Convergence of fixed-point iteration

Theorem

Convergence of fixed-point iteration

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Statement

Let \((X,d)\) be a non-empty complete metric space and let \(g:X\to X\) be a contraction, i.e. there exists a constant \(L\) with \(0 \le L \lt 1\) such that \(d(g(x),g(y)) \le L\,d(x,y)\) for all \(x,y \in X\). Then: (i) \(g\) has a unique fixed point \(x^\ast \in X\) with \(g(x^\ast)=x^\ast\); (ii) for any starting point \(x_0 \in X\), the iteration \(x_{n+1} = g(x_n)\) converges to \(x^\ast\), i.e. \(x_n \to x^\ast\) as \(n \to \infty\); (iii) the convergence is at least linear (geometric), with the a priori bound \(d(x_n,x^\ast) \le \dfrac{L^n}{1-L}\,d(x_1,x_0)\) and the a posteriori bound \(d(x_n,x^\ast) \le \dfrac{L}{1-L}\,d(x_n,x_{n-1})\), for all \(n \ge 1\).

Why it matters

This is the Banach fixed-point theorem, specialised to the setting most useful in numerical analysis: it is the single result that certifies, in advance and without knowing the answer, that an iterative scheme \(x_{n+1}=g(x_n)\) will converge, and it hands you an explicit, computable error bound as it does so. Almost every fixed-point-style solver used in practice — Newton's method (locally, via its own contraction constant), the Jacobi and Gauss–Seidel iterations for linear systems written as \(x = Bx+c\), Picard iteration for ODE existence, and fixed-point schemes for nonlinear equations \(x=g(x)\) — is, at bottom, an appeal to this theorem or a close relative of it.

What makes the result genuinely useful rather than merely elegant is that it converts a qualitative existence statement into a quantitative stopping criterion: the a posteriori bound lets you certify, from two consecutive computed iterates alone, how close you already are to the true root, which is exactly what a numerical routine needs to decide when to stop.

Hypotheses

Completeness of \((X,d)\).Take \(X=\mathbb{Q}\) with the usual metric (incomplete) and \(g(x) = x/2 + 1/x\), whose only fixed point in \(\mathbb{R}\) is \(x^\ast=\sqrt2 \notin \mathbb{Q}\); the Cauchy sequence of iterates has no limit inside \(X\), so "convergence" fails for lack of a target to converge to, even though the contraction estimate still holds on suitable subsets of \(\mathbb{Q}\).

\(g\) maps \(X\) into itself.If \(g\) is only defined and contractive on a subset \(X\) but iterates can leave \(X\), the sequence \(x_n\) may exit the space where the contraction estimate applies; e.g. \(g(x)=1-2x\) restricted to \([0,1]\) sends \(x=0.9\) to \(-0.8 \notin [0,1]\), so \(g\) is not self-mapping on \([0,1]\) and the argument breaks down immediately at \(n=1\).

Uniform contraction constant \(L \lt 1\) (not merely \(d(g(x),g(y)) \lt d(x,y)\) pointwise).Take \(X=[1,\infty)\) with \(g(x) = x + 1/x\). Then \(g(x)-g(y) = (x-y)\left(1 - \frac{1}{xy}\right)\), so \(|g(x)-g(y)| \lt |x-y|\) strictly whenever \(x\ne y\), yet \(\sup\) of the ratio over all pairs is \(1\) (attained in the limit as \(x,y\to\infty\)), so no single \(L\lt1\) works uniformly; indeed \(g(x)\to\infty\) and the iteration diverges from every starting point.

Non-emptiness of \(X\).Vacuous but necessary for the existence claim to be meaningful: if \(X=\varnothing\) there is trivially no fixed point, so the statement "\(g\) has a fixed point" would be false as an existence claim (though vacuously any universally-quantified convergence statement about "all starting points" holds trivially).

Proof

\text{Define the iterates } x_0 \in X,\quad x_{n+1} = g(x_n) \text{ for } n \ge 0.

Well-defined because \(g:X\to X\), so \(x_n \in X\) for every \(n\) by induction. A

d(x_{n+1},x_n) \le L\, d(x_n,x_{n-1}) \quad \text{for all } n \ge 1.

Apply the contraction hypothesis directly with \(x=x_n,\,y=x_{n-1}\): \(d(g(x_n),g(x_{n-1})) \le L\,d(x_n,x_{n-1})\), and \(g(x_n)=x_{n+1}\), \(g(x_{n-1})=x_n\). A

d(x_{n+1},x_n) \le L^n\, d(x_1,x_0) \quad \text{for all } n \ge 0.

Induction on \(n\) using Step 2 as the inductive step; base case \(n=0\) is an equality. A

\text{For } m \gt n:\quad d(x_m,x_n) \le \sum_{k=n}^{m-1} d(x_{k+1},x_k) \le \sum_{k=n}^{m-1} L^k d(x_1,x_0) \le \frac{L^n}{1-L}\,d(x_1,x_0).

First inequality is the triangle inequality applied repeatedly along the chain \(x_n,x_{n+1},\dots,x_m\); second substitutes Step 3; third bounds the finite geometric sum by the infinite tail \(\sum_{k=n}^{\infty}L^k = L^n/(1-L)\), valid since \(0\le L\lt1\). B

(x_n)_{n\ge0} \text{ is a Cauchy sequence in } (X,d).

Given \(\varepsilon\gt0\), since \(L^n \to 0\) (as \(0\le L\lt1\)), choose \(N\) with \(\frac{L^N}{1-L}d(x_1,x_0) \lt \varepsilon\); then Step 4 gives \(d(x_m,x_n)\le\frac{L^n}{1-L}d(x_1,x_0)\lt\varepsilon\) for all \(m\gt n\ge N\), which is precisely the Cauchy condition. (If \(d(x_1,x_0)=0\) the sequence is constant and trivially Cauchy.) C

\exists\, x^\ast \in X \text{ such that } x_n \to x^\ast.

By completeness of \((X,d)\), every Cauchy sequence in \(X\) converges to a limit in \(X\); apply this to the Cauchy sequence from Step 5. This is exactly where completeness is used, and nowhere else in the proof. B

g(x^\ast) = x^\ast.

\(g\) is Lipschitz continuous (constant \(L\)), hence continuous, so \(g(x_n) \to g(x^\ast)\). But \(g(x_n) = x_{n+1} \to x^\ast\) also (a subsequence, in fact a shift, of a convergent sequence has the same limit). Limits in a metric space are unique, so \(g(x^\ast)=x^\ast\). B

\text{Uniqueness: if } g(y^\ast)=y^\ast \text{ also, then } y^\ast = x^\ast.

\(d(x^\ast,y^\ast) = d(g(x^\ast),g(y^\ast)) \le L\,d(x^\ast,y^\ast)\), so \((1-L)\,d(x^\ast,y^\ast)\le 0\); since \(1-L\gt0\) and \(d\ge0\), this forces \(d(x^\ast,y^\ast)=0\), i.e. \(x^\ast=y^\ast\) by the metric axiom (\(d(a,b)=0 \iff a=b\)). A

d(x_n,x^\ast) \le \frac{L^n}{1-L}\,d(x_1,x_0) \quad (\text{a priori bound}).

Let \(m\to\infty\) in Step 4 (with the fixed lower index \(n\)); the right-hand side is independent of \(m\), and \(d(\cdot,x_n)\) is continuous, so \(d(x_m,x_n)\to d(x^\ast,x_n)\) by Step 6, giving the stated bound in the limit. C

d(x_n,x^\ast) \le \frac{L}{1-L}\,d(x_n,x_{n-1}) \quad (\text{a posteriori bound}).

Apply the a priori bound (Step 9) with starting index shifted to \(x_{n-1}\) in place of \(x_0\): the tail sequence \((x_{n-1},x_n,x_{n+1},\dots)\) is itself generated by \(g\) from \(x_{n-1}\), so \(d(x_n, x^\ast) \le \frac{L^1}{1-L} d(x_n,x_{n-1})\), i.e. the case "one step in" of Step 9 applied to the restarted iteration. B

Result

g \text{ contraction on complete } (X,d) \;\Longrightarrow\; \exists! \, x^\ast=g(x^\ast), \quad x_n=g^n(x_0)\to x^\ast, \quad d(x_n,x^\ast)\le \frac{L^n}{1-L}d(x_1,x_0)

Reading. If repeatedly applying a map always shrinks distances by at least a fixed factor \(L\lt1\), and the space has no "holes" for a limit to fall through, then repeated application from any starting guess homes in on one and only one fixed point, and it does so at a guaranteed geometric rate you can bound without ever knowing the fixed point itself.

Scope. Applies to any complete metric space — finite-dimensional \(\mathbb{R}^n\) with any equivalent norm, Banach spaces of functions (giving Picard–Lindelöf existence for ODEs), or discrete iteration schemes in numerics. It does not require differentiability of \(g\), linearity, or that \(X\) be a vector space — only a metric and completeness. It gives global convergence from any \(x_0\in X\); this is stronger than, and should not be confused with, purely local convergence results such as Newton's method's quadratic convergence, which only apply once \(x_0\) is already close to \(x^\ast\).

Corollaries & converses

Rate corollary. The convergence is at least linear with asymptotic error constant \(L\): from Step 9, \(\limsup_{n\to\infty} d(x_n,x^\ast)^{1/n} \le L\), so smaller \(L\) means faster guaranteed convergence.
Stopping-criterion corollary. The a posteriori bound (Step 10) is used directly in practice: once \(\frac{L}{1-L}d(x_n,x_{n-1}) \lt \text{tol}\), you may certify \(d(x_n,x^\ast)\lt\text{tol}\) without ever computing \(x^\ast\).
Differentiable corollary (mean value form). If \(X\subseteq\mathbb{R}\) is a closed interval and \(g\) is differentiable with \(\sup_{x\in X}|g'(x)| = L \lt 1\), then \(g\) is a contraction with that constant \(L\), by the Mean Value Theorem applied to \(g\) on the segment between any \(x,y\in X\).
Converse fails. A map can have a unique, attracting fixed point that iteration converges to, without being a contraction in any metric compatible with the topology on a natural domain — e.g. \(g(x)=x^2\) on \([0,1)\) has fixed point \(0\) attracting all points, but \(g\) is not a contraction on all of \([0,1)\) since \(|g(x)-g(0)|/|x-0| = x \to 1\) as \(x\to1\), and completeness of \([0,1)\) also fails. So convergence of an iteration does not imply the hypotheses of this theorem held; the theorem gives a sufficient, not necessary, condition.
Local converse partly holds. If \(g\) is \(C^1\) near a fixed point \(x^\ast\) with \(|g'(x^\ast)|\lt1\), then \(g\) is a contraction on some sufficiently small closed neighbourhood of \(x^\ast\) (by continuity of \(g'\)), recovering local applicability of the theorem even when \(g\) is not a global contraction.

Fails without

Drop \(L\lt1\) (allow \(L=1\), i.e. only \(d(g(x),g(y))\le d(x,y)\)): take \(X=\mathbb{R}\), \(g(x)=x+1\). Then \(d(g(x),g(y))=d(x,y)\) exactly (an isometry, so the weak inequality holds with \(L=1\)), \(g\) has no fixed point at all, and the iterates \(x_n = x_0+n \to \infty\), diverging.
Drop completeness of \(X\): take \(X=(0,1]\) (not complete, missing its limit point \(0\)) with \(g(x)=x/2\), which satisfies \(d(g(x),g(y)) = \frac12 d(x,y)\), a genuine contraction with \(L=1/2\). The iterates \(x_n = x_0/2^n \to 0\), but \(0\notin X\), so the "fixed point" the sequence is converging to does not exist inside the space — existence of \(x^\ast\in X\) fails even though the Cauchy estimate (Steps 1–5) is unaffected.
Drop self-mapping \(g:X\to X\): on \(X=[0,1]\), let \(g(x) = 2x \bmod{\text{(defined only for }x\le1/2)}\), i.e. \(g\) is only defined on \([0,1/2]\) but \(x_0=0.9\) is not in the domain at all — the iteration is not even well-posed, so "convergence" is meaningless before contraction is ever invoked.

Common errors

Checking \(|g(x)-g(y)| \lt |x-y|\) at isolated test points (or even "for all \(x\ne y\) pointwise") and treating that as verifying a contraction, instead of bounding \(\sup |g'|\) or the Lipschitz ratio uniformly over the whole domain — this is exactly the trap in the \(g(x)=x+1/x\) counterexample above.
Using the a priori error bound with the wrong base distance, e.g. writing \(\frac{L^n}{1-L}d(x_n,x_{n-1})\) instead of \(d(x_1,x_0)\) — conflating the a priori bound (Step 9, uses the very first step) with the a posteriori bound (Step 10, uses the latest step).
Forgetting to verify \(g\) maps the candidate domain into itself before checking the contraction ratio, so the iteration silently leaves the region where the Lipschitz bound was established.
Assuming any \(L \lt 1\) found on a subinterval automatically applies globally, when \(g'\) may exceed 1 elsewhere on the full stated domain \(X\).
Confusing this global, any-starting-point convergence with the local quadratic convergence of Newton's method, then being surprised that a "fixed-point" reformulation of Newton's method converges only linearly and only near the root unless the contraction hypothesis is separately re-verified for it.

Discussion

The result is usually named for Stefan Banach, who proved it in his 1922 thesis in the abstract setting of complete normed (later: metric) spaces, but the underlying iterative idea — successive approximation — long predates that abstraction, appearing in Picard's 1890 existence proof for ordinary differential equations and, in disguised numerical form, in far older root-finding recipes. Banach's contribution was to isolate exactly the right hypotheses (completeness plus a uniform contraction ratio) so that the same three-line argument proves existence, uniqueness, and a constructive convergence rate all at once — a rare combination in existence theorems, most of which give you no algorithm.

In numerical analysis the theorem is less often invoked by its full metric-space name and more often used through its \(\mathbb{R}^n\) or \(C^1\) corollary: to solve \(f(x)=0\), one rewrites it as \(x=g(x)\) for some cleverly chosen \(g\) (there are infinitely many rewritings of the same root equation), and the entire practical question is choosing \(g\) so that \(|g'(x^\ast)|\lt1\) near the root — this is precisely why naive fixed-point iteration on \(x=\cos x\) rewritten badly can diverge while a rescaled version converges rapidly: the algebra of the equation is unchanged, only the contraction constant of the chosen \(g\) changes.

There is a genuine conceptual link to Newton's method: Newton iteration for \(f\) is fixed-point iteration for \(g(x) = x - f(x)/f'(x)\), and one computes \(g'(x^\ast) = f(x^\ast)f''(x^\ast)/f'(x^\ast)^2 = 0\) whenever \(x^\ast\) is a simple root, which is exactly why Newton's method achieves quadratic (not merely the linear rate guaranteed by this theorem) convergence locally — this theorem only certifies that Newton's method converges at all once you are inside a suitable ball, not the faster rate, which needs a separate, sharper Taylor-remainder argument.

Common misconception: that "the map looks like it's shrinking distances" is enough — informally checking a few sample points, or even checking \(|g'|\lt1\) at the fixed point alone (rather than uniformly on the whole relevant domain, or at least on a closed neighbourhood as in the local converse corollary), is not a valid verification of the hypothesis and is the single most common source of false claims of convergence in student work.

Worked examples

\text{Solve } x = \cos x \text{ by fixed-point iteration, } x_0 = 0.5, \; X=[0.5,0.8].

Set \(g(x)=\cos x\). Check self-mapping: \(g\) is decreasing on \([0.5,0.8]\) (since \(\sin x\gt0\) there), with \(g(0.5)=\cos(0.5)\approx0.8776\) and \(g(0.8)=\cos(0.8)\approx0.6967\), both in \([0.5,0.8]\) (in fact in \([0.69,0.88]\subset[0.5,0.9]\); restrict to \(X=[0.5,0.9]\) to be safe), so \(g:X\to X\). A

|g'(x)| = |\!-\!\sin x| \le \sin(0.9) \approx 0.7833 = L \lt 1 \quad \text{on } X=[0.5,0.9].

\(g'(x)=-\sin x\), and \(\sin\) is increasing on \([0.5,0.9]\subset[0,\pi/2]\), so its supremum on \(X\) is at \(x=0.9\); apply the differentiable corollary (Mean Value Theorem) to conclude \(g\) is a contraction with this \(L\). B

x_0=0.5,\; x_1=\cos(0.5)\approx0.8776,\; x_2=\cos(0.8776)\approx0.6390,\; x_3\approx0.8026,\; x_4\approx0.6947,\dots

Direct computation of \(x_{n+1}=g(x_n)\); \([0.5,0.9]\) with the sup-norm is complete (closed subset of \(\mathbb{R}\)), so the theorem applies and guarantees convergence to the unique fixed point in \(X\). A

d(x_4,x^\ast) \le \frac{L}{1-L}\,|x_4-x_3| \approx \frac{0.7833}{0.2167}\times|0.6947-0.8026| \approx 3.615 \times 0.1079 \approx 0.390.

A posteriori bound (Step 10 of the proof), computed purely from the two most recent iterates, no knowledge of \(x^\ast\) required. (The bound is loose here because \(L\) is not small; more iterations or a better-chosen \(g\) tighten it rapidly since the error itself shrinks geometrically even though this particular bound is conservative.) A

x^\ast \approx 0.7391 \text{ (the Dottie number)}, \quad x_n \to x^\ast \text{ geometrically with ratio} \le 0.783

\text{Solve } Ax=b \text{ via Jacobi iteration for } A=\begin{pmatrix}4&1\\1&3\end{pmatrix},\; b=\begin{pmatrix}5\\4\end{pmatrix}.

Write \(A=D+R\) with \(D=\mathrm{diag}(4,3)\); Jacobi rewrites \(Ax=b\) as the fixed-point form \(x = -D^{-1}Rx + D^{-1}b =: g(x)\), i.e. \(g(x)=Bx+c\) with \(B=-D^{-1}R=\begin{pmatrix}0&-1/4\\-1/3&0\end{pmatrix}\), \(c=(5/4,\,4/3)^{\mathsf T}\). A

\(g\) is affine, so it is Lipschitz with constant equal to the induced operator norm \(\|B\|_\infty\) (max absolute row sum), which is exactly the standard sufficient condition for Jacobi convergence: strict diagonal dominance of \(A\) (here \(4\gt1\) and \(3\gt1\)) forces \(\|B\|_\infty\lt1\). B

(\mathbb{R}^2,\|\cdot\|_\infty) \text{ is complete, and } g:\mathbb{R}^2\to\mathbb{R}^2 \text{ trivially, so all hypotheses hold with } L=\tfrac13.

Finite-dimensional normed spaces are always complete (standard fact from analysis: Cauchy sequences converge componentwise by completeness of \(\mathbb{R}\)); \(g\) affine on all of \(\mathbb{R}^2\) is automatically self-mapping. A

x_0=(0,0)^{\mathsf T}: \; x_1=(1.25,\,1.333),\; x_2=(0.9167,\,0.9167),\; x_3=(1.0208,\,1.0278),\dots \to x^\ast=(1,1)^{\mathsf T}.

Iterate \(x_{n+1}=Bx_n+c\); check directly that \(x^\ast=(1,1)\) solves \(Ax=b\): \(4(1)+1(1)=5\), \(1(1)+3(1)=4\). Convergence is guaranteed a priori by the theorem before ever computing a single iterate. A

x^\ast = (1,1)^{\mathsf T}, \quad \|x_n - x^\ast\|_\infty \le \frac{(1/3)^n}{2/3}\,\|x_1-x_0\|_\infty \to 0 \text{ geometrically}

Problems

Let \(X=[1,2]\) and \(g(x) = \frac{1}{2}\left(x + \frac{2}{x}\right)\). Show \(g:X\to X\) and that \(g\) is a contraction on \(X\); find the smallest \(L\) your argument gives.
Solution
\(g'(x) = \frac12 - \frac{1}{x^2}\). On \([1,2]\), \(g'(1)=\frac12-1=-\frac12\) and \(g'(2)=\frac12-\frac14=\frac14\); since \(g'\) is increasing on \([1,2]\) (as \(g''(x)=2/x^3\gt0\)), \(g'\) ranges continuously over \([-\frac12,\frac14]\), so \(|g'(x)|\le\frac12\) for all \(x\in[1,2]\). By the Mean Value Theorem corollary, \(g\) is a contraction with \(L=\frac12\). For self-mapping: \(g\) has a minimum where \(g'(x)=0\), i.e. \(x=\sqrt2\), giving \(g(\sqrt2)=\sqrt2\approx1.414\); endpoints give \(g(1)=1.5\), \(g(2)=1.5\); since \(g\) is decreasing then increasing on \([1,2]\) with minimum \(\sqrt2\gt1\) and both endpoint values \(1.5\le2\), we get \(g(X)\subseteq[\sqrt2,1.5]\subset[1,2]=X\). Both hypotheses hold with \(L=\frac12\). (This is Newton's method for \(\sqrt2\), and indeed \(x^\ast=\sqrt2\).)
For the map in Problem 1, starting from \(x_0=2\), compute \(x_1,x_2\) and use the a priori bound to guarantee \(d(x_4,x^\ast)\) is below what tolerance?
Solution
\(x_1 = \frac12(2+1)=1.5\). \(x_2=\frac12(1.5+2/1.5)=\frac12(1.5+1.3333)=1.41667\). A priori bound: \(d(x_n,x^\ast)\le \frac{L^n}{1-L}d(x_1,x_0)\), with \(L=\frac12\), \(d(x_1,x_0)=|1.5-2|=0.5\). For \(n=4\): \(\frac{(1/2)^4}{1/2}\times0.5 = \frac{1/16}{1/2}\times0.5 = \frac18\times0.5=0.0625\). So \(d(x_4,x^\ast)\le0.0625\) is guaranteed (in fact actual convergence is much faster here since this is Newton's method with quadratic local rate, but the theorem only certifies the linear bound).
Show that \(g(x) = 4x(1-x)\) (the logistic map at parameter 4) is not a contraction on \(X=[0,1]\), and explain why this does not by itself prove the iteration diverges.
Solution
\(g'(x)=4-8x\), so \(|g'(x)|\) ranges up to \(|g'(0)|=4\) and \(|g'(1)|=4\), far exceeding 1; more directly, take \(x=0.01,\,y=0\): \(g(x)-g(y) = 4(0.01)(0.99) - 0 = 0.0396\), while \(x-y=0.01\), giving ratio \(3.96\), and this ratio is unbounded as points near \(x=0,1\) are chosen closer together with one fixed at an endpoint... more carefully, near \(x=1/2\) also \(g'(1/2)=0\), so the ratio is not uniformly bounded below 1 anywhere on \([0,1]\); hence \(g\) fails the contraction hypothesis on all of \(X\). This does not prove divergence because the theorem only gives a sufficient condition for convergence (see Corollaries: converse fails); indeed the logistic map at parameter 4 is a classical example of chaotic, non-convergent behaviour for generic starting points, consistent with the hypothesis failing, but the failure of the hypothesis alone is not a proof of chaos — that requires separate argument (e.g. conjugacy to the tent map).
Let \(X=\mathbb{R}^2\) with the Euclidean norm and \(g(x) = Bx\) where \(B\) is a \(2\times2\) matrix. State a necessary and sufficient condition on \(B\) for \(g\) to be a contraction on all of \(X\) in the \(\ell^2\) operator norm, and relate it to the theorem's conclusion for solving \(x=Bx+c\).
Solution
\(g\) is a contraction in the \(\ell^2\)-induced norm iff \(\|B\|_2 \lt 1\), where \(\|B\|_2\) is the largest singular value of \(B\) (equivalently \(\sqrt{\lambda_{\max}(B^{\mathsf T}B)}\)); this is because \(d(g(x),g(y))=\|B(x-y)\|_2 \le \|B\|_2\|x-y\|_2\) with equality attained for some direction, so \(\|B\|_2\) is exactly the smallest valid Lipschitz constant. Since \((\mathbb{R}^2,\|\cdot\|_2)\) is complete and \(g\) is affine hence self-mapping on all of \(\mathbb{R}^2\), the theorem then guarantees: a unique \(x^\ast\) with \(x^\ast=Bx^\ast+c\) (equivalently \((I-B)x^\ast=c\) has a unique solution, consistent with \(1\) not being an eigenvalue of \(B\) since \(\|B\|_2\lt1\) bounds all eigenvalues' moduli below 1), and that \(x_n=B^nx_0 + (\text{terms in }c) \to x^\ast\) from every starting point, geometrically with ratio \(\|B\|_2\).
Construct an explicit example of a map \(g\) on a complete space \(X\) that has a unique fixed point which globally attracts every orbit, yet \(g\) is not a contraction with respect to any metric giving the same convergent behaviour on all of \(X\) — i.e. verify the converse genuinely fails, and explain what a numerical analyst should do given only convergent-looking numerics with no such certificate.
Solution
Take \(X=[0,\infty)\) (complete, closed subset of \(\mathbb{R}\)) and \(g(x) = \frac{x}{1+x}\). Then \(g(0)=0\) is the unique fixed point (solve \(x=x/(1+x) \Rightarrow x(1+x)=x \Rightarrow x^2=0 \Rightarrow x=0\)), and for any \(x_0\gt0\), \(x_{n+1}=x_n/(1+x_n) \lt x_n\), a decreasing sequence bounded below by \(0\), hence convergent, and one checks the limit \(\ell\) satisfies \(\ell=\ell/(1+\ell)\Rightarrow\ell=0\); so every orbit converges to \(0\). But \(g'(x) = 1/(1+x)^2\), which \(\to1\) as \(x\to0^+\), so \(\sup_{x\in X}|g'(x)|=1\) (approached, not exceeded, but not bounded below 1 either), meaning no \(L\lt1\) works uniformly on all of \(X\) — indeed \(d(g(x),g(0))/d(x,0) = 1/(1+x) \to 1\) as \(x\to0\), so the ratio gets arbitrarily close to \(1\) near the fixed point itself, the theorem's hypothesis genuinely fails. Given only numerical evidence that an iteration "looks convergent," an analyst cannot conclude a contraction certificate exists; the correct response is either to prove convergence by a different route (here, monotone-bounded-sequence argument, or Lyapunov-type methods) or restrict to a subdomain bounded away from the point where the ratio approaches 1 (e.g. \(X'=[\delta,\infty)\) for fixed \(\delta\gt0\), on which \(\sup|g'|=1/(1+\delta)^2\lt1\) does give a valid local contraction certificate).