{
  "index": "2012-B-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Prove that, for any two bounded functions $g_1, g_2: \\RR \\to [1, \\infty)$,\nthere exist functions $h_1, h_2: \\RR \\to \\RR$ such that, for every $x \\in \\RR$,\n\\[\n\\sup_{s \\in \\RR} (g_1(s)^x g_2(s))  = \\max_{t \\in \\RR} (x h_1(t) + h_2(t)).\n\\]",
  "solution": "Define the function\n\\[\nf(x) = \\sup_{s \\in \\RR} \\{x \\log g_1(s) + \\log g_2(s)\\}.\n\\]\nAs a function of $x$, $f$ is the supremum of a collection of affine functions, so it is convex.\nThe function $e^{f(x)}$ is then also convex, as may be checked directly from the definition:\nfor $x_1, x_2 \\in \\RR$ and $t \\in [0,1]$, by the weighted AM-GM inequality\n\\begin{align*}\nt e^{f(x_1)} + (1-t) e^{f(x_2)}&\\geq e^{t f(x_1) + (1-t)f(x_2)} \\\\\n&\\geq e^{f(t x_1 + (1-t)x_2)}.\n\\end{align*}\nFor each $t \\in \\RR$, draw a supporting line to the graph of $e^{f(x)}$ at $x=t$;\nit has the form $y = x h_1(t) + h_2(t)$ for some $h_1(t), h_2(t) \\in \\RR$. For all $x$, we then have\n\\[\n\\sup_{s \\in \\RR} \\{g_1(s)^x g_2(s) \\} \\geq x h_1(t) + h_2(t)\n\\]\nwith equality for $x = t$. This proves the desired equality (including the fact that the maximum on the right side is achieved).\n\n\\noindent\n\\textbf{Remark.}\nThis problem demonstrates an example of \\emph{duality} for convex functions.",
  "vars": [
    "x",
    "s",
    "t",
    "f",
    "y"
  ],
  "params": [
    "g_1",
    "g_2",
    "h_1",
    "h_2"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variable",
        "s": "argument",
        "t": "parameter",
        "f": "convexfunction",
        "y": "ordinate",
        "g_1": "firstbounded",
        "g_2": "secondbounded",
        "h_1": "firstsupport",
        "h_2": "secondsupport"
      },
      "question": "Prove that, for any two bounded functions $firstbounded, secondbounded: \\RR \\to [1, \\infty)$, there exist functions $firstsupport, secondsupport: \\RR \\to \\RR$ such that, for every $variable \\in \\RR$,\\[\n\\sup_{argument \\in \\RR} (firstbounded(argument)^{variable} secondbounded(argument))  = \\max_{parameter \\in \\RR} (variable firstsupport(parameter) + secondsupport(parameter)).\n\\]",
      "solution": "Define the function\n\\[\nconvexfunction(variable) = \\sup_{argument \\in \\RR} \\{variable \\log firstbounded(argument) + \\log secondbounded(argument)\\}.\n\\]\nAs a function of $variable$, $convexfunction$ is the supremum of a collection of affine functions, so it is convex.\nThe function $e^{convexfunction(variable)}$ is then also convex, as may be checked directly from the definition:\nfor $variable_{1}, variable_{2} \\in \\RR$ and $parameter \\in [0,1]$, by the weighted AM-GM inequality\n\\begin{align*}\nparameter e^{convexfunction(variable_{1})} + (1-parameter) e^{convexfunction(variable_{2})}&\\geq e^{parameter \\, convexfunction(variable_{1}) + (1-parameter)\\,convexfunction(variable_{2})} \\\\\n&\\geq e^{convexfunction(parameter \\, variable_{1} + (1-parameter)\\,variable_{2})}.\n\\end{align*}\nFor each $parameter \\in \\RR$, draw a supporting line to the graph of $e^{convexfunction(variable)}$ at $variable=parameter$;\nit has the form $ordinate = variable firstsupport(parameter) + secondsupport(parameter)$ for some $firstsupport(parameter), secondsupport(parameter) \\in \\RR$. For all $variable$, we then have\n\\[\n\\sup_{argument \\in \\RR} \\{firstbounded(argument)^{variable} secondbounded(argument) \\} \\geq variable firstsupport(parameter) + secondsupport(parameter)\n\\]\nwith equality for $variable = parameter$. This proves the desired equality (including the fact that the maximum on the right side is achieved).\n\n\\noindent\n\\textbf{Remark.}\nThis problem demonstrates an example of \\emph{duality} for convex functions."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sandcastle",
        "s": "lanternfish",
        "t": "parchment",
        "f": "drumbeat",
        "y": "silhouette",
        "g_1": "waterfall",
        "g_2": "moonlight",
        "h_1": "arrowhead",
        "h_2": "windchime"
      },
      "question": "Prove that, for any two bounded functions $waterfall, moonlight: \\RR \\to [1, \\infty)$,\nthere exist functions $arrowhead, windchime: \\RR \\to \\RR$ such that, for every $sandcastle \\in \\RR$,\n\\[\n\\sup_{lanternfish \\in \\RR} (waterfall(lanternfish)^{sandcastle} moonlight(lanternfish))  = \\max_{parchment \\in \\RR} (sandcastle \\, arrowhead(parchment) + windchime(parchment)).\n\\]",
      "solution": "Define the function\n\\[\ndrumbeat(sandcastle) = \\sup_{lanternfish \\in \\RR} \\{sandcastle \\log waterfall(lanternfish) + \\log moonlight(lanternfish)\\}.\n\\]\nAs a function of $sandcastle$, $drumbeat$ is the supremum of a collection of affine functions, so it is convex.\nThe function $e^{drumbeat(sandcastle)}$ is then also convex, as may be checked directly from the definition:\nfor $sandcastle_1, sandcastle_2 \\in \\RR$ and $parchment \\in [0,1]$, by the weighted AM–GM inequality\n\\begin{align*}\nparchment \\, e^{drumbeat(sandcastle_1)} + (1-parchment) \\, e^{drumbeat(sandcastle_2)}&\\ge e^{parchment \\, drumbeat(sandcastle_1) + (1-parchment)drumbeat(sandcastle_2)} \\\\\n&\\ge e^{drumbeat(parchment \\, sandcastle_1 + (1-parchment)sandcastle_2)}.\n\\end{align*}\nFor each $parchment \\in \\RR$, draw a supporting line to the graph of $e^{drumbeat(sandcastle)}$ at $sandcastle=parchment$;\nit has the form $\\silhouette = sandcastle \\, arrowhead(parchment) + windchime(parchment)$ for some $arrowhead(parchment), windchime(parchment) \\in \\RR$. For all $sandcastle$, we then have\n\\[\n\\sup_{lanternfish \\in \\RR} \\{waterfall(lanternfish)^{sandcastle} moonlight(lanternfish) \\} \\ge sandcastle \\, arrowhead(parchment) + windchime(parchment)\n\\]\nwith equality for $sandcastle = parchment$. This proves the desired equality (including the fact that the maximum on the right side is achieved).\n\n\\noindent\n\\textbf{Remark.}\nThis problem demonstrates an example of \\emph{duality} for convex functions."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "immutable",
        "s": "steadfast",
        "t": "timeless",
        "f": "constant",
        "y": "grounded",
        "g_1": "shrinkerone",
        "g_2": "shrinkertwo",
        "h_1": "undermineone",
        "h_2": "underminetwo"
      },
      "question": "Prove that, for any two bounded functions $shrinkerone, shrinkertwo: \\RR \\to [1, \\infty)$,\nthere exist functions $undermineone, underminetwo: \\RR \\to \\RR$ such that, for every $immutable \\in \\RR$,\n\\[\n\\sup_{steadfast \\in \\RR} (shrinkerone(steadfast)^{immutable} shrinkertwo(steadfast))  = \\max_{timeless \\in \\RR} (immutable\\; undermineone(timeless) + underminetwo(timeless)).\n\\]",
      "solution": "Define the function\n\\[\nconstant(immutable) = \\sup_{steadfast \\in \\RR} \\{ immutable \\log shrinkerone(steadfast) + \\log shrinkertwo(steadfast) \\}.\n\\]\nAs a function of $immutable$, $constant$ is the supremum of a collection of affine functions, so it is convex.\nThe function $e^{constant(immutable)}$ is then also convex, as may be checked directly from the definition:\nfor $x_1, x_2 \\in \\RR$ and $timeless \\in [0,1]$, by the weighted AM-GM inequality\n\\begin{align*}\n timeless e^{constant(x_1)} + (1-timeless) e^{constant(x_2)} &\\geq e^{timeless constant(x_1) + (1-timeless)constant(x_2)} \\\\\n &\\geq e^{constant(timeless x_1 + (1-timeless)x_2)}.\n\\end{align*}\nFor each $timeless \\in \\RR$, draw a supporting line to the graph of $e^{constant(immutable)}$ at $immutable=timeless$;\nit has the form $grounded = immutable\\; undermineone(timeless) + underminetwo(timeless)$ for some $undermineone(timeless),\\; underminetwo(timeless) \\in \\RR$. For all $immutable$, we then have\n\\[\n\\sup_{steadfast \\in \\RR} \\{ shrinkerone(steadfast)^{immutable} shrinkertwo(steadfast) \\} \\geq immutable\\; undermineone(timeless) + underminetwo(timeless)\n\\]\nwith equality for $immutable = timeless$. This proves the desired equality (including the fact that the maximum on the right side is achieved).\n\n\\noindent\n\\textbf{Remark.}\nThis problem demonstrates an example of \\emph{duality} for convex functions."
    },
    "garbled_string": {
      "map": {
        "x": "klmqpers",
        "s": "fdjsklwe",
        "t": "ghrplmno",
        "f": "bvnzxcqe",
        "y": "lskdjfue",
        "g_1": "qzxwvtnp",
        "g_2": "hjgtrsla",
        "h_1": "mnbvcxzq",
        "h_2": "plmoknij"
      },
      "question": "Prove that, for any two bounded functions $qzxwvtnp, hjgtrsla: \\RR \\to [1, \\infty)$,\nthere exist functions $mnbvcxzq, plmoknij: \\RR \\to \\RR$ such that, for every $klmqpers \\in \\RR$,\n\\[\n\\sup_{fdjsklwe \\in \\RR} (qzxwvtnp(fdjsklwe)^{klmqpers} hjgtrsla(fdjsklwe))  = \\max_{ghrplmno \\in \\RR} (klmqpers mnbvcxzq(ghrplmno) + plmoknij(ghrplmno)).\n\\]",
      "solution": "Define the function\n\\[\nbvnzxcqe(klmqpers) = \\sup_{fdjsklwe \\in \\RR} \\{klmqpers \\log qzxwvtnp(fdjsklwe) + \\log hjgtrsla(fdjsklwe)\\}.\n\\]\nAs a function of $klmqpers$, $bvnzxcqe$ is the supremum of a collection of affine functions, so it is convex.\nThe function $e^{bvnzxcqe(klmqpers)}$ is then also convex, as may be checked directly from the definition:\nfor $klmqpers_1, klmqpers_2 \\in \\RR$ and $ghrplmno \\in [0,1]$, by the weighted AM-GM inequality\n\\begin{align*}\nghrplmno e^{bvnzxcqe(klmqpers_1)} + (1-ghrplmno) e^{bvnzxcqe(klmqpers_2)}&\\geq e^{ghrplmno bvnzxcqe(klmqpers_1) + (1-ghrplmno)bvnzxcqe(klmqpers_2)} \\\\\n&\\geq e^{bvnzxcqe(ghrplmno klmqpers_1 + (1-ghrplmno)klmqpers_2)}.\n\\end{align*}\nFor each $ghrplmno \\in \\RR$, draw a supporting line to the graph of $e^{bvnzxcqe(klmqpers)}$ at $klmqpers=ghrplmno$;\nit has the form $lskdjfue = klmqpers mnbvcxzq(ghrplmno) + plmoknij(ghrplmno)$ for some $mnbvcxzq(ghrplmno), plmoknij(ghrplmno) \\in \\RR$. For all $klmqpers$, we then have\n\\[\n\\sup_{fdjsklwe \\in \\RR} \\{qzxwvtnp(fdjsklwe)^{klmqpers} hjgtrsla(fdjsklwe) \\} \\geq klmqpers mnbvcxzq(ghrplmno) + plmoknij(ghrplmno)\n\\]\nwith equality for $klmqpers = ghrplmno$. This proves the desired equality (including the fact that the maximum on the right side is achieved).\n\n\\noindent\n\\textbf{Remark.}\nThis problem demonstrates an example of \\emph{duality} for convex functions."
    },
    "kernel_variant": {
      "question": "Let k ≥ 2 and d be positive integers and let A be a k × d real matrix of full row–rank.  \nAssume that the linear system A s = c is consistent, i.e. the affine sub–space  \n\n  S := {s ∈ ℝᵈ : A s = c}                                                                                                          (0)\n\nis non–empty.  \nLet  \n\n  g₀,…,g_k : ℝᵈ → [2,1000]\n\nbe arbitrary bounded Borel maps and, for x = (x₁,…,x_k) ∈ ℝᵏ, define  \n\n  F(x) := sup_{s∈S} g₀(s)·g₁(s)^{x₁}⋯g_k(s)^{x_k},                                          (★)  \n  f := log F.\n\nFor s ∈ S introduce the logarithmic coordinates  \n\n  α(s) := (log g₁(s),…,log g_k(s)) ∈ ℝᵏ, β(s) := log g₀(s) ∈ ℝ.                  (1)\n\nBecause 2 ≤ g_j ≤ 1000 we have  \n\n  ‖α(s)‖₂ ≤ √k log 1000, |β(s)| ≤ log 1000 =: L.                            (2)\n\nProblem  \n\n(i)  Show that f is finite–valued, convex and globally Lipschitz on ℝᵏ.  \n     (Give an explicit Lipschitz constant.)  \n\n(ii) Put   T := cl{(α(s),β(s)): s∈S} ⊂ ℝ^{k+1}.                                       (3)\n\n     Prove that T is compact and that  \n\n  f(x) = max_{(a,b)∈T} (a·x + b) for every x ∈ ℝᵏ.                        (†)\n\n(iii) (Uniform polyhedral approximation)  \n      For every compact K ⊂ ℝᵏ and ε > 0 there exists a finite set  \n      T_{K,ε} ⊂ T such that  \n\n  0 ≤ f(x) − max_{t∈T_{K,ε}}(a(t)·x + b(t)) ≤ ε for all x∈K.            (4)\n\n(iv) (Quantitative version of (iii))  \n      Let  \n\n  M := sup_{s∈S}‖α(s)‖₂ ≤ √k L, diam K := sup_{x,y∈K}‖x−y‖₂,\n\n      and let C_k be a constant depending only on k.  \n      Then one can choose T_{K,ε} so that  \n\n  |T_{K,ε}| ≤ C_k ( M·diam K / ε )^k.                                         (5)\n\n(v) (Polyhedrality criterion)  Denote by conv T the closed convex hull of T.  \n    The following statements are equivalent.  \n\n (a) There exists a finite subset T₀ ⊂ T with  \n        f(x) = max_{t∈T₀}(a(t)·x + b(t)) for all x∈ℝᵏ.  \n\n (b) conv T is a polytope, i.e. the convex hull of finitely many points in ℝ^{k+1}.  \n\n    Moreover, a sufficient (but not necessary) condition for (a)–(b) is  \n\n (c) There exists a finite subset S₀ ⊂ S such that  \n        conv T = conv{(α(s),β(s)) : s∈S₀}.  \n\nTasks (iii)–(v) require both quantitative and qualitative tools from convex geometry.",
      "solution": "Throughout abbreviate  \n\n M := sup_{s∈S}‖α(s)‖₂ ≤ √k L.                                                        (6)\n\n\n\nStep 1. Logarithmic rewriting.  \nTaking logarithms in (★) gives  \n\n f(x) = sup_{s∈S} (x·α(s) + β(s)),                                             (7)\n\nso f is the point–wise supremum of affine functions of x.\n\n\n\n2. Proof of (i).\n\nFiniteness. For every x∈ℝᵏ and s∈S, by (2)\n\n |x·α(s)+β(s)| ≤ ‖x‖₂‖α(s)‖₂ + |β(s)| ≤ ‖x‖₂ M + L,\n\nhence −‖x‖₂ M − L ≤ f(x) ≤ ‖x‖₂ M + L, so f is finite everywhere.\n\nConvexity. A supremum of affine functions is convex.\n\nLipschitzness. For x,y∈ℝᵏ,\n\n |f(x)−f(y)| ≤ sup_{s∈S}|(x−y)·α(s)| ≤ M‖x−y‖₂,                         (8)\n\nthus f is globally M–Lipschitz.\n\n\n\n3. Compact parameter set and representation (†).\n\nBoundedness of α(s),β(s) from (2) implies boundedness of {(α(s),β(s)) : s∈S}; its closure T is therefore compact.\n\nFor a fixed x the map (a,b)↦a·x+b is continuous, so\n\n f(x)=max_{(a,b)∈T}(a·x+b)                                                       (by (7)),\n\nestablishing (†).\n\n\n\n4. Uniform polyhedral approximation – proof of (iii).\n\nLet K ⊂ ℝᵏ be compact and ε>0.  \nPut δ := ε/(2M).  Because K is compact, finitely many closed balls  \nB(x^i, δ) (i=1,…,N) cover K.  For each centre x^i pick (a^i,b^i)∈T such that  \n\n f(x^i)=a^i·x^i+b^i.                                                            (9)\n\nSet T_{K,ε}:={(a^i,b^i):1≤i≤N}.  \nFix x∈K and choose i with ‖x−x^i‖₂≤δ.  Using (8) and (9):\n\n f(x)−(a^i·x+b^i)  \n = [f(x)−f(x^i)] + [a^i·x^i+b^i−a^i·x−b^i]  \n ≤ M‖x−x^i‖₂ + ‖a^i‖₂‖x−x^i‖₂  \n ≤ Mδ + Mδ = 2Mδ = ε.                                                         (10)\n\nBecause every affine form on the right–hand side of (4) is one used in (7), we always have  \n\n max_{t∈T_{K,ε}}(a(t)·x+b(t)) ≤ f(x).  \n\nCombining with (10) proves (4).\n\n\n\n5. Quantitative bound – proof of (iv).\n\nStandard volumetric packing/covering shows that the minimal number N of closed balls of radius δ needed to cover a k–dimensional set of diameter diam K satisfies  \n\n N ≤ C_k (diam K / δ)^k.                                                       (11)\n\nWith δ = ε/(2M) inequality (11) yields  \n\n |T_{K,ε}| = N ≤ C_k ( M·diam K / ε )^k,\n\nwhich is precisely (5).\n\n\n\n6. Polyhedrality criterion – proof of (v).\n\nIntroduce the support functions\n\n σ_T(x) := max_{t∈T} t·(x,1), σ_{T₀}(x) := max_{t∈T₀} t·(x,1).              (12)\n\nObserve that t=(a,b) acts on (x,1) as a·x + b, so condition (a) is exactly σ_T = σ_{T₀}.\n\n(a) ⇒ (b). The function σ_T is the support function of the closed convex set conv T.  \nSince σ_T = σ_{T₀} and T₀ is finite, conv T₀ is a polytope.  \nClosed convex sets with the same support function coincide, hence  \n\n conv T = conv T₀,\n\nso conv T is a polytope.\n\n(b) ⇒ (a). Assume conv T = conv{t¹,…,t^m}.  \nPut T₀ := {t¹,…,t^m}.  For every x ∈ ℝᵏ we have\n\n f(x) = σ_T(x) = σ_{T₀}(x) = max_{1≤j≤m} t^j·(x,1),\n\nyielding (a).\n\nSufficiency of (c). If there exists S₀ with conv T = conv{(α(s),β(s)) : s∈S₀} then, setting  \nT₀ := {(α(s),β(s)) : s∈S₀}, we obtain conv T = conv T₀, hence σ_T = σ_{T₀} and (a) follows, so (b) holds as well.\n\nRemark. The converse (b) ⇒ (c) is false in general:  take k = 1, d = 2, A = (1 0), c = 0, so S = {(0,s₂):s₂∈ℝ}.  \nLet g₀≡1 and g₁(0,s₂)=2+e^{-|s₂|}.  Then T = [log 2, log 3] × {0}.  \nHere conv T = T is a line segment (a polytope), yet no finite S₀ can satisfy (c).\n\n\n\n7. Differentiability remark.\n\nAt every point x where f is differentiable the maximiser in (†) is unique and equals  \n\n (∇f(x), f(x) − ∇f(x)·x).\n\n∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.824736",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension & more variables:  The exponent now contains a k-tuple x (k ≥ 2) instead of a single real; the domain of maximisation is the affine subspace {s : A s=c} of ℝᵈ (d may be arbitrarily large).\n\n2.  Extra constraints:  The maximisation is not over all s but over those satisfying A s=c, forcing the solver to cope with an external linear restriction.\n\n3.  Deeper theory:  The solution must invoke\n   •  vector–valued convex analysis (support functions of compact sets in ℝ^{k+1});  \n   •  topological arguments guaranteeing attainment of suprema via compactification;  \n   •  measurable selection theorems and δ–net (or Carathéodory–type) arguments to extract finite determining sets.\n\n4.  Multiple interacting concepts:  The proof intertwines boundedness, compact closures, supporting hyperplanes, subgradients, Lipschitz estimates, and finite covering arguments.\n\n5.  Additional conclusion (iii):  Producing a finite determining set on each compact K is absent from the original problem and forces the contestant to deploy a quantitative compactness argument beyond the basic supporting-line construction.\n\nAltogether these extensions raise the problem well above the original in both technical breadth and conceptual depth; simple pattern matching no longer suffices."
      }
    },
    "original_kernel_variant": {
      "question": "Let k ≥ 2 and d be positive integers and let A be a k × d real matrix of full row–rank.  \nAssume that the linear system A s = c is consistent, i.e. the affine sub–space  \n\n  S := {s ∈ ℝᵈ : A s = c}                                                                                                          (0)\n\nis non–empty.  \nLet  \n\n  g₀,…,g_k : ℝᵈ → [2,1000]\n\nbe arbitrary bounded Borel maps and, for x = (x₁,…,x_k) ∈ ℝᵏ, define  \n\n  F(x) := sup_{s∈S} g₀(s)·g₁(s)^{x₁}⋯g_k(s)^{x_k},                                          (★)  \n  f := log F.\n\nFor s ∈ S introduce the logarithmic coordinates  \n\n  α(s) := (log g₁(s),…,log g_k(s)) ∈ ℝᵏ, β(s) := log g₀(s) ∈ ℝ.                  (1)\n\nBecause 2 ≤ g_j ≤ 1000 we have  \n\n  ‖α(s)‖₂ ≤ √k log 1000, |β(s)| ≤ log 1000 =: L.                            (2)\n\nProblem  \n\n(i)  Show that f is finite–valued, convex and globally Lipschitz on ℝᵏ.  \n     (Give an explicit Lipschitz constant.)  \n\n(ii) Put   T := cl{(α(s),β(s)): s∈S} ⊂ ℝ^{k+1}.                                       (3)\n\n     Prove that T is compact and that  \n\n  f(x) = max_{(a,b)∈T} (a·x + b) for every x ∈ ℝᵏ.                        (†)\n\n(iii) (Uniform polyhedral approximation)  \n      For every compact K ⊂ ℝᵏ and ε > 0 there exists a finite set  \n      T_{K,ε} ⊂ T such that  \n\n  0 ≤ f(x) − max_{t∈T_{K,ε}}(a(t)·x + b(t)) ≤ ε for all x∈K.            (4)\n\n(iv) (Quantitative version of (iii))  \n      Let  \n\n  M := sup_{s∈S}‖α(s)‖₂ ≤ √k L, diam K := sup_{x,y∈K}‖x−y‖₂,\n\n      and let C_k be a constant depending only on k.  \n      Then one can choose T_{K,ε} so that  \n\n  |T_{K,ε}| ≤ C_k ( M·diam K / ε )^k.                                         (5)\n\n(v) (Polyhedrality criterion)  Denote by conv T the closed convex hull of T.  \n    The following statements are equivalent.  \n\n (a) There exists a finite subset T₀ ⊂ T with  \n        f(x) = max_{t∈T₀}(a(t)·x + b(t)) for all x∈ℝᵏ.  \n\n (b) conv T is a polytope, i.e. the convex hull of finitely many points in ℝ^{k+1}.  \n\n    Moreover, a sufficient (but not necessary) condition for (a)–(b) is  \n\n (c) There exists a finite subset S₀ ⊂ S such that  \n        conv T = conv{(α(s),β(s)) : s∈S₀}.  \n\nTasks (iii)–(v) require both quantitative and qualitative tools from convex geometry.",
      "solution": "Throughout abbreviate  \n\n M := sup_{s∈S}‖α(s)‖₂ ≤ √k L.                                                        (6)\n\n\n\nStep 1. Logarithmic rewriting.  \nTaking logarithms in (★) gives  \n\n f(x) = sup_{s∈S} (x·α(s) + β(s)),                                             (7)\n\nso f is the point–wise supremum of affine functions of x.\n\n\n\n2. Proof of (i).\n\nFiniteness. For every x∈ℝᵏ and s∈S, by (2)\n\n |x·α(s)+β(s)| ≤ ‖x‖₂‖α(s)‖₂ + |β(s)| ≤ ‖x‖₂ M + L,\n\nhence −‖x‖₂ M − L ≤ f(x) ≤ ‖x‖₂ M + L, so f is finite everywhere.\n\nConvexity. A supremum of affine functions is convex.\n\nLipschitzness. For x,y∈ℝᵏ,\n\n |f(x)−f(y)| ≤ sup_{s∈S}|(x−y)·α(s)| ≤ M‖x−y‖₂,                         (8)\n\nthus f is globally M–Lipschitz.\n\n\n\n3. Compact parameter set and representation (†).\n\nBoundedness of α(s),β(s) from (2) implies boundedness of {(α(s),β(s)) : s∈S}; its closure T is therefore compact.\n\nFor a fixed x the map (a,b)↦a·x+b is continuous, so\n\n f(x)=max_{(a,b)∈T}(a·x+b)                                                       (by (7)),\n\nestablishing (†).\n\n\n\n4. Uniform polyhedral approximation – proof of (iii).\n\nLet K ⊂ ℝᵏ be compact and ε>0.  \nPut δ := ε/(2M).  Because K is compact, finitely many closed balls  \nB(x^i, δ) (i=1,…,N) cover K.  For each centre x^i pick (a^i,b^i)∈T such that  \n\n f(x^i)=a^i·x^i+b^i.                                                            (9)\n\nSet T_{K,ε}:={(a^i,b^i):1≤i≤N}.  \nFix x∈K and choose i with ‖x−x^i‖₂≤δ.  Using (8) and (9):\n\n f(x)−(a^i·x+b^i)  \n = [f(x)−f(x^i)] + [a^i·x^i+b^i−a^i·x−b^i]  \n ≤ M‖x−x^i‖₂ + ‖a^i‖₂‖x−x^i‖₂  \n ≤ Mδ + Mδ = 2Mδ = ε.                                                         (10)\n\nBecause every affine form on the right–hand side of (4) is one used in (7), we always have  \n\n max_{t∈T_{K,ε}}(a(t)·x+b(t)) ≤ f(x).  \n\nCombining with (10) proves (4).\n\n\n\n5. Quantitative bound – proof of (iv).\n\nStandard volumetric packing/covering shows that the minimal number N of closed balls of radius δ needed to cover a k–dimensional set of diameter diam K satisfies  \n\n N ≤ C_k (diam K / δ)^k.                                                       (11)\n\nWith δ = ε/(2M) inequality (11) yields  \n\n |T_{K,ε}| = N ≤ C_k ( M·diam K / ε )^k,\n\nwhich is precisely (5).\n\n\n\n6. Polyhedrality criterion – proof of (v).\n\nIntroduce the support functions\n\n σ_T(x) := max_{t∈T} t·(x,1), σ_{T₀}(x) := max_{t∈T₀} t·(x,1).              (12)\n\nObserve that t=(a,b) acts on (x,1) as a·x + b, so condition (a) is exactly σ_T = σ_{T₀}.\n\n(a) ⇒ (b). The function σ_T is the support function of the closed convex set conv T.  \nSince σ_T = σ_{T₀} and T₀ is finite, conv T₀ is a polytope.  \nClosed convex sets with the same support function coincide, hence  \n\n conv T = conv T₀,\n\nso conv T is a polytope.\n\n(b) ⇒ (a). Assume conv T = conv{t¹,…,t^m}.  \nPut T₀ := {t¹,…,t^m}.  For every x ∈ ℝᵏ we have\n\n f(x) = σ_T(x) = σ_{T₀}(x) = max_{1≤j≤m} t^j·(x,1),\n\nyielding (a).\n\nSufficiency of (c). If there exists S₀ with conv T = conv{(α(s),β(s)) : s∈S₀} then, setting  \nT₀ := {(α(s),β(s)) : s∈S₀}, we obtain conv T = conv T₀, hence σ_T = σ_{T₀} and (a) follows, so (b) holds as well.\n\nRemark. The converse (b) ⇒ (c) is false in general:  take k = 1, d = 2, A = (1 0), c = 0, so S = {(0,s₂):s₂∈ℝ}.  \nLet g₀≡1 and g₁(0,s₂)=2+e^{-|s₂|}.  Then T = [log 2, log 3] × {0}.  \nHere conv T = T is a line segment (a polytope), yet no finite S₀ can satisfy (c).\n\n\n\n7. Differentiability remark.\n\nAt every point x where f is differentiable the maximiser in (†) is unique and equals  \n\n (∇f(x), f(x) − ∇f(x)·x).\n\n∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.630781",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension & more variables:  The exponent now contains a k-tuple x (k ≥ 2) instead of a single real; the domain of maximisation is the affine subspace {s : A s=c} of ℝᵈ (d may be arbitrarily large).\n\n2.  Extra constraints:  The maximisation is not over all s but over those satisfying A s=c, forcing the solver to cope with an external linear restriction.\n\n3.  Deeper theory:  The solution must invoke\n   •  vector–valued convex analysis (support functions of compact sets in ℝ^{k+1});  \n   •  topological arguments guaranteeing attainment of suprema via compactification;  \n   •  measurable selection theorems and δ–net (or Carathéodory–type) arguments to extract finite determining sets.\n\n4.  Multiple interacting concepts:  The proof intertwines boundedness, compact closures, supporting hyperplanes, subgradients, Lipschitz estimates, and finite covering arguments.\n\n5.  Additional conclusion (iii):  Producing a finite determining set on each compact K is absent from the original problem and forces the contestant to deploy a quantitative compactness argument beyond the basic supporting-line construction.\n\nAltogether these extensions raise the problem well above the original in both technical breadth and conceptual depth; simple pattern matching no longer suffices."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}