{
  "index": "2010-A-3",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Suppose that the function $h:\\mathbb{R}^2\\to \\mathbb{R}$ has continuous partial\nderivatives and satisfies the equation\n\\[\nh(x,y) = a \\frac{\\partial h}{\\partial x}(x,y) +\nb \\frac{\\partial h}{\\partial y}(x,y)\n\\]\nfor some constants $a,b$. Prove that if there is a constant $M$ such that\n$|h(x,y)|\\leq M$ for all $(x,y) \\in \\mathbb{R}^2$, then $h$ is identically zero.",
  "solution": "If $a=b=0$, then the desired result holds trivially, so we assume that at least one of $a,b$ is nonzero.\nPick any point $(a_0, b_0) \\in \\mathbb{R}^2$, and let $L$ be the line given by the parametric equation\n$L(t) = (a_0,b_0) + (a,b) t$ for $t\\in \\mathbb{R}$. By the chain rule and the given equation, we have $\\frac{d}{dt}(h\\circ L) = h\\circ L$. If we write $f = h\\circ L:\\mathbb{R} \\to \\mathbb{R}$, then $f'(t) = f(t)$ for all $t$. It follows that $f(t) = Ce^t$ for some constant $C$. Since $|f(t)| \\leq M$ for all $t$, we must have $C=0$.\nIt follows that $h(a_0,b_0) = 0$; since $(a_0,b_0)$ was an arbitrary point,\n$h$ is identically $0$ over all of $\\mathbb{R}^2$.",
  "vars": [
    "f",
    "h",
    "L",
    "t",
    "x",
    "y"
  ],
  "params": [
    "a",
    "a_0",
    "b",
    "b_0",
    "C",
    "M",
    "R"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "curveval",
        "h": "planefun",
        "L": "straightpath",
        "t": "paramval",
        "x": "horizvar",
        "y": "vertvar",
        "a": "coeffax",
        "a_0": "originax",
        "b": "coeffy",
        "b_0": "originby",
        "C": "expconst",
        "M": "boundmax",
        "R": "realset"
      },
      "question": "Suppose that the function $planefun:\\mathbb{realset}^2\\to \\mathbb{realset}$ has continuous partial\nderivatives and satisfies the equation\n\\[\nplanefun(horizvar,vertvar) = coeffax \\frac{\\partial planefun}{\\partial horizvar}(horizvar,vertvar) +\ncoeffy \\frac{\\partial planefun}{\\partial vertvar}(horizvar,vertvar)\n\\]\nfor some constants $coeffax,coeffy$. Prove that if there is a constant $boundmax$ such that\n$|planefun(horizvar,vertvar)|\\leq boundmax$ for all $(horizvar,vertvar) \\in \\mathbb{realset}^2$, then $planefun$ is identically zero.",
      "solution": "If $coeffax=coeffy=0$, then the desired result holds trivially, so we assume that at least one of $coeffax,coeffy$ is nonzero.\nPick any point $(originax, originby) \\in \\mathbb{realset}^2$, and let $straightpath$ be the line given by the parametric equation\n$straightpath(paramval) = (originax,originby) + (coeffax,coeffy) paramval$ for $paramval\\in \\mathbb{realset}$. By the chain rule and the given equation, we have $\\frac{d}{dparamval}(planefun\\circ straightpath) = planefun\\circ straightpath$. If we write $curveval = planefun\\circ straightpath:\\mathbb{realset} \\to \\mathbb{realset}$, then $curveval'(paramval) = curveval(paramval)$ for all $paramval$. It follows that $curveval(paramval) = expconst e^{paramval}$ for some constant $expconst$. Since $|curveval(paramval)| \\leq boundmax$ for all $paramval$, we must have $expconst=0$.\nIt follows that $planefun(originax,originby) = 0$; since $(originax,originby)$ was an arbitrary point,\n$planefun$ is identically $0$ over all of $\\mathbb{realset}^2$."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "blueberry",
        "h": "sandstone",
        "L": "lighthouse",
        "t": "marigolds",
        "x": "pendulum",
        "y": "gemstone",
        "a": "sailplane",
        "a_0": "raincloud",
        "b": "campfire",
        "b_0": "waterfall",
        "C": "hummingbird",
        "M": "blacksmith",
        "R": "windswept"
      },
      "question": "Suppose that the function $sandstone:\\mathbb{R}^2\\to \\mathbb{R}$ has continuous partial\nderivatives and satisfies the equation\n\\[\nsandstone(pendulum,gemstone) = sailplane \\frac{\\partial sandstone}{\\partial pendulum}(pendulum,gemstone) +\ncampfire \\frac{\\partial sandstone}{\\partial gemstone}(pendulum,gemstone)\n\\]\nfor some constants $sailplane,campfire$. Prove that if there is a constant $blacksmith$ such that\n$|sandstone(pendulum,gemstone)|\\leq blacksmith$ for all $(pendulum,gemstone) \\in \\mathbb{R}^2$, then $sandstone$ is identically zero.",
      "solution": "If $sailplane=campfire=0$, then the desired result holds trivially, so we assume that at least one of $sailplane,campfire$ is nonzero.\nPick any point $(raincloud, waterfall) \\in \\mathbb{R}^2$, and let $lighthouse$ be the line given by the parametric equation\n$lighthouse(marigolds) = (raincloud,waterfall) + (sailplane,campfire) \\, marigolds$ for $marigolds\\in \\mathbb{R}$. By the chain rule and the given equation, we have $\\frac{d}{d marigolds}(sandstone\\circ lighthouse) = sandstone\\circ lighthouse$. If we write $blueberry = sandstone\\circ lighthouse:\\mathbb{R} \\to \\mathbb{R}$, then $blueberry'(marigolds) = blueberry(marigolds)$ for all $marigolds$. It follows that $blueberry(marigolds) = hummingbird e^{marigolds}$ for some constant $hummingbird$. Since $|blueberry(marigolds)| \\leq blacksmith$ for all $marigolds$, we must have $hummingbird=0$.\nIt follows that $sandstone(raincloud,waterfall) = 0$; since $(raincloud,waterfall)$ was an arbitrary point,\n$sandstone$ is identically $0$ over all of $\\mathbb{R}^2$.}"
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "constantval",
        "h": "deepness",
        "L": "curvature",
        "t": "spaceval",
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "a": "variable",
        "a_0": "variableone",
        "b": "dynamism",
        "b_0": "dynamismone",
        "C": "changing",
        "M": "unbounded",
        "R": "imaginary"
      },
      "question": "Suppose that the function $deepness:\\mathbb{R}^2\\to \\mathbb{R}$ has continuous partial\nderivatives and satisfies the equation\n\\[\ndeepness(verticalaxis,horizontalaxis) = variable \\frac{\\partial deepness}{\\partial verticalaxis}(verticalaxis,horizontalaxis) +\ndynamism \\frac{\\partial deepness}{\\partial horizontalaxis}(verticalaxis,horizontalaxis)\n\\]\nfor some constants $variable,dynamism$. Prove that if there is a constant $unbounded$ such that\n$|deepness(verticalaxis,horizontalaxis)|\\leq unbounded$ for all $(verticalaxis,horizontalaxis) \\in \\mathbb{R}^2$, then $deepness$ is identically zero.",
      "solution": "If $variable=dynamism=0$, then the desired result holds trivially, so we assume that at least one of $variable,dynamism$ is nonzero.\nPick any point $(variableone, dynamismone) \\in \\mathbb{R}^2$, and let $curvature$ be the line given by the parametric equation\n$curvature(spaceval) = (variableone,dynamismone) + (variable,dynamism) spaceval$ for $spaceval\\in \\mathbb{R}$. By the chain rule and the given equation, we have $\\frac{d}{d spaceval}(deepness\\circ curvature) = deepness\\circ curvature$. If we write $constantval = deepness\\circ curvature:\\mathbb{R} \\to \\mathbb{R}$, then $constantval'(spaceval) = constantval(spaceval)$ for all $spaceval$. It follows that $constantval(spaceval) = changing e^{spaceval}$ for some constant $changing$. Since $|constantval(spaceval)| \\leq unbounded$ for all $spaceval$, we must have $changing=0$.\nIt follows that $deepness(variableone,dynamismone) = 0$; since $(variableone,dynamismone)$ was an arbitrary point,\n$deepness$ is identically $0$ over all of $\\mathbb{R}^2$. "
    },
    "garbled_string": {
      "map": {
        "f": "qzxwvtnp",
        "h": "hjgrksla",
        "L": "klmnbvcx",
        "t": "pqowieur",
        "x": "zmcnvbqw",
        "y": "lkjhgfdp",
        "a": "asdfghjk",
        "a_0": "plmoknij",
        "b": "zxcvbnml",
        "b_0": "qazwsxed",
        "C": "mnbvcxzq",
        "M": "poiuytre",
        "R": "lkjqwert"
      },
      "question": "Suppose that the function $hjgrksla:\\mathbb{R}^2\\to \\mathbb{R}$ has continuous partial derivatives and satisfies the equation\n\\[\nhjgrksla(zmcnvbqw,lkjhgfdp) = asdfghjk \\frac{\\partial hjgrksla}{\\partial zmcnvbqw}(zmcnvbqw,lkjhgfdp) +\nzxcvbnml \\frac{\\partial hjgrksla}{\\partial lkjhgfdp}(zmcnvbqw,lkjhgfdp)\n\\]\nfor some constants $asdfghjk,zxcvbnml$. Prove that if there is a constant $poiuytre$ such that\n$|hjgrksla(zmcnvbqw,lkjhgfdp)|\\leq poiuytre$ for all $(zmcnvbqw,lkjhgfdp) \\in \\mathbb{R}^2$, then $hjgrksla$ is identically zero.",
      "solution": "If $asdfghjk=zxcvbnml=0$, then the desired result holds trivially, so we assume that at least one of $asdfghjk,zxcvbnml$ is nonzero. Pick any point $(plmoknij, qazwsxed) \\in \\mathbb{R}^2$, and let $klmnbvcx$ be the line given by the parametric equation $klmnbvcx(pqowieur) = (plmoknij,qazwsxed) + (asdfghjk,zxcvbnml) pqowieur$ for $pqowieur\\in \\mathbb{R}$. By the chain rule and the given equation, we have $\\frac{d}{d pqowieur}(hjgrksla\\circ klmnbvcx) = hjgrksla\\circ klmnbvcx$. If we write $qzxwvtnp = hjgrksla\\circ klmnbvcx:\\mathbb{R} \\to \\mathbb{R}$, then $qzxwvtnp'(pqowieur) = qzxwvtnp(pqowieur)$ for all $pqowieur$. It follows that $qzxwvtnp(pqowieur) = mnbvcxzq e^{pqowieur}$ for some constant $mnbvcxzq$. Since $|qzxwvtnp(pqowieur)| \\leq poiuytre$ for all $pqowieur$, we must have $mnbvcxzq=0$. It follows that $hjgrksla(plmoknij,qazwsxed) = 0$; since $(plmoknij,qazwsxed)$ was an arbitrary point, $hjgrksla$ is identically $0$ over all of \\mathbb{R}^2$."
    },
    "kernel_variant": {
      "question": "Let H be a real, separable Hilbert space with inner product ⟨·,·⟩ and norm ‖·‖.  \nLet A : D(A) ⊂ H → H be a (possibly unbounded) densely-defined, closed linear operator that generates an exponentially stable C₀-semigroup S(t)=e^{tA} (t ≥ 0); i.e. there are constants  \n\n  C ≥ 1  and  γ > 0  such that  ‖S(t)‖ ≤ C e^{−γ t} for every t ≥ 0.               (0)\n\nA function h : H → ℝ is called Fréchet differentiable if for every x ∈ H there exists a bounded linear functional Dh(x) satisfying  \n\n  lim_{‖v‖→0} ( h(x+v) − h(x) − Dh(x)[v] ) / ‖v‖ = 0.\n\nVia the Riesz representation theorem we identify Dh(x) with the gradient ∇h(x) ∈ H so that Dh(x)[v]=⟨∇h(x),v⟩ for all v∈H.\n\nAssume\n(1)  h is bounded: sup_{y∈H}|h(y)| < ∞;  \n(2)  h is Fréchet differentiable and the map x ↦ ∇h(x) is continuous on H (h is therefore continuous as well);  \n(3)  h satisfies the first–order functional identity on the (possibly proper) domain D(A):\n  h(x) = ⟨∇h(x), Ax⟩  for every x ∈ D(A).                                                            (★)\n\nProve that h is identically zero on the whole Hilbert space H.",
      "solution": "We write S(t):=e^{tA}, note that S(t)D(A)⊂D(A) for all t≥0 (standard semigroup theory), and keep the constants C,γ of (0).  Fix M:=sup_{y∈H}|h(y)|<∞.\n\nStep 1 – Absolute continuity of  φ_x(t):=h(S(t)x)  when x ∈ D(A).  \nFor x∈D(A) the map t↦S(t)x is C¹ in H and satisfies d/dt S(t)x=AS(t)x.  \nBecause h is C¹ (Fréchet) and ∇h is continuous, the classical chain rule in Banach spaces gives that φ_x is absolutely continuous on every compact interval and\n\n  φ_x′(t)=⟨∇h(S(t)x),AS(t)x⟩         a.e. t≥0.                             (1)\n\nA short derivation: write Δh = h(S(t+h)x)−h(S(t)x) and use the first-order Taylor formula with remainder  \n Δh = ⟨∇h(S(t)x),S(t+h)x−S(t)x⟩ + r(t,h),  r(t,h)=o(‖S(t+h)x−S(t)x‖).  \nSince S(t+h)x−S(t)x = ∫₀^h S(t+s)Ax ds, divide by h and let h→0; boundedness of ∇h on bounded sets together with (0) provides a dominating L¹-function in s ∈ (0,h), justifying the limit by dominated convergence.\n\nStep 2 – The orbit ODE.  \nBecause S(t)x∈D(A) for every t and because (★) holds on D(A),\n\n φ_x′(t)=⟨∇h(S(t)x),AS(t)x⟩ = h(S(t)x)=φ_x(t) a.e. t≥0.\n\nThus φ_x solves φ′=φ, hence\n\n  φ_x(t)=h(x)e^{t}  for all t≥0.                                             (2)\n\nStep 3 – Exponential stability forces vanishing on D(A).  \nThe semigroup estimate (0) gives ‖S(t)x‖ ≤ C‖x‖e^{−γ t} → 0   as t→∞.  \nBecause h is bounded,\n\n |h(x)| e^{t}=|φ_x(t)|=|h(S(t)x)| ≤ M  ∀ t≥0.\n\nLetting t→∞ yields h(x)=0 for every x∈D(A).                               (3)\n\nStep 4 – Extension from D(A) to all of H.  \nSince A is densely defined, D(A) is dense in H.  \nBecause h is continuous on H, its zero set contains the dense subset D(A) and is therefore the whole space:\n\n ∀x∈H,  ∃x_n∈D(A) with x_n→x ⇒ h(x)=lim_{n→∞}h(x_n)=0.\n\nThus h≡0 on H.\n\n∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.816413",
        "was_fixed": false,
        "difficulty_analysis": "1. Infinite-dimensional setting  \n   • The domain is an arbitrary separable Hilbert space, not ℝⁿ, so standard finite-dimensional tools (determinants, coordinates, compactness of spheres) no longer apply.  \n   • One must manipulate bounded linear operators and their functional calculus (the semigroup e^{t𝑇}) instead of simple vectors.\n\n2. Spectral theory and semigroup estimates  \n   • The hypothesis s(𝑇)<0 invokes the spectrum of 𝑇 and requires knowing that e^{t𝑇} forms a strongly continuous semigroup with exponential decay.  \n   • Handling e^{t𝑇} demands familiarity with the Hille–Yosida theory or equivalent functional-analytic results.\n\n3. Fréchet differentiability and the Riesz identification  \n   • Recasting the Fréchet derivative as a gradient via the Riesz map introduces an additional layer of abstraction not present in the original finite-dimensional gradient.\n\n4. Method of characteristics in Hilbert spaces  \n   • The solution traces the flow of the vector field x↦𝑇x and derives an ODE along infinite-dimensional integral curves, a non-trivial extension of the finite-dimensional characteristic method.\n\n5. Subtle use of boundedness  \n   • Boundedness is combined with exponential decay of the orbit to force vanishing of the coefficient in an exponential ODE, a step absent from the original problem.\n\nBecause of these added layers – spectral analysis, semigroup theory, infinite-dimensional calculus, and a more delicate exploitation of boundedness – the enhanced variant demands considerably deeper theoretical insight and a longer chain of advanced arguments than both the original problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let H be a real, separable Hilbert space with inner product ⟨·,·⟩ and norm ‖·‖.  \nLet A : D(A) ⊂ H → H be a (possibly unbounded) densely-defined, closed linear operator that generates an exponentially stable C₀-semigroup S(t)=e^{tA} (t ≥ 0); i.e. there are constants  \n\n  C ≥ 1  and  γ > 0  such that  ‖S(t)‖ ≤ C e^{−γ t} for every t ≥ 0.               (0)\n\nA function h : H → ℝ is called Fréchet differentiable if for every x ∈ H there exists a bounded linear functional Dh(x) satisfying  \n\n  lim_{‖v‖→0} ( h(x+v) − h(x) − Dh(x)[v] ) / ‖v‖ = 0.\n\nVia the Riesz representation theorem we identify Dh(x) with the gradient ∇h(x) ∈ H so that Dh(x)[v]=⟨∇h(x),v⟩ for all v∈H.\n\nAssume\n(1)  h is bounded: sup_{y∈H}|h(y)| < ∞;  \n(2)  h is Fréchet differentiable and the map x ↦ ∇h(x) is continuous on H (h is therefore continuous as well);  \n(3)  h satisfies the first–order functional identity on the (possibly proper) domain D(A):\n  h(x) = ⟨∇h(x), Ax⟩  for every x ∈ D(A).                                                            (★)\n\nProve that h is identically zero on the whole Hilbert space H.",
      "solution": "We write S(t):=e^{tA}, note that S(t)D(A)⊂D(A) for all t≥0 (standard semigroup theory), and keep the constants C,γ of (0).  Fix M:=sup_{y∈H}|h(y)|<∞.\n\nStep 1 – Absolute continuity of  φ_x(t):=h(S(t)x)  when x ∈ D(A).  \nFor x∈D(A) the map t↦S(t)x is C¹ in H and satisfies d/dt S(t)x=AS(t)x.  \nBecause h is C¹ (Fréchet) and ∇h is continuous, the classical chain rule in Banach spaces gives that φ_x is absolutely continuous on every compact interval and\n\n  φ_x′(t)=⟨∇h(S(t)x),AS(t)x⟩         a.e. t≥0.                             (1)\n\nA short derivation: write Δh = h(S(t+h)x)−h(S(t)x) and use the first-order Taylor formula with remainder  \n Δh = ⟨∇h(S(t)x),S(t+h)x−S(t)x⟩ + r(t,h),  r(t,h)=o(‖S(t+h)x−S(t)x‖).  \nSince S(t+h)x−S(t)x = ∫₀^h S(t+s)Ax ds, divide by h and let h→0; boundedness of ∇h on bounded sets together with (0) provides a dominating L¹-function in s ∈ (0,h), justifying the limit by dominated convergence.\n\nStep 2 – The orbit ODE.  \nBecause S(t)x∈D(A) for every t and because (★) holds on D(A),\n\n φ_x′(t)=⟨∇h(S(t)x),AS(t)x⟩ = h(S(t)x)=φ_x(t) a.e. t≥0.\n\nThus φ_x solves φ′=φ, hence\n\n  φ_x(t)=h(x)e^{t}  for all t≥0.                                             (2)\n\nStep 3 – Exponential stability forces vanishing on D(A).  \nThe semigroup estimate (0) gives ‖S(t)x‖ ≤ C‖x‖e^{−γ t} → 0   as t→∞.  \nBecause h is bounded,\n\n |h(x)| e^{t}=|φ_x(t)|=|h(S(t)x)| ≤ M  ∀ t≥0.\n\nLetting t→∞ yields h(x)=0 for every x∈D(A).                               (3)\n\nStep 4 – Extension from D(A) to all of H.  \nSince A is densely defined, D(A) is dense in H.  \nBecause h is continuous on H, its zero set contains the dense subset D(A) and is therefore the whole space:\n\n ∀x∈H,  ∃x_n∈D(A) with x_n→x ⇒ h(x)=lim_{n→∞}h(x_n)=0.\n\nThus h≡0 on H.\n\n∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.624866",
        "was_fixed": false,
        "difficulty_analysis": "1. Infinite-dimensional setting  \n   • The domain is an arbitrary separable Hilbert space, not ℝⁿ, so standard finite-dimensional tools (determinants, coordinates, compactness of spheres) no longer apply.  \n   • One must manipulate bounded linear operators and their functional calculus (the semigroup e^{t𝑇}) instead of simple vectors.\n\n2. Spectral theory and semigroup estimates  \n   • The hypothesis s(𝑇)<0 invokes the spectrum of 𝑇 and requires knowing that e^{t𝑇} forms a strongly continuous semigroup with exponential decay.  \n   • Handling e^{t𝑇} demands familiarity with the Hille–Yosida theory or equivalent functional-analytic results.\n\n3. Fréchet differentiability and the Riesz identification  \n   • Recasting the Fréchet derivative as a gradient via the Riesz map introduces an additional layer of abstraction not present in the original finite-dimensional gradient.\n\n4. Method of characteristics in Hilbert spaces  \n   • The solution traces the flow of the vector field x↦𝑇x and derives an ODE along infinite-dimensional integral curves, a non-trivial extension of the finite-dimensional characteristic method.\n\n5. Subtle use of boundedness  \n   • Boundedness is combined with exponential decay of the orbit to force vanishing of the coefficient in an exponential ODE, a step absent from the original problem.\n\nBecause of these added layers – spectral analysis, semigroup theory, infinite-dimensional calculus, and a more delicate exploitation of boundedness – the enhanced variant demands considerably deeper theoretical insight and a longer chain of advanced arguments than both the original problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}