Problems mainly based on a composite function $f(f(x))={f(x)}^3$

Question

For $f(x)$ which is differentiable over all $\mathbb{R}$, $f(x)$ meets four conditions

(a) $f(4) < 1 $

(b) For all $\mathbb{R}$, $f(f(x))={f(x)}^3$

(c) if $p < q$, $0\le f(p) \le f(q)$

(d). if $p < q$, $f'(p) + \cfrac{3}{4}p \le f'(q) + \cfrac{3}{4}q$

for all $f(x)$ that meets above conditions, $M(x)$ is the function where $f(4)$ become largest number. (maximum)

Find the value of $$\int_{-2}^{2} M(x) dx$$ Source: A Korean hs problem

I was struggling to understand what (b) means, and I couldn't figure it out from the start.

(detailed questions after some comments)

I was mainly curious about the condition (b).

Simply thought, $f(x)$ is $0$, $1$ or $x^3$

But it is not likely, because of the condition (a)

I think the problem wanted me to think about the domain and range of $f(x)$

But at that point, I wasn't able to move forward.

I don't know how to connect that thought with other conditions.

Answer 1

Some observations:

1. Rearranging condition 'd' , we find: $ \frac{f'(p) - f'(q)}{p-q} \leq \frac{3}{4}$, take the limit as $ p \to q$ with $p>q$ on LHS. Take the upper side limit to avoid problems: $$ \lim_{ p \to q^+} \frac{ f'(p) - f'(q)}{q-p} \leq \frac34$$

We find that: $ -f''(x) \leq \frac34$(*)

3. Equation b, we find that if $f(x)=0$ at any $ x \in \mathbb{R}$ , then it must be that $x=0$ is a zero for the function as well.

4. The c. condition can be written as: $f'(x) \geq 0$

5. If $f(4)<1$, then it must be that $f(f(4))<f(4)$, by the facts that $f(4)<4$ and $f(x)$ is an increasing function.

We can also prove by induction on the above fact that upon setting the number of compositions to infinity, the function goes to zero. Since $0 \leq f(x)$, we find the lower bound is zero.

6. (From Lutz Lehman) , if a function is continuous for all $x \in \mathbb{R}$, then it must be that the range can be written as an open interval $(0,b)$. Further, we can identify a polynomial equation which the upper bound must satisfy.

Let $b$ be the upper bound, if $b>1$ then $f(b) \geq b$, then $b^3 \geq b$, this is only without contradiction if $b^3 = b$.

If $b<1$, then we find that $1 \leq b$

Hence, we arrive at the conclusion of Lutz Lehman on the bounds:

$$0 \leq b <1$$

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Example function:

As noted by Lutz Lehman, $f(x)=0$ seems to be a function satisfying the criteria in the question + the ones derived above.

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*: I assumed existence of second derivative to make this work, otherwise there is no reason for upper limit and lower limit to agree.

**: I assumed $f'(x) \neq 0$