Here is Prob. 7, Sec. 28, in the book Topology by James R. Munkres, 2nd edition:
Let $(X, d)$ be a metric space. If $f$ satisfies the condition $$ d\big( f(x), f(y) \big) < d(x, y) $$ for all $x, y \in X$ with $x \neq y$, then $f$ is called a shrinking map. If there is a number $\alpha < 1$ such that $$ d \big( f(x), f(y) \big) \leq \alpha d(x, y) $$ for all $x, y \in X$, then $f$ is called a contraction. A fixed point of $f$ is a point $x$ such that $f(x) = x$.
(a) If $f$ is a contraction and $X$ is compact, show $f$ has a unique fixed point. [Hint: Define $f^1 = f$ and $f^{n+1} = f \circ f^n$. Consider the intersection $A$ of the sets $A_n = f^n(X)$.]
(b) Show more generally that if $f$ is a shrinking map and $X$ is compact, then $f$ has a unique fixed point. [Hint: Let $A$ be as before. Given $x \in A$, choose $x_n$ so that $x = f^{n+1}\left(x_n\right)$. If $a$ is the limit of some subsequence of the sequence $y_n = f^n \left( x_n \right)$, show that $a \in A$ and $f(a) = x$. Conclude that $A = f(A)$, so that $\mathrm{diam}\, A = 0$.]
(c) Let $X = [0, 1]$. Show that $f(x) = x - x^2/2$ maps $X$ into $X$ and is a shrinking map that is not a contraction. [Hint: Use the mean-value theorem of calculus.]
(d) The result in (a) holds if $X$ is a complete metric space, such as $\mathbb{R}$; see the exercises of Sec. 43. The result in (b) does not: Show that the map $f \colon \mathbb{R} \to \mathbb{R}$ given by $f(x) = \left[ x + \left( x^2 + 1 \right)^{1/2} \right]/2$ is a shrinking map that is not a contraction and has no fixed point.
Here is my Math SE post on Prob. 7 (a).
Here is my Math SE post on Prob. 7 (b).
And, here is my Math SE post on Prob. 7 (c).
Here I'll only be attempting Part (d).
My Attempt:
Part (d):
Here $$ f(x) = \frac{x + \sqrt{ x^2 + 1} }{ 2 } \ \mbox{ for all } \ x\in \mathbb{R}. \tag{0} $$ So we obtain $$ f^\prime(x) = \frac{ 1 + \frac{x}{\sqrt{x^2+1}} }{ 2} = \frac{ \sqrt{x^2 + 1} + x }{ 2 \sqrt{ x^2 + 1 } } = \frac{1}{2} \left( 1 + \frac{x}{ \sqrt{x^2+1} } \right) > 0 \ \mbox{ for all } \ x \in \mathbb{R}. \tag{1} $$ Thus $f$ is strictly increasing.
Now suppose that there exists a real number $\alpha \in [0, 1)$ such that $$ \big\lvert f(x) - f(y) \big\rvert \leq \alpha \lvert x - y \rvert \ \mbox{ for all } \ x, y \in \mathbb{R}. \tag{A} $$
Let $x, y \in \mathbb{R}$ such that $x < y$. Then $f(x) < f(y)$, and so $$ f(y) - f(x) = \big\lvert f(x) - f(y) \big\rvert \leq \alpha \lvert x - y \rvert = \alpha (y-x), $$ and so $$ \begin{align} \alpha &\geq \frac{ f(y) - f(x) }{ y-x} \\ &= \frac{ \frac{y + \sqrt{ y^2 + 1} }{ 2 } - \frac{x + \sqrt{ x^2 + 1} }{ 2 } }{ y-x } \\ &= \frac{1}{2} + \frac{ \sqrt{y^2+1} - \sqrt{x^2+1} }{ 2(y-x) } \\ &= \frac{1}{2} + \frac{ y^2 - x^2 }{ 2(y-x) \left( \sqrt{ y^2+1} + \sqrt{x^2+1} \right) } \\ &= \frac{1}{2} + \frac{ y + x }{ 2 \left( \sqrt{ y^2+1} + \sqrt{x^2+1} \right) }. \end{align} $$ Thus we find that $$ \alpha \geq \frac{1}{2} \left( 1 + \frac{ y + x }{ \sqrt{ y^2+1} + \sqrt{x^2+1} } \right) $$ for all $x, y \in \mathbb{R}$ such that $x < y$. Interchanging the roles of $x$ and $y$ in the preceding relation, we obtain the same relation for all $x, y \in \mathbb{R}$ such that $y < x$. Thus we can conclude that $$ \alpha \geq \frac{1}{2} \left( 1 + \frac{ y + x }{ \sqrt{ y^2+1} + \sqrt{x^2+1} } \right) \tag{2} $$ for all $x, y \in \mathbb{R}$ such that $x \neq y$.
If in (2) we take the limit as $y \to +\infty$ keeping $x \in \mathbb{R}$ constant, we obtain $$ \alpha \geq \frac{1}{2} \left( 1 + 1 \right) = 1, $$ which contradicts our supposition that $\alpha \in [0, 1)$. Hence our $f$ is not a contraction.
Is my reasoning correct?
We find that, for all $x \in \mathbb{R}$, $$ -1 < \frac{ x }{ \sqrt{x^2+1} } < 1, $$ and so $$ 0 < 1 + \frac{ x }{ \sqrt{x^2+1} } < 2, $$ and once again from (1) above, we have $$ 0 < f^\prime(x) < \frac{1}{2} \left( 1 + 1 \right) = 1. \tag{2} $$
Now let $x$ and $y$ be any real numbers such that $x \neq y$.
First let us suppose that $x < y$. Then by the Lagrange's mean-value theorem there exists a real number $c \in (x, y)$ such that $$ f(y) - f(x) = f^\prime(c) (y-x), $$ and since $f$ is strictly increasing, therefore (2) yields $$ \big\lvert f(x) - f(y) \big\rvert = f(y) - f(x) = f^\prime(c) (y-x) < (y-x) = \lvert x- y \rvert. $$ And, interchanging the roles of $x$ and $y$ in the above relation, we obtain the same relation if $y < x$. Therefore we can conclude $$ \big\lvert f(x) - f(y) \big\rvert < \lvert x- y \rvert $$ for all $x, y \in \mathbb{R}$ such that $x \neq y$. Hence $f$ is a shrinking map.
Is my reasoning correct?
Is each of my two proofs correct and clear enough in each and every details? Or, are there any lacunas?
