{
  "index": "2011-B-3",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "interval containing $0$, with $g$ nonzero and continuous at $0$.  If $fg$\nand $f/g$ are differentiable at $0$, must $f$ be differentiable at 0?",
  "solution": "Yes, it follows that $f$ is differentiable.\n\n\\noindent\n\\textbf{First solution.}\nNote first that at $0$, $f/g$ and $g$ are both continuous, as then is their product $f$.\nIf $f(0) \\neq 0$, then in some neighborhood of $0$, $f$ is either always positive or always\nnegative. We can thus choose $\\epsilon \\in \\{\\pm 1\\}$ so that\n$\\epsilon f$ is the composition of the differentiable function\n$(fg)\\cdot (f/g)$ with the square root function. By the chain rule, $f$ is differentiable at $0$.\n\nIf $f(0) = 0$, then $(f/g)(0) = 0$, so we have\n\\[\n(f/g)'(0) = \\lim_{x \\to 0}\n\\frac{f(x)}{x g(x)}.\n\\]\nSince $g$ is continuous at 0, we may multiply limits to deduce that $\\lim_{x \\to 0} f(x)/x$ exists.\n\n\\noindent\n\\textbf{Second solution.}\nChoose a neighborhood $N$ of $0$ on which $g(x) \\neq 0$.\nDefine the following functions on $N \\setminus \\{0\\}$:\n$h_1(x) = \\frac{f(x)g(x)-f(0)g(0)}{x}$;\n$h_2(x) = \\frac{f(x)g(0)-f(0)g(x)}{xg(0)g(x)}$;\n$h_3(x) = g(0)g(x)$;\n$h_4(x) = \\frac{1}{g(x)+g(0)}$. Then\nby assumption, $h_1,h_2,h_3,h_4$ all have limits as $x \\to 0$. On the\nother hand,\n\\[\n\\frac{f(x)-f(0)}{x} = (h_1(x)+h_2(x)h_3(x))h_4(x),\n\\]\nand it follows that $\\lim_{x\\to 0} \\frac{f(x)-f(0)}{x}$ exists, as desired.",
  "vars": [
    "f",
    "g",
    "x",
    "h_1",
    "h_2",
    "h_3",
    "h_4"
  ],
  "params": [
    "\\\\epsilon",
    "N"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "function",
        "g": "generator",
        "x": "matvarx",
        "h_1": "helpera",
        "h_2": "helperb",
        "h_3": "helperc",
        "h_4": "helperd",
        "\\epsilon": "limavar",
        "N": "neighbr"
      },
      "question": "interval containing $0$, with $generator$ nonzero and continuous at $0$.  If $functiongenerator$ and $function/generator$ are differentiable at $0$, must $function$ be differentiable at 0?",
      "solution": "Yes, it follows that $function$ is differentiable.\n\n\\noindent\n\\textbf{First solution.}\nNote first that at $0$, $function/generator$ and $generator$ are both continuous, as then is their product $function$.\nIf $function(0) \\neq 0$, then in some neighborhood of $0$, $function$ is either always positive or always\nnegative. We can thus choose $limavar \\in \\{\\pm 1\\}$ so that\n$limavar function$ is the composition of the differentiable function\n$(functiongenerator)\\cdot (function/generator)$ with the square root function. By the chain rule, $function$ is differentiable at $0$.\n\nIf $function(0) = 0$, then $(function/generator)(0) = 0$, so we have\n\\[\n(function/generator)'(0) = \\lim_{matvarx \\to 0}\n\\frac{function(matvarx)}{matvarx\\, generator(matvarx)}.\n\\]\nSince $generator$ is continuous at 0, we may multiply limits to deduce that $\\lim_{matvarx \\to 0} function(matvarx)/matvarx$ exists.\n\n\\noindent\n\\textbf{Second solution.}\nChoose a neighborhood $neighbr$ of $0$ on which $generator(matvarx) \\neq 0$.\nDefine the following functions on $neighbr \\setminus \\{0\\}$:\n$helpera(matvarx) = \\frac{function(matvarx)generator(matvarx)-function(0)generator(0)}{matvarx}$;\n$helperb(matvarx) = \\frac{function(matvarx)generator(0)-function(0)generator(matvarx)}{matvarx\\, generator(0)\\, generator(matvarx)}$;\n$helperc(matvarx) = generator(0)generator(matvarx)$;\n$helperd(matvarx) = \\frac{1}{generator(matvarx)+generator(0)}$. Then\nby assumption, $helpera,helperb,helperc,helperd$ all have limits as $matvarx \\to 0$. On the\nother hand,\n\\[\n\\frac{function(matvarx)-function(0)}{matvarx} = (helpera(matvarx)+helperb(matvarx)helperc(matvarx))helperd(matvarx),\n\\]\nand it follows that $\\lim_{matvarx\\to 0} \\frac{function(matvarx)-function(0)}{matvarx}$ exists, as desired."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "blueberry",
        "g": "calendar",
        "x": "lantern",
        "h_1": "teapotage",
        "h_2": "paperclips",
        "h_3": "snowballer",
        "h_4": "thunderous",
        "\\epsilon": "marigolds",
        "N": "waterfall"
      },
      "question": "interval containing $0$, with $calendar$ nonzero and continuous at $0$.  If $blueberry\\,calendar$ and $blueberry/calendar$ are differentiable at $0$, must $blueberry$ be differentiable at 0?",
      "solution": "Yes, it follows that $blueberry$ is differentiable.\n\n\\noindent\n\\textbf{First solution.}\nNote first that at $0$, $blueberry/calendar$ and $calendar$ are both continuous, as then is their product $blueberry$.\nIf $blueberry(0) \\neq 0$, then in some neighborhood of $0$, $blueberry$ is either always positive or always\nnegative. We can thus choose $marigolds \\in \\{\\pm 1\\}$ so that\n$marigolds\\,blueberry$ is the composition of the differentiable function\n$(blueberry\\,calendar)\\cdot (blueberry/calendar)$ with the square root function. By the chain rule, $blueberry$ is differentiable at $0$.\n\nIf $blueberry(0) = 0$, then $(blueberry/calendar)(0) = 0$, so we have\n\\[\n(blueberry/calendar)'(0) = \\lim_{lantern \\to 0}\n\\frac{blueberry(lantern)}{lantern\\, calendar(lantern)}.\n\\]\nSince $calendar$ is continuous at 0, we may multiply limits to deduce that $\\lim_{lantern \\to 0} blueberry(lantern)/lantern$ exists.\n\n\\noindent\n\\textbf{Second solution.}\nChoose a neighborhood $waterfall$ of $0$ on which $calendar(lantern) \\neq 0$.\nDefine the following functions on $waterfall \\setminus \\{0\\}$:\n$teapotage(lantern) = \\frac{blueberry(lantern)calendar(lantern)-blueberry(0)calendar(0)}{lantern}$;\n$paperclips(lantern) = \\frac{blueberry(lantern)calendar(0)-blueberry(0)calendar(lantern)}{lantern calendar(0)calendar(lantern)}$;\n$snowballer(lantern) = calendar(0)calendar(lantern)$;\n$thunderous(lantern) = \\frac{1}{calendar(lantern)+calendar(0)}$. Then\nby assumption, $teapotage,paperclips,snowballer,thunderous$ all have limits as $lantern \\to 0$. On the\nother hand,\n\\[\n\\frac{blueberry(lantern)-blueberry(0)}{lantern} = (teapotage(lantern)+paperclips(lantern)snowballer(lantern))thunderous(lantern),\n\\]\nand it follows that $\\lim_{lantern\\to 0} \\frac{blueberry(lantern)-blueberry(0)}{lantern}$ exists, as desired."
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "nonfunction",
        "g": "disfunction",
        "x": "constantval",
        "h_1": "hindrancea",
        "h_2": "hindranceb",
        "h_3": "hindrancec",
        "h_4": "hindranced",
        "\\epsilon": "megavalue",
        "N": "remotezone"
      },
      "question": "interval containing $0$, with $disfunction$ nonzero and continuous at $0$.  If $nonfunction disfunction$ and $nonfunction/disfunction$ are differentiable at $0$, must $nonfunction$ be differentiable at 0?",
      "solution": "Yes, it follows that $nonfunction$ is differentiable.\n\n\\noindent\n\\textbf{First solution.}\nNote first that at $0$, $nonfunction/disfunction$ and $disfunction$ are both continuous, as then is their product $nonfunction$.\nIf $nonfunction(0) \\neq 0$, then in some neighborhood of $0$, $nonfunction$ is either always positive or always\nnegative. We can thus choose $megavalue \\in \\{\\pm 1\\}$ so that\n$megavalue nonfunction$ is the composition of the differentiable function\n$(nonfunction disfunction)\\cdot (nonfunction/disfunction)$ with the square root function. By the chain rule, $nonfunction$ is differentiable at $0$.\n\nIf $nonfunction(0) = 0$, then $(nonfunction/disfunction)(0) = 0$, so we have\n\\[\n(nonfunction/disfunction)'(0) = \\lim_{constantval \\to 0}\n\\frac{nonfunction(constantval)}{constantval disfunction(constantval)}.\n\\]\nSince $disfunction$ is continuous at 0, we may multiply limits to deduce that $\\lim_{constantval \\to 0} nonfunction(constantval)/constantval$ exists.\n\n\\noindent\n\\textbf{Second solution.}\nChoose a neighborhood $remotezone$ of $0$ on which $disfunction(constantval) \\neq 0$.\nDefine the following functions on $remotezone \\setminus \\{0\\}$:\n$hindrancea(constantval) = \\frac{nonfunction(constantval)disfunction(constantval)-nonfunction(0)disfunction(0)}{constantval}$;\n$hindranceb(constantval) = \\frac{nonfunction(constantval)disfunction(0)-nonfunction(0)disfunction(constantval)}{constantval disfunction(0)disfunction(constantval)}$;\n$hindrancec(constantval) = disfunction(0)disfunction(constantval)$;\n$hindranced(constantval) = \\frac{1}{disfunction(constantval)+disfunction(0)}$. Then\nby assumption, $hindrancea,hindranceb,hindrancec,hindranced$ all have limits as $constantval \\to 0$. On the\nother hand,\n\\[\n\\frac{nonfunction(constantval)-nonfunction(0)}{constantval} = (hindrancea(constantval)+hindranceb(constantval)hindrancec(constantval))hindranced(constantval),\n\\]\nand it follows that $\\lim_{constantval\\to 0} \\frac{nonfunction(constantval)-nonfunction(0)}{constantval}$ exists, as desired."
    },
    "garbled_string": {
      "map": {
        "f": "qzxwvtnp",
        "g": "hjgrksla",
        "x": "mplqodnq",
        "h_1": "anfldtkg",
        "h_2": "revsopjn",
        "h_3": "uvxtshkp",
        "h_4": "bklyfrma",
        "\\epsilon": "kowrjzmx",
        "N": "sudbclwa"
      },
      "question": "interval containing $0$, with $hjgrksla$ nonzero and continuous at $0$.  If $qzxwvtnp hjgrksla$\nand $qzxwvtnp/hjgrksla$ are differentiable at $0$, must $qzxwvtnp$ be differentiable at 0?",
      "solution": "Yes, it follows that $qzxwvtnp$ is differentiable.\n\n\\noindent\n\\textbf{First solution.}\nNote first that at $0$, $qzxwvtnp/hjgrksla$ and $hjgrksla$ are both continuous, as then is their product $qzxwvtnp$.\nIf $qzxwvtnp(0) \\neq 0$, then in some neighborhood of $0$, $qzxwvtnp$ is either always positive or always\nnegative. We can thus choose $kowrjzmx \\in \\{\\pm 1\\}$ so that\n$kowrjzmx qzxwvtnp$ is the composition of the differentiable function\n$(qzxwvtnp hjgrksla)\\cdot (qzxwvtnp/hjgrksla)$ with the square root function. By the chain rule, $qzxwvtnp$ is differentiable at $0$.\n\nIf $qzxwvtnp(0) = 0$, then $(qzxwvtnp/hjgrksla)(0) = 0$, so we have\n\\[\n(qzxwvtnp/hjgrksla)'(0) = \\lim_{mplqodnq \\to 0}\n\\frac{qzxwvtnp(mplqodnq)}{mplqodnq\\, hjgrksla(mplqodnq)}.\n\\]\nSince $hjgrksla$ is continuous at 0, we may multiply limits to deduce that $\\lim_{mplqodnq \\to 0} qzxwvtnp(mplqodnq)/mplqodnq$ exists.\n\n\\noindent\n\\textbf{Second solution.}\nChoose a neighborhood $sudbclwa$ of $0$ on which $hjgrksla(mplqodnq) \\neq 0$.\nDefine the following functions on $sudbclwa \\setminus \\{0\\}$:\n$anfldtkg(mplqodnq) = \\frac{qzxwvtnp(mplqodnq)hjgrksla(mplqodnq)-qzxwvtnp(0)hjgrksla(0)}{mplqodnq}$;\n$revsopjn(mplqodnq) = \\frac{qzxwvtnp(mplqodnq)hjgrksla(0)-qzxwvtnp(0)hjgrksla(mplqodnq)}{mplqodnq hjgrksla(0)hjgrksla(mplqodnq)}$;\n$uvxtshkp(mplqodnq) = hjgrksla(0)hjgrksla(mplqodnq)$;\n$bklyfrma(mplqodnq) = \\frac{1}{hjgrksla(mplqodnq)+hjgrksla(0)}$. Then\nby assumption, $anfldtkg,revsopjn,uvxtshkp,bklyfrma$ all have limits as $mplqodnq \\to 0$. On the\nother hand,\n\\[\n\\frac{qzxwvtnp(mplqodnq)-qzxwvtnp(0)}{mplqodnq} = (anfldtkg(mplqodnq)+revsopjn(mplqodnq)uvxtshkp(mplqodnq))bklyfrma(mplqodnq),\n\\]\nand it follows that $\\lim_{mplqodnq\\to 0} \\frac{qzxwvtnp(mplqodnq)-qzxwvtnp(0)}{mplqodnq}$ exists, as desired."
    },
    "kernel_variant": {
      "question": "Let E be a real Banach space and A a unital (possibly non-commutative) Banach algebra with identity 1.  \nLet U ⊂ E be an open neighbourhood of 0 and  \n\n  f , g : U → A  \n\ntwo mappings that satisfy  \n\n(H1) g is Fréchet-differentiable at 0; put G₀ := g(0).  \n   G₀ is invertible and—after shrinking U—g(x) is invertible for every x ∈ U.  \n   Write G(x) := g(x) and G⁻¹(x) := g(x)⁻¹.  \n\n(H2) The product F(x) := f(x) G(x) and the “quotient’’  \n   Q(x) := G⁻¹(x) f(x) are both Fréchet-differentiable at 0.\n\nProve the following statements.\n\n(a) f is Fréchet-differentiable at 0.\n\n(b) For every h ∈ E one has  \n  Df(0) h = ½[ DF(0) h ⋅ G₀⁻¹ + G₀ ⋅ DQ(0) h ] + ½[ DG(0) h ⋅ G₀⁻¹ , f(0) ],  (*)\n\n  where [X,Y] := XY − YX denotes the commutator in A.\n\n(c) If either A is commutative or f(0) lies in the centre of A, the commutator vanishes and (*) reduces to  \n\n  Df(0) h = ½[ DF(0) h ⋅ G₀⁻¹ + G₀ ⋅ DQ(0) h ].\n\n(d) If, for some k ∈ ℕ ∪ {∞}, the map g itself is of class Cᵏ on a neighbourhood of 0 and, in addition, F and Q are of class Cᵏ on the same neighbourhood, then f is of class Cᵏ there as well.",
      "solution": "Throughout ‖·‖ denotes the norm in the Banach space or Banach algebra that is currently used;  \no(‖x‖) means a quantity whose norm divided by ‖x‖ tends to 0 as x → 0.  \nPut f₀ := f(0).\n\nStep 1. Linear expansions of G and G⁻¹.  \nBecause g is Fréchet-differentiable at 0 we have  \n\n  G(x) = G₀ + DG(0) x + R_G(x),  ‖R_G(x)‖ = o(‖x‖).  (1)\n\nThe inversion map ι : G ↦ G⁻¹ is C¹ on the open set of invertible elements, and  \n\n  Dι(G₀)(H) = − G₀⁻¹ H G₀⁻¹.  \n\nCombining this with (1) yields  \n\n  G⁻¹(x) = G₀⁻¹ − G₀⁻¹ DG(0) x G₀⁻¹ + R_inv(x),  ‖R_inv(x)‖ = o(‖x‖).  (2)\n\nStep 2. Remainders for F and Q.  \nThe differentiability of F and Q at 0 gives  \n\n  F(x) = F(0) + DF(0) x + R_F(x),  ‖R_F(x)‖ = o(‖x‖),  (3)  \n  Q(x) = Q(0) + DQ(0) x + R_Q(x),  ‖R_Q(x)‖ = o(‖x‖).  (4)\n\nNote F(0) = f₀ G₀ and Q(0) = G₀⁻¹ f₀.\n\nStep 3. A fundamental identity for f.  \nFor every x with ‖x‖ small enough,\n\n  f(x) = ½[ F(x) G⁻¹(x) + G(x) Q(x) ].  (5)\n\nIndeed each summand equals f(x).\n\nStep 4. Extraction of the first-order term.  \nInsert (1)–(4) into (5) and retain only the constant and linear contributions.\n\n• Constant part.  \n ½[ F(0) G₀⁻¹ + G₀ Q(0) ] = ½[ f₀ G₀ G₀⁻¹ + G₀ G₀⁻¹ f₀ ] = f₀.\n\n• Linear part.  \n\nF(x) G⁻¹(x) contributes  DF(0) x G₀⁻¹  and  F(0) Dι(G₀)(DG(0) x)  \n             = − f₀ DG(0) x G₀⁻¹;  \n\nG(x) Q(x) contributes  DG(0) x Q(0) + G₀ DQ(0) x  \n             = DG(0) x G₀⁻¹ f₀ + G₀ DQ(0) x.\n\nHence the entire linear term equals  \n\n ½[ DF(0) x G₀⁻¹ + G₀ DQ(0) x ] + ½[ DG(0) x G₀⁻¹ , f₀ ].  (6)\n\nDefine  \n\n L(h) := ½[ DF(0) h G₀⁻¹ + G₀ DQ(0) h ] + ½[ DG(0) h G₀⁻¹ , f₀ ].\n\nStep 5. Quadratic remainders.  \nSubtract f₀ + L(x) from (5).  Every remaining monomial contains at least one factor R_G, R_inv, R_F or R_Q; by (1)–(4) and (2),\n\n ‖R_F(x)‖ + ‖R_Q(x)‖ + ‖R_G(x)‖ + ‖R_inv(x)‖ = o(‖x‖).\n\nBecause multiplication in a Banach algebra is continuous, each product that survives after the subtraction has norm o(‖x‖). Consequently  \n\n ‖f(x) − f(0) − L(x)‖ = o(‖x‖),  x → 0.\n\nThus f is Fréchet-differentiable at 0 and Df(0) = L, which proves (a) and formula (*) in (b).\n\nStep 6. Simplifications under commutativity/centrality.  \nIf A is commutative, or merely if f₀ lies in the centre of A, the commutator in (*) vanishes, giving the simpler expression announced in (c).\n\nStep 7. Higher regularity.  \nAssume now that, for some k ∈ ℕ ∪ {∞}, g, F and Q are Cᵏ on a neighbourhood of 0.  \nBecause inversion is a C^∞ map on the group of units of A, the maps  \n\n x ↦ G(x) and x ↦ G⁻¹(x)  \n\nare also of class Cᵏ.  Formula (5) combines the Cᵏ-maps  \n\n F, G⁻¹, G, Q  \n\nvia continuous bilinear operations and an averaging; the resulting map f is therefore of class Cᵏ.  This establishes (d). ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.821076",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher-dimensional framework: the problem is set in an arbitrary Banach space (domain) and a Banach algebra (range), far beyond the 1-dimensional scalar situation of the original.  \n•  Fréchet differentiability replaces elementary one-variable limits, forcing the solver to control operator norms and linear maps.  \n•  The algebraic structure is non-commutative; one must carefully justify manipulations such as (4) and control non-commuting factors, which are absent in the scalar case.  \n•  The derivative must be produced explicitly in terms of the (unknown) derivatives of two composite maps, requiring a delicate decomposition (Step 2) and a norm estimate (Step 4) instead of simple scalar limits.  \n•  Part (c) links differentiability of various orders, introducing smooth Banach-manifold techniques and demanding knowledge of the calculus of Banach-space valued mappings.\n\nAll these ingredients—operator-norm estimates, non-commutativity, Fréchet calculus, and higher-order smoothness—make the enhanced variant substantially more sophisticated and technically involved than both the original problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let E be a real Banach space and A a unital (possibly non-commutative) Banach algebra with identity 1.  \nLet U ⊂ E be an open neighbourhood of 0 and  \n\n  f , g : U → A  \n\ntwo mappings that satisfy  \n\n(H1) g is Fréchet-differentiable at 0; put G₀ := g(0).  \n   G₀ is invertible and—after shrinking U—g(x) is invertible for every x ∈ U.  \n   Write G(x) := g(x) and G⁻¹(x) := g(x)⁻¹.  \n\n(H2) The product F(x) := f(x) G(x) and the “quotient’’  \n   Q(x) := G⁻¹(x) f(x) are both Fréchet-differentiable at 0.\n\nProve the following statements.\n\n(a) f is Fréchet-differentiable at 0.\n\n(b) For every h ∈ E one has  \n  Df(0) h = ½[ DF(0) h ⋅ G₀⁻¹ + G₀ ⋅ DQ(0) h ] + ½[ DG(0) h ⋅ G₀⁻¹ , f(0) ],  (*)\n\n  where [X,Y] := XY − YX denotes the commutator in A.\n\n(c) If either A is commutative or f(0) lies in the centre of A, the commutator vanishes and (*) reduces to  \n\n  Df(0) h = ½[ DF(0) h ⋅ G₀⁻¹ + G₀ ⋅ DQ(0) h ].\n\n(d) If, for some k ∈ ℕ ∪ {∞}, the map g itself is of class Cᵏ on a neighbourhood of 0 and, in addition, F and Q are of class Cᵏ on the same neighbourhood, then f is of class Cᵏ there as well.",
      "solution": "Throughout ‖·‖ denotes the norm in the Banach space or Banach algebra that is currently used;  \no(‖x‖) means a quantity whose norm divided by ‖x‖ tends to 0 as x → 0.  \nPut f₀ := f(0).\n\nStep 1. Linear expansions of G and G⁻¹.  \nBecause g is Fréchet-differentiable at 0 we have  \n\n  G(x) = G₀ + DG(0) x + R_G(x),  ‖R_G(x)‖ = o(‖x‖).  (1)\n\nThe inversion map ι : G ↦ G⁻¹ is C¹ on the open set of invertible elements, and  \n\n  Dι(G₀)(H) = − G₀⁻¹ H G₀⁻¹.  \n\nCombining this with (1) yields  \n\n  G⁻¹(x) = G₀⁻¹ − G₀⁻¹ DG(0) x G₀⁻¹ + R_inv(x),  ‖R_inv(x)‖ = o(‖x‖).  (2)\n\nStep 2. Remainders for F and Q.  \nThe differentiability of F and Q at 0 gives  \n\n  F(x) = F(0) + DF(0) x + R_F(x),  ‖R_F(x)‖ = o(‖x‖),  (3)  \n  Q(x) = Q(0) + DQ(0) x + R_Q(x),  ‖R_Q(x)‖ = o(‖x‖).  (4)\n\nNote F(0) = f₀ G₀ and Q(0) = G₀⁻¹ f₀.\n\nStep 3. A fundamental identity for f.  \nFor every x with ‖x‖ small enough,\n\n  f(x) = ½[ F(x) G⁻¹(x) + G(x) Q(x) ].  (5)\n\nIndeed each summand equals f(x).\n\nStep 4. Extraction of the first-order term.  \nInsert (1)–(4) into (5) and retain only the constant and linear contributions.\n\n• Constant part.  \n ½[ F(0) G₀⁻¹ + G₀ Q(0) ] = ½[ f₀ G₀ G₀⁻¹ + G₀ G₀⁻¹ f₀ ] = f₀.\n\n• Linear part.  \n\nF(x) G⁻¹(x) contributes  DF(0) x G₀⁻¹  and  F(0) Dι(G₀)(DG(0) x)  \n             = − f₀ DG(0) x G₀⁻¹;  \n\nG(x) Q(x) contributes  DG(0) x Q(0) + G₀ DQ(0) x  \n             = DG(0) x G₀⁻¹ f₀ + G₀ DQ(0) x.\n\nHence the entire linear term equals  \n\n ½[ DF(0) x G₀⁻¹ + G₀ DQ(0) x ] + ½[ DG(0) x G₀⁻¹ , f₀ ].  (6)\n\nDefine  \n\n L(h) := ½[ DF(0) h G₀⁻¹ + G₀ DQ(0) h ] + ½[ DG(0) h G₀⁻¹ , f₀ ].\n\nStep 5. Quadratic remainders.  \nSubtract f₀ + L(x) from (5).  Every remaining monomial contains at least one factor R_G, R_inv, R_F or R_Q; by (1)–(4) and (2),\n\n ‖R_F(x)‖ + ‖R_Q(x)‖ + ‖R_G(x)‖ + ‖R_inv(x)‖ = o(‖x‖).\n\nBecause multiplication in a Banach algebra is continuous, each product that survives after the subtraction has norm o(‖x‖). Consequently  \n\n ‖f(x) − f(0) − L(x)‖ = o(‖x‖),  x → 0.\n\nThus f is Fréchet-differentiable at 0 and Df(0) = L, which proves (a) and formula (*) in (b).\n\nStep 6. Simplifications under commutativity/centrality.  \nIf A is commutative, or merely if f₀ lies in the centre of A, the commutator in (*) vanishes, giving the simpler expression announced in (c).\n\nStep 7. Higher regularity.  \nAssume now that, for some k ∈ ℕ ∪ {∞}, g, F and Q are Cᵏ on a neighbourhood of 0.  \nBecause inversion is a C^∞ map on the group of units of A, the maps  \n\n x ↦ G(x) and x ↦ G⁻¹(x)  \n\nare also of class Cᵏ.  Formula (5) combines the Cᵏ-maps  \n\n F, G⁻¹, G, Q  \n\nvia continuous bilinear operations and an averaging; the resulting map f is therefore of class Cᵏ.  This establishes (d). ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.628261",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher-dimensional framework: the problem is set in an arbitrary Banach space (domain) and a Banach algebra (range), far beyond the 1-dimensional scalar situation of the original.  \n•  Fréchet differentiability replaces elementary one-variable limits, forcing the solver to control operator norms and linear maps.  \n•  The algebraic structure is non-commutative; one must carefully justify manipulations such as (4) and control non-commuting factors, which are absent in the scalar case.  \n•  The derivative must be produced explicitly in terms of the (unknown) derivatives of two composite maps, requiring a delicate decomposition (Step 2) and a norm estimate (Step 4) instead of simple scalar limits.  \n•  Part (c) links differentiability of various orders, introducing smooth Banach-manifold techniques and demanding knowledge of the calculus of Banach-space valued mappings.\n\nAll these ingredients—operator-norm estimates, non-commutativity, Fréchet calculus, and higher-order smoothness—make the enhanced variant substantially more sophisticated and technically involved than both the original problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}