Some Final Problems - #6

Hey everyone, Mark said that there would be at least one problem like #6,7,8 using the MVT on the Final Exam, so I figured I would show my solution to #6.
The problem reads: Suppose that $f$ is differentiable on an interval $I$ and there is a number $r$ in $(0,1)$ with $|f'(x)|\leq r$ for all $x\in I$. Use the MVT to prove that $f$ is a contraction on $I$. Recall that $f$ is called a contraction on $I$ if there is a number $r$ with $0< r <1$ and $$|f(y)-f(x)| \leq r|y-x|$$ for all $x,y\in I$

Proof:

Let $f$ be differentiable on an interval $I$, where $\exists r\in (0,1)$ with $|f'(x)|\leq r$ $\forall x\in I$. Next let $x,y\in I$ and without loss of generality, assume $x < y$.
Then by the MVT $\exists c\in I$, where $x < c < y$ such that
$$f'(c)=\frac{f(y)-f(x)}{y-x}.$$
But then
$$\left| \frac{f(y)-f(x)}{y-x} \right| = |f'(c)| \leq r.$$
Hence
$$|f(y)-f(x)|\leq r|y-x|.$$
Now since $x$ and $y$ were chosen arbitrarily, $f$ is a contraction on $I$.

7 on the Review Sheet- Exercise 5.3.5:

Show that if $f$ is differentiable on an interval with $f'(x) \ne 1$, then $f$ can have at most one fixed point.

Proof:
Using proof by contradiction, assume that $f$ has two distinct fixed points $a_1$ and $a_2$. Thus, $f(a_1) = a_1$ and $f(a_2) = a_2$ and the MVT implies that there must exist a $c$ where
$$ f'(c) = \frac{f(a_1) - f(a_2)}{ a_1 - a_2} = \frac{a_1 - a_2}{a_1 - a_2} = 1.$$
But this does not hold true and $f'(c) \ne 1.$ Thus, $f$ can have at most one fixed point.

Hey @lhouse1 I would be careful to point out in your preamble that $a_2 < a_1$. Also, you have shown here that $f$ cannot have $2$ fixed points, but what about more than two? Or prove that we can have less than two? Other than that it looks great.

I think what your getting at is covered. @lhouse1 showed it doesn't work in the case of 2 arbitrary fixed points. even if we had $f(a_{1}) = a_{1}, f(a_{2}) = a_{2}, ..., f(a_{n}) = a_{n}$ We would have $$\frac{f(a_{1}) - f(a_{2})-...-f(a_{n})}{a_{1} - a_{2} - ...-a_{n}} = \frac{a_{1} - a_{2} - ...-a_{n}}{a_{1} - a_{2} - ...-a_{n}} = 1.$$
Since $a_{1} - a_{2} - ...-a_{n} \in\mathbb{R}$.

Have anyone done 5.b? I understand that g'(0)=1 and that the functions has infinitely many "swings" due to theh sin function but I can't think of a way to prove it formally.

I agree if anyone has 5, i just think the $sin(1/x)$ is throwing me off.