(a) and (b) Step 1. If $f$ is a shrinking map then there is at most one fixed point. Suppose there are two fixed points $x$ and $y$ . Then $d(x,y)=d(f(x),f(y))<d(x,y)$ .Step 2. We construct a closed compact set $C$ that will be proved to have a fixed point. Let $C_0=X$ and $C_n=f^n(X)$ . $f$ is a continuous map from a compact space to Hausdorff space, therefore, it is a closed map. Then, by induction, every set in the sequence is closed. Moreover, since the image of a subset lies within the image of a superset, the sequence is a nested sequence of nonempty closed sets that has a nonempty closed and compact intersection $C$ .Step 3. We show that $f(C)\subseteq C$ . Indeed, suppose that for $z\in C$ : $f(z)\notin C$ . Hence, there is $n\ge 1$ such that $f(z)\in C_n-C_{n+1}$ . This implies that $z\notin C_n$ which contradicts the fact that $c\in C$ .Step 4. We show that $C$ has only one point, this implies that the point is the fixed point. If $f$ is a contraction we obtain the result immediately. Indeed, the diameter of $C_n$ decreases exponentially and there cannot be more than 1 point in the intersection. More generally, we want to show that any $z\in C$ is the image of some point in $C$ . For each $n$ : $z\in C_n$ , i.e. there exists $x_n\in X$ such that $f^n(x_n)=z$ . A compact metric space is sequentially compact, hence, there is a subsequence of $\{y_n=f^{n-1}(x_n)\}$ that converges to some point $a$ . Any neighborhood of $a$ contains infinitely many members of the sequence, therefore, it is in the closure of every $C_n$ , but $C_n$ ’s are closed, therefore, $a\in C$ . Moreover, since $f(y_n)=z$ for all $n$ and $f$ is continuous, $f(a)=z$ . We conclude that $C\subseteq f(C)\subseteq C$ , i.e. $C=f(C)$ . The distance on $C$ is a continuous function from the compact product $C\times C$ to the ordered set $\mathbb{R}$ . Therefore, there is a pair of points $x,y\in C$ with the maximum distance between them. Let $x=f(a),y=f(b)$ where $a,b\in C$ . Then if $x\neq y$ : $d(x,y)=d(f(a),f(b))<d(a,b)\le d(x,y)$ . This contradiction implies that there is only one point in $C$ , the fixed point of $f$ .(c) $f$ is strictly increasing from 0 to 1/2. $f(y)-f(x)=(y-x)(1-(y+x)/2)$ implies it is a shrinking map (the unique fixed point is 0). $f(1)-f(x)=(1-x)^2/2$ implies it is not a contraction.(d) $f$ is strictly increasing, moreover $f(x)>(x+|x|)/2\ge x$ for all $x$ . So, there is no fixed point. $(f(y)-f(x))/(y-x)=[1+(y+x)/(\sqrt{y^2+1}+\sqrt{x^2+1})]/2<1$ but can be made as close to 1 as desired when $x,y\to+\infty$ . Therefore, it is a shrinking map which is not contraction and has no fixed point.