{
  "index": "1993-B-4",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "The function $K(x,y)$ is positive and continuous for $0 \\leq x \\leq 1, 0\n\\leq y \\leq 1$, and the functions $f(x)$ and $g(x)$ are positive and\ncontinuous for $0 \\leq x \\leq 1$. Suppose that for all $x$, $0 \\leq x \\leq 1$,\n\\[\n\\int_0^1 f(y)K(x,y)\\,dy = g(x)\n\\]\nand\n\\[\n\\int_0^1 g(y)K(x,y)\\,dy = f(x).\n\\]\nShow that $f(x) = g(x)$ for $0 \\leq x \\leq 1$.",
  "solution": "Solution. For convenience of notation, define the linear operator \\( T \\) by\n\\[\n(T h)(x)=\\int_{0}^{1} h(y) K(x, y) d y\n\\]\nfor any continuous function \\( h \\) on \\( [0,1] \\). Then \\( T f=g \\) and \\( T g=f \\).\nBy the Extreme Value Theorem [Ap1, Section 3.16], the continuous function \\( f(x) / g(x) \\) attains a minimum value on \\( [0,1] \\), say \\( r \\). Thus \\( f(x)-r g(x) \\geq 0 \\) on \\( [0,1] \\), with equality somewhere. Suppose that \\( f-r g \\) is not identically zero on \\( [0,1] \\). Then by continuity \\( f-r g \\) is positive on some interval, so \\( T(f-r g) \\) is positive on \\( [0,1] \\), and so is \\( T^{2}(f-r g) \\). But by linearity, \\( T^{2}(f-r g)=T(g-r f)=f-r g \\), which is zero somewhere. This contradiction shows that \\( f-r g=0 \\). Thus \\( T g=r g \\), and \\( g=T^{2} g=r^{2} g \\), so \\( r^{2}=1 \\). Also \\( r \\geq 0 \\), so \\( r=1 \\). Hence \\( f=g \\).",
  "vars": [
    "x",
    "y",
    "h"
  ],
  "params": [
    "f",
    "g",
    "K",
    "T",
    "r"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "indepx",
        "y": "indepy",
        "h": "auxfun",
        "f": "funone",
        "g": "funtwo",
        "K": "kernel",
        "T": "operator",
        "r": "minratio"
      },
      "question": "The function $kernel(indepx,indepy)$ is positive and continuous for $0 \\leq indepx \\leq 1, 0\\leq indepy \\leq 1$, and the functions $funone(indepx)$ and $funtwo(indepx)$ are positive and continuous for $0 \\leq indepx \\leq 1$. Suppose that for all $indepx$, $0 \\leq indepx \\leq 1$,\\[\\int_0^1 funone(indepy)kernel(indepx,indepy)\\,dindepy = funtwo(indepx)\\]and\\[\\int_0^1 funtwo(indepy)kernel(indepx,indepy)\\,dindepy = funone(indepx).\\]Show that $funone(indepx) = funtwo(indepx)$ for $0 \\leq indepx \\leq 1$.",
      "solution": "Solution. For convenience of notation, define the linear operator \\( operator \\) by\\[ (operator\\,auxfun)(indepx)=\\int_{0}^{1} auxfun(indepy) \\; kernel(indepx, indepy) \\, dindepy \\]for any continuous function \\( auxfun \\) on \\( [0,1] \\). Then \\( operator\\,funone = funtwo \\) and \\( operator\\,funtwo = funone \\). By the Extreme Value Theorem [Ap1, Section 3.16], the continuous function \\( funone(indepx) / funtwo(indepx) \\) attains a minimum value on \\( [0,1] \\), say \\( minratio \\). Thus \\( funone(indepx)-minratio\\,funtwo(indepx) \\geq 0 \\) on \\( [0,1] \\), with equality somewhere. Suppose that \\( funone-minratio\\,funtwo \\) is not identically zero on \\( [0,1] \\). Then by continuity \\( funone-minratio\\,funtwo \\) is positive on some interval, so \\( operator(funone-minratio\\,funtwo) \\) is positive on \\( [0,1] \\), and so is \\( operator^{2}(funone-minratio\\,funtwo) \\). But by linearity, \\( operator^{2}(funone-minratio\\,funtwo)=operator(funtwo-minratio\\,funone)=funone-minratio\\,funtwo \\), which is zero somewhere. This contradiction shows that \\( funone-minratio\\,funtwo=0 \\). Thus \\( operator\\,funtwo=minratio\\,funtwo \\), and \\( funtwo=operator^{2}\\,funtwo=minratio^{2}\\,funtwo \\), so \\( minratio^{2}=1 \\). Also \\( minratio \\geq 0 \\), so \\( minratio=1 \\). Hence \\( funone=funtwo \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sandcastle",
        "y": "lanternfish",
        "h": "marshmallow",
        "f": "tapestry",
        "g": "whirlpool",
        "K": "pendulum",
        "T": "crossbow",
        "r": "gemstone"
      },
      "question": "The function $pendulum(sandcastle,lanternfish)$ is positive and continuous for $0 \\leq sandcastle \\leq 1, 0\\n\\leq lanternfish \\leq 1$, and the functions $tapestry(sandcastle)$ and $whirlpool(sandcastle)$ are positive and\\ncontinuous for $0 \\leq sandcastle \\leq 1$. Suppose that for all $sandcastle$, $0 \\leq sandcastle \\leq 1$,\\n\\[\\n\\int_0^1 tapestry(lanternfish) \\, pendulum(sandcastle,lanternfish)\\,d\\!lanternfish = whirlpool(sandcastle)\\n\\]\\nand\\n\\[\\n\\int_0^1 whirlpool(lanternfish) \\, pendulum(sandcastle,lanternfish)\\,d\\!lanternfish = tapestry(sandcastle).\\n\\]\\nShow that $tapestry(sandcastle) = whirlpool(sandcastle)$ for $0 \\leq sandcastle \\leq 1$.",
      "solution": "Solution. For convenience of notation, define the linear operator \\( crossbow \\) by\\n\\[\\n(crossbow\\, marshmallow)(sandcastle)=\\int_{0}^{1} marshmallow(lanternfish)\\, pendulum(sandcastle, lanternfish)\\, d\\!lanternfish\\n\\]\\nfor any continuous function \\( marshmallow \\) on \\( [0,1] \\). Then \\( crossbow\\, tapestry=whirlpool \\) and \\( crossbow\\, whirlpool=tapestry \\).\\nBy the Extreme Value Theorem [Ap1, Section 3.16], the continuous function \\( tapestry(sandcastle) / whirlpool(sandcastle) \\) attains a minimum value on \\( [0,1] \\), say \\( gemstone \\). Thus \\( tapestry(sandcastle)-gemstone\\, whirlpool(sandcastle) \\geq 0 \\) on \\( [0,1] \\), with equality somewhere. Suppose that \\( tapestry-gemstone\\,whirlpool \\) is not identically zero on \\( [0,1] \\). Then by continuity \\( tapestry-gemstone\\,whirlpool \\) is positive on some interval, so \\( crossbow(tapestry-gemstone\\,whirlpool) \\) is positive on \\( [0,1] \\), and so is \\( crossbow^{2}(tapestry-gemstone\\,whirlpool) \\). But by linearity, \\( crossbow^{2}(tapestry-gemstone\\,whirlpool)=crossbow(whirlpool-gemstone\\,tapestry)=tapestry-gemstone\\,whirlpool \\), which is zero somewhere. This contradiction shows that \\( tapestry-gemstone\\,whirlpool=0 \\). Thus \\( crossbow\\, whirlpool=gemstone\\,whirlpool \\), and \\( whirlpool=crossbow^{2} whirlpool=gemstone^{2} whirlpool \\), so \\( gemstone^{2}=1 \\). Also \\( gemstone \\geq 0 \\), so \\( gemstone=1 \\). Hence \\( tapestry=whirlpool \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "outerplace",
        "y": "innerplace",
        "h": "constantfun",
        "f": "negativefunction",
        "g": "zerofunction",
        "K": "nullkernel",
        "T": "nonlinear",
        "r": "maximumvalue"
      },
      "question": "The function $nullkernel(outerplace,innerplace)$ is positive and continuous for $0 \\leq outerplace \\leq 1, 0 \\leq innerplace \\leq 1$, and the functions $negativefunction(outerplace)$ and $zerofunction(outerplace)$ are positive and continuous for $0 \\leq outerplace \\leq 1$. Suppose that for all $outerplace$, $0 \\leq outerplace \\leq 1$,\\[\n\\int_0^1 negativefunction(innerplace)nullkernel(outerplace,innerplace)\\,dinnerplace = zerofunction(outerplace)\n\\]and\\[\n\\int_0^1 zerofunction(innerplace)nullkernel(outerplace,innerplace)\\,dinnerplace = negativefunction(outerplace).\n\\]Show that $negativefunction(outerplace) = zerofunction(outerplace)$ for $0 \\leq outerplace \\leq 1$.",
      "solution": "Solution. For convenience of notation, define the linear operator \\( nonlinear \\) by\n\\[\n(nonlinear\\, constantfun)(outerplace)=\\int_{0}^{1} constantfun(innerplace) nullkernel(outerplace, innerplace) d innerplace\n\\]\nfor any continuous function \\( constantfun \\) on \\( [0,1] \\). Then \\( nonlinear\\, negativefunction=zerofunction \\) and \\( nonlinear\\, zerofunction=negativefunction \\).\nBy the Extreme Value Theorem [Ap1, Section 3.16], the continuous function \\( negativefunction(outerplace) / zerofunction(outerplace) \\) attains a minimum value on \\( [0,1] \\), say \\( maximumvalue \\). Thus \\( negativefunction(outerplace)-maximumvalue \\, zerofunction(outerplace) \\geq 0 \\) on \\( [0,1] \\), with equality somewhere. Suppose that \\( negativefunction-maximumvalue \\, zerofunction \\) is not identically zero on \\( [0,1] \\). Then by continuity \\( negativefunction-maximumvalue \\, zerofunction \\) is positive on some interval, so \\( nonlinear(negativefunction-maximumvalue \\, zerofunction) \\) is positive on \\( [0,1] \\), and so is \\( nonlinear^{2}(negativefunction-maximumvalue \\, zerofunction) \\). But by linearity, \\( nonlinear^{2}(negativefunction-maximumvalue \\, zerofunction)=nonlinear(zerofunction-maximumvalue \\, negativefunction)=negativefunction-maximumvalue \\, zerofunction \\), which is zero somewhere. This contradiction shows that \\( negativefunction-maximumvalue \\, zerofunction=0 \\). Thus \\( nonlinear\\, zerofunction=maximumvalue \\, zerofunction \\), and \\( zerofunction=nonlinear^{2} \\, zerofunction=maximumvalue^{2} \\, zerofunction \\), so \\( maximumvalue^{2}=1 \\). Also \\( maximumvalue \\geq 0 \\), so \\( maximumvalue=1 \\). Hence \\( negativefunction=zerofunction \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "h": "vcmptlqe",
        "f": "iksdhpor",
        "g": "zmbqltna",
        "K": "rnfovdae",
        "T": "dawumepy",
        "r": "ybznkqse"
      },
      "question": "The function $rnfovdae(qzxwvtnp,hjgrksla)$ is positive and continuous for $0 \\leq qzxwvtnp \\leq 1, 0\n\\leq hjgrksla \\leq 1$, and the functions $iksdhpor(qzxwvtnp)$ and $zmbqltna(qzxwvtnp)$ are positive and\ncontinuous for $0 \\leq qzxwvtnp \\leq 1$. Suppose that for all $qzxwvtnp$, $0 \\leq qzxwvtnp \\leq 1$,\n\\[\n\\int_0^1 iksdhpor(hjgrksla) rnfovdae(qzxwvtnp,hjgrksla)\\,d hjgrksla = zmbqltna(qzxwvtnp)\n\\]\nand\n\\[\n\\int_0^1 zmbqltna(hjgrksla) rnfovdae(qzxwvtnp,hjgrksla)\\,d hjgrksla = iksdhpor(qzxwvtnp).\n\\]\nShow that $iksdhpor(qzxwvtnp) = zmbqltna(qzxwvtnp)$ for $0 \\leq qzxwvtnp \\leq 1$.",
      "solution": "Solution. For convenience of notation, define the linear operator \\( dawumepy \\) by\n\\[\n(dawumepy \\, vcmptlqe)(qzxwvtnp)=\\int_{0}^{1} vcmptlqe(hjgrksla)\\, rnfovdae(qzxwvtnp, hjgrksla)\\,d hjgrksla\n\\]\nfor any continuous function \\( vcmptlqe \\) on \\( [0,1] \\). Then \\( dawumepy \\, iksdhpor = zmbqltna \\) and \\( dawumepy \\, zmbqltna = iksdhpor \\).\nBy the Extreme Value Theorem [Ap1, Section 3.16], the continuous function \\( iksdhpor(qzxwvtnp) / zmbqltna(qzxwvtnp) \\) attains a minimum value on \\( [0,1] \\), say \\( ybznkqse \\). Thus \\( iksdhpor(qzxwvtnp)- ybznkqse\\, zmbqltna(qzxwvtnp) \\geq 0 \\) on \\( [0,1] \\), with equality somewhere. Suppose that \\( iksdhpor - ybznkqse\\, zmbqltna \\) is not identically zero on \\( [0,1] \\). Then by continuity \\( iksdhpor - ybznkqse\\, zmbqltna \\) is positive on some interval, so \\( dawumepy( iksdhpor - ybznkqse\\, zmbqltna) \\) is positive on \\( [0,1] \\), and so is \\( dawumepy^{2}( iksdhpor - ybznkqse\\, zmbqltna) \\). But by linearity, \\( dawumepy^{2}( iksdhpor - ybznkqse\\, zmbqltna)= dawumepy( zmbqltna - ybznkqse\\, iksdhpor)= iksdhpor - ybznkqse\\, zmbqltna \\), which is zero somewhere. This contradiction shows that \\( iksdhpor - ybznkqse\\, zmbqltna = 0 \\). Thus \\( dawumepy \\, zmbqltna = ybznkqse \\, zmbqltna \\), and \\( zmbqltna = dawumepy^{2} \\, zmbqltna = ybznkqse^{2} \\, zmbqltna \\), so \\( ybznkqse^{2}=1 \\). Also \\( ybznkqse \\geq 0 \\), so \\( ybznkqse = 1 \\). Hence \\( iksdhpor = zmbqltna \\)."
    },
    "kernel_variant": {
      "question": "Let an integer n ≥ 2 be fixed and set  \n Ω := [−1,2]ⁿ ⊂ ℝⁿ, equipped with the n–dimensional Lebesgue measure dx.  \n\nAssume  \n\n(1)  K : Ω × Ω → ℝ is of class C², strictly positive and symmetric, i.e.  \n  K(x,y)=K(y,x)>0 for every x,y∈Ω.  \n\n(2)  The integral operator  \n  (Th)(x) := ∫_Ω K(x,y) h(y) dy, x∈Ω,  \nacts on C(Ω̄); thus T ∈ 𝓛(C(Ω̄)).  \n\nTwo functions f , g ∈ C²(Ω̄) satisfy  \n\n (a)  Coupled Fredholm relations Tf = g and Tg = f in Ω;  \n\n (b)  Homogeneous Dirichlet boundary conditions f|_{∂Ω}=g|_{∂Ω}=0;  \n\n (c)  A common Helmholtz law −Δf = λf and −Δg = λg in Ω  \n   for some real constant λ > 0;  \n\n (d)  Strict interior positivity f(x)>0 , g(x)>0 for all x∈Ω.  \n\nProve that  \n\n(i)  f(x)=g(x) for every x∈Ω;  \n\n(ii) λ equals the first (smallest) Dirichlet eigenvalue λ₁(Ω) of −Δ on Ω.  \n\nConsequently, f=g coincides—up to a positive multiplicative factor—with the unique positive first Dirichlet eigenfunction of −Δ on Ω.",
      "solution": "All Banach–lattice notions (positivity, spectral radius, etc.) are considered in the space C(Ω̄) endowed with ‖·‖_∞.  An arrow “≫” denotes strict positivity.\n\n────────────────────────────────────────────────────────\nStep 1.  Compactness and strong positivity of T and T².  \nBecause K∈C²(Ω̄×Ω̄) and Ω̄×Ω̄ is compact, the kernel is uniformly continuous; hence T: C(Ω̄)→C(Ω̄) is linear, bounded and compact (Arzelà–Ascoli).  Since K>0,  \n\n h ≥ 0, h ≢ 0 ⇒ (Th)(x) > 0 ∀x∈Ω                             (1.1)\n\nholds, i.e. T is strongly positive, and so is T².  \nBy the Kreĭn–Rutman theorem the spectral radius r(T²)>0 is an algebraically simple eigenvalue; the corresponding eigenfunction ψ satisfies ψ≫0 and is unique up to positive scaling.\n\n────────────────────────────────────────────────────────\nStep 2.  Spectrum of T²; first comparison of f and g.  \nUsing the relations (a) one computes\n\n T²f = T(Tf) = T(g) = f,  T²g = T(Tg) = T(f) = g.        (2.1)\n\nThus f and g are positive eigenfunctions of T² with eigenvalue 1, so\n\n 1 = r(T²).                                                  (2.2)\n\nBy the simplicity of r(T²) there is a constant α>0 such that\n\n f = α g.                                                     (2.3)\n\n────────────────────────────────────────────────────────\nStep 3.  The proportionality constant α satisfies α=1.  \nInsert f=αg into Tf=g:\n\n Tf = T(αg) = α Tg = α f = α² g.                            (3.1)\n\nBut Tf=g by hypothesis, hence g=α² g.  Strict positivity of g forces\n\n α² = 1 ⇒ α = 1 (α>0 by construction).                      (3.2)\n\nConsequently\n\n f≡g on Ω̄.                                                  (3.3)\n\nAssertion (i) is proved.\n\n────────────────────────────────────────────────────────\nStep 4.  Reduction to a single unknown.  \nSet u:=f=g.  Then\n\n Tu=u, −Δu=λu in Ω, u=0 on ∂Ω, u>0 in Ω.                   (4.1)\n\n────────────────────────────────────────────────────────\nStep 5.  Identification of λ.  \nFirst recall two classical facts.\n\n(5.1)  The variational characterisation:  \n\n λ₁(Ω)=min_{v∈H₀¹(Ω)\\{0}} ∫|∇v|² / ∫v² .                    (5.2)\n\n(5.2)  Courant’s nodal domain theorem: any Dirichlet eigenfunction for\n−Δ having eigenvalue strictly larger than λ₁(Ω) possesses at least two nodal domains.\n\nBecause u solves −Δu=λu and u>0 in Ω, u has exactly one nodal domain.  Therefore λ cannot exceed λ₁(Ω).  On the other hand, the Rayleigh quotient of u equals λ by (4.1) and by definition is bounded below by λ₁(Ω).  Hence\n\n λ₁(Ω) ≤ λ ≤ λ₁(Ω) ⇒ λ = λ₁(Ω).                             (5.3)\n\nThis proves assertion (ii).\n\n────────────────────────────────────────────────────────\nStep 6.  Uniqueness up to scaling.  \nSince λ=λ₁(Ω) and the first eigenspace is one–dimensional, u must be a positive multiple of the (normalised) first eigenfunction φ₁.  Linearity of T and Tu=u show that every c φ₁ (c>0) satisfies the full coupled system, and no further solutions exist.  The family\n\n { c φ₁ : c>0 }                                              (6.1)\n\ntherefore exhausts all pairs (f,g). ▢",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.733419",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensional setting: the domain is Ω=[−1,2]ⁿ with arbitrary n≥2 instead of an interval, so both elliptic PDE theory and multi-variable integration are required.\n\n2. Additional structures: the problem mixes three distinct operators—\n   • a compact positive integral operator T,  \n   • the Laplacian −Δ with Dirichlet boundary conditions,  \n   • the boundary trace operator—whose eigenstructures have to be related.\n\n3. Extra constraints: the simultaneous satisfaction of Tf=g, Tg=f, vanishing boundary\n   data, and −Δf=λf, −Δg=λg forces the solver to coordinate tools from integral-operator\n   theory (Kreĭn–Rutman), elliptic regularity, the strong maximum principle, Hopf’s lemma,\n   and the variational characterization of Laplacian eigenvalues.\n\n4. Deeper insights: establishing f=g is no longer sufficient; one must also identify\n   the specific eigenvalue λ and show that no other positive solution can appear.\n   This requires a spectral comparison between T and −Δ and a delicate use of sign\n   properties of higher Laplacian eigenfunctions.\n\n5. More steps: the solution now proceeds through six substantial stages (positivity,\n   ratio argument, spectral radius of T, expansion in Laplacian eigenbasis, variational\n   minimization, uniqueness), each invoking advanced results absent from the original\n   exercise.\n\nAll these additions make the variant significantly more intricate than either the\noriginal problem or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let an integer n ≥ 2 be fixed and set  \n Ω := [−1,2]ⁿ ⊂ ℝⁿ, equipped with the n–dimensional Lebesgue measure dx.  \n\nAssume  \n\n(1)  K : Ω × Ω → ℝ is of class C², strictly positive and symmetric, i.e.  \n  K(x,y)=K(y,x)>0 for every x,y∈Ω.  \n\n(2)  The integral operator  \n  (Th)(x) := ∫_Ω K(x,y) h(y) dy, x∈Ω,  \nacts on C(Ω̄); thus T ∈ 𝓛(C(Ω̄)).  \n\nTwo functions f , g ∈ C²(Ω̄) satisfy  \n\n (a)  Coupled Fredholm relations Tf = g and Tg = f in Ω;  \n\n (b)  Homogeneous Dirichlet boundary conditions f|_{∂Ω}=g|_{∂Ω}=0;  \n\n (c)  A common Helmholtz law −Δf = λf and −Δg = λg in Ω  \n   for some real constant λ > 0;  \n\n (d)  Strict interior positivity f(x)>0 , g(x)>0 for all x∈Ω.  \n\nProve that  \n\n(i)  f(x)=g(x) for every x∈Ω;  \n\n(ii) λ equals the first (smallest) Dirichlet eigenvalue λ₁(Ω) of −Δ on Ω.  \n\nConsequently, f=g coincides—up to a positive multiplicative factor—with the unique positive first Dirichlet eigenfunction of −Δ on Ω.",
      "solution": "All Banach–lattice notions (positivity, spectral radius, etc.) are considered in the space C(Ω̄) endowed with ‖·‖_∞.  An arrow “≫” denotes strict positivity.\n\n────────────────────────────────────────────────────────\nStep 1.  Compactness and strong positivity of T and T².  \nBecause K∈C²(Ω̄×Ω̄) and Ω̄×Ω̄ is compact, the kernel is uniformly continuous; hence T: C(Ω̄)→C(Ω̄) is linear, bounded and compact (Arzelà–Ascoli).  Since K>0,  \n\n h ≥ 0, h ≢ 0 ⇒ (Th)(x) > 0 ∀x∈Ω                             (1.1)\n\nholds, i.e. T is strongly positive, and so is T².  \nBy the Kreĭn–Rutman theorem the spectral radius r(T²)>0 is an algebraically simple eigenvalue; the corresponding eigenfunction ψ satisfies ψ≫0 and is unique up to positive scaling.\n\n────────────────────────────────────────────────────────\nStep 2.  Spectrum of T²; first comparison of f and g.  \nUsing the relations (a) one computes\n\n T²f = T(Tf) = T(g) = f,  T²g = T(Tg) = T(f) = g.        (2.1)\n\nThus f and g are positive eigenfunctions of T² with eigenvalue 1, so\n\n 1 = r(T²).                                                  (2.2)\n\nBy the simplicity of r(T²) there is a constant α>0 such that\n\n f = α g.                                                     (2.3)\n\n────────────────────────────────────────────────────────\nStep 3.  The proportionality constant α satisfies α=1.  \nInsert f=αg into Tf=g:\n\n Tf = T(αg) = α Tg = α f = α² g.                            (3.1)\n\nBut Tf=g by hypothesis, hence g=α² g.  Strict positivity of g forces\n\n α² = 1 ⇒ α = 1 (α>0 by construction).                      (3.2)\n\nConsequently\n\n f≡g on Ω̄.                                                  (3.3)\n\nAssertion (i) is proved.\n\n────────────────────────────────────────────────────────\nStep 4.  Reduction to a single unknown.  \nSet u:=f=g.  Then\n\n Tu=u, −Δu=λu in Ω, u=0 on ∂Ω, u>0 in Ω.                   (4.1)\n\n────────────────────────────────────────────────────────\nStep 5.  Identification of λ.  \nFirst recall two classical facts.\n\n(5.1)  The variational characterisation:  \n\n λ₁(Ω)=min_{v∈H₀¹(Ω)\\{0}} ∫|∇v|² / ∫v² .                    (5.2)\n\n(5.2)  Courant’s nodal domain theorem: any Dirichlet eigenfunction for\n−Δ having eigenvalue strictly larger than λ₁(Ω) possesses at least two nodal domains.\n\nBecause u solves −Δu=λu and u>0 in Ω, u has exactly one nodal domain.  Therefore λ cannot exceed λ₁(Ω).  On the other hand, the Rayleigh quotient of u equals λ by (4.1) and by definition is bounded below by λ₁(Ω).  Hence\n\n λ₁(Ω) ≤ λ ≤ λ₁(Ω) ⇒ λ = λ₁(Ω).                             (5.3)\n\nThis proves assertion (ii).\n\n────────────────────────────────────────────────────────\nStep 6.  Uniqueness up to scaling.  \nSince λ=λ₁(Ω) and the first eigenspace is one–dimensional, u must be a positive multiple of the (normalised) first eigenfunction φ₁.  Linearity of T and Tu=u show that every c φ₁ (c>0) satisfies the full coupled system, and no further solutions exist.  The family\n\n { c φ₁ : c>0 }                                              (6.1)\n\ntherefore exhausts all pairs (f,g). ▢",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.568544",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensional setting: the domain is Ω=[−1,2]ⁿ with arbitrary n≥2 instead of an interval, so both elliptic PDE theory and multi-variable integration are required.\n\n2. Additional structures: the problem mixes three distinct operators—\n   • a compact positive integral operator T,  \n   • the Laplacian −Δ with Dirichlet boundary conditions,  \n   • the boundary trace operator—whose eigenstructures have to be related.\n\n3. Extra constraints: the simultaneous satisfaction of Tf=g, Tg=f, vanishing boundary\n   data, and −Δf=λf, −Δg=λg forces the solver to coordinate tools from integral-operator\n   theory (Kreĭn–Rutman), elliptic regularity, the strong maximum principle, Hopf’s lemma,\n   and the variational characterization of Laplacian eigenvalues.\n\n4. Deeper insights: establishing f=g is no longer sufficient; one must also identify\n   the specific eigenvalue λ and show that no other positive solution can appear.\n   This requires a spectral comparison between T and −Δ and a delicate use of sign\n   properties of higher Laplacian eigenfunctions.\n\n5. More steps: the solution now proceeds through six substantial stages (positivity,\n   ratio argument, spectral radius of T, expansion in Laplacian eigenbasis, variational\n   minimization, uniqueness), each invoking advanced results absent from the original\n   exercise.\n\nAll these additions make the variant significantly more intricate than either the\noriginal problem or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}