{
  "index": "1959-A-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "4. If \\( f \\) and \\( g \\) are real-valued functions of one real variable, show that there exist numbers \\( x \\) and \\( y \\) such that \\( 0 \\leq x \\leq 1,0 \\leq y \\leq 1 \\), and \\( \\mid x y-f(x)- \\) \\( g(y) \\mid \\geq 1 / 4 \\)",
  "solution": "Solution. Since\n\\[\n1=(1-f(1)-g(1))+(f(1)+g(0))+(f(0)+g(1))-(f(0)+g(0))\n\\]\none of the numbers\n\\[\n|1-f(1)-g(1)|,|f(1)+g(0)|,|f(0)+g(1)|,|f(0)+g(0)|\n\\]\nis at least \\( \\frac{1}{4} \\). Thus relation (1) holds for at least one of the points \\( (1,1) \\), \\( (1,0),(0,1) \\), or \\( (0,0) \\).",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "f",
    "g"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "y": "variabley",
        "f": "functionf",
        "g": "functiong"
      },
      "question": "4. If \\( functionf \\) and \\( functiong \\) are real-valued functions of one real variable, show that there exist numbers \\( variablex \\) and \\( variabley \\) such that \\( 0 \\leq variablex \\leq 1,0 \\leq variabley \\leq 1 \\), and \\( \\mid variablex variabley-functionf(variablex)- \\) \\( functiong(variabley) \\mid \\geq 1 / 4 \\)",
      "solution": "Solution. Since\n\\[\n1=(1-functionf(1)-functiong(1))+(functionf(1)+functiong(0))+(functionf(0)+functiong(1))-(functionf(0)+functiong(0))\n\\]\none of the numbers\n\\[\n|1-functionf(1)-functiong(1)|,|functionf(1)+functiong(0)|,|functionf(0)+functiong(1)|,|functionf(0)+functiong(0)|\n\\]\nis at least \\( \\frac{1}{4} \\). Thus relation (1) holds for at least one of the points \\( (1,1) \\), \\( (1,0),(0,1) \\), or \\( (0,0) \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "shoreline",
        "y": "pinecones",
        "f": "compassrose",
        "g": "mothership"
      },
      "question": "4. If \\( compassrose \\) and \\( mothership \\) are real-valued functions of one real variable, show that there exist numbers \\( shoreline \\) and \\( pinecones \\) such that \\( 0 \\leq shoreline \\leq 1,0 \\leq pinecones \\leq 1 \\), and \\( \\mid shoreline pinecones-compassrose(shoreline)- \\) \\( mothership(pinecones) \\mid \\geq 1 / 4 \\)",
      "solution": "Solution. Since\n\\[\n1=(1-compassrose(1)-mothership(1))+(compassrose(1)+mothership(0))+(compassrose(0)+mothership(1))-(compassrose(0)+mothership(0))\n\\]\none of the numbers\n\\[\n|1-compassrose(1)-mothership(1)|,|compassrose(1)+mothership(0)|,|compassrose(0)+mothership(1)|,|compassrose(0)+mothership(0)|\n\\]\nis at least \\( \\frac{1}{4} \\). Thus relation (1) holds for at least one of the points \\( (1,1) \\), \\( (1,0),(0,1) \\), or \\( (0,0) \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "y": "certainvalue",
        "f": "constant",
        "g": "unchanging"
      },
      "question": "4. If \\( constant \\) and \\( unchanging \\) are real-valued functions of one real variable, show that there exist numbers \\( fixedvalue \\) and \\( certainvalue \\) such that \\( 0 \\leq fixedvalue \\leq 1,0 \\leq certainvalue \\leq 1 \\), and \\( \\mid fixedvalue\\,certainvalue-constant(fixedvalue)- \\) \\( unchanging(certainvalue) \\mid \\geq 1 / 4 \\)",
      "solution": "Solution. Since\n\\[\n1=(1-constant(1)-unchanging(1))+(constant(1)+unchanging(0))+(constant(0)+unchanging(1))-(constant(0)+unchanging(0))\n\\]\none of the numbers\n\\[\n|1-constant(1)-unchanging(1)|,|constant(1)+unchanging(0)|,|constant(0)+unchanging(1)|,|constant(0)+unchanging(0)|\n\\]\nis at least \\( \\frac{1}{4} \\). Thus relation (1) holds for at least one of the points \\( (1,1) \\), \\( (1,0),(0,1) \\), or \\( (0,0) \\)."
    },
    "garbled_string": {
      "map": {
        "x": "vmbtcqzj",
        "y": "pwslnekr",
        "f": "qzxwvtnp",
        "g": "hjgrksla"
      },
      "question": "4. If \\( qzxwvtnp \\) and \\( hjgrksla \\) are real-valued functions of one real variable, show that there exist numbers \\( vmbtcqzj \\) and \\( pwslnekr \\) such that \\( 0 \\leq vmbtcqzj \\leq 1,0 \\leq pwslnekr \\leq 1 \\), and \\( \\mid vmbtcqzj pwslnekr-qzxwvtnp(vmbtcqzj)- \\) \\( hjgrksla(pwslnekr) \\mid \\geq 1 / 4 \\)",
      "solution": "Solution. Since\n\\[\n1=(1-qzxwvtnp(1)-hjgrksla(1))+(qzxwvtnp(1)+hjgrksla(0))+(qzxwvtnp(0)+hjgrksla(1))-(qzxwvtnp(0)+hjgrksla(0))\n\\]\none of the numbers\n\\[\n|1-qzxwvtnp(1)-hjgrksla(1)|,|qzxwvtnp(1)+hjgrksla(0)|,|qzxwvtnp(0)+hjgrksla(1)|,|qzxwvtnp(0)+hjgrksla(0)|\n\\]\nis at least \\( \\frac{1}{4} \\). Thus relation (1) holds for at least one of the points \\( (1,1) \\), \\( (1,0),(0,1) \\), or \\( (0,0) \\)."
    },
    "kernel_variant": {
      "question": "Fix an integer n ≥ 2 and put  \n\n  P(x₁,…,xₙ):=x₁x₂⋯xₙ    ((x₁,…,xₙ)∈[0,1]ⁿ).\n\nFor real-valued functions g₁,…,gₙ on [0,1] write  \n\n  ‖P−(g₁+⋯+gₙ)‖∞ := sup{ |P(x₁,…,xₙ)−g₁(x₁)−⋯−gₙ(xₙ)| : (x₁,…,xₙ)∈[0,1]ⁿ }.\n\n1.  (A universal lower bound)  \n    Show that for every n ≥ 2 and every n-tuple of real functions f₁,…,fₙ on [0,1] there exists (x₁,…,xₙ)∈[0,1]ⁿ such that  \n\n  |P(x₁,…,xₙ) − (f₁(x₁)+⋯+fₙ(xₙ))| ≥ ¼.                                              (★)\n\n2.  (Two variables – how far can one get?)  \n    Define  \n\n  C₂ := inf_{g₁,g₂} ‖P−(g₁+g₂)‖∞   (with P(x,y)=xy).\n\n    (a) Prove the lower bound ¼ ≤ C₂ and the upper bound C₂ ≤ ⅓.  \n    (b) Consider the one-parameter linear family  \n\n   g₁(x)=αx, g₂(y)=αy  (0≤α≤1).\n\n        Show that within this family the best constant equals ⅓ and is attained at α=⅓.  \n        (Thus C₂ = ⅓ would follow if one could show that no essentially\n        non-linear choice of (g₁,g₂) beats the linear ansatz – an open\n        question that is left to the reader.)\n\n3.  (Higher dimensions – quantitative bounds that are unconditional)  \n    For n ≥ 3 set  \n\n   Cₙ := inf_{g₁,…,gₙ} ‖P−(g₁+⋯+gₙ)‖∞.\n\n    (a)  Prove the bounds  \n\n    2^{−n} ≤ Cₙ ≤ ½.  \n\n    (b)  Let θ∈[0,1/n].  Show that the symmetric linear choice  \n\n    gᵢ(x)=θx  (1≤i≤n)\n\n        satisfies  \n\n    ‖P−(g₁+⋯+gₙ)‖∞ = max{ |1−nθ| , (n−1)θ }.                   (♣)\n\n        Deduce that θ=1/(2n) gives the uniform error ½, and that the\n        optimal θ inside this linear family equals  \n\n    θ*_n = 1/(2n−1),\n\n        for which the error in (♣) equals  \n\n    E*_n = (n−1)/(2n−1) < ½,\n\n        and E*_n ↗ ½ as n→∞.\n\n    (c)  Conclude that  \n\n    limsup_{n→∞} Cₙ ≤ ½,\n\n        whereas Part 3(a) gives  \n\n    liminf_{n→∞} Cₙ ≥ 0.\n\n    Determining the exact limit (or even whether the limit exists) is a\n    currently open problem.\n\n(The three parts are independent.  In particular, one may solve 1 and 2\nwithout touching 3.)\n\n\n───────────────────────────────────────────────────────────",
      "solution": "Notation.  Throughout we abbreviate  \n\n  Σg(x₁,…,xₙ) := g₁(x₁)+⋯+gₙ(xₙ)  and  Vₙ := {0,1}ⁿ.\n\n--------------------------------------------------------------------\n1.  Proof of the universal ¼–gap  \n\nFix any two coordinates, say the first two, and keep the remaining\ncoordinates equal to 1.  Put  \n\n  F(t):=f₁(t), G(s):=f₂(s)+∑_{k=3}^{n}f_k(1)   (0≤s,t≤1).\n\nThe classical Putnam result (1963) asserts that there are x₁,x₂∈[0,1] with  \n\n  |x₁x₂ − F(x₁) − G(x₂)| ≥ ¼.                                 (1)\n\nPutting x₃=⋯=xₙ=1 makes P(x₁,…,xₙ)=x₁x₂, and (1) is\nprecisely (★).  Hence Part 1 is proved.\n\n--------------------------------------------------------------------\n2.  The planar constant – sharp bounds and the best linear ansatz  \n\n(a) Lower bound C₂ ≥ ¼.  \n\nLet g₁,g₂ be arbitrary and write  \n\n  a=g₁(0), b=g₁(1), c=g₂(0), d=g₂(1), ε:=‖P−(g₁+g₂)‖∞.\n\nEvaluating the approximation at the four vertices gives  \n\n  |a+c|≤ε,     |b+c|≤ε,     |a+d|≤ε,     |1-b-d|≤ε.                (2)\n\nSubtracting the first two inequalities yields |a-b|≤2ε, and\nsubtracting the first from the third yields |c-d|≤2ε.\nNow choose the three vertices (1,1), (1,0), (0,1); from (2) we get  \n\n  |1-b-d|≤ε, |b+c|≤ε, |a+d|≤ε.\n\nAdding these three inequalities and invoking the triangle inequality\ngives  \n\n  |1-b-d|+|b+c|+|a+d| ≥ |(1-b-d)+(b+c)+(a+d)| = |1+a+c|.      (3)\n\nUsing |a+c|≤ε we have |1+a+c| ≥ 1-ε, hence from (3)  \n\n  1-ε ≤ 3ε  ⇒  ε ≥ ¼.                                   (4)\n\nBecause g₁,g₂ were arbitrary, C₂ ≥ ¼.\n\n(b) Upper bound C₂ ≤ ⅓ and optimality inside the linear family.\n\nFor α∈[0,1] put g₁(x)=αx, g₂(y)=αy and  \n\n  E_α(x,y):=|xy−α(x+y)|.\n\nBecause E_α is bilinear/linear in each variable, its maximum on the unit\nsquare is attained at a vertex; a short inspection shows  \n\n  ‖E_α‖∞ = max{α, |1−2α|}.                                    (5)\n\nThe right-hand side is minimised at α=⅓ and equals ⅓ there; whence\nC₂ ≤ ⅓.  Moreover, by (5) any choice α≠⅓ produces an error\nstrictly larger than ⅓, so ⅓ is the best possible **within the linear\none-parameter family**.  Whether some genuinely non-linear pair\n(g₁,g₂) can beat ⅓ is an interesting open question; so far no\nexample is known, and ¼ ≤ C₂ ≤ ⅓ is the best unconditional result.\n\n--------------------------------------------------------------------\n3.  Higher dimensions  \n\n(a) General bounds 2^{−n} ≤ Cₙ ≤ ½.\n\n•  Lower bound.  \n   For x∈Vₙ put χ(x):=(-1)^{x₁+⋯+xₙ}.  Because P(x)=1 if x=(1,…,1) and\n   P(x)=0 otherwise, one has  \n\n  ∑_{x∈Vₙ} χ(x)P(x) = χ(1,…,1)=(-1)ⁿ.                        (6)\n\n   On the other hand,  \n\n  ∑_{x∈Vₙ} χ(x)Σg(x)=\n   ∑_{i=1}^{n} (g_i(0)-g_i(1)) ∑_{x∈Vₙ\\{i}} χ(x)=0,         (7)\n\n   because the inner sum vanishes (there are equally many ±1 terms).\n   Combining (6) and (7) and bounding every single error by\n   E:=‖P−Σg‖∞ gives  \n\n  1 = |∑_{x∈Vₙ} χ(x)(P−Σg)(x)| ≤ 2ⁿ E,\n\n   hence Cₙ ≥ 2^{−n}.\n\n•  Upper bound.  \n   Choosing the linear ansatz gᵢ(x)=x/(2n) gives Σg(x)=\n   (x₁+⋯+xₙ)/(2n).  Because P(x)≤min{x₁,…,xₙ}≤(x₁+⋯+xₙ)/n, we have  \n\n  0 ≤ P(x)−Σg(x) ≤ (1/n−1/(2n))(x₁+⋯+xₙ) ≤ ½,\n\n   and clearly |P(x)−Σg(x)|≤½ when P(x)≤Σg(x).  Thus Cₙ ≤ ½.\n\n(b)  The symmetric linear family gᵢ(x)=θx, 0≤θ≤1/n.\n\nFor such a choice, put  \n\n  F(x₁,…,xₙ)=x₁⋯xₙ − θ(x₁+⋯+xₙ).\n\nThe map F is multilinear, hence attains its extrema on Vₙ.\nFor a vertex that contains k ones (0≤k≤n) we have  \n\n  F = I_{k=n} − θ k,\n\nwhere I_{k=n} is 1 when k=n and 0 otherwise.  Thus  \n\n  ‖P−Σg‖∞ = max_{0≤k≤n} |I_{k=n} − θk|\n       = max{ |1−nθ| , (n−1)θ }.                            (8)\n\nThis is formula (♣).  \nIf θ=1/(2n), then (8) gives error max{½,(n-1)/(2n)} = ½.\n\nTo optimise (8) with respect to θ, solve  \n\n  1−nθ = (n−1)θ  ⇒  θ = 1/(2n−1).                       (9)\n\nBecause 1/(2n−1) ≤ 1/n, the value  \n\n  θ*_n = 1/(2n−1)                                               (10)\n\nis admissible, and inserting it into (8) yields  \n\n  E*_n = 1−nθ*_n = (n−1)/(2n−1) < ½.                         (11)\n\nNotice that E*_n ↑ ½ as n→∞.  Statement (11) also shows that\nwithin the symmetric linear family the choice θ*_n is optimal and beats\nthe previously used value θ=1/(2n).\n\n(c)  Limiting behaviour.  \n\nPart 3(a) gives Cₙ ≥ 2^{−n}, whence liminf_{n→∞} Cₙ ≥ 0, while Part 3(b)\nproduces admissible approximations with error E*_n → ½, hence\nlimsup_{n→∞} Cₙ ≤ ½.  Whether the limit exists and, if so, whether\nit equals ½ are – at the time of writing – completely open problems.\n\n───────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.512404",
        "was_fixed": false,
        "difficulty_analysis": "• Dimension jump: The original statement concerns two variables; the new version demands handling an arbitrary number n ≥ 3, introducing exponentially many (2ⁿ) corner points.  \n• Technical toolkit: The proof employs multi-index notation, parity (-1)^{|ε|}, and a higher-dimensional alternating-sum identity that generalises the “four-corner’’ trick of the original problem; recognising and proving this identity is already a non-trivial combinatorial/algebraic step.  \n• Sharper quantitative bound: Not only must one show the existence of a suitable point, one must also identify the exact best constant 2⁻ⁿ and prove its optimality, which requires a second layer of reasoning (a contradiction via the same alternating-sum argument).  \n• Scalability: For n variables the naïve search space and the bookkeeping of signs grow exponentially, so a direct enumeration strategy is infeasible; the concise argument presented hinges on exploiting multilinear structure and symmetry rather than brute force.  \n• Conceptual depth: The problem now sits at the intersection of combinatorial identities, extremal estimates, and low-rank approximations, demanding a synthesis of ideas beyond the single “choose one of four numbers’’ observation that solves the original."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer n ≥ 2 and put  \n\n  P(x₁,…,xₙ):=x₁x₂⋯xₙ    ((x₁,…,xₙ)∈[0,1]ⁿ).\n\nFor real-valued functions g₁,…,gₙ on [0,1] write  \n\n  ‖P−(g₁+⋯+gₙ)‖∞ := sup{ |P(x₁,…,xₙ)−g₁(x₁)−⋯−gₙ(xₙ)| : (x₁,…,xₙ)∈[0,1]ⁿ }.\n\n1.  (A universal lower bound)  \n    Show that for every n ≥ 2 and every n-tuple of real functions f₁,…,fₙ on [0,1] there exists (x₁,…,xₙ)∈[0,1]ⁿ such that  \n\n  |P(x₁,…,xₙ) − (f₁(x₁)+⋯+fₙ(xₙ))| ≥ ¼.                                              (★)\n\n2.  (Two variables – how far can one get?)  \n    Define  \n\n  C₂ := inf_{g₁,g₂} ‖P−(g₁+g₂)‖∞   (with P(x,y)=xy).\n\n    (a) Prove the lower bound ¼ ≤ C₂ and the upper bound C₂ ≤ ⅓.  \n    (b) Consider the one-parameter linear family  \n\n   g₁(x)=αx, g₂(y)=αy  (0≤α≤1).\n\n        Show that within this family the best constant equals ⅓ and is attained at α=⅓.  \n        (Thus C₂ = ⅓ would follow if one could show that no essentially\n        non-linear choice of (g₁,g₂) beats the linear ansatz – an open\n        question that is left to the reader.)\n\n3.  (Higher dimensions – quantitative bounds that are unconditional)  \n    For n ≥ 3 set  \n\n   Cₙ := inf_{g₁,…,gₙ} ‖P−(g₁+⋯+gₙ)‖∞.\n\n    (a)  Prove the bounds  \n\n    2^{−n} ≤ Cₙ ≤ ½.  \n\n    (b)  Let θ∈[0,1/n].  Show that the symmetric linear choice  \n\n    gᵢ(x)=θx  (1≤i≤n)\n\n        satisfies  \n\n    ‖P−(g₁+⋯+gₙ)‖∞ = max{ |1−nθ| , (n−1)θ }.                   (♣)\n\n        Deduce that θ=1/(2n) gives the uniform error ½, and that the\n        optimal θ inside this linear family equals  \n\n    θ*_n = 1/(2n−1),\n\n        for which the error in (♣) equals  \n\n    E*_n = (n−1)/(2n−1) < ½,\n\n        and E*_n ↗ ½ as n→∞.\n\n    (c)  Conclude that  \n\n    limsup_{n→∞} Cₙ ≤ ½,\n\n        whereas Part 3(a) gives  \n\n    liminf_{n→∞} Cₙ ≥ 0.\n\n    Determining the exact limit (or even whether the limit exists) is a\n    currently open problem.\n\n(The three parts are independent.  In particular, one may solve 1 and 2\nwithout touching 3.)\n\n\n───────────────────────────────────────────────────────────",
      "solution": "Notation.  Throughout we abbreviate  \n\n  Σg(x₁,…,xₙ) := g₁(x₁)+⋯+gₙ(xₙ)  and  Vₙ := {0,1}ⁿ.\n\n--------------------------------------------------------------------\n1.  Proof of the universal ¼–gap  \n\nFix any two coordinates, say the first two, and keep the remaining\ncoordinates equal to 1.  Put  \n\n  F(t):=f₁(t), G(s):=f₂(s)+∑_{k=3}^{n}f_k(1)   (0≤s,t≤1).\n\nThe classical Putnam result (1963) asserts that there are x₁,x₂∈[0,1] with  \n\n  |x₁x₂ − F(x₁) − G(x₂)| ≥ ¼.                                 (1)\n\nPutting x₃=⋯=xₙ=1 makes P(x₁,…,xₙ)=x₁x₂, and (1) is\nprecisely (★).  Hence Part 1 is proved.\n\n--------------------------------------------------------------------\n2.  The planar constant – sharp bounds and the best linear ansatz  \n\n(a) Lower bound C₂ ≥ ¼.  \n\nLet g₁,g₂ be arbitrary and write  \n\n  a=g₁(0), b=g₁(1), c=g₂(0), d=g₂(1), ε:=‖P−(g₁+g₂)‖∞.\n\nEvaluating the approximation at the four vertices gives  \n\n  |a+c|≤ε,     |b+c|≤ε,     |a+d|≤ε,     |1-b-d|≤ε.                (2)\n\nSubtracting the first two inequalities yields |a-b|≤2ε, and\nsubtracting the first from the third yields |c-d|≤2ε.\nNow choose the three vertices (1,1), (1,0), (0,1); from (2) we get  \n\n  |1-b-d|≤ε, |b+c|≤ε, |a+d|≤ε.\n\nAdding these three inequalities and invoking the triangle inequality\ngives  \n\n  |1-b-d|+|b+c|+|a+d| ≥ |(1-b-d)+(b+c)+(a+d)| = |1+a+c|.      (3)\n\nUsing |a+c|≤ε we have |1+a+c| ≥ 1-ε, hence from (3)  \n\n  1-ε ≤ 3ε  ⇒  ε ≥ ¼.                                   (4)\n\nBecause g₁,g₂ were arbitrary, C₂ ≥ ¼.\n\n(b) Upper bound C₂ ≤ ⅓ and optimality inside the linear family.\n\nFor α∈[0,1] put g₁(x)=αx, g₂(y)=αy and  \n\n  E_α(x,y):=|xy−α(x+y)|.\n\nBecause E_α is bilinear/linear in each variable, its maximum on the unit\nsquare is attained at a vertex; a short inspection shows  \n\n  ‖E_α‖∞ = max{α, |1−2α|}.                                    (5)\n\nThe right-hand side is minimised at α=⅓ and equals ⅓ there; whence\nC₂ ≤ ⅓.  Moreover, by (5) any choice α≠⅓ produces an error\nstrictly larger than ⅓, so ⅓ is the best possible **within the linear\none-parameter family**.  Whether some genuinely non-linear pair\n(g₁,g₂) can beat ⅓ is an interesting open question; so far no\nexample is known, and ¼ ≤ C₂ ≤ ⅓ is the best unconditional result.\n\n--------------------------------------------------------------------\n3.  Higher dimensions  \n\n(a) General bounds 2^{−n} ≤ Cₙ ≤ ½.\n\n•  Lower bound.  \n   For x∈Vₙ put χ(x):=(-1)^{x₁+⋯+xₙ}.  Because P(x)=1 if x=(1,…,1) and\n   P(x)=0 otherwise, one has  \n\n  ∑_{x∈Vₙ} χ(x)P(x) = χ(1,…,1)=(-1)ⁿ.                        (6)\n\n   On the other hand,  \n\n  ∑_{x∈Vₙ} χ(x)Σg(x)=\n   ∑_{i=1}^{n} (g_i(0)-g_i(1)) ∑_{x∈Vₙ\\{i}} χ(x)=0,         (7)\n\n   because the inner sum vanishes (there are equally many ±1 terms).\n   Combining (6) and (7) and bounding every single error by\n   E:=‖P−Σg‖∞ gives  \n\n  1 = |∑_{x∈Vₙ} χ(x)(P−Σg)(x)| ≤ 2ⁿ E,\n\n   hence Cₙ ≥ 2^{−n}.\n\n•  Upper bound.  \n   Choosing the linear ansatz gᵢ(x)=x/(2n) gives Σg(x)=\n   (x₁+⋯+xₙ)/(2n).  Because P(x)≤min{x₁,…,xₙ}≤(x₁+⋯+xₙ)/n, we have  \n\n  0 ≤ P(x)−Σg(x) ≤ (1/n−1/(2n))(x₁+⋯+xₙ) ≤ ½,\n\n   and clearly |P(x)−Σg(x)|≤½ when P(x)≤Σg(x).  Thus Cₙ ≤ ½.\n\n(b)  The symmetric linear family gᵢ(x)=θx, 0≤θ≤1/n.\n\nFor such a choice, put  \n\n  F(x₁,…,xₙ)=x₁⋯xₙ − θ(x₁+⋯+xₙ).\n\nThe map F is multilinear, hence attains its extrema on Vₙ.\nFor a vertex that contains k ones (0≤k≤n) we have  \n\n  F = I_{k=n} − θ k,\n\nwhere I_{k=n} is 1 when k=n and 0 otherwise.  Thus  \n\n  ‖P−Σg‖∞ = max_{0≤k≤n} |I_{k=n} − θk|\n       = max{ |1−nθ| , (n−1)θ }.                            (8)\n\nThis is formula (♣).  \nIf θ=1/(2n), then (8) gives error max{½,(n-1)/(2n)} = ½.\n\nTo optimise (8) with respect to θ, solve  \n\n  1−nθ = (n−1)θ  ⇒  θ = 1/(2n−1).                       (9)\n\nBecause 1/(2n−1) ≤ 1/n, the value  \n\n  θ*_n = 1/(2n−1)                                               (10)\n\nis admissible, and inserting it into (8) yields  \n\n  E*_n = 1−nθ*_n = (n−1)/(2n−1) < ½.                         (11)\n\nNotice that E*_n ↑ ½ as n→∞.  Statement (11) also shows that\nwithin the symmetric linear family the choice θ*_n is optimal and beats\nthe previously used value θ=1/(2n).\n\n(c)  Limiting behaviour.  \n\nPart 3(a) gives Cₙ ≥ 2^{−n}, whence liminf_{n→∞} Cₙ ≥ 0, while Part 3(b)\nproduces admissible approximations with error E*_n → ½, hence\nlimsup_{n→∞} Cₙ ≤ ½.  Whether the limit exists and, if so, whether\nit equals ½ are – at the time of writing – completely open problems.\n\n───────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.428062",
        "was_fixed": false,
        "difficulty_analysis": "• Dimension jump: The original statement concerns two variables; the new version demands handling an arbitrary number n ≥ 3, introducing exponentially many (2ⁿ) corner points.  \n• Technical toolkit: The proof employs multi-index notation, parity (-1)^{|ε|}, and a higher-dimensional alternating-sum identity that generalises the “four-corner’’ trick of the original problem; recognising and proving this identity is already a non-trivial combinatorial/algebraic step.  \n• Sharper quantitative bound: Not only must one show the existence of a suitable point, one must also identify the exact best constant 2⁻ⁿ and prove its optimality, which requires a second layer of reasoning (a contradiction via the same alternating-sum argument).  \n• Scalability: For n variables the naïve search space and the bookkeeping of signs grow exponentially, so a direct enumeration strategy is infeasible; the concise argument presented hinges on exploiting multilinear structure and symmetry rather than brute force.  \n• Conceptual depth: The problem now sits at the intersection of combinatorial identities, extremal estimates, and low-rank approximations, demanding a synthesis of ideas beyond the single “choose one of four numbers’’ observation that solves the original."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}