ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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

Example

Let

$φ : [1, 2] \to R, φ (x) : = \frac{1}{2} (x + \frac{2}{x})$

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

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

No derivative is used, and none is available at this point in the reading order. The usual argument that $φ$ maps $[1, 2]$ into itself computes the minimum of $φ$ 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.

Facts & Assumptions

Given: The interval $[1, 2]$ and the function $φ (x) = \frac{1}{2} (x + 2 / x)$ on it.

[L1]

Algebra of continuous functions: the identity and constants are continuous, sums and scalar multiples of continuous functions are continuous, and the reciprocal of a continuous nowhere-vanishing function is continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L2]

One-dimensional fixed point theorem: a continuous $φ : [a, b] \to R$ with $a \leq b$ and $φ (x) \in [a, b]$ for all $x \in [a, b]$ has a fixed point in $[a, b]$ (Every continuous map of a closed bounded interval into itself has a fixed point, Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ).

[L3]

Reciprocals and order: for $1 \leq x \leq 2$ one has $0 < x$ , hence $1 / 2 \leq 1 / x \leq 1$ and so $1 \leq 2 / x \leq 2$ (Inverses of positives are positive, and reciprocation reverses order, Ordered field).

[L4]

Square roots: for $a \geq 0$ there is a unique $s \geq 0$ with $s^{2} = a$ , written $a^{1 / 2} = \sqrt{a}$ ; and $t \mapsto t^{2}$ is strictly increasing on the nonnegative reals (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , Integer powers $a^{m}$ ).

[L5]

Ordered-field arithmetic: $1 \leq 2$ ; halving preserves order; and $1^{2} = 1 \leq 2 \leq 4 = 2^{2}$ (Ordered field, Complete ordered field (least-upper-bound property), Integer powers $a^{m}$ ).

Verification

technique · direct

1.1

Claim 1. On $[1, 2]$ the identity is continuous and does not vanish, since $x \geq 1 > 0$ ; so $x \mapsto 2 / x$ is continuous there by [L1], and $φ = \frac{1}{2} (i d + 2 / i d)$ is continuous on $[1, 2]$ as a scalar multiple of a sum of continuous functions.

L1L3

1.2

Claim 2. Let $x \in [1, 2]$ . By [L3] we have $1 \leq 2 / x \leq 2$ , and by hypothesis $1 \leq x \leq 2$ ; adding, $2 \leq x + 2 / x \leq 4$ , and halving gives $1 \leq φ (x) \leq 2$ by [L5]. So $φ (x) \in [1, 2]$ .

L3L5

2.1

Claim 3. By [L2], applied with $a : = 1$ , $b : = 2$ and the map $φ$ , which is continuous by step 1.1 and maps $[1, 2]$ into itself by step 1.2, there is $c \in [1, 2]$ with $φ (c) = c$ .

step 1.1step 1.2L2choose

3.1

Every fixed point squares to $2$ . Let $c \in [1, 2]$ satisfy $φ (c) = c$ . Then $c \geq 1 > 0$ , so multiplying $\frac{1}{2} (c + 2 / c) = c$ by $2 c$ gives $c^{2} + 2 = 2 c^{2}$ , that is $c^{2} = 2$ .

step 2.1L3L5

4.1

Claim 4. By [L4] there is exactly one nonnegative real whose square is $2$ , namely $\sqrt{2}$ ; since every fixed point $c$ is $\geq 1 > 0$ and satisfies $c^{2} = 2$ by step 3.1, the fixed point is unique and equals $\sqrt{2}$ . And $\sqrt{2}$ does lie in $[1, 2]$ : from $1^{2} = 1 \leq 2 \leq 4 = 2^{2}$ and the strict monotonicity of $t \mapsto t^{2}$ on the nonnegative reals ([L4], [L5]) one gets $1 \leq \sqrt{2} \leq 2$ .

step 3.1L4L5∎

Remarks

A sharper bound, from a square. For $x > 0$ one has $(\sqrt{x} - \sqrt{2 / x})^{2} \geq 0$ , and expanding gives $x + 2 / x \geq 2 \sqrt{2}$ , so $φ (x) \geq \sqrt{2}$ 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 $φ$ 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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

53 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Fixed point (mathematics) (Wikipedia) (standard reference, not scraped)
Methods of computing square roots (Wikipedia) (standard reference, not scraped)
Intermediate value theorem (Wikipedia) (standard reference, not scraped)

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