Problem 5. Let $(\mathcal{X},\mathrm{d}_\mathcal{X})$ be a non-empty complete metric space. Suppose that $f,g:\mathcal{X}\longrightarrow\mathcal{X}$ are two Banach's contractions of $\mathcal{X}$. Prove that there always exists a unique point $x_0\in\mathcal{X}$ such that $f\big(g(x)\big)=x_0$.
Dear all, I am revising for my exam in two weeks by doing old exams. This was a problem in an old exam, but I don't know how to prove this. I guess that f and g need to have the same fixed point in order for this to be true, so I tried to prove this, but I didn't come far. Could anyone please help me with solving this problem? Thanks in advance!
They do not need to have the same fixed point—the composition is also a contraction!
I.e., if there exists $λ , μ \in [0,1)$ such that, for all $x,y \in \mathcal X$,
$$d\big(f(x),f(y)\big) \leq λ d\big(f(x),f(y)\big)$$
and
$$d\big(g(x),g(y)\big) \leq μ d\big(g(x),g(y)\big),$$
then, for all $x,y \in \mathcal X$,
$$d\big(fg(x),fg(y)\big) \leq λ \,d\big(g(x),g(y)\big) \leq λ μ \,d\big(x,y\big) $$
This should suffice! I.e., apply the CMT after composing, not before.
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