{
  "index": "1958-2-A-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "6. Let \\( a(x) \\) and \\( b(x) \\) be continuous functions on \\( 0 \\leq x \\leq 1 \\) and let \\( 0 \\leq \\) \\( a(x) \\leq a<1 \\) on that range. Under what other conditions (if any) is the solution of the equation for \\( u \\),\n\\[\nu=\\underset{0 \\leq x \\leq 1}{\\operatorname{maximum}}[b(x)+a(x) \\cdot u],\n\\]\ngiven by\n\\[\nu=\\operatorname{maximum}_{0 \\leq x \\leq 1}\\left[\\frac{b(x)}{1-a(x)}\\right] ?\n\\]",
  "solution": "Solution. Since the functions involved are continuous, the maximum values must be attained, and since we are given that \\( a(x) \\) is bounded and less than unity, we see that the following statements are all equivalent:\n(1) \\( u=\\max _{0 \\leq x \\leq 1}[b(x)+a(x) u] \\)\n(2) ( \\( \\forall x) \\quad u \\geq b(x)+a(x) \\cdot u \\) with equality for at least one value of \\( x \\).\n(3) ( \\( \\forall x) \\quad(1-a(x)) u \\geq b(x) \\quad \\) with equality for at least one value of \\( x \\).\n(4) \\( (\\forall x) \\quad u \\geq \\frac{b(x)}{1-a(x)} \\) with equality for at least one value of \\( x \\).\n(5) \\( u=\\max _{0 \\leq x \\leq 1} \\frac{b(x)}{1-a(x)} \\).\n\nThus no additional conditions are required. Equation (1) always has a unique solution for \\( u \\), given by (5).",
  "vars": [
    "x",
    "u"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "u": "unknownu",
        "a": "functiona",
        "b": "functionb"
      },
      "question": "6. Let \\( functiona(variablex) \\) and \\( functionb(variablex) \\) be continuous functions on \\( 0 \\leq variablex \\leq 1 \\) and let \\( 0 \\leq functiona(variablex) \\leq functiona<1 \\) on that range. Under what other conditions (if any) is the solution of the equation for \\( unknownu \\),\n\\[\nunknownu=\\underset{0 \\leq variablex \\leq 1}{\\operatorname{maximum}}[functionb(variablex)+functiona(variablex) \\cdot unknownu],\n\\]\ngiven by\n\\[\nunknownu=\\operatorname{maximum}_{0 \\leq variablex \\leq 1}\\left[\\frac{functionb(variablex)}{1-functiona(variablex)}\\right] ?\n\\]",
      "solution": "Solution. Since the functions involved are continuous, the maximum values must be attained, and since we are given that \\( functiona(variablex) \\) is bounded and less than unity, we see that the following statements are all equivalent:\n(1) \\( unknownu=\\max _{0 \\leq variablex \\leq 1}[functionb(variablex)+functiona(variablex) unknownu] \\)\n(2) ( \\( \\forall variablex) \\quad unknownu \\geq functionb(variablex)+functiona(variablex) \\cdot unknownu \\) with equality for at least one value of \\( variablex \\).\n(3) ( \\( \\forall variablex) \\quad(1-functiona(variablex)) unknownu \\geq functionb(variablex) \\quad \\) with equality for at least one value of \\( variablex \\).\n(4) \\( (\\forall variablex) \\quad unknownu \\geq \\frac{functionb(variablex)}{1-functiona(variablex)} \\) with equality for at least one value of \\( variablex \\).\n(5) \\( unknownu=\\max _{0 \\leq variablex \\leq 1} \\frac{functionb(variablex)}{1-functiona(variablex)} \\).\n\nThus no additional conditions are required. Equation (1) always has a unique solution for \\( unknownu \\), given by (5)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineapple",
        "u": "shoelaces",
        "a": "lighthouse",
        "b": "squirrels"
      },
      "question": "6. Let \\( lighthouse(pineapple) \\) and \\( squirrels(pineapple) \\) be continuous functions on \\( 0 \\leq pineapple \\leq 1 \\) and let \\( 0 \\leq \\) \\( lighthouse(pineapple) \\leq lighthouse<1 \\) on that range. Under what other conditions (if any) is the solution of the equation for \\( shoelaces \\),\n\\[\nshoelaces=\\underset{0 \\leq pineapple \\leq 1}{\\operatorname{maximum}}[squirrels(pineapple)+lighthouse(pineapple) \\cdot shoelaces],\n\\]\ngiven by\n\\[\nshoelaces=\\operatorname{maximum}_{0 \\leq pineapple \\leq 1}\\left[\\frac{squirrels(pineapple)}{1-lighthouse(pineapple)}\\right] ?\n\\]",
      "solution": "Solution. Since the functions involved are continuous, the maximum values must be attained, and since we are given that \\( lighthouse(pineapple) \\) is bounded and less than unity, we see that the following statements are all equivalent:\n(1) \\( shoelaces=\\max _{0 \\leq pineapple \\leq 1}[squirrels(pineapple)+lighthouse(pineapple) shoelaces] \\)\n(2) \\( (\\forall pineapple) \\quad shoelaces \\geq squirrels(pineapple)+lighthouse(pineapple) \\cdot shoelaces \\) with equality for at least one value of \\( pineapple \\).\n(3) \\( (\\forall pineapple) \\quad(1-lighthouse(pineapple)) shoelaces \\geq squirrels(pineapple) \\quad \\) with equality for at least one value of \\( pineapple \\).\n(4) \\( (\\forall pineapple) \\quad shoelaces \\geq \\frac{squirrels(pineapple)}{1-lighthouse(pineapple)} \\) with equality for at least one value of \\( pineapple \\).\n(5) \\( shoelaces=\\max _{0 \\leq pineapple \\leq 1} \\frac{squirrels(pineapple)}{1-lighthouse(pineapple)} \\).\n\nThus no additional conditions are required. Equation (1) always has a unique solution for \\( shoelaces \\), given by (5)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "u": "minimumvalue",
        "a": "inflationfactor",
        "b": "subtractive"
      },
      "question": "6. Let \\( inflationfactor(fixedvalue) \\) and \\( subtractive(fixedvalue) \\) be continuous functions on \\( 0 \\leq fixedvalue \\leq 1 \\) and let \\( 0 \\leq inflationfactor(fixedvalue) \\leq inflationfactor<1 \\) on that range. Under what other conditions (if any) is the solution of the equation for \\( minimumvalue \\),\n\\[\nminimumvalue=\\underset{0 \\leq fixedvalue \\leq 1}{\\operatorname{maximum}}[subtractive(fixedvalue)+inflationfactor(fixedvalue) \\cdot minimumvalue],\n\\]\n given by\n\\[\nminimumvalue=\\operatorname{maximum}_{0 \\leq fixedvalue \\leq 1}\\left[\\frac{subtractive(fixedvalue)}{1-inflationfactor(fixedvalue)}\\right] ?\n\\]",
      "solution": "Solution. Since the functions involved are continuous, the maximum values must be attained, and since we are given that \\( inflationfactor(fixedvalue) \\) is bounded and less than unity, we see that the following statements are all equivalent:\n(1) \\( minimumvalue=\\max _{0 \\leq fixedvalue \\leq 1}[subtractive(fixedvalue)+inflationfactor(fixedvalue) minimumvalue] \\)\n(2) ( \\( \\forall fixedvalue) \\quad minimumvalue \\geq subtractive(fixedvalue)+inflationfactor(fixedvalue) \\cdot minimumvalue \\) with equality for at least one value of \\( fixedvalue \\).\n(3) ( \\( \\forall fixedvalue) \\quad(1-inflationfactor(fixedvalue)) minimumvalue \\geq subtractive(fixedvalue) \\quad \\) with equality for at least one value of \\( fixedvalue \\).\n(4) \\( (\\forall fixedvalue) \\quad minimumvalue \\geq \\frac{subtractive(fixedvalue)}{1-inflationfactor(fixedvalue)} \\) with equality for at least one value of \\( fixedvalue \\).\n(5) \\( minimumvalue=\\max _{0 \\leq fixedvalue \\leq 1} \\frac{subtractive(fixedvalue)}{1-inflationfactor(fixedvalue)} \\).\n\nThus no additional conditions are required. Equation (1) always has a unique solution for \\( minimumvalue \\), given by (5)."
    },
    "garbled_string": {
      "map": {
        "x": "mrqslpva",
        "u": "zpdnxwqo",
        "a": "fkvhceut",
        "b": "tjdwplra"
      },
      "question": "6. Let \\( fkvhceut(mrqslpva) \\) and \\( tjdwplra(mrqslpva) \\) be continuous functions on \\( 0 \\leq mrqslpva \\leq 1 \\) and let \\( 0 \\leq \\) \\( fkvhceut(mrqslpva) \\leq fkvhceut<1 \\) on that range. Under what other conditions (if any) is the solution of the equation for \\( zpdnxwqo \\),\n\\[\nzpdnxwqo=\\underset{0 \\leq mrqslpva \\leq 1}{\\operatorname{maximum}}[tjdwplra(mrqslpva)+fkvhceut(mrqslpva) \\cdot zpdnxwqo],\n\\]\ngiven by\n\\[\nzpdnxwqo=\\operatorname{maximum}_{0 \\leq mrqslpva \\leq 1}\\left[\\frac{tjdwplra(mrqslpva)}{1-fkvhceut(mrqslpva)}\\right] ?\n\\]",
      "solution": "Solution. Since the functions involved are continuous, the maximum values must be attained, and since we are given that \\( fkvhceut(mrqslpva) \\) is bounded and less than unity, we see that the following statements are all equivalent:\n(1) \\( zpdnxwqo=\\max _{0 \\leq mrqslpva \\leq 1}[tjdwplra(mrqslpva)+fkvhceut(mrqslpva) zpdnxwqo] \\)\n(2) ( \\( \\forall mrqslpva) \\quad zpdnxwqo \\geq tjdwplra(mrqslpva)+fkvhceut(mrqslpva) \\cdot zpdnxwqo \\) with equality for at least one value of \\( mrqslpva \\).\n(3) ( \\( \\forall mrqslpva) \\quad(1-fkvhceut(mrqslpva)) zpdnxwqo \\geq tjdwplra(mrqslpva) \\quad \\) with equality for at least one value of \\( mrqslpva \\).\n(4) \\( (\\forall mrqslpva) \\quad zpdnxwqo \\geq \\frac{tjdwplra(mrqslpva)}{1-fkvhceut(mrqslpva)} \\) with equality for at least one value of \\( mrqslpva \\).\n(5) \\( zpdnxwqo=\\max _{0 \\leq mrqslpva \\leq 1} \\frac{tjdwplra(mrqslpva)}{1-fkvhceut(mrqslpva)} \\).\n\nThus no additional conditions are required. Equation (1) always has a unique solution for \\( zpdnxwqo \\), given by (5)."
    },
    "kernel_variant": {
      "question": "Let m ≥ 2, k ≥ 1 and denote 𝔇 := [−2,2]^k.  \nFor every t ∈ 𝔇 we are given continuous data  \n\n• a non–negative m×m–matrix A(t) whose row–sums satisfy  \n  ‖A(t)‖_∞ = max_{1≤i≤m} Σ_{j=1}^{m} A_{ij}(t) ≤ 0.8;  \n\n• a vector B(t) ∈ ℝ^{m}.  \n\nFix a constant λ with 0.8 < λ ≤ 1.5.\n\nAdditional structure (simultaneous positive monomial diagonalisation)  \n(SD)  There exists an invertible positive monomial matrix  \n  S = P diag(s_1,…,s_m)  (s_i>0, P a permutation matrix)                                  (♠)  \nsuch that, for every t ∈ 𝔇,  \n\n  D(t) := S^{−1} A(t) S = diag( a_1(t), … , a_m(t) )\n\nis diagonal.  (Hence 0 ≤ a_i(t) ≤ 0.8 for all i and t.)\n\nFor a family {v(t)}_{t∈𝔇} ⊂ ℝ^{m} the componentwise maximum is defined by  \n (max_{t∈𝔇} v(t))_i := max_{t∈𝔇} v_i(t).\n\nConsider the nonlinear system\n\n(★)  λ u = max_{t∈𝔇} [ B(t) + A(t) u ] ,  u ∈ ℝ^{m}.  \n\n(a) Prove that (★) possesses exactly one solution u ∈ ℝ^{m}.  \n\n(b) Put  B̃(t) := S^{−1} B(t) and y := S^{−1} u.  \n Show that the coordinates of y are given explicitly by  \n   y_i = max_{t∈𝔇}  B̃_i(t) / (λ − a_i(t))  (1 ≤ i ≤ m),  \n and hence that the unique solution of (★) admits the closed form  \n   u = S  max_{t∈𝔇} (λ I − D(t))^{−1} B̃(t)                    (♣)  \n equivalently  \n   u = S  max_{t∈𝔇} (λ I − S^{−1} A(t) S)^{−1} S^{−1} B(t).  \n\n(c) Establish the a-priori bound  \n   ‖u‖_∞ ≤ 1/(λ − 0.8) · max_{t∈𝔇} ‖B(t)‖_∞.",
      "solution": "Step 0.  Preparatory facts  \nBecause ‖A(t)‖_∞ ≤ 0.8 < λ, all matrices  \n\n  M(t) := λ I − A(t)   \n\nare nonsingular M–matrices.  Hence M(t)^{−1} ≥ 0 and  \n\n  ‖M(t)^{−1}‖_∞ ≤ 1/(λ − 0.8). (0.1)\n\nDefine the order–preserving operator  \n\n  F(u) := max_{t∈𝔇}[ B(t)+A(t)u ] (u ∈ ℝ^{m}). (0.2)\n\n\n\nStep 1.  Existence and uniqueness – task (a)  \nFor any u,v ∈ ℝ^{m} and every t,  \n\n  ‖A(t)u − A(t)v‖_∞ ≤ 0.8‖u−v‖_∞.\n\nHence  \n\n  ‖F(u) − F(v)‖_∞ ≤ 0.8‖u−v‖_∞.\n\nSet G(u) := (1/λ)F(u).  Then  \n\n  ‖G(u) − G(v)‖_∞ ≤ 0.8/λ · ‖u−v‖_∞ < ‖u−v‖_∞,\n\nso G is a contraction on the complete metric space (ℝ^{m},‖·‖_∞).  By the Banach fixed-point theorem there exists a unique u with u = G(u), i.e. (★) has exactly one solution.  \n\n\n\nStep 2.  Passing to diagonal coordinates  \nBecause S in (♠) is a positive monomial matrix, both S and S^{−1} are non-negative and, crucially,\n\n  S^{−1}(max_{t∈𝔇} v(t)) = max_{t∈𝔇} (S^{−1} v(t))  (0.3)\n\nfor every family {v(t)}.  Indeed, writing S^{−1}=diag(d_1,…,d_m)P^{T} with d_i>0,  \n  (S^{−1}w)_i = d_i w_{σ(i)}, σ a permutation.  \nHence the i-th component of the left–hand side of (0.3) equals  \n  d_i max_t v_{σ(i)}(t)  \nwhile that of the right-hand side equals  \n  max_t d_i v_{σ(i)}(t)=d_i max_t v_{σ(i)}(t).  \n\nConsequently we may left-multiply (★) by S^{−1} and push S^{−1} through the componentwise maximum:\n\n  λ y = max_{t∈𝔇}[ B̃(t)+D(t) y ],   y := S^{−1}u, B̃(t):=S^{−1}B(t). (2.1)\n\n\n\nStep 3.  Solving the decoupled scalar equations – task (b)  \nBecause each D(t) is diagonal, (2.1) splits into m independent scalar relations:\n\n  λ y_i = max_{t∈𝔇}[ B̃_i(t)+a_i(t) y_i ]  (1≤i≤m). (2.2)\n\nFix an index i.  For any t,\n\n  B̃_i(t)+a_i(t) y_i ≤ max_{s∈𝔇} B̃_i(s) + 0.8 y_i,\n\nso from (2.2),\n\n  λ y_i ≤ max_{s∈𝔇} B̃_i(s) + 0.8 y_i → (λ-0.8) y_i ≤ max_{s∈𝔇} B̃_i(s),\n\nhence y_i is finite.\n\nConversely define  \n\n  y_i^* := max_{t∈𝔇} B̃_i(t)/(λ - a_i(t)). (2.3)\n\nFor every t,  \n\n  λ y_i^* ≥ B̃_i(t)+a_i(t) y_i^*,  \n\nand taking the maximum over t yields λ y_i^* ≥ RHS of (2.2).  \nAt a point t* attaining the maximum in (2.3) we have equality, so λ y_i^* equals that maximum.  Therefore y_i^* satisfies (2.2).  Uniqueness from Step 1 forces y_i = y_i^*.\n\nCollecting the coordinates,\n\n  y = max_{t∈𝔇}(λ I − D(t))^{−1} B̃(t),\n\nand returning to u = S y we obtain the explicit representation (♣).  \n\n\n\nStep 4.  A-priori bound – task (c)  \nFrom (★),\n\n  λ‖u‖_∞ = ‖F(u)‖_∞  \n      ≤ max_{t∈𝔇}[ ‖B(t)‖_∞ + ‖A(t)‖_∞‖u‖_∞ ]  \n      ≤ max_{t∈𝔇} ‖B(t)‖_∞ + 0.8‖u‖_∞.\n\nHence (λ−0.8)‖u‖_∞ ≤ max_{t∈𝔇} ‖B(t)‖_∞, i.e.\n\n  ‖u‖_∞ ≤ 1/(λ − 0.8) · max_{t∈𝔇} ‖B(t)‖_∞.\n\nThis completes the proof of all three tasks.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.495423",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: The unknown is now an m–dimensional vector; matrices A(t) enter instead of scalars.  \n• Additional structure: non-negative matrices with spectral‐radius control, M–matrix theory, Neumann series estimates.  \n• Deeper theory: Uses Banach’s fixed-point theorem, properties of M–matrices, spectral radius arguments, and matrix norms.  \n• Multiple interacting concepts: continuity on a higher-dimensional domain, component-wise order, contraction mapping, and explicit inversion of parameter‐dependent matrices.  \n• More steps: establishment of positivity/invertibility, two-sided comparison leading to an equality, Lipschitz-type a-priori bound.\n\nThese layers of matrix analysis and functional analysis make the variant substantially more intricate than the scalar original and the current kernel problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let m ≥ 2, k ≥ 1 and denote 𝔇 := [−2,2]^k.  \nFor every t ∈ 𝔇 we are given continuous data  \n\n• a non–negative m×m–matrix A(t) whose row–sums satisfy  \n  ‖A(t)‖_∞ = max_{1≤i≤m} Σ_{j=1}^{m} A_{ij}(t) ≤ 0.8;  \n\n• a vector B(t) ∈ ℝ^{m}.  \n\nFix a constant λ with 0.8 < λ ≤ 1.5.\n\nAdditional structure (simultaneous positive monomial diagonalisation)  \n(SD)  There exists an invertible positive monomial matrix  \n  S = P diag(s_1,…,s_m)  (s_i>0, P a permutation matrix)                                  (♠)  \nsuch that, for every t ∈ 𝔇,  \n\n  D(t) := S^{−1} A(t) S = diag( a_1(t), … , a_m(t) )\n\nis diagonal.  (Hence 0 ≤ a_i(t) ≤ 0.8 for all i and t.)\n\nFor a family {v(t)}_{t∈𝔇} ⊂ ℝ^{m} the componentwise maximum is defined by  \n (max_{t∈𝔇} v(t))_i := max_{t∈𝔇} v_i(t).\n\nConsider the nonlinear system\n\n(★)  λ u = max_{t∈𝔇} [ B(t) + A(t) u ] ,  u ∈ ℝ^{m}.  \n\n(a) Prove that (★) possesses exactly one solution u ∈ ℝ^{m}.  \n\n(b) Put  B̃(t) := S^{−1} B(t) and y := S^{−1} u.  \n Show that the coordinates of y are given explicitly by  \n   y_i = max_{t∈𝔇}  B̃_i(t) / (λ − a_i(t))  (1 ≤ i ≤ m),  \n and hence that the unique solution of (★) admits the closed form  \n   u = S  max_{t∈𝔇} (λ I − D(t))^{−1} B̃(t)                    (♣)  \n equivalently  \n   u = S  max_{t∈𝔇} (λ I − S^{−1} A(t) S)^{−1} S^{−1} B(t).  \n\n(c) Establish the a-priori bound  \n   ‖u‖_∞ ≤ 1/(λ − 0.8) · max_{t∈𝔇} ‖B(t)‖_∞.",
      "solution": "Step 0.  Preparatory facts  \nBecause ‖A(t)‖_∞ ≤ 0.8 < λ, all matrices  \n\n  M(t) := λ I − A(t)   \n\nare nonsingular M–matrices.  Hence M(t)^{−1} ≥ 0 and  \n\n  ‖M(t)^{−1}‖_∞ ≤ 1/(λ − 0.8). (0.1)\n\nDefine the order–preserving operator  \n\n  F(u) := max_{t∈𝔇}[ B(t)+A(t)u ] (u ∈ ℝ^{m}). (0.2)\n\n\n\nStep 1.  Existence and uniqueness – task (a)  \nFor any u,v ∈ ℝ^{m} and every t,  \n\n  ‖A(t)u − A(t)v‖_∞ ≤ 0.8‖u−v‖_∞.\n\nHence  \n\n  ‖F(u) − F(v)‖_∞ ≤ 0.8‖u−v‖_∞.\n\nSet G(u) := (1/λ)F(u).  Then  \n\n  ‖G(u) − G(v)‖_∞ ≤ 0.8/λ · ‖u−v‖_∞ < ‖u−v‖_∞,\n\nso G is a contraction on the complete metric space (ℝ^{m},‖·‖_∞).  By the Banach fixed-point theorem there exists a unique u with u = G(u), i.e. (★) has exactly one solution.  \n\n\n\nStep 2.  Passing to diagonal coordinates  \nBecause S in (♠) is a positive monomial matrix, both S and S^{−1} are non-negative and, crucially,\n\n  S^{−1}(max_{t∈𝔇} v(t)) = max_{t∈𝔇} (S^{−1} v(t))  (0.3)\n\nfor every family {v(t)}.  Indeed, writing S^{−1}=diag(d_1,…,d_m)P^{T} with d_i>0,  \n  (S^{−1}w)_i = d_i w_{σ(i)}, σ a permutation.  \nHence the i-th component of the left–hand side of (0.3) equals  \n  d_i max_t v_{σ(i)}(t)  \nwhile that of the right-hand side equals  \n  max_t d_i v_{σ(i)}(t)=d_i max_t v_{σ(i)}(t).  \n\nConsequently we may left-multiply (★) by S^{−1} and push S^{−1} through the componentwise maximum:\n\n  λ y = max_{t∈𝔇}[ B̃(t)+D(t) y ],   y := S^{−1}u, B̃(t):=S^{−1}B(t). (2.1)\n\n\n\nStep 3.  Solving the decoupled scalar equations – task (b)  \nBecause each D(t) is diagonal, (2.1) splits into m independent scalar relations:\n\n  λ y_i = max_{t∈𝔇}[ B̃_i(t)+a_i(t) y_i ]  (1≤i≤m). (2.2)\n\nFix an index i.  For any t,\n\n  B̃_i(t)+a_i(t) y_i ≤ max_{s∈𝔇} B̃_i(s) + 0.8 y_i,\n\nso from (2.2),\n\n  λ y_i ≤ max_{s∈𝔇} B̃_i(s) + 0.8 y_i → (λ-0.8) y_i ≤ max_{s∈𝔇} B̃_i(s),\n\nhence y_i is finite.\n\nConversely define  \n\n  y_i^* := max_{t∈𝔇} B̃_i(t)/(λ - a_i(t)). (2.3)\n\nFor every t,  \n\n  λ y_i^* ≥ B̃_i(t)+a_i(t) y_i^*,  \n\nand taking the maximum over t yields λ y_i^* ≥ RHS of (2.2).  \nAt a point t* attaining the maximum in (2.3) we have equality, so λ y_i^* equals that maximum.  Therefore y_i^* satisfies (2.2).  Uniqueness from Step 1 forces y_i = y_i^*.\n\nCollecting the coordinates,\n\n  y = max_{t∈𝔇}(λ I − D(t))^{−1} B̃(t),\n\nand returning to u = S y we obtain the explicit representation (♣).  \n\n\n\nStep 4.  A-priori bound – task (c)  \nFrom (★),\n\n  λ‖u‖_∞ = ‖F(u)‖_∞  \n      ≤ max_{t∈𝔇}[ ‖B(t)‖_∞ + ‖A(t)‖_∞‖u‖_∞ ]  \n      ≤ max_{t∈𝔇} ‖B(t)‖_∞ + 0.8‖u‖_∞.\n\nHence (λ−0.8)‖u‖_∞ ≤ max_{t∈𝔇} ‖B(t)‖_∞, i.e.\n\n  ‖u‖_∞ ≤ 1/(λ − 0.8) · max_{t∈𝔇} ‖B(t)‖_∞.\n\nThis completes the proof of all three tasks.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.414617",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: The unknown is now an m–dimensional vector; matrices A(t) enter instead of scalars.  \n• Additional structure: non-negative matrices with spectral‐radius control, M–matrix theory, Neumann series estimates.  \n• Deeper theory: Uses Banach’s fixed-point theorem, properties of M–matrices, spectral radius arguments, and matrix norms.  \n• Multiple interacting concepts: continuity on a higher-dimensional domain, component-wise order, contraction mapping, and explicit inversion of parameter‐dependent matrices.  \n• More steps: establishment of positivity/invertibility, two-sided comparison leading to an equality, Lipschitz-type a-priori bound.\n\nThese layers of matrix analysis and functional analysis make the variant substantially more intricate than the scalar original and the current kernel problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}