Are There Any Solutions?

Calculus Level 4

True or False?

There exists a continuous function $f: \mathbb{R} \to \mathbb{R}$ such that for all $x\in \mathbb R$ , $f\big(f(x)\big) = \cos x$ .

True False

3 solutions

Mark Hennings
Oct 31, 2016

Let $x_0$ be the unique solution. of $\cos x = x$ . Since $f(\cos x) = f(f(f(x))) = \cos(f(x))$ , we see that $f(x_0) = \cos(f(x_0))$ , and so $f(x_0) = x_0$ .

If $0 \le x,y \le \tfrac12\pi$ and $f(x) = f(y)$ , then $\cos x = f(f(x)) = f(f(y)) = \cos y$ , and hence $x=y$ . Thus $f$ is injective on $[0,\tfrac12\pi]$ . By determined use of the Intermediate Value Theorem, this means that $f$ is either strictly increasing or strictly decreasing on $[0,\tfrac12\pi]$ . This means that there will be an interval $I$ around $x_0$ so that $0 \le f(x) \le \tfrac12\pi$ for all $x \in I$ , which implies that $f \circ f = \cos$ is increasing on $I$ , which is not the case.

Thus no continuous function $f$ exists with $f \circ f = \cos$ .

Can you explain why $\cos f( x_0 ) = x_0$ ? I thought all we had was $\cos x_0 = x_0$ , and it's not clear to me why $f(x_0) = x_0$ too.

Calvin Lin Staff - 4 years, 7 months ago

Happy with $f(\cos x) = f(f(f(x))) = \cos(f(x))$ ? Then $f(x_0) = f(\cos x_0) = \cos(f(x_0))$ , and so $f(x_0)$ is a solution of $\cos y = y$ , and hence $f(x_0) = x_0$ .

Mark Hennings - 4 years, 7 months ago

If any function $F:A \to A$ is such that $F\circ F$ has a unique fixed point $\alpha$ , then $\alpha$ is the unique fixed point of $F$ .

Proof :

Since $\alpha = F(F(\alpha))$ , we have $F(\alpha) = F(F(F(\alpha)))$ . Thus $F(\alpha)$ is a fixed point of $F\circ F$ , and so $F(\alpha) = \alpha$ , so $\alpha$ is a fixed point of $F$ .

If $\beta$ is a fixed point of $F$ , then $F(F(\beta)) = F(\beta) = \beta$ , and so $\beta$ is a fixed point of $F\circ F$ , and hence $\beta = \alpha$ .

A generalisation of this technique is part of the proof that sufficiently well-behaved differential equations have unique solutions. The equation $y'(x) \; = \; f(x,y) \hspace{2cm} y(0) = a$ is equivalent to the integral equation $y(x) \; = \; I[y](x) \; = \; a + \int_0^x f(u,y(u))\,du$ Under suitable conditions on the function $f$ , the operator $I$ is such that some $n$ -fold composite $I \circ I \circ \cdots \circ I$ is a contraction mapping on $C[0,X]$ , and hence has a unique fixed point. This unique fixed point is then the unique fixed point of $I$ , and hence the unique solution of the differential equation on the interval $[0,X]$ .

Mark Hennings - 4 years, 7 months ago

Aditya Narayan Sharma
Oct 30, 2016

Note that $\displaystyle f(f(0))=1,f(f(\frac{\pi}{2}))=0$ and since $\displaystyle f(f(f(x)))=f(\cos x) \implies \cos(f(x))=f(\cos x)$

Now since $\cos x$ is a bounded function and is decreasing or increasing as per the choice of intervals $f$ must be so.

Now if we take for granted that $f$ is decreasing then , $\displaystyle \frac{\pi}{2}>0 \implies f(\frac{\pi}{2})<f(0)\implies f(f(\frac{\pi}{2}))>f(f(0))\implies 0>1$ which is absurd.

Again if $f$ would have been increasing then, $\displaystyle \frac{\pi}{2}>0 \implies f(\frac{\pi}{2})>f(0) \implies f(f(\frac{\pi}{2}))>f(f(0)) \implies 0>1$ which is again a contradiction .

Thus there exists no such function.

Why must the function be increasing in the (restricted) domain? Why can't it bounce up and down?

Calvin Lin Staff - 4 years, 7 months ago

Since the function is continuous & non-constant so if it bounces up and down, the entire graph can be subdivided into intervals where the function is either increasing or decreasing. But it is neither increasing nor decreasing as I showed for $\forall x$

Aditya Narayan Sharma - 4 years, 7 months ago

No, you needed "If $f$ is decreasing in the entire region from 0 to $\frac{ \pi}{2}$ ", since you applied it at the comparison at the points $0, \frac{ \pi}{2}$ .

I do not see how your solution applies to "If $f$ is decreasing in the region 0 to $a$ , increasing in the region from $a$ to $b$ , decreasing in the region from $b$ to $c$ , ...". In particular, it could be decreasing from 0 to 1, increasing from 1 to 1.1, decreasing from 1 to $\frac{ \pi}{2}$ , and we still have $f(\frac{\pi}{2} ) > f(0)$ (because the increasing region had a huge increase).

Calvin Lin Staff - 4 years, 7 months ago

@Calvin Lin – Okk, So how can we improve this solution ?

Aditya Narayan Sharma - 4 years, 7 months ago

@Aditya Narayan Sharma – I don't think there is an easy fix to this approach. But I've not thought about it much.

Calvin Lin Staff - 4 years, 7 months ago

@Calvin Lin – Ah, Mark Henning's solution includes an explanation for why the function is injective and hence strictly increasing/decreasing in the region we care about.

You can use that and help patch up your solution.

Calvin Lin Staff - 4 years, 7 months ago

Kushal Bose
Nov 2, 2016

I have thought in this way.I m not sure whether I m right or wrong.

$f(f(x))= cosx$ . Here $f(x)$ has domain in $R$ and range is also $R$ .So, $f(f(x))$ has domain which is the range of $f(x)$ i.e. $R$ .Now range of $f(f(x))$ has range in $R$ .But it is equal to $cosx$ which is bounded between $[-1,1]$ .So it contradicts.

That's why, there is no such function.

That's not correct. For example, there are many solutions to $f(f(x) ) = 0$ , like for example $f(x) = 0$ or $f(x) = \{ x \}$ (fractional part).

Your "domain-range" argument would still hold, which contradicts the existence of solutions. There is something deeper going on.

Calvin Lin Staff - 4 years, 7 months ago

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