)|$.

Let's note that since $f$ is differentiable, then according to Lagrange's theorem, we can write:

$$\exists c\in (x,y): |f(x)-f(y)|=\frac{|f(x)-f(y)|}{|x-y|}|x-y|=|f'(c)||x-y|=|-a\sin(ac)||x-y|.$$

Since we want $f$ to be a contracting map, then we want this expression to be less than $\alpha|x-y|$ for some $\alpha \in (0,1)$.

Therefore, we want the following:

$$\exists \alpha \in (0,1): |-a\sin(ac)||x-y|< \alpha |x-y|.$$

Let's note that $|\sin(ac)|\leq 1$, and therefore, $|-a\sin(ac)||x-y|\leq |-a||x-y|=|a||x-y|$.

Therefore, we want $|a|\in (0,1)$, in order to ensure that $\exists \alpha \in (0,1): |a||x-y|<\alpha |x-y|$.

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The map is also a contraction when $a = 0$. –  Pete L. Clark Apr 28 '13 at 19:14
thank you for that! how would I do a contraction mapping proof? |sin b - sin a|<_ |b-a| i need to prove that for all a,b E R that the above statement is true. could i use the mean value theorem for this? which kind of proof technique could i use? –  Anona anon Apr 28 '13 at 19:35