{
  "index": "1954-A-5",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "5. If \\( f(x) \\) is a real-valued function defined for \\( 0<x<1 \\), then the formula \\( f(x)=o(x) \\) is an abbreviation for the statement that\n\\[\n\\frac{f(x)}{x} \\rightarrow 0 \\quad \\text { as } x \\rightarrow 0\n\\]\n\nKeeping this in mind, prove the following: if\n\\[\n\\lim _{x \\rightarrow 0} f(x)=0 \\text { and } f(x)-f\\left(\\frac{x}{2}\\right)=o(x)\n\\]\nthen \\( f(x)=o(x) \\).",
  "solution": "Solution. Let \\( \\epsilon>0 \\) be given. Choose \\( \\delta \\) so that for all \\( x \\) satisfying \\( 0<x<\\delta \\)\n\\[\n\\left|\\frac{1}{x}[f(x)-f(x / 2)]\\right|<\\frac{1}{2} \\epsilon .\n\\]\n\nNow fix \\( y, 0<y<\\delta \\). Then\n\\[\n\\begin{array}{l} \nf(y)=\\left[f(y)-f\\left(\\frac{y}{2}\\right)\\right]+\\left[f\\left(\\frac{y}{2}\\right)-f\\left(\\frac{y}{4}\\right)\\right] \\\\\n+\\cdots+\\left[f\\left(\\frac{y}{2^{n-1}}\\right)-f\\left(\\frac{y}{2^{n}}\\right)\\right]+f\\left(\\frac{y}{2^{n}}\\right)\n\\end{array}\n\\]\n\nSo\n\\[\n\\begin{array}{l}\n|f(y)| \\leq \\sum_{i=1}^{n} \\left\\lvert\\, f\\left(\\frac{y}{2^{i-1}}\\right)\\right.-f\\left(\\frac{y}{2^{i}}\\right)\\left|+\\left|f\\left(\\frac{y}{2^{n}}\\right)\\right|\\right. \\\\\n\\leq \\sum_{i=1}^{n} \\frac{y}{2^{i}} \\epsilon+\\left|f\\left(\\frac{y}{2^{n}}\\right)\\right|=y \\epsilon\\left(1-\\frac{1}{2^{n}}\\right)+\\left|f\\left(\\frac{y}{2^{n}}\\right)\\right|\n\\end{array}\n\\]\nusing (1).\nLetting \\( n \\rightarrow \\infty \\) we have\n\\[\n|f(y)| \\leq \\epsilon y\n\\]\nsince \\( f(x)-0 \\) as \\( x-0 \\).\nThus we have proved: For all \\( \\epsilon>0 \\), there is a \\( \\delta>0 \\) such that \\( |f(y) / y| \\) \\( \\leq \\epsilon \\) provided \\( 0<y<\\cdot \\delta \\). But, by definition, this is\n\\[\nf(x)=o(x) .\n\\]",
  "vars": [
    "x",
    "y",
    "n",
    "i",
    "f"
  ],
  "params": [
    "\\\\epsilon",
    "\\\\delta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "y": "variabley",
        "n": "indexnn",
        "i": "indexii",
        "f": "function",
        "\\epsilon": "smallconst",
        "\\delta": "smallrange"
      },
      "question": "5. If \\( function(variablex) \\) is a real-valued function defined for \\( 0<variablex<1 \\), then the formula \\( function(variablex)=o(variablex) \\) is an abbreviation for the statement that\n\\[\n\\frac{function(variablex)}{variablex} \\rightarrow 0 \\quad \\text { as } variablex \\rightarrow 0\n\\]\n\nKeeping this in mind, prove the following: if\n\\[\n\\lim _{variablex \\rightarrow 0} function(variablex)=0 \\text { and } function(variablex)-function\\left(\\frac{variablex}{2}\\right)=o(variablex)\n\\]\nthen \\( function(variablex)=o(variablex) \\).",
      "solution": "Solution. Let \\( smallconst>0 \\) be given. Choose \\( smallrange \\) so that for all \\( variablex \\) satisfying \\( 0<variablex<smallrange \\)\n\\[\n\\left|\\frac{1}{variablex}[function(variablex)-function(variablex / 2)]\\right|<\\frac{1}{2} smallconst .\n\\]\n\nNow fix \\( variabley, 0<variabley<smallrange \\). Then\n\\[\n\\begin{array}{l} \nfunction(variabley)=\\left[function(variabley)-function\\left(\\frac{variabley}{2}\\right)\\right]+\\left[function\\left(\\frac{variabley}{2}\\right)-function\\left(\\frac{variabley}{4}\\right)\\right] \\\\\n+\\cdots+\\left[function\\left(\\frac{variabley}{2^{indexnn-1}}\\right)-function\\left(\\frac{variabley}{2^{indexnn}}\\right)\\right]+function\\left(\\frac{variabley}{2^{indexnn}}\\right)\n\\end{array}\n\\]\n\nSo\n\\[\n\\begin{array}{l}\n|function(variabley)| \\leq \\sum_{indexii=1}^{indexnn} \\left\\lvert\\, function\\left(\\frac{variabley}{2^{indexii-1}}\\right)\\right.-function\\left(\\frac{variabley}{2^{indexii}}\\right)\\left|+\\left|function\\left(\\frac{variabley}{2^{indexnn}}\\right)\\right|\\right. \\\\\n\\leq \\sum_{indexii=1}^{indexnn} \\frac{variabley}{2^{indexii}} smallconst+\\left|function\\left(\\frac{variabley}{2^{indexnn}}\\right)\\right|=variabley smallconst\\left(1-\\frac{1}{2^{indexnn}}\\right)+\\left|function\\left(\\frac{variabley}{2^{indexnn}}\\right)\\right|\n\\end{array}\n\\]\nusing (1).\nLetting \\( indexnn \\rightarrow \\infty \\) we have\n\\[\n|function(variabley)| \\leq smallconst variabley\n\\]\nsince \\( function(variablex)\\to0 \\) as \\( variablex\\to0 \\).\nThus we have proved: For all \\( smallconst>0 \\), there is a \\( smallrange>0 \\) such that \\( |function(variabley) / variabley| \\leq smallconst \\) provided \\( 0<variabley<smallrange \\). But, by definition, this is\n\\[\nfunction(variablex)=o(variablex) .\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "gazeboair",
        "y": "lanternglow",
        "n": "orchardmix",
        "i": "planktonic",
        "f": "meadowspur",
        "\\epsilon": "sandpiper",
        "\\delta": "kingfisher"
      },
      "question": "5. If \\( meadowspur(gazeboair) \\) is a real-valued function defined for \\( 0<gazeboair<1 \\), then the formula \\( meadowspur(gazeboair)=o(gazeboair) \\) is an abbreviation for the statement that\n\\[\n\\frac{meadowspur(gazeboair)}{gazeboair} \\rightarrow 0 \\quad \\text { as } gazeboair \\rightarrow 0\n\\]\n\nKeeping this in mind, prove the following: if\n\\[\n\\lim _{gazeboair \\rightarrow 0} meadowspur(gazeboair)=0 \\text { and } meadowspur(gazeboair)-meadowspur\\left(\\frac{gazeboair}{2}\\right)=o(gazeboair)\n\\]\nthen \\( meadowspur(gazeboair)=o(gazeboair) \\).",
      "solution": "Solution. Let \\( sandpiper>0 \\) be given. Choose \\( kingfisher \\) so that for all \\( gazeboair \\) satisfying \\( 0<gazeboair<kingfisher \\)\n\\[\n\\left|\\frac{1}{gazeboair}[meadowspur(gazeboair)-meadowspur(gazeboair / 2)]\\right|<\\frac{1}{2} sandpiper .\n\\]\n\nNow fix \\( lanternglow, 0<lanternglow<kingfisher \\). Then\n\\[\n\\begin{array}{l} \nmeadowspur(lanternglow)=\\left[meadowspur(lanternglow)-meadowspur\\left(\\frac{lanternglow}{2}\\right)\\right]+\\left[meadowspur\\left(\\frac{lanternglow}{2}\\right)-meadowspur\\left(\\frac{lanternglow}{4}\\right)\\right] \\\\\n+\\cdots+\\left[meadowspur\\left(\\frac{lanternglow}{2^{orchardmix-1}}\\right)-meadowspur\\left(\\frac{lanternglow}{2^{orchardmix}}\\right)\\right]+meadowspur\\left(\\frac{lanternglow}{2^{orchardmix}}\\right)\n\\end{array}\n\\]\n\nSo\n\\[\n\\begin{array}{l}\n|meadowspur(lanternglow)| \\leq \\sum_{planktonic=1}^{orchardmix} \\left\\lvert\\, meadowspur\\left(\\frac{lanternglow}{2^{planktonic-1}}\\right)\\right.-meadowspur\\left(\\frac{lanternglow}{2^{planktonic}}\\right)\\left|+\\left|meadowspur\\left(\\frac{lanternglow}{2^{orchardmix}}\\right)\\right|\\right. \\\\\n\\leq \\sum_{planktonic=1}^{orchardmix} \\frac{lanternglow}{2^{planktonic}} sandpiper+\\left|meadowspur\\left(\\frac{lanternglow}{2^{orchardmix}}\\right)\\right|=lanternglow sandpiper\\left(1-\\frac{1}{2^{orchardmix}}\\right)+\\left|meadowspur\\left(\\frac{lanternglow}{2^{orchardmix}}\\right)\\right|\n\\end{array}\n\\]\nusing (1).\nLetting \\( orchardmix \\rightarrow \\infty \\) we have\n\\[\n|meadowspur(lanternglow)| \\leq sandpiper lanternglow\n\\]\nsince \\( meadowspur(gazeboair)\\to 0 \\) as \\( gazeboair\\to 0 \\).\nThus we have proved: For all \\( sandpiper>0 \\), there is a \\( kingfisher>0 \\) such that \\( |meadowspur(lanternglow) / lanternglow| \\) \\( \\leq sandpiper \\) provided \\( 0<lanternglow<\\cdot kingfisher \\). But, by definition, this is\n\\[\nmeadowspur(gazeboair)=o(gazeboair) .\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "vastvalue",
        "y": "colossal",
        "n": "minuscule",
        "i": "gigantic",
        "f": "malfunction",
        "\\epsilon": "enormity",
        "\\delta": "distance"
      },
      "question": "5. If \\( malfunction(vastvalue) \\) is a real-valued function defined for \\( 0<vastvalue<1 \\), then the formula \\( malfunction(vastvalue)=o(vastvalue) \\) is an abbreviation for the statement that\n\\[\n\\frac{malfunction(vastvalue)}{vastvalue} \\rightarrow 0 \\quad \\text { as } vastvalue \\rightarrow 0\n\\]\n\nKeeping this in mind, prove the following: if\n\\[\n\\lim _{vastvalue \\rightarrow 0} malfunction(vastvalue)=0 \\text { and } malfunction(vastvalue)-malfunction\\left(\\frac{vastvalue}{2}\\right)=o(vastvalue)\n\\]\nthen \\( malfunction(vastvalue)=o(vastvalue) \\).",
      "solution": "Solution. Let \\( enormity>0 \\) be given. Choose \\( distance \\) so that for all \\( vastvalue \\) satisfying \\( 0<vastvalue<distance \\)\n\\[\n\\left|\\frac{1}{vastvalue}[malfunction(vastvalue)-malfunction(vastvalue / 2)]\\right|<\\frac{1}{2} enormity .\n\\]\n\nNow fix \\( colossal, 0<colossal<distance \\). Then\n\\[\n\\begin{array}{l} \nmalfunction(colossal)=\\left[malfunction(colossal)-malfunction\\left(\\frac{colossal}{2}\\right)\\right]+\\left[malfunction\\left(\\frac{colossal}{2}\\right)-malfunction\\left(\\frac{colossal}{4}\\right)\\right] \\\\\n+\\cdots+\\left[malfunction\\left(\\frac{colossal}{2^{minuscule-1}}\\right)-malfunction\\left(\\frac{colossal}{2^{minuscule}}\\right)\\right]+malfunction\\left(\\frac{colossal}{2^{minuscule}}\\right)\n\\end{array}\n\\]\n\nSo\n\\[\n\\begin{array}{l}\n|malfunction(colossal)| \\leq \\sum_{gigantic=1}^{minuscule} \\left\\lvert\\, malfunction\\left(\\frac{colossal}{2^{gigantic-1}}\\right)\\right.-malfunction\\left(\\frac{colossal}{2^{gigantic}}\\right)\\left|+\\left|malfunction\\left(\\frac{colossal}{2^{minuscule}}\\right)\\right|\\right. \\\\\n\\leq \\sum_{gigantic=1}^{minuscule} \\frac{colossal}{2^{gigantic}} enormity+\\left|malfunction\\left(\\frac{colossal}{2^{minuscule}}\\right)\\right|=colossal\\ enormity\\left(1-\\frac{1}{2^{minuscule}}\\right)+\\left|malfunction\\left(\\frac{colossal}{2^{minuscule}}\\right)\\right|\n\\end{array}\n\\]\nusing (1).\nLetting \\( minuscule \\rightarrow \\infty \\) we have\n\\[\n|malfunction(colossal)| \\leq enormity\\ colossal\n\\]\nsince \\( malfunction(vastvalue)-0 \\) as \\( vastvalue-0 \\).\nThus we have proved: For all \\( enormity>0 \\), there is a \\( distance>0 \\) such that \\( |malfunction(colossal) / colossal| \\leq enormity \\) provided \\( 0<colossal<\\cdot distance \\). But, by definition, this is\n\\[\nmalfunction(vastvalue)=o(vastvalue) .\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "n": "vckmroqe",
        "i": "zpjdfyal",
        "f": "bnchwguo",
        "\\epsilon": "uvwqrjst",
        "\\delta": "nfbciyma"
      },
      "question": "5. If \\( bnchwguo(qzxwvtnp) \\) is a real-valued function defined for \\( 0<qzxwvtnp<1 \\), then the formula \\( bnchwguo(qzxwvtnp)=o(qzxwvtnp) \\) is an abbreviation for the statement that\n\\[\n\\frac{bnchwguo(qzxwvtnp)}{qzxwvtnp} \\rightarrow 0 \\quad \\text { as } qzxwvtnp \\rightarrow 0\n\\]\n\nKeeping this in mind, prove the following: if\n\\[\n\\lim _{qzxwvtnp \\rightarrow 0} bnchwguo(qzxwvtnp)=0 \\text { and } bnchwguo(qzxwvtnp)-bnchwguo\\left(\\frac{qzxwvtnp}{2}\\right)=o(qzxwvtnp)\n\\]\nthen \\( bnchwguo(qzxwvtnp)=o(qzxwvtnp) \\).",
      "solution": "Solution. Let \\( uvwqrjst>0 \\) be given. Choose nfbciyma so that for all qzxwvtnp satisfying \\( 0<qzxwvtnp<nfbciyma \\)\n\\[\n\\left|\\frac{1}{qzxwvtnp}[bnchwguo(qzxwvtnp)-bnchwguo(qzxwvtnp / 2)]\\right|<\\frac{1}{2} uvwqrjst .\n\\]\n\nNow fix hjgrksla, \\( 0<hjgrksla<nfbciyma \\). Then\n\\[\n\\begin{array}{l} \nbnchwguo(hjgrksla)=\\left[bnchwguo(hjgrksla)-bnchwguo\\left(\\frac{hjgrksla}{2}\\right)\\right]+\\left[bnchwguo\\left(\\frac{hjgrksla}{2}\\right)-bnchwguo\\left(\\frac{hjgrksla}{4}\\right)\\right] \\\\\n+\\cdots+\\left[bnchwguo\\left(\\frac{hjgrksla}{2^{vckmroqe-1}}\\right)-bnchwguo\\left(\\frac{hjgrksla}{2^{vckmroqe}}\\right)\\right]+bnchwguo\\left(\\frac{hjgrksla}{2^{vckmroqe}}\\right)\n\\end{array}\n\\]\n\nSo\n\\[\n\\begin{array}{l}\n|bnchwguo(hjgrksla)| \\leq \\sum_{zpjdfyal=1}^{vckmroqe} \\left\\lvert\\, bnchwguo\\left(\\frac{hjgrksla}{2^{zpjdfyal-1}}\\right)\\right.-bnchwguo\\left(\\frac{hjgrksla}{2^{zpjdfyal}}\\right)\\left|+\\left|bnchwguo\\left(\\frac{hjgrksla}{2^{vckmroqe}}\\right)\\right|\\right. \\\\\n\\leq \\sum_{zpjdfyal=1}^{vckmroqe} \\frac{hjgrksla}{2^{zpjdfyal}} uvwqrjst+\\left|bnchwguo\\left(\\frac{hjgrksla}{2^{vckmroqe}}\\right)\\right|=hjgrksla uvwqrjst\\left(1-\\frac{1}{2^{vckmroqe}}\\right)+\\left|bnchwguo\\left(\\frac{hjgrksla}{2^{vckmroqe}}\\right)\\right|\n\\end{array}\n\\]\nusing (1).\nLetting \\( vckmroqe \\rightarrow \\infty \\) we have\n\\[\n|bnchwguo(hjgrksla)| \\leq uvwqrjst hjgrksla\n\\]\nsince \\( bnchwguo(qzxwvtnp)-0 \\) as \\( qzxwvtnp-0 \\).\nThus we have proved: For all \\( uvwqrjst>0 \\), there is a nfbciyma>0 such that \\( |bnchwguo(hjgrksla) / hjgrksla| \\) \\( \\leq uvwqrjst \\) provided \\( 0<hjgrksla<\\cdot nfbciyma \\). But, by definition, this is\n\\[\nbnchwguo(qzxwvtnp)=o(qzxwvtnp) .\n\\]"
    },
    "kernel_variant": {
      "question": "Let n ≥ 2 and fix an induced norm ‖·‖ on ℝⁿ.  \nLet f : ℝⁿ \\ {0} → ℝ be defined in a punctured neighbourhood of the origin and assume  \n\n(i) lim_{x→0} f(x)=0.  \n\nChoose an integer p ≥ 2 and invertible matrices A₁,…,A_p ∈ ℝ^{n×n} such that  \n\n  ρ := max_{1≤j≤p}‖A_j‖ < 1.  \n\nLet positive weights c₁,…,c_p satisfy ∑_{j=1}^p c_j=1.  \nAssume that, as x→0,\n\n  f(x) − ∑_{j=1}^p c_j f(A_j x) = o(‖x‖).                                (*)\n\nProve simultaneously that  \n\n(1) f(x)=o(‖x‖) as x→0, and  \n\n(2) the ratio f(x)/‖x‖ tends to 0 uniformly in every direction, i.e.  \n\n  lim_{t→0⁺} sup_{‖u‖=1} |f(tu)|/t = 0.",
      "solution": "Step 0.  Notation.  \nFor a multi–index α=(α₁,…,α_k) with entries in {1,…,p} set\n\n  A_α := A_{α₁}A_{α₂}⋯A_{α_k},  c_α := c_{α₁}c_{α₂}⋯c_{α_k},  |α|=k.\n\nBecause ‖A_j‖≤ρ<1 we have ‖A_α‖≤ρ^{|α|}.  \nFor α=∅ (the empty word) put A_∅=I, c_∅=1.\n\nStep 1.  Converting (*) into a quantitative estimate.  \nBy definition of the little–o term, there exists δ₀>0 such that\n\n  |f(x)−∑_{j=1}^p c_j f(A_j x)| ≤ ε₀ (1−ρ)‖x‖              (1)\n\nwhenever 0<‖x‖<δ₀, where ε₀>0 is arbitrary (we shall later send ε₀ to 0).  \nThe factor (1−ρ) is inserted for convenience.\n\nDenote the error term in (1) by R₁(x):  \n  R₁(x)=f(x)−∑_{j=1}^p c_j f(A_j x),  |R₁(x)|≤ε₀(1−ρ)‖x‖.    (2)\n\nStep 2.  First iteration.  \nInsert the identity f(A_j x)=∑_{k=1}^p c_k f(A_kA_j x)+R₁(A_j x) coming from (2) into the right-hand side of f(x)=∑_{j=1}^p c_j f(A_j x)+R₁(x).  We obtain\n\n  f(x)=∑_{|α|=2} c_α f(A_α x)+∑_{j=1}^p c_j R₁(A_j x)+R₁(x).    (3)\n\nStep 3.  k-th iteration.  \nProceeding inductively, after k steps we have\n\n  f(x)=∑_{|α|=k} c_α f(A_α x)+∑_{m=0}^{k-1} ∑_{|α|=m} c_α R₁(A_α x).    (4)\n\n(The second sum contains the remainders produced at every level m.)\n\nStep 4.  Passage to the limit as k→∞.  \nFix x with 0<‖x‖<δ₀.  Because ‖A_α‖≤ρ^{|α|}, the norm of A_α x does not exceed ρ^{k}‖x‖ when |α|=k.  Hence\n\n  lim_{k→∞} sup_{|α|=k}|f(A_α x)| = 0 (5)\n\nby (i).  Therefore the first term on the right-hand side of (4) vanishes as k→∞.\n\nFor the remainder part we use (2):\n\n|R₁(A_α x)| ≤ ε₀(1−ρ)‖A_α x‖ ≤ ε₀(1−ρ)ρ^{|α|}‖x‖.\n\nSumming over all words of length m gives ∑_{|α|=m} c_α=1.  Hence\n\n  ∑_{|α|=m} c_α|R₁(A_α x)|≤ε₀(1−ρ)ρ^{m}‖x‖.               (6)\n\nInsert (6) into (4) and let k→∞:\n\n|f(x)|≤∑_{m=0}^{∞} ε₀(1−ρ)ρ^{m}‖x‖  \n    = ε₀‖x‖.                                                              (7)\n\nStep 5.  Uniform estimate and conclusion.  \nInequality (7) holds for every x with 0<‖x‖<δ₀ and for the arbitrarily chosen ε₀.  Hence for any ε>0 we may pick ε₀=ε to obtain\n\n  sup_{0<‖x‖<δ₀} |f(x)|/‖x‖ ≤ ε.                             (8)\n\nLetting ε→0 proves\n\n  lim_{x→0} f(x)/‖x‖ = 0,                                     (9)\n\nwhich is statement (1).\n\nBecause the right-hand side of (7) depends only on ‖x‖, the bound is radial; taking the supremum over all unit vectors u yields\n\n  sup_{‖u‖=1}|f(tu)|/t ≤ ε  (0<t<δ₀).                        (10)\n\nAgain ε is arbitrary, so the left-hand side tends to 0 as t→0⁺, proving (2).\n\n∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.465789",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: The variable is now x ∈ ℝⁿ with an arbitrary induced norm; scalar-valued functions of one real variable no longer suffice.\n\n2. Multiple contractions: Instead of a single scaling x↦x/3, the problem involves p ≥ 2 different invertible contraction matrices.  The iteration therefore grows on a branching tree rather than a single chain, and one must manage multi-indices and countably many remainder terms.\n\n3. Weighted refinement equation: The relation (*) mixes the values of f at p different points with positive weights, so simple telescoping fails.  One needs an averaging argument plus control of the total weight at each depth.\n\n4. Uniform limit:  Beyond showing f(x)=o(‖x‖), the solver must establish uniform convergence of f(x)/‖x‖ over all directions, adding a layer of subtlety absent in the original problem.\n\n5. Advanced techniques:  The proof uses operator norms, estimates on products of matrices (ρ^{|α|}), geometric-series bounds over an infinite rooted tree of compositions, and uniformity arguments—tools well beyond the elementary one-dimensional telescoping sum employed in the original solution.\n\nThese additions collectively raise the conceptual and technical load, making the enhanced variant significantly more challenging than both the original exercise and the preceding kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let n ≥ 2 and fix an induced norm ‖·‖ on ℝⁿ.  \nLet f : ℝⁿ \\ {0} → ℝ be defined in a punctured neighbourhood of the origin and assume  \n\n(i) lim_{x→0} f(x)=0.  \n\nChoose an integer p ≥ 2 and invertible matrices A₁,…,A_p ∈ ℝ^{n×n} such that  \n\n  ρ := max_{1≤j≤p}‖A_j‖ < 1.  \n\nLet positive weights c₁,…,c_p satisfy ∑_{j=1}^p c_j=1.  \nAssume that, as x→0,\n\n  f(x) − ∑_{j=1}^p c_j f(A_j x) = o(‖x‖).                                (*)\n\nProve simultaneously that  \n\n(1) f(x)=o(‖x‖) as x→0, and  \n\n(2) the ratio f(x)/‖x‖ tends to 0 uniformly in every direction, i.e.  \n\n  lim_{t→0⁺} sup_{‖u‖=1} |f(tu)|/t = 0.",
      "solution": "Step 0.  Notation.  \nFor a multi–index α=(α₁,…,α_k) with entries in {1,…,p} set\n\n  A_α := A_{α₁}A_{α₂}⋯A_{α_k},  c_α := c_{α₁}c_{α₂}⋯c_{α_k},  |α|=k.\n\nBecause ‖A_j‖≤ρ<1 we have ‖A_α‖≤ρ^{|α|}.  \nFor α=∅ (the empty word) put A_∅=I, c_∅=1.\n\nStep 1.  Converting (*) into a quantitative estimate.  \nBy definition of the little–o term, there exists δ₀>0 such that\n\n  |f(x)−∑_{j=1}^p c_j f(A_j x)| ≤ ε₀ (1−ρ)‖x‖              (1)\n\nwhenever 0<‖x‖<δ₀, where ε₀>0 is arbitrary (we shall later send ε₀ to 0).  \nThe factor (1−ρ) is inserted for convenience.\n\nDenote the error term in (1) by R₁(x):  \n  R₁(x)=f(x)−∑_{j=1}^p c_j f(A_j x),  |R₁(x)|≤ε₀(1−ρ)‖x‖.    (2)\n\nStep 2.  First iteration.  \nInsert the identity f(A_j x)=∑_{k=1}^p c_k f(A_kA_j x)+R₁(A_j x) coming from (2) into the right-hand side of f(x)=∑_{j=1}^p c_j f(A_j x)+R₁(x).  We obtain\n\n  f(x)=∑_{|α|=2} c_α f(A_α x)+∑_{j=1}^p c_j R₁(A_j x)+R₁(x).    (3)\n\nStep 3.  k-th iteration.  \nProceeding inductively, after k steps we have\n\n  f(x)=∑_{|α|=k} c_α f(A_α x)+∑_{m=0}^{k-1} ∑_{|α|=m} c_α R₁(A_α x).    (4)\n\n(The second sum contains the remainders produced at every level m.)\n\nStep 4.  Passage to the limit as k→∞.  \nFix x with 0<‖x‖<δ₀.  Because ‖A_α‖≤ρ^{|α|}, the norm of A_α x does not exceed ρ^{k}‖x‖ when |α|=k.  Hence\n\n  lim_{k→∞} sup_{|α|=k}|f(A_α x)| = 0 (5)\n\nby (i).  Therefore the first term on the right-hand side of (4) vanishes as k→∞.\n\nFor the remainder part we use (2):\n\n|R₁(A_α x)| ≤ ε₀(1−ρ)‖A_α x‖ ≤ ε₀(1−ρ)ρ^{|α|}‖x‖.\n\nSumming over all words of length m gives ∑_{|α|=m} c_α=1.  Hence\n\n  ∑_{|α|=m} c_α|R₁(A_α x)|≤ε₀(1−ρ)ρ^{m}‖x‖.               (6)\n\nInsert (6) into (4) and let k→∞:\n\n|f(x)|≤∑_{m=0}^{∞} ε₀(1−ρ)ρ^{m}‖x‖  \n    = ε₀‖x‖.                                                              (7)\n\nStep 5.  Uniform estimate and conclusion.  \nInequality (7) holds for every x with 0<‖x‖<δ₀ and for the arbitrarily chosen ε₀.  Hence for any ε>0 we may pick ε₀=ε to obtain\n\n  sup_{0<‖x‖<δ₀} |f(x)|/‖x‖ ≤ ε.                             (8)\n\nLetting ε→0 proves\n\n  lim_{x→0} f(x)/‖x‖ = 0,                                     (9)\n\nwhich is statement (1).\n\nBecause the right-hand side of (7) depends only on ‖x‖, the bound is radial; taking the supremum over all unit vectors u yields\n\n  sup_{‖u‖=1}|f(tu)|/t ≤ ε  (0<t<δ₀).                        (10)\n\nAgain ε is arbitrary, so the left-hand side tends to 0 as t→0⁺, proving (2).\n\n∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.393242",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: The variable is now x ∈ ℝⁿ with an arbitrary induced norm; scalar-valued functions of one real variable no longer suffice.\n\n2. Multiple contractions: Instead of a single scaling x↦x/3, the problem involves p ≥ 2 different invertible contraction matrices.  The iteration therefore grows on a branching tree rather than a single chain, and one must manage multi-indices and countably many remainder terms.\n\n3. Weighted refinement equation: The relation (*) mixes the values of f at p different points with positive weights, so simple telescoping fails.  One needs an averaging argument plus control of the total weight at each depth.\n\n4. Uniform limit:  Beyond showing f(x)=o(‖x‖), the solver must establish uniform convergence of f(x)/‖x‖ over all directions, adding a layer of subtlety absent in the original problem.\n\n5. Advanced techniques:  The proof uses operator norms, estimates on products of matrices (ρ^{|α|}), geometric-series bounds over an infinite rooted tree of compositions, and uniformity arguments—tools well beyond the elementary one-dimensional telescoping sum employed in the original solution.\n\nThese additions collectively raise the conceptual and technical load, making the enhanced variant significantly more challenging than both the original exercise and the preceding kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}