Pick out the true statements:
a. $|\cos^2 x − \cos^2 y| \le|x − y|$ for all $x, y \in \mathbb{R}$.
b. If $f : \mathbb{R} \to \mathbb{R}$ satisfies $|f(x) − f(y)|\le|x − y|^{\sqrt{2}}$for all $x, y \in \mathbb{R}$ then $f$ must be a constant function.
c. Let$f : \mathbb{R} \to \mathbb{R}$ be continuously differentiable and such that $|f'(x)|  \le4/5$ for all $x \in \mathbb{R}$. Then, there exists a unique $x \in \mathbb{R}$ such that $f(x) = x$.

(a) using Lagranges theorem I get that this is true.
But I have no idea about the others.Can I get some help please?

  • $\begingroup$ "Lagranges theorem" = "Mean Value Theorem"? $\endgroup$
    – Braindead
    Commented Jan 10, 2014 at 7:12

2 Answers 2


For part c), here is an alternate solution for the existence of a fixed point on $f$. If $|f'(x)| \le a$ for $0<a<1$, we have:

$-a \le f'(x) \le a$, which maybe integrated to $-a x + b \le f(x) \le a x + b$, and so

$$(-a-1)x + b \le f(x) - x \le (a-1)x + b$$

Since $(a-1)$ is not zero, one can find a sufficiently large $x$ so that $f(x) - x$ is negative.

Since $(-a-1)$ is not zero, one can find a sufficiently small $x$ so that $f(x) - x$ is positive.

Intermediate value theorem guarantees that $f(x) - x = 0$ for some $x$.

Intuitive picture: (two lines, opposite slopes, absolute value less than one; the line $y=x$. The three lines form a triangle. There is a part of $f$ that is trapped in the triangle, and it has no choice but to go through $y=x$.)


Note that since $\sqrt{2}>1$, then we have $$ 0\leq \lim\limits_{y\to x}\frac{\vert f(x)-f(y)\vert}{\vert x-y\vert}\leq \lim\limits_{y\to x} \vert x-y\vert^{\sqrt{2}-1}=0 $$ Therefore, the limit in the middle exist and is equal to zero. What does this tell you about $f$ and its derivative? For the last point, note that this implies that $\vert f'(x)\vert <1$. If there would exist two points $y$ and $x$ suche that $f(y)=y$ and $f(x)=x$, what would implie the Lagrange theorem that you used in a)?

Since for every $x, y$ and for some $c$ between them we have $$ \vert f(x)-f(y)\vert=\vert f'(c)\vert\vert (x-y)\vert \leq\frac{4}{5} \vert x-y\vert $$ Then, $f$ is a contracting mapping and we can use the Banach fixed point theorem to show the existence of such an $x$

  • $\begingroup$ How would we know if this $x$ and $y$ actually exists? $\endgroup$
    – Lemon
    Commented Jan 10, 2014 at 7:00
  • $\begingroup$ No, the argument is to show that there exist at most one point $x$ such that $f(x)=x$ (those points are called fixed points}. I forgot to suggesr how to shoe the existence. I will fix it $\endgroup$
    – Brandon
    Commented Jan 10, 2014 at 7:04
  • $\begingroup$ Oh I see, you were going for some sort of contradiction. $\endgroup$
    – Lemon
    Commented Jan 10, 2014 at 7:07
  • $\begingroup$ For existence, you can integrate the inequality $-4/5 \le f'(x) \le 4/5$. $\endgroup$
    – Braindead
    Commented Jan 10, 2014 at 7:09
  • $\begingroup$ Ah okay, thanks. I up vote you $\endgroup$
    – Lemon
    Commented Jan 10, 2014 at 7:11

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