A worked fixed point on $[1,2]$ for the map $x \mapsto (x + 2/x)/2$, from the one-dimensional fixed point theorem

Let

$\varphi : [1,2] \to \mathbb{R}, \qquad \varphi(x) := \frac{1}{2}\Bigl(x + \frac{2}{x}\Bigr)$

(Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). Then:

1. $\varphi$ is continuous on $[1,2]$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point);
2. $\varphi(x) \in [1,2]$ for every $x \in [1,2]$;
3. by Every continuous map of a closed bounded interval into itself has a fixed point, $\varphi$ has a fixed point in $[1,2]$; and
4. that fixed point is unique and equals $\sqrt{2} = 2^{1/2}$ (Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$).

What the example is for. It is the smallest nontrivial instance of the one-dimensional fixed point theorem in which the fixed point can be named, and it shows that the theorem, which asserts existence only, may be combined with an algebraic identity to pin the point down. The identity is elementary: $\varphi(c) = c$ says $c + 2/c = 2c$, that is $c^{2} = 2$.

No derivative is used, and none is available at this point in the reading order. The usual argument that $\varphi$ maps $[1,2]$ into itself computes the minimum of $\varphi$ by differentiation; the two-line order estimate of step 1.2 below replaces it. The same map is treated as a contraction of $[1,2]$ in The map $x \mapsto (x + 2/x)/2$ is a contraction of $[1,2]$ with fixed point $\sqrt{2}$, and the a priori bound gives the error after $n$ steps, where the Banach fixed point theorem gives the same point together with an error bound after $n$ iterations; that route needs completeness of the metric subspace, this one needs only the intermediate value theorem.

• A sharper bound, from a square. For $x > 0$ one has $\bigl(\sqrt{x} - \sqrt{2/x}\bigr)^{2} \ge 0$, and expanding gives $x + 2/x \ge 2\sqrt{2}$, so $\varphi(x) \ge \sqrt{2}$ for every $x > 0$. That is the same identity The map $x \mapsto (x + 2/x)/2$ is a contraction of $[1,2]$ with fixed point $\sqrt{2}$, and the a priori bound gives the error after $n$ steps uses, and it shows that $\varphi$ maps $(0,\infty)$ into $[\sqrt{2}, \infty)$; the crude estimate of step 1.2 is all that claim 2 needs, and it avoids square roots entirely.

• Existence and identification are separate steps. Every continuous map of a closed bounded interval into itself has a fixed point gives claim 3 with no information about where the point is; claim 4 is pure algebra and would be equally valid if no fixed point existed, since it only says which number a fixed point must be. It is the combination that names $\sqrt{2}$.

• The interval matters. On $[-2,-1]$ the same formula has the fixed point $-\sqrt{2}$, and on an interval straddling $0$ the map is not even defined. Choosing $[1,2]$ is what makes claim 2 true and isolates the positive root.