Proving some statements, given that $|f(x)-f(y)| \leq M|x-y|^{\alpha}$

Question

Let $\alpha \in (0,1)$, $M>0$ and $f:\mathbb{R} \to \mathbb{R}$ such that $$|f(x)-f(y)| \leq M|x-y|^{\alpha}, \: \forall x,y \in \mathbb{R}$$
a) Prove that there is $p \in (0,\infty)$ such that $f([-p,p])\subset [-p,p]$
b) Prove that there are $m,n \in \mathbb{R}$ such that $f(m)f(n)+mn=0$

Obviously, $f$ is continuous from the inequality given. I tried a proof by contradiction: suppose that $\forall p>0$, there is $x_p \in [-p,p]$ such that $|f(x_p)|>p$. Then, I tried to make $p$ go to $0$ and obtain some sort of chain inequalities that contradict the one given in the statement, but I didn't succeed. The fact that $\alpha$ is in $(0,1)$ really gives me a hard time.

Edit: I think I found a solution for b)

If there is a $m \in \mathbb{R}$ such that $f(m)=0$, then take $n=0$ and the conclusion follows.
Suppose now that there is no $x$ such that $f(x)=0$, i.e. $f$ maintains a constant sign on $\mathbb{R}$. Without loss of generality, let $f(x)>0$
Let's assume that $f(x)>x, \forall x>0$. Then, going back to the inequality in the statement with $y=0$ we get $$Mx^{\alpha}=M|x|^{\alpha} \geq |f(x)-f(0)| \geq |f(x)|-|f(0)|=f(x)-f(0)>x-f(0)$$ Taking limits here as $x \to \infty$ yields $0 \geq \infty$, because $\alpha<1$, which is impossible. Thus, there must be a $n>0$ such that $f(n) \leq n$

Now consider $p>0$ from a). Hence we know that $ f(p) \geq -p$ and $f(-p) \leq p$. Define the continuous function $g:[-p,p]\to \mathbb{R}$ with $g(x)=f(x)f(n)+xn$. Then $$g(-p)=f(-p)f(n)-pn \leq p(f(n)-n) \leq 0$$ and $$g(p)=f(p)f(n)+pn \geq p(n-f(n))\geq 0$$ Now we can conclude that there must be $m \in [-p,p]$ such that $g(m)=0$ which means that $$f(m)f(n)+mn=0$$

Answer 1

Continuing from your work, we have the chain of inequalities $$p-|f(0)| < |f(x_p)| - |f(0)| \le |f(x_p) - f(0)| \le M p^\alpha,$$ which leads to a contradiction when taking $p \to \infty$.

Answer 2

For $a)$: Consider $f([-p,p])$, and suppose that $f([-p,p])\subset[-p,p]$. Even though this isn't a proof, this scratch work will lead to the conditions we need to prove.

Let's start by considering $p\in[-p,p]$. For the inclusion to hold, we need $|f(p)|\leq p$. We know that $$ |f(p)|=|f(p)-f(0)+f(0)|\leq |f(p)-f(0)|+|f(0)|. $$
By assumption, it then follows that $$ |f(p)|\leq M|p-0|^\alpha+|f(0)|=Mp^\alpha+|f(0)|. $$ To make $|f(p)|\leq p$, we assume that $$ Mp^\alpha+|f(0)|\leq p. $$ This inequality is true if $$ |f(0)|\leq p^\alpha(p^{1-\alpha}-M). $$ If $p$ is sufficiently large, $p^{1-\alpha}-M\geq 1$ and $p^\alpha\geq |f(0)|.$ Both of these conditions can be found explicitly. Now, we have only found the appropriate conditions for $p$, but if $|x|<p$, only one more inequality is needed in the proof $(|x|^\alpha\leq|p|^\alpha)$.