Exact supremum error on a three-point domain

Goal and scope

Prove g_n(a)=1/(n+2),g_n(b)=-4/(n+2),g_n(c)=2/(n+2) converges uniformly to zero.

A complete finite domain and rational tails for all n≥0; N must not depend on x.

Givens and symbols

• algebra
• epsilon
• definition_here

• G1 — D={a,b,c}, g=0; the three formulas define all of g_n:D to R for every n≥0.

• E_n — whole-domain supremum error
• N — common tail index

Rules and sources

Some rule statements have reader clarifications in this revision: they are paraphrases, not verbatim quotations. Referenced original source bytes and hashes are unchanged.

• mko_definition · cited — Use a finite domain and symbolic rational tails. The maximum of all pointwise witnesses gives a uniform witness on this complete finite domain, not on arbitrary infinite domains. Check exact supremum gaps and the all-n tail; a finite prefix is insufficient. · mko-uniform-convergence-of-functions
• algebra · prerequisite — Use real-field identities; division requires a nonzero denominator and multiplication by a positive number preserves inequalities.
• epsilon · prerequisite — Convergence requires for every positive epsilon an integer N controlling every n≥N by a strict error bound; use the Archimedean property of the reals.
• definition_here · prerequisite — A nonempty finite set of real numbers has its maximum as its supremum; uniform convergence requires, for each positive epsilon, one common N making the error smaller than epsilon for every n≥N and every x.

Main derivation

n\ge0

n+2>0

(|g_n(a)|,|g_n(b)|,|g_n(c)|)

(1,4,2)/(n+2)

E_n=\sup_{x\in D}|g_n(x)-g(x)|

E_n=\max\{1,4,2\}/(n+2)=4/(n+2)

\varepsilon>0,\quad N=\lfloor4/\varepsilon\rfloor+1

N>4/\varepsilon

n\ge N

|g_n(x)|\le E_n=4/(n+2)<\varepsilon\quad\forall x\in D

\forall\varepsilon>0\ \exists N\ \forall n\ge N\ \forall x\in D

g_n\to0\text{ uniformly on }D

N_x=\lfloor |c_x|/\varepsilon\rfloor+1,\quad(c_a,c_b,c_c)=(1,-4,2)

\max_{x\in D}N_x=\lfloor4/\varepsilon\rfloor+1=N

\varepsilon=1/10,\quad N=41

E_n\le4/43<1/10\quad(n\ge41)

Substeps and lemmas

No substeps in this guide.

Reverse audits

• S08 · premise_audit · S04, S05 — This numerical audit does not replace the proof for every epsilon.
• S07 · premise_audit · G1 — The maximum exists because the entire domain is finite.
• S06 · premise_audit · S05 — Choosing N only after x would not prove general uniform convergence.
• S05 · premise_audit · S03, S04 — The statement is for all n≥N, not only n=N.
• S04 · premise_audit · S03 — The formula for N contains no x.
• S03 · premise_audit · S02 — No sampled maximum is used to estimate an unknown infinite-domain supremum.
• S02 · inverse_equivalence · G1, S01 — The maximum signed value is not the maximum absolute error.
• S01 · premise_audit · G1 — The common index and domain remain fixed.

Error controls and negative controls

Absolute values first

List absolute errors before taking their maximum.

Common index

N depends only on epsilon and controls every later n.

• The supremum error is 2/(n+2). → reject — It misses b's absolute error.
• N may depend on x with no common bound. → reject — That is only the pointwise quantifier order.
• A finite-grid supremum must equal the supremum over the whole continuous interval. → reject — For h(x)=x*(1-x), grid {0,1} has maximum zero, while x=1/2 in [0,1] has value 1/4.

Result and limitations

The supremum error is exactly 4/(n+2); common N=floor(4/epsilon)+1 proves uniform convergence.

Verification and source binding

1 declared checks for this guide; only the listed expressions are covered.

• source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
• adopted source SHA-256: 1a3da85dbb40310b278c7800673a12807c5d786870d07ca8dc757999c5f695d4
• original source SHA-256: 729e76f868ad91fc9a80212ab0f92d3ee8fb3af4d052e2d03b2201388c4fd253
• Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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