I have the following question:
Let $\phi: M \to M$ to be an operator such that $\phi^3 = \phi \circ \phi \circ \phi: M \to M$ is a contraction mapping, i.e., $$\|\phi^3(u) -\phi^3(v)\| \leq k\|u-v\| \quad \forall u,v \in M,$$ where $M$ is a closed subset of a Banach Space $X$ and $k \in (0,1)$. Show that $\phi$ has a unique fixed point.
Let $\phi$ to be as the exercise description. Then, it follows that $\phi^3$ has a unique fixed point, say, $p$. In other words, we have $$\phi^3(p)=p.$$ If we apply $\phi$ to the previous equation, we get: $$\phi(\phi^3(p)) = \phi^3(\phi(p)) = \phi(p),$$ that is, $\phi(p)$ is also a fixed point. By uniqueness, it follows that $\phi(p) = p $ and therefore, $\phi$ has a fixed point. It remains to show that $p$ is unique. Say there exists another fixed point, $r$ such that $\phi(r) =r$. Then, it follows that: $$0 \leq \|p-r\| = \|\phi(p) -\phi(r)\| = \|\phi^3(\phi(p)) -\phi^3(\phi(r))\| \leq k\|\phi(p) -\phi(r)\| \leq k\|p -r\|,$$ but $k \in (0,1)$, so $\|p-r\| = 0,$ and therefore $p = r$. Thus, $\phi$ has a unique fixed point.
No, just because $\phi^3$ is a contraction it doesn't follow that $\phi$ is a contraction. For example, on $\mathbb R^2$, the linear transformation $A$ corresponding to the matrix $$ \pmatrix{1 & -7\cr 1/4 & -3/2\cr}$$ is not a contraction, but $A^3 = I/8$ is.
Hint: if $p$ is a fixed point of $\phi^3$, then so is $\phi(p)$.