{
  "index": "1974-B-4",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "B-4. In the standard definition, a real -valued function of two real variables \\( g: R^{2} \\rightarrow R^{1} \\) is continuous if, for every point \\( \\left(x_{0}, y_{0}\\right) \\in R^{2} \\) and every \\( \\varepsilon>0 \\), there is a corresponding \\( \\delta>0 \\) such that \\( \\left[\\left(x-x_{0}\\right)^{2}+\\left(y-y_{0}\\right)^{2}\\right]^{1 / 2}<\\delta \\) implies \\( \\left|g(x, y)-g\\left(x_{0}, y_{0}\\right)\\right|<\\varepsilon \\).\n\nBy contrast, \\( f: R^{\\mathbf{2}} \\rightarrow R^{\\prime} \\) is said to be continuous in each variable separately if, for each fixed value \\( y_{0} \\) of \\( y \\), the function \\( f\\left(x, y_{0}\\right) \\) is continuous in the usual sense as a function of \\( x \\), and similarly \\( f\\left(x_{0}, y\\right) \\) is continuous as a function of \\( y \\) for each fixed \\( x_{0} \\).\n\nLet \\( f: R^{\\mathbf{2}} \\rightarrow R^{\\prime} \\) be continuous in each variable separately. Show that there exists a sequence of continuous functions \\( g_{n}: R^{2} \\rightarrow R^{\\prime} \\) such that\n\\[\nf(x, y)=\\lim _{n \\rightarrow 0} g_{n}(x, y) \\text { for all }(x, y) \\in R^{2} .\n\\]",
  "solution": "B-4.\nFor each \\( n \\), we construct the function \\( g_{n}(x, y) \\) as follows: First divide the \\( x y \\)-plane into vertical strips of width \\( 1 / n \\) separated by the lines \\( \\{x=m / n\\}, m \\) an integer. Now set \\( g_{n}(x, y)=f(x, y) \\) along each vertical line \\( x=m / n \\), and interpolate linearly (holding \\( y \\) fixed and letting \\( x \\) vary) in between. Then \\( g_{n}(x, y) \\) is continuous because \\( f\\left(x_{0}, y\\right) \\) is continuous in \\( y ; g_{n}(x, y) \\rightarrow f(x, y) \\) because \\( f\\left(x, y_{0}\\right) \\) is continuous in \\( x \\).\n\nRemarks. This result has two interesting consequences for functions which are continuous in each variable separately:\n(i) Such functions are Borel measurable.\n(ii) They are continuous (in the usual sense) except on a set of points of the first Baire category. (In particular, there is no function which is continuous in each variable separately and yet discontinuous at every point.)",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "g",
    "x_0",
    "y_0",
    "\\\\varepsilon",
    "\\\\delta",
    "f",
    "g_n",
    "n",
    "m",
    "R"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "varxcoord",
        "y": "varycoord",
        "g": "basefunc",
        "x_0": "xorigin",
        "y_0": "yorigin",
        "\\varepsilon": "epsilonsym",
        "\\delta": "deltasymb",
        "f": "targetfunc",
        "g_n": "approxfunc",
        "n": "indexvar",
        "m": "stripindex",
        "R": "realspace"
      },
      "question": "B-4. In the standard definition, a real-valued function of two real variables \\( basefunc: realspace^{2} \\rightarrow realspace^{1} \\) is continuous if, for every point \\( \\left(xorigin, yorigin\\right) \\in realspace^{2} \\) and every \\( epsilonsym>0 \\), there is a corresponding \\( deltasymb>0 \\) such that \\( \\left[\\left(varxcoord-xorigin\\right)^{2}+\\left(varycoord-yorigin\\right)^{2}\\right]^{1 / 2}<deltasymb \\) implies \\( \\left|basefunc(varxcoord, varycoord)-basefunc\\left(xorigin, yorigin\\right)\\right|<epsilonsym \\).\n\nBy contrast, \\( targetfunc: realspace^{\\mathbf{2}} \\rightarrow realspace^{\\prime} \\) is said to be continuous in each variable separately if, for each fixed value \\( yorigin \\) of \\( varycoord \\), the function \\( targetfunc\\left(varxcoord, yorigin\\right) \\) is continuous in the usual sense as a function of \\( varxcoord \\), and similarly \\( targetfunc\\left(xorigin, varycoord\\right) \\) is continuous as a function of \\( varycoord \\) for each fixed \\( xorigin \\).\n\nLet \\( targetfunc: realspace^{\\mathbf{2}} \\rightarrow realspace^{\\prime} \\) be continuous in each variable separately. Show that there exists a sequence of continuous functions \\( approxfunc: realspace^{2} \\rightarrow realspace^{\\prime} \\) such that\n\\[\ntargetfunc(varxcoord, varycoord)=\\lim _{indexvar \\rightarrow 0} approxfunc(varxcoord, varycoord) \\text { for all }(varxcoord, varycoord) \\in realspace^{2} .\n\\]",
      "solution": "B-4.\nFor each \\( indexvar \\), we construct the function \\( approxfunc(varxcoord, varycoord) \\) as follows: First divide the \\( varxcoord varycoord \\)-plane into vertical strips of width \\( 1 / indexvar \\) separated by the lines \\( \\{varxcoord=stripindex / indexvar\\}, stripindex \\) an integer. Now set \\( approxfunc(varxcoord, varycoord)=targetfunc(varxcoord, varycoord) \\) along each vertical line \\( varxcoord=stripindex / indexvar \\), and interpolate linearly (holding \\( varycoord \\) fixed and letting \\( varxcoord \\) vary) in between. Then \\( approxfunc(varxcoord, varycoord) \\) is continuous because \\( targetfunc\\left(xorigin, varycoord\\right) \\) is continuous in \\( varycoord ; approxfunc(varxcoord, varycoord) \\rightarrow targetfunc(varxcoord, varycoord) \\) because \\( targetfunc\\left(varxcoord, yorigin\\right) \\) is continuous in \\( varxcoord \\).\n\nRemarks. This result has two interesting consequences for functions which are continuous in each variable separately:\n(i) Such functions are Borel measurable.\n(ii) They are continuous (in the usual sense) except on a set of points of the first Baire category. (In particular, there is no function which is continuous in each variable separately and yet discontinuous at every point.)"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "ambergris",
        "y": "calliope",
        "g": "snowflake",
        "x_0": "lemonade",
        "y_0": "kingfisher",
        "\\varepsilon": "goldenrod",
        "\\delta": "hummingbird",
        "f": "starfruit",
        "g_n": "marshmallow",
        "n": "tangerine",
        "m": "pineapple",
        "R": "universe"
      },
      "question": "B-4. In the standard definition, a real -valued function of two real variables \\( snowflake: universe^{2} \\rightarrow universe^{1} \\) is continuous if, for every point \\( \\left(lemonade, kingfisher\\right) \\in universe^{2} \\) and every \\( goldenrod>0 \\), there is a corresponding \\( hummingbird>0 \\) such that \\( \\left[\\left(ambergris-lemonade\\right)^{2}+\\left(calliope-kingfisher\\right)^{2}\\right]^{1 / 2}<hummingbird \\) implies \\( \\left|snowflake(ambergris, calliope)-snowflake\\left(lemonade, kingfisher\\right)\\right|<goldenrod \\).\n\nBy contrast, \\( starfruit: universe^{\\mathbf{2}} \\rightarrow universe^{\\prime} \\) is said to be continuous in each variable separately if, for each fixed value \\( kingfisher \\) of \\( calliope \\), the function \\( starfruit\\left(ambergris, kingfisher\\right) \\) is continuous in the usual sense as a function of \\( ambergris \\), and similarly \\( starfruit\\left(lemonade, calliope\\right) \\) is continuous as a function of \\( calliope \\) for each fixed \\( lemonade \\).\n\nLet \\( starfruit: universe^{\\mathbf{2}} \\rightarrow universe^{\\prime} \\) be continuous in each variable separately. Show that there exists a sequence of continuous functions \\( marshmallow: universe^{2} \\rightarrow universe^{\\prime} \\) such that\n\\[\nstarfruit(ambergris, calliope)=\\lim _{tangerine \\rightarrow 0} marshmallow(ambergris, calliope) \\text { for all }(ambergris, calliope) \\in universe^{2} .\n\\]",
      "solution": "B-4.\nFor each \\( tangerine \\), we construct the function \\( marshmallow(ambergris, calliope) \\) as follows: First divide the \\( ambergris calliope \\)-plane into vertical strips of width \\( 1 / tangerine \\) separated by the lines \\( \\{ambergris=pineapple / tangerine\\}, pineapple \\) an integer. Now set \\( marshmallow(ambergris, calliope)=starfruit(ambergris, calliope) \\) along each vertical line \\( ambergris=pineapple / tangerine \\), and interpolate linearly (holding \\( calliope \\) fixed and letting \\( ambergris \\) vary) in between. Then \\( marshmallow(ambergris, calliope) \\) is continuous because \\( starfruit\\left(lemonade, calliope\\right) \\) is continuous in \\( calliope ; marshmallow(ambergris, calliope) \\rightarrow starfruit(ambergris, calliope) \\) because \\( starfruit\\left(ambergris, kingfisher\\right) \\) is continuous in \\( ambergris \\).\n\nRemarks. This result has two interesting consequences for functions which are continuous in each variable separately:\n(i) Such functions are Borel measurable.\n(ii) They are continuous (in the usual sense) except on a set of points of the first Baire category. (In particular, there is no function which is continuous in each variable separately and yet discontinuous at every point.)"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "immobilept",
        "y": "frozenaxis",
        "g": "staticelem",
        "x_0": "infinitepoint",
        "y_0": "unboundedaxis",
        "\\\\varepsilon": "giganticval",
        "\\\\delta": "colossalrad",
        "f": "steadystate",
        "g_n": "steadystaticseq",
        "n": "continuousindex",
        "m": "irrationalnum",
        "R": "imaginaryset"
      },
      "question": "B-4. In the standard definition, a real -valued function of two real variables \\( staticelem: imaginaryset^{2} \\rightarrow imaginaryset^{1} \\) is continuous if, for every point \\( \\left(infinitepoint, unboundedaxis\\right) \\in imaginaryset^{2} \\) and every \\( giganticval>0 \\), there is a corresponding \\( colossalrad>0 \\) such that \\( \\left[\\left(immobilept-infinitepoint\\right)^{2}+\\left(frozenaxis-unboundedaxis\\right)^{2}\\right]^{1 / 2}<colossalrad \\) implies \\( \\left|staticelem(immobilept, frozenaxis)-staticelem\\left(infinitepoint, unboundedaxis\\right)\\right|<giganticval \\).\n\nBy contrast, \\( steadystate: imaginaryset^{\\mathbf{2}} \\rightarrow imaginaryset^{\\prime} \\) is said to be continuous in each variable separately if, for each fixed value \\( unboundedaxis \\) of \\( frozenaxis \\), the function \\( steadystate\\left(immobilept, unboundedaxis\\right) \\) is continuous in the usual sense as a function of \\( immobilept \\), and similarly \\( steadystate\\left(infinitepoint, frozenaxis\\right) \\) is continuous as a function of \\( frozenaxis \\) for each fixed \\( infinitepoint \\).\n\nLet \\( steadystate: imaginaryset^{\\mathbf{2}} \\rightarrow imaginaryset^{\\prime} \\) be continuous in each variable separately. Show that there exists a sequence of continuous functions \\( steadystaticseq: imaginaryset^{2} \\rightarrow imaginaryset^{\\prime} \\) such that\n\\[\nsteadystate(immobilept, frozenaxis)=\\lim _{continuousindex \\rightarrow 0} steadystaticseq(immobilept, frozenaxis) \\text { for all }(immobilept, frozenaxis) \\in imaginaryset^{2} .\n\\]",
      "solution": "B-4.\nFor each \\( continuousindex \\), we construct the function \\( steadystaticseq(immobilept, frozenaxis) \\) as follows: First divide the immobilept frozenaxis-plane into vertical strips of width \\( 1 / continuousindex \\) separated by the lines \\( \\{immobilept=irrationalnum / continuousindex\\}, irrationalnum \\) an integer. Now set \\( steadystaticseq(immobilept, frozenaxis)=steadystate(immobilept, frozenaxis) \\) along each vertical line \\( immobilept=irrationalnum / continuousindex \\), and interpolate linearly (holding \\( frozenaxis \\) fixed and letting \\( immobilept \\) vary) in between. Then \\( steadystaticseq(immobilept, frozenaxis) \\) is continuous because \\( steadystate\\left(infinitepoint, frozenaxis\\right) \\) is continuous in \\( frozenaxis ; steadystaticseq(immobilept, frozenaxis) \\rightarrow steadystate(immobilept, frozenaxis) \\) because \\( steadystate\\left(immobilept, unboundedaxis\\right) \\) is continuous in \\( immobilept \\).\n\nRemarks. This result has two interesting consequences for functions which are continuous in each variable separately:\n(i) Such functions are Borel measurable.\n(ii) They are continuous (in the usual sense) except on a set of points of the first Baire category. (In particular, there is no function which is continuous in each variable separately and yet discontinuous at every point.)"
    },
    "garbled_string": {
      "map": {
        "x": "uqvmzkle",
        "y": "porhtnsw",
        "g": "djmwqzsh",
        "x_0": "vxsqeaml",
        "y_0": "zrywgfuj",
        "\\varepsilon": "kzjxvbqe",
        "\\delta": "pytrmnhs",
        "f": "zcyrplkd",
        "g_n": "xhqskmdl",
        "n": "vbclzqjh",
        "m": "tpsnwdke",
        "R": "nyfqtzso"
      },
      "question": "B-4. In the standard definition, a real -valued function of two real variables \\( djmwqzsh: nyfqtzso^{2} \\rightarrow nyfqtzso^{1} \\) is continuous if, for every point \\( \\left(vxsqeaml, zrywgfuj\\right) \\in nyfqtzso^{2} \\) and every \\( kzjxvbqe>0 \\), there is a corresponding \\( pytrmnhs>0 \\) such that \\( \\left[\\left(uqvmzkle-vxsqeaml\\right)^{2}+\\left(porhtnsw-zrywgfuj\\right)^{2}\\right]^{1 / 2}<pytrmnhs \\) implies \\( \\left|djmwqzsh(uqvmzkle, porhtnsw)-djmwqzsh\\left(vxsqeaml, zrywgfuj\\right)\\right|<kzjxvbqe \\).\n\nBy contrast, \\( zcyrplkd: nyfqtzso^{\\mathbf{2}} \\rightarrow nyfqtzso^{\\prime} \\) is said to be continuous in each variable separately if, for each fixed value \\( zrywgfuj \\) of \\( porhtnsw \\), the function \\( zcyrplkd\\left(uqvmzkle, zrywgfuj\\right) \\) is continuous in the usual sense as a function of \\( uqvmzkle \\), and similarly \\( zcyrplkd\\left(vxsqeaml, porhtnsw\\right) \\) is continuous as a function of \\( porhtnsw \\) for each fixed \\( vxsqeaml \\).\n\nLet \\( zcyrplkd: nyfqtzso^{\\mathbf{2}} \\rightarrow nyfqtzso^{\\prime} \\) be continuous in each variable separately. Show that there exists a sequence of continuous functions \\( xhqskmdl: nyfqtzso^{2} \\rightarrow nyfqtzso^{\\prime} \\) such that\n\\[\nzcyrplkd(uqvmzkle, porhtnsw)=\\lim _{vbclzqjh \\rightarrow 0} xhqskmdl(uqvmzkle, porhtnsw) \\text { for all }(uqvmzkle, porhtnsw) \\in nyfqtzso^{2} .\n\\]",
      "solution": "B-4.\nFor each \\( vbclzqjh \\), we construct the function \\( xhqskmdl(uqvmzkle, porhtnsw) \\) as follows: First divide the \\( uqvmzkle porhtnsw \\)-plane into vertical strips of width \\( 1 / vbclzqjh \\) separated by the lines \\( \\{uqvmzkle=tpsnwdke / vbclzqjh\\}, tpsnwdke \\) an integer. Now set \\( xhqskmdl(uqvmzkle, porhtnsw)=zcyrplkd(uqvmzkle, porhtnsw) \\) along each vertical line \\( uqvmzkle=tpsnwdke / vbclzqjh \\), and interpolate linearly (holding \\( porhtnsw \\) fixed and letting \\( uqvmzkle \\) vary) in between. Then \\( xhqskmdl(uqvmzkle, porhtnsw) \\) is continuous because \\( zcyrplkd\\left(vxsqeaml, porhtnsw\\right) \\) is continuous in \\( porhtnsw ; xhqskmdl(uqvmzkle, porhtnsw) \\rightarrow zcyrplkd(uqvmzkle, porhtnsw) \\) because \\( zcyrplkd\\left(uqvmzkle, zrywgfuj\\right) \\) is continuous in \\( uqvmzkle \\).\n\nRemarks. This result has two interesting consequences for functions which are continuous in each variable separately:\n(i) Such functions are Borel measurable.\n(ii) They are continuous (in the usual sense) except on a set of points of the first Baire category. (In particular, there is no function which is continuous in each variable separately and yet discontinuous at every point.)"
    },
    "kernel_variant": {
      "question": "\\[\n\\textbf{Dyadic interpolation, quantitative smooth approximation, and quasi–uniform convergence outside the discontinuity set}\n\\]\n\nFix an integer $k\\ge 2$.  \nGiven a function $u:\\mathbb R^{k}\\longrightarrow\\mathbb R$, write  \n\\[\n\\operatorname{Lip}(u):=\\sup_{x\\neq y}\\frac{|u(x)-u(y)|}{|x-y|}\n\\quad(\\text{Euclidean distance }|\\cdot|),\n\\]\nand, for an open non–empty set $U\\subset\\mathbb R^{k}$ and $R>0$, put  \n\\[\n\\operatorname{Lip}\\!\\bigl(u\\!\\mid_{U}\\bigr)\n :=\\sup_{\\substack{x,y\\in U\\\\x\\neq y}}\n       \\frac{|u(x)-u(y)|}{|x-y|},\n \\qquad  \n M_{U,R}:=\\sup_{\\operatorname{dist}(x,U)\\le R}|u(x)|.\n\\]\n\nLet  \n\\[\nf:\\mathbb R^{k}\\longrightarrow \\mathbb R\n\\]\nbe \\emph{separately continuous} (continuous in each coordinate) and \\emph{locally bounded}.  \nFor $n\\ge 1$ set $\\Delta_{n}:=2^{-n}$ and introduce the dyadic lattice  \n\\[\n\\Lambda_{n}:=(\\Delta_{n}\\mathbb Z)^{k}\\subset\\mathbb R^{k}.\n\\]\n\n(A) Show that there exists a sequence $\\bigl(g_{n}\\bigr)_{n\\ge 1}\\subset C^{\\infty}(\\mathbb R^{k})$ such that, for every $n\\ge 1$,\n\\begin{itemize}\n\\item[(i)] $g_{n}(v)=f(v)$ for all lattice points $v\\in\\Lambda_{n}$;\n\\item[(ii)] for every bounded open set $U\\subset\\mathbb R^{k}$  \n      \\[\n      \\operatorname{Lip}\\!\\bigl(g_{n}\\!\\mid_{U}\\bigr)\n            \\le C(k)\\,M_{U,\\sqrt{k}}\\;2^{\\,n},\n      \\tag{$\\ast$}\n      \\]\n      where $C(k)>0$ depends only on $k$;\n\\item[(iii)] $g_{n}(x)\\longrightarrow f(x)$ for every $x\\in\\mathbb R^{k}$.\n\\end{itemize}\n\n(B) Let $K\\subset\\mathbb R^{k}$ be compact and write $D(f)$ for the discontinuity set of $f$.  \nProve that there exists an \\emph{increasing} sequence of closed sets  \n\\[\nF_{1}\\subset F_{2}\\subset\\cdots\\subset K\\setminus D(f),\n\\qquad\n\\bigcup_{m=1}^{\\infty}F_{m}=K\\setminus D(f),\n\\]\nsuch that the convergence in (iii) is \\emph{uniform on each} $F_{m}$ (hence quasi–uniform on $K\\setminus D(f)$).\n\n(C) Deduce that $D(f)$ is an $F_{\\sigma}$–set of first (Baire) category.\n\n(No assertion is required concerning Lebesgue measure.)\n\n\\bigskip",
      "solution": "Throughout, $C_{*}(k)>0$ denotes a numerical constant depending only on $k$ whose value may change from line to line.  \nAll radii we need are $\\le\\sqrt{k}$, so it is convenient to abbreviate  \n\\[\nM_{U}:=M_{U,\\sqrt{k}}\n      =\\sup_{\\operatorname{dist}(x,U)\\le\\sqrt{k}}|f(x)|.\n\\]\n\nThe proof is divided into seven steps.\n\n\\bigskip\n\\textbf{Step 1 – Multilinear interpolation on a dyadic cube.}\n\nFix $n\\ge 1$ and let  \n\\[\nQ=[a_{1},a_{1}+\\Delta_{n}]\\times\\cdots\\times[a_{k},a_{k}+\\Delta_{n}]\n\\]\nbe a closed dyadic cube of side $\\Delta_{n}$ that meets a bounded open set $U$.  \nIndex its $2^{k}$ vertices by $\\varepsilon\\in\\{0,1\\}^{k}$:\n\\[\nv^{\\varepsilon}_{j}=a_{j}+\\varepsilon_{j}\\Delta_{n},\n\\qquad\nj=1,\\dots,k .\n\\]\n\nFor $x\\in Q$ set  \n\\[\n\\lambda_{j}(x):=\\frac{x_{j}-a_{j}}{\\Delta_{n}}\\in[0,1],\n\\quad\n1-\\lambda_{j}(x)=\\frac{a_{j}+\\Delta_{n}-x_{j}}{\\Delta_{n}},\n\\]\nand define the tensor–product multilinear interpolant\n\\[\nI_{Q}(x):=\\sum_{\\varepsilon\\in\\{0,1\\}^{k}}\n           \\Bigl(\\prod_{j=1}^{k}\n           \\bigl(\\varepsilon_{j}\\lambda_{j}(x)+(1-\\varepsilon_{j})(1-\\lambda_{j}(x))\\bigr)\\Bigr)\\;\n           f\\!\\bigl(v^{\\varepsilon}\\bigr).\n\\tag{1.1}\n\\]\n\nBecause $I_{Q}$ is affine in each coordinate,  \n\\[\n\\frac{\\partial I_{Q}}{\\partial x_{j}}(x)\n=\\frac1{\\Delta_{n}}\n  \\sum_{\\varepsilon\\in\\{0,1\\}^{k}}\n  (-1)^{\\varepsilon_{j}}\n  \\Bigl(\\prod_{i\\neq j}\n     \\bigl(\\varepsilon_{i}\\lambda_{i}(x)+(1-\\varepsilon_{i})(1-\\lambda_{i}(x))\\bigr)\\Bigr)\\,\n     f(v^{\\varepsilon}).\n\\]\n\nAll coefficients have absolute value $\\le 1$, hence  \n\\[\n\\Bigl|\\frac{\\partial I_{Q}}{\\partial x_{j}}(x)\\Bigr|\n      \\le\\frac{2^{k}}{\\Delta_{n}}\\,\n           \\max_{\\varepsilon}|f(v^{\\varepsilon})|.\n\\]\n\nSince $\\operatorname{diam}Q=\\sqrt{k}\\,\\Delta_{n}\\le\\sqrt{k}$, every\nvertex satisfies $\\operatorname{dist}(v^{\\varepsilon},U)\\le\\sqrt{k}$, so\n$|f(v^{\\varepsilon})|\\le M_{U}$.  Consequently  \n\\[\n\\Bigl|\\frac{\\partial I_{Q}}{\\partial x_{j}}(x)\\Bigr|\n        \\le C_{0}(k)\\,M_{U}\\,2^{\\,n},\n\\tag{1.2}\n\\]\nwhence\n\\[\n\\operatorname{Lip}\\!\\bigl(I_{Q}\\!\\mid_{U\\cap Q}\\bigr)\n        \\le C_{0}(k)\\,M_{U}\\,2^{\\,n}.\n\\tag{1.3}\n\\]\n\n\\bigskip\n\\textbf{Step 2 – A global continuous interpolant $g_{n}^{\\,0}$.}\n\nDefine $g_{n}^{\\,0}$ by prescribing $g_{n}^{\\,0}\\!\\mid_{Q}:=I_{Q}$ on each dyadic\ncube $Q$ of side $\\Delta_{n}$.  \nFormulas coincide on common faces, hence $g_{n}^{\\,0}\\in C(\\mathbb R^{k})$ and\n$g_{n}^{\\,0}=f$ on $\\Lambda_{n}$.  \nBecause each point of $U$ belongs to at most $3^{k}$ cubes that intersect $U$,\nestimate (1.3) yields  \n\\[\n\\operatorname{Lip}\\!\\bigl(g_{n}^{\\,0}\\!\\mid_{U}\\bigr)\n      \\le C_{1}(k)\\,M_{U}\\,2^{\\,n}.\n\\tag{2.1}\n\\]\n\n\\bigskip\n\\textbf{Step 3 – Small–scale mollification.}\n\nFix a standard mollifier $\\rho\\in C^{\\infty}_{c}(\\mathbb R^{k})$ supported in\n$\\overline{B(0,1)}$, $\\rho\\ge 0$, $\\int\\rho=1$, and set\n$\\rho_{\\eta}(x)=\\eta^{-k}\\rho(x/\\eta)$.  \nChoose  \n\\[\n\\eta_{n}:=\\Delta_{n}^{2}=2^{-2n},\\qquad h_{n}:=g_{n}^{\\,0}*\\rho_{\\eta_{n}}.\n\\]\n\nIf $x,y\\in U$,\n\\[\n|h_{n}(x)-h_{n}(y)|\n   \\le\\operatorname{Lip}\\!\\bigl(g_{n}^{\\,0}\\!\\mid_{U^{+\\eta_{n}}}\\bigr)\\,|x-y|\n   \\stackrel{(2.1)}{\\le} C_{1}(k)\\,M_{U}\\,2^{\\,n}|x-y|,\n\\]\nhence  \n\\[\n\\operatorname{Lip}\\!\\bigl(h_{n}\\!\\mid_{U}\\bigr)\n     \\le C_{1}(k)\\,M_{U}\\,2^{\\,n}.\n\\tag{3.1}\n\\]\n\n\\bigskip\n\\textbf{Step 4 – Pointwise correction on the lattice.}\n\nPut  \n\\[\n\\delta_{n}(v):=h_{n}(v)-f(v),\\qquad v\\in\\Lambda_{n}.\n\\]\n\nUsing (5.1) below we shall have\n\\[\n|\\delta_{n}(v)|\\le C_{5}(k)\\,M_{U}\\,2^{-n}.\n\\]\nPick $\\theta\\in C^{\\infty}_{c}(\\mathbb R^{k})$, $\\theta\\ge 0$,\n$\\operatorname{supp}\\theta\\subset\\overline{B(0,1)}$, $\\theta(0)=1$,\n$\\operatorname{Lip}(\\theta)\\le 2$, and set  \n\\[\nr_{n}:=\\frac{\\Delta_{n}}{8},\\qquad\n\\theta_{v}(x):=\\theta\\!\\Bigl(\\frac{x-v}{r_{n}}\\Bigr).\n\\]\n\nThe balls $B(v,r_{n})$ are disjoint, so  \n\\[\nb_{n}(x):=\\sum_{v\\in\\Lambda_{n}}\\delta_{n}(v)\\,\\theta_{v}(x)\n\\]\ncontains at most one non–zero term and lies in $C^{\\infty}(\\mathbb R^{k})$.  \nDefine  \n\\[\ng_{n}:=h_{n}-b_{n}.\n\\tag{4.1}\n\\]\n\nThen $g_{n}(v)=f(v)$ for $v\\in\\Lambda_{n}$.\n\nFor $x,y\\in U$, at most $C_{2}(k)$ of the $\\theta_{v}$’s differ between $x$\nand $y$, and  \n$|\\delta_{n}(v)|\\le C_{5}(k)M_{U}2^{-n}$, so  \n\\[\n\\operatorname{Lip}\\!\\bigl(b_{n}\\!\\mid_{U}\\bigr)\\le C_{3}(k)\\,M_{U}.\n\\tag{4.2}\n\\]\n\nCombining (3.1) and (4.2) gives  \n\\[\n\\operatorname{Lip}\\!\\bigl(g_{n}\\!\\mid_{U}\\bigr)\n      \\le C_{4}(k)\\,M_{U}\\,2^{\\,n},\n\\]\nthat is, property $(\\ast)$ in (A)(ii).\n\n\\bigskip\n\\textbf{Step 5 – Pointwise estimate for $|g_{n}-f|$.}\n\nFor $x\\in\\mathbb R^{k}$ and $\\delta>0$ define the coordinatewise modulus  \n\\[\n\\omega_{j}(x;\\delta):=\n   \\sup_{|s-x_{j}|\\le\\delta}\n     \\bigl|f(x_{1},\\dots,x_{j-1},s,x_{j+1},\\dots,x_{k})-f(x)\\bigr|,\n   \\quad j=1,\\dots,k .\n\\]\n\nChoose a bounded open $U\\ni x$.  \n\n\\smallskip\n(a)  As $\\eta_{n}\\le\\Delta_{n}$,\n\\[\n|h_{n}(x)-g_{n}^{\\,0}(x)|\n   \\le\\operatorname{Lip}\\!\\bigl(g_{n}^{\\,0}\\!\\mid_{U^{+\\eta_{n}}}\\bigr)\\,\\eta_{n}\n   \\stackrel{(2.1)}{\\le}C_{1}(k)\\,M_{U}\\,2^{-n}.\n\\tag{5.1}\n\\]\n\n(b)  From formula (1.1),  \n\\[\n|g_{n}^{\\,0}(x)-f(x)|\n     \\le\\sum_{j=1}^{k}\\omega_{j}(x;\\Delta_{n}).\n\\tag{5.2}\n\\]\n\n(c)  At most one bump is active at $x$, so  \n\\[\n|b_{n}(x)|\\le C_{5}(k)\\,M_{U}\\,2^{-n}.\n\\tag{5.3}\n\\]\n\nCollecting (4.1) and (5.1)–(5.3) we find  \n\\[\n|g_{n}(x)-f(x)|\n   \\le\\sum_{j=1}^{k}\\omega_{j}(x;\\Delta_{n})+C_{6}(k)\\,M_{U}\\,2^{-n}.\n\\tag{5.4}\n\\]\n\nSince $f$ is separately continuous, each $\\omega_{j}(x;\\Delta_{n})\\to 0$ as\n$n\\to\\infty$, so $g_{n}(x)\\to f(x)$ pointwise, proving (A)(iii).\n\n\\bigskip\n\\textbf{Step 6 – Quasi–uniform convergence on a nested family of closed sets.}\n\nFix a compact set $K\\subset\\mathbb R^{k}$.  \nFor $p\\in\\mathbb N$ set  \n\\[\nF_{p}:=\n   \\Bigl\\{x\\in K\\setminus D(f):\n          \\max_{1\\le j\\le k}\\omega_{j}\\bigl(x;2^{-p}\\bigr)\\le2^{-p}\\Bigr\\}.\n\\tag{6.1}\n\\]\n\n\\emph{$F_{p}$ is closed.}  \nThe map $x\\mapsto\\omega_{j}(x;2^{-p})$ is upper semicontinuous (being the\nmaximum of continuous maps), hence the sub-level set (6.1) is closed.  \nBecause $f$ is continuous at each $x\\notin D(f)$, for every such $x$ there\nexists $p$ with $x\\in F_{p}$.  Thus  \n\\[\nK\\setminus D(f)=\\bigcup_{p=1}^{\\infty}F_{p},\n\\quad\nF_{p}\\subset F_{p+1}.\n\\tag{6.2}\n\\]\n\n\\emph{Uniform convergence on $F_{p}$.}  \nFix $p$ and let $U$ be a bounded open set with $K\\subset U$.  \nFor $x\\in F_{p}$, inequality (5.4) gives  \n\\[\n|g_{n}(x)-f(x)|\n   \\le k\\cdot 2^{-p}+C_{6}(k)\\,M_{U}\\,2^{-n}.\n\\]\n\nHence  \n\\[\n\\sup_{x\\in F_{p}}|g_{n}(x)-f(x)|\n     \\le k\\cdot 2^{-p}+C_{6}(k)\\,M_{U}\\,2^{-n}.\n\\]\n\nChoose $n\\ge n(p):=p+\\lceil\\log_{2}(2kC_{6}(k)M_{U})\\rceil$; then  \n\\[\n\\sup_{x\\in F_{p}}|g_{n}(x)-f(x)|\\le 2^{1-p}.\n\\]\nTherefore $\\bigl(g_{n}\\bigr)$ converges uniformly to $f$ on every $F_{p}$,\nestablishing Part (B).\n\n\\bigskip\n\\textbf{Step 7 – The discontinuity set is an $F_{\\sigma}$ and meagre.}\n\n\\medskip\n\\emph{(i) $D(f)$ is an $F_{\\sigma}$.}  \nFor $m\\in\\mathbb N$ put\n\\[\nE_{m}:=\\Bigl\\{x\\in\\mathbb R^{k}:\n              \\limsup_{y\\to x}|f(y)-f(x)|\\ge\\tfrac1m\\Bigr\\}.\n\\]\nWrite $\\omega_{f}(x):=\\displaystyle\\limsup_{y\\to x}|f(y)-f(x)|$; the map\n$\\omega_{f}$ is \\emph{upper semicontinuous}\nbecause\n\\[\n\\omega_{f}(x)=\\inf_{r>0}\\sup_{\\substack{y\\in\\mathbb R^{k}\\\\|y-x|\\le r}}\n                  |f(y)-f(x)|\n\\]\nis an infimum of continuous functions.  Consequently each $E_{m}$ is closed and\n\\[\nD(f)=\\bigcup_{m=1}^{\\infty}E_{m}.\n\\tag{7.1}\n\\]\n\n\\medskip\n\\emph{(ii) Each $E_{m}$ is nowhere dense.}  \nLet $O\\subset\\mathbb R^{k}$ be a non–empty open set.  \nWe show $O\\setminus E_{m}\\neq\\varnothing$.  \nChoose a closed axis–parallel cube\n\\[\nQ^{(0)}=[a_{1},b_{1}]\\times\\cdots\\times[a_{k},b_{k}]\\subset O.\n\\]\nWe shall construct by induction a sequence of closed cubes\n$Q^{(0)}\\supset Q^{(1)}\\supset Q^{(2)}\\supset\\cdots$ such that\n\\begin{equation}\n\\operatorname{diam}Q^{(n)}\\longrightarrow 0\n\\quad\\text{and}\\quad\n\\operatorname{osc}_{f}(Q^{(n)}):=\n     \\sup_{x,y\\in Q^{(n)}}|f(x)-f(y)|\\le\\frac1m.\n\\tag{7.2}\n\\end{equation}\n\nAssuming $Q^{(n)}$ has been chosen, set\n$Q^{(n)}=I_{1}\\times\\cdots\\times I_{k}$ with intervals $I_{j}$.  \nLet $j\\equiv n\\pmod{k}$; fix the other $(k-1)$ coordinates and use\ncontinuity of $s\\mapsto f(x_{1},\\dots,x_{j-1},s,x_{j+1},\\dots,x_{k})$ in\ns to obtain a subdivision point $c\\in I_{j}$ such that for\n\\[\nI'_{j}:=[a_{j},c],\\qquad\nI''_{j}:=[c,b_{j}],\n\\]\nwe have\n\\[\n\\operatorname{osc}_{f}\\bigl(\n  I'_{j}\\times\\prod_{i\\neq j}I_{i}\\bigr)\n\\le\\operatorname{osc}_{f}(Q^{(n)})-\\frac1{2m}\n\\quad\\text{or}\\quad\n\\operatorname{osc}_{f}\\bigl(\n  I''_{j}\\times\\prod_{i\\neq j}I_{i}\\bigr)\n\\le\\operatorname{osc}_{f}(Q^{(n)})-\\frac1{2m}.\n\\]\nSelect the sub–cube $Q^{(n+1)}$ for which the latter inequality holds.\nInduction decreases the oscillation by at least $1/(2m)$ every $k$\nsteps; consequently the oscillation becomes $\\le 1/m$ after finitely many\nsteps, while the diameters tend to $0$.  (We also keep a factor $1/2$–shrink of\neach edge so diameters decrease geometrically.)\n\nProperty (7.2) implies that the singleton\n\\[\n\\{x_{*}\\}:=\\bigcap_{n=0}^{\\infty}Q^{(n)}\n\\]\nis contained in $O$ and satisfies\n$\\displaystyle\\limsup_{y\\to x_{*}}|f(y)-f(x_{*})|\\le 1/m$;\nhence $x_{*}\\notin E_{m}$.  Therefore $O\\setminus E_{m}\\neq\\varnothing$,\nso $E_{m}$ has empty interior in $O$ and is nowhere dense.\n\n\\medskip\n\\emph{(iii) Conclusion.}  \nBecause $D(f)=\\bigcup_{m}E_{m}$ with each $E_{m}$ closed nowhere dense,\n$D(f)$ is an $F_{\\sigma}$–set of first (Baire) category, proving (C).\n\n\\hfill$\\square$\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.615498",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensional setting: the problem is lifted from ℝ² to arbitrary\n   ℝᵏ, forcing the solver to manage k–dimensional dyadic lattices and\n   multilinear interpolation.\n\n2. Stronger approximation: the task is no longer merely to obtain\n   continuous approximants but smooth (C^∞) ones with an explicit global\n   Lipschitz control that worsens at a prescribed rate.\n\n3. Exact interpolation constraint: the approximants must coincide with f\n   on an exponentially dense lattice – this severely restricts the\n   admissible smoothing procedures and forces a delicate combination of\n   Whitney–type extension and mollification.\n\n4. Uniform convergence on compacta away from D(f) must be demonstrated,\n   introducing an interplay between pointwise estimates, moduli of\n   continuity, and compactness arguments.\n\n5. Additional conclusions about the structure of the discontinuity set\n   (F_σ, first category, measure 0) require both Baire-category theory\n   and measure-theoretic covering arguments, far beyond the scope of the\n   original exercise.\n\nCollectively these enhancements demand mastery of interpolation theory,\nmollifiers, Lipschitz estimates, Baire category, and measure theory,\nmaking the variant significantly harder than both the original problem\nand the earlier kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "\\[\n\\textbf{Dyadic interpolation, quantitative smooth approximation, and the size of the discontinuity set}\n\\]\n\nFix an integer $k\\ge 2$.  \nFor a function $u:\\mathbb R^{k}\\longrightarrow\\mathbb R$ put  \n\n\\[\n\\operatorname{Lip}(u):=\\sup_{x\\ne y}\\frac{|u(x)-u(y)|}{|x-y|}\n\\quad(\\text{Euclidean distance }|\\cdot|),\n\\]\nand, for an open non–empty set $U\\subset\\mathbb R^{k}$ and $R>0$, write  \n\n\\[\n\\operatorname{Lip}\\!\\bigl(u\\!\\mid_{U}\\bigr)\n :=\\sup_{\\substack{x,y\\in U \\\\ x\\ne y}}\\frac{|u(x)-u(y)|}{|x-y|},\n \\qquad  \n M_{U,R}:=\\sup_{\\operatorname{dist}(x,U)\\le R}|u(x)|.\n\\]\n\nLet  \n\n\\[\nf:\\mathbb R^{k}\\longrightarrow \\mathbb R\n\\]\n\nbe \\emph{separately continuous} (i.e. continuous in each coordinate variable) and \\emph{locally bounded}.  \nFor $n\\ge 1$ put $\\Delta_{n}:=2^{-n}$ and denote by  \n\n\\[\n\\Lambda_{n}:=(\\Delta_{n}\\mathbb Z)^{k}\\subset\\mathbb R^{k}\n\\]\n\nthe $k$–dimensional dyadic lattice of mesh $\\Delta_{n}$.\n\n(A) Prove that there exists a sequence $\\bigl(g_{n}\\bigr)_{n\\ge 1}$ of $C^{\\infty}$–functions on $\\mathbb R^{k}$ such that, for every $n\\ge 1$,\n\n(i) $g_{n}(v)=f(v)$ for every lattice point $v\\in\\Lambda_{n}$;  \n\n(ii) for every bounded open set $U\\subset\\mathbb R^{k}$  \n\n\\[\n\\operatorname{Lip}\\!\\bigl(g_{n}\\!\\mid_{U}\\bigr)\n      \\le C(k)\\,M_{U,\\sqrt{k}}\\;2^{\\,n},\n\\tag{$\\ast$}\n\\]\n\nwhere $C(k)>0$ depends only on $k$;  \n\n(iii) $g_{n}(x)\\longrightarrow f(x)$ for every $x\\in\\mathbb R^{k}$.\n\n(B) Show that for every compact set $K\\subset\\mathbb R^{k}$ the convergence in (iii) is \\emph{uniform} on $K\\setminus D(f)$, where $D(f)$ denotes the discontinuity set of $f$.\n\n(C) Deduce that $D(f)$ is an $F_{\\sigma}$–set of first (Baire) category.  \n(No statement is required about Lebesgue measure.)",
      "solution": "Throughout the proof $\\;C_{*}(k)\\!>\\!0$ denotes a constant depending only on $k$ whose value may change from line to line.  \nAll radii that appear will not exceed $\\sqrt{k}$, so every estimate involves the single quantity  \n\n\\[\nM_{U}:=M_{U,\\sqrt{k}}\n      =\\sup_{\\operatorname{dist}(x,U)\\le\\sqrt{k}}|f(x)|.\n\\]\n\nThe argument is divided into seven steps.\n\n\\bigskip\n\\textbf{Step 1 – Multilinear interpolation on one dyadic cube.}\n\nFix $n\\ge 1$ and let $Q=[a_{1},a_{1}+\\Delta_{n}]\\times\\dots\\times[a_{k},a_{k}+\\Delta_{n}]$ be a \\emph{closed} dyadic cube of side $\\Delta_{n}$ which intersects a bounded open set $U$.  \nIts $2^{k}$ vertices are indexed by $\\varepsilon\\in\\{0,1\\}^{k}$:\n\n\\[\nv^{\\varepsilon}_{j}=a_{j}+\\varepsilon_{j}\\Delta_{n},\\qquad\nj=1,\\dots,k .\n\\]\n\nFor $x\\in Q$ put  \n\n\\[\n\\lambda_{j}(x):=\\frac{x_{j}-a_{j}}{\\Delta_{n}}\\in[0,1],\n\\qquad\n1-\\lambda_{j}(x)=\\frac{a_{j}+\\Delta_{n}-x_{j}}{\\Delta_{n}},\n\\]\nand define the standard tensor–product interpolant\n\n\\[\nI_{Q}(x):=\\sum_{\\varepsilon\\in\\{0,1\\}^{k}}\n           \\Bigl(\\prod_{j=1}^{k}\n           \\bigl(\\varepsilon_{j}\\lambda_{j}(x)+(1-\\varepsilon_{j})(1-\\lambda_{j}(x))\\bigr)\\Bigr)\\;\n           f\\!\\bigl(v^{\\varepsilon}\\bigr).\n\\tag{1.1}\n\\]\n\n\\emph{Derivative estimate.}  \nSince $I_{Q}$ is affine in each coordinate,\n\\[\n\\frac{\\partial I_{Q}}{\\partial x_{j}}(x)\n   =\\frac1{\\Delta_{n}}\n     \\sum_{\\varepsilon\\in\\{0,1\\}^{k}}\n     (-1)^{\\varepsilon_{j}}\n     \\Bigl(\\prod_{i\\ne j}\n     \\bigl(\\varepsilon_{i}\\lambda_{i}(x)+(1-\\varepsilon_{i})(1-\\lambda_{i}(x))\\bigr)\\Bigr)\\,\n     f\\!\\bigl(v^{\\varepsilon}\\bigr).\n\\]\n\nAll coefficients in the sum have absolute value at most $1$, hence  \n\n\\[\n\\Bigl|\\frac{\\partial I_{Q}}{\\partial x_{j}}(x)\\Bigr|\n   \\le \\frac{2^{k}}{\\Delta_{n}}\n        \\max_{\\varepsilon}|f(v^{\\varepsilon})|.\n\\]\n\nBecause $x\\in U$ and every vertex satisfies\n\\[\n\\operatorname{dist}\\!\\bigl(v^{\\varepsilon},U\\bigr)\n      \\le \\operatorname{diam}Q\n      =\\sqrt{k}\\,\\Delta_{n}\\le\\sqrt{k},\n\\]\nwe have $|f(v^{\\varepsilon})|\\le M_{U}$. Therefore  \n\n\\[\n\\Bigl|\\frac{\\partial I_{Q}}{\\partial x_{j}}(x)\\Bigr|\n          \\le C_{0}(k)\\,M_{U}\\,2^{\\,n},\n\\tag{1.2}\n\\]\nand\n\n\\[\n\\operatorname{Lip}\\!\\bigl(I_{Q}\\!\\mid_{U\\cap Q}\\bigr)\n          \\le C_{0}(k)\\,M_{U}\\,2^{\\,n}.\n\\tag{1.3}\n\\]\n\n\\bigskip\n\\textbf{Step 2 – A global continuous interpolant $g_{n}^{\\,0}$.}\n\nDefine $g_{n}^{\\,0}$ by setting $g_{n}^{\\,0}\\!\\mid_{Q}:=I_{Q}$ for every dyadic cube $Q$ of side $\\Delta_{n}$.  \nThe formulas on neighbouring cubes coincide on common faces, hence $g_{n}^{\\,0}\\in C(\\mathbb R^{k})$ and $g_{n}^{\\,0}=f$ on $\\Lambda_{n}$.  \nEach $x\\in U$ is contained in at most $3^{k}$ cubes intersecting $U$, so from (1.3)\n\n\\[\n\\operatorname{Lip}\\!\\bigl(g_{n}^{\\,0}\\!\\mid_{U}\\bigr)\n      \\le C_{1}(k)\\,M_{U}\\,2^{\\,n}.\n\\tag{2.1}\n\\]\n\n\\bigskip\n\\textbf{Step 3 – Small–scale mollification.}\n\nLet $\\rho\\in C^{\\infty}_{c}(\\mathbb R^{k})$ be non–negative, radial, supported in $\\overline{B(0,1)}$ and satisfy $\\int\\rho=1$.  \nFor $\\eta>0$ put $\\rho_{\\eta}(x):=\\eta^{-k}\\rho(x/\\eta)$ and choose  \n\n\\[\n\\eta_{n}:=\\Delta_{n}^{2}=2^{-2n}.\n\\]\n\nDefine\n\\[\nh_{n}:=g_{n}^{\\,0}*\\rho_{\\eta_{n}}.\n\\]\n\nIf $x,y\\in U$, then\n\n\\[\n\\begin{aligned}\n|h_{n}(x)-h_{n}(y)|\n  &\\le\\int |g_{n}^{\\,0}(x-z)-g_{n}^{\\,0}(y-z)|\n          \\rho_{\\eta_{n}}(z)\\,dz      \\\\\n  &\\le \\operatorname{Lip}\\!\\bigl(g_{n}^{\\,0}\\!\\mid_{U^{+\\eta_{n}}}\\bigr)\\,|x-y|.\n\\end{aligned}\n\\]\n\nBecause $\\eta_{n}<1\\le\\sqrt{k}$ and $U^{+\\eta_{n}}\\subset U^{+\\sqrt{k}}$,  \n(2.1) yields\n\n\\[\n\\operatorname{Lip}\\!\\bigl(h_{n}\\!\\mid_{U}\\bigr)\n     \\le C_{1}(k)\\,M_{U}\\,2^{\\,n}.\n\\tag{3.1}\n\\]\n\n\\bigskip\n\\textbf{Step 4 – Local bump corrections.}\n\nSet  \n\n\\[\n\\delta_{n}(v):=h_{n}(v)-f(v),\\qquad v\\in\\Lambda_{n}.\n\\]\n\nChoose a non–negative bump $\\theta\\in C^{\\infty}_{c}(\\mathbb R^{k})$ supported in $\\overline{B(0,1)}$ with $\\theta(0)=1$ and $\\operatorname{Lip}(\\theta)\\le 2$.  \nLet  \n\n\\[\nr_{n}:=\\frac{\\Delta_{n}}{8},\\qquad\n\\theta_{v}(x):=\\theta\\!\\Bigl(\\frac{x-v}{r_{n}}\\Bigr).\n\\]\n\nThe balls $B(v,r_{n})$ are disjoint, hence  \n\n\\[\nb_{n}(x):=\\sum_{v\\in\\Lambda_{n}}\\delta_{n}(v)\\,\\theta_{v}(x)\n\\]\nhas at most one non–zero summand and belongs to $C^{\\infty}(\\mathbb R^{k})$.  \nPut  \n\n\\[\ng_{n}:=h_{n}-b_{n}.\n\\tag{4.1}\n\\]\n\n\\emph{Property (A)(i).}  \nFor $v\\in\\Lambda_{n}$, $\\theta_{v}(v)=1$ and $\\theta_{w}(v)=0$ ($w\\ne v$), so $g_{n}(v)=f(v)$.\n\n\\emph{Lipschitz estimate for the bump term.}  \nWrite $U_{n}:=\\{v\\in\\Lambda_{n}:B(v,r_{n})\\cap U\\ne\\varnothing\\}$.  \nFor $x,y\\in U$ there exist at most $C_{2}(k)$ points $v\\in U_{n}$ such that at least one of $x,y$ lies in $B(v,r_{n})$.  \nBecause $|\\delta_{n}(v)|\\le 2M_{U}$ (vertices lie within $r_{n}<1$ of $U$),\n\n\\[\n\\begin{aligned}\n|b_{n}(x)-b_{n}(y)|\n   &\\le \\sum_{v\\in U_{n}} |\\delta_{n}(v)|\\;\n        |\\theta_{v}(x)-\\theta_{v}(y)|   \\\\\n   &\\le \\sum_{v\\in U_{n}} 2M_{U}\\;\\frac{2}{r_{n}}\\;|x-y|\n   \\le C_{3}(k)\\,M_{U}\\,2^{\\,n}\\,|x-y|.\n\\end{aligned}\n\\]\nConsequently  \n\n\\[\n\\operatorname{Lip}\\!\\bigl(b_{n}\\!\\mid_{U}\\bigr)\\le C_{3}(k)\\,M_{U}\\,2^{\\,n}.\n\\tag{4.2}\n\\]\n\nCombining (3.1) and (4.2):\n\n\\[\n\\operatorname{Lip}\\!\\bigl(g_{n}\\!\\mid_{U}\\bigr)\n  \\le \\bigl(C_{1}(k)+C_{3}(k)\\bigr)\\,M_{U}\\,2^{\\,n}\n  =:C_{4}(k)\\,M_{U}\\,2^{\\,n},\n\\]\nwhich is precisely (\\*) of part (A)(ii).\n\n\\bigskip\n\\textbf{Step 5 – Pointwise control of $|g_{n}-f|$.}\n\nFor $x\\in\\mathbb R^{k}$ and $\\delta>0$ define the \\emph{one–sided modulus}\n\n\\[\n\\omega_{j}(x;\\delta)\n   :=\\sup_{|s-x_{j}|\\le\\delta}\n     \\Bigl|f(x_{1},\\dots,x_{j-1},s,x_{j+1},\\dots,x_{k})-f(x)\\Bigr|,\n   \\qquad j=1,\\dots,k .\n\\]\n\nChoose $U\\subset\\mathbb R^{k}$ bounded with $x\\in U$.  \n\n\\smallskip\n\\emph{(a) Difference $h_{n}-g_{n}^{\\,0}$.}  \nSince $\\eta_{n}\\le\\Delta_{n}$,\n\n\\[\n|h_{n}(x)-g_{n}^{\\,0}(x)|\n   \\le \\operatorname{Lip}\\!\\bigl(g_{n}^{\\,0}\\!\\mid_{U^{+\\eta_{n}}}\\bigr)\\,\\eta_{n}\n   \\stackrel{(2.1)}{\\le} C_{1}(k)\\,M_{U}\\,2^{\\,n}\\eta_{n}\n   = C_{5}(k)\\,M_{U}\\,2^{-n}.\n\\tag{5.1}\n\\]\n\n\\smallskip\n\\emph{(b) Difference $g_{n}^{\\,0}-f$.}  \nInside the cube $Q_{x}$ of side $\\Delta_{n}$ that contains $x$,  \nformula (1.1) expresses $g_{n}^{\\,0}(x)$ as a convex combination of the $2^{k}$ values $f(v^{\\varepsilon})$.  \nChanging one coordinate at a time and using the triangle inequality,\n\n\\[\n|g_{n}^{\\,0}(x)-f(x)|\n   \\le \\sum_{j=1}^{k}\\omega_{j}(x;\\Delta_{n}).\n\\tag{5.2}\n\\]\n\n\\smallskip\n\\emph{(c) The bump term.}  \nAt most one $\\theta_{v}$ is non–zero at $x$, and $|\\delta_{n}(v)|\\le 2M_{U}$, so  \n\n\\[\n|b_{n}(x)|\\le 2M_{U}.\n\\tag{5.3}\n\\]\n\n\\smallskip\nCollecting (4.1) and (5.1)–(5.3):\n\n\\[\n|g_{n}(x)-f(x)|\n  \\le \\sum_{j=1}^{k}\\omega_{j}(x;\\Delta_{n})\n       + C_{6}(k)\\,M_{U}\\,2^{-n}.\n\\tag{5.4}\n\\]\n\nBecause $f$ is separately continuous, $\\omega_{j}(x;\\Delta_{n})\\to 0$ ($n\\to\\infty$) for every $j$, proving $g_{n}(x)\\to f(x)$ and finishing (A)(iii).\n\n\\bigskip\n\\textbf{Step 6 – Uniform convergence on $K\\setminus D(f)$.}\n\nLet $K\\subset\\mathbb R^{k}$ be compact and $\\widehat K:=K\\setminus D(f)$.  \nOn the closed set $\\widehat K$ the function $f$ is jointly continuous, hence has a uniform modulus of continuity $\\Omega$:  \n\n\\[\n|f(x)-f(y)|\\le\\Omega(|x-y|),\\qquad x,y\\in\\widehat K,\n\\quad\\text{with}\\quad\\Omega(t)\\xrightarrow[t\\to0]{}0.\n\\]\n\nTaking $U$ bounded open with $K\\subset U$ and applying (5.4):\n\n\\[\n\\sup_{x\\in\\widehat K}|g_{n}(x)-f(x)|\n   \\le k\\,\\Omega(\\Delta_{n})\n        +C_{6}(k)\\,M_{U}\\,2^{-n}\n   \\xrightarrow[n\\to\\infty]{}0,\n\\]\nestablishing part (B).\n\n\\bigskip\n\\textbf{Step 7 – The discontinuity set is $F_{\\sigma}$ and meagre.}\n\nFix $m\\in\\mathbb N$ and define  \n\n\\[\nE_{m}:=\n   \\Bigl\\{x\\in\\mathbb R^{k}:\n          \\limsup_{y\\to x}|f(y)-f(x)|\\ge\\tfrac1m\\Bigr\\}.\n\\]\n\nThe function $x\\mapsto\\limsup_{y\\to x}|f(y)-f(x)|$ is upper semicontinuous, so each $E_{m}$ is closed and  \n\n\\[\nD(f)=\\bigcup_{m=1}^{\\infty}E_{m}.\n\\]\n\nTo prove that every $E_{m}$ is nowhere dense let $O\\subset\\mathbb R^{k}$ be non–empty and open.  \nFor $n\\ge 1$ set  \n\n\\[\nO_{n}:=\\Bigl\\{x\\in O:\\;|g_{n}(x)-f(x)|<\\tfrac1{6m}\\Bigr\\},\n\\]\nwhich are open and satisfy $\\bigcup_{n\\ge1}O_{n}=O$ (by $g_{n}\\to f$ pointwise).  \nChoose $n_{0}$ with $O_{n_{0}}\\ne\\varnothing$ and pick $x_{0}\\in O_{n_{0}}$.  \nBecause $O_{n_{0}}$ is open there exists $r_{0}>0$ such that  \n\n\\[\nB_{0}:=B(x_{0},r_{0})\\subset O_{n_{0}}\\subset O .\n\\tag{7.1}\n\\]\n\nPut $U:=B(x_{0},1)$ and $L:=\\operatorname{Lip}\\bigl(g_{n_{0}}\\!\\mid_{U}\\bigr)\\le C_{4}(k)\\,M_{U}\\,2^{\\,n_{0}}$.  \nDefine  \n\n\\[\nr:=\\min\\!\\Bigl\\{\\tfrac{r_{0}}{2},\\,\\frac1{6mL}\\Bigr\\}>0\n\\quad\\text{and}\\quad\nB:=B(x_{0},r)\\subset B_{0}\\subset O .\n\\]\n\nFor any $y\\in B$ we have  \n\n\\[\n|f(y)-f(x_{0})|\n   \\le |f(y)-g_{n_{0}}(y)|\n       +|g_{n_{0}}(y)-g_{n_{0}}(x_{0})|\n       +|g_{n_{0}}(x_{0})-f(x_{0})|.\n\\]\n\nThe first and third terms are each $<\\tfrac1{6m}$ by $y,x_{0}\\in O_{n_{0}}$,  \nwhile the middle term is bounded by $L|y-x_{0}|\\le Lr\\le\\tfrac1{6m}$.  \nHence $|f(y)-f(x_{0})|<\\tfrac1m$ for every $y\\in B$, so the oscillation of $f$ at every $x\\in B$ is $<\\tfrac1m$.  \nThus $B\\subset O\\setminus E_{m}$ and $E_{m}$ has empty interior in $O$; that is, $E_{m}$ is nowhere dense.\n\nTherefore $D(f)$, being the countable union of closed nowhere–dense sets $(E_{m})_{m\\ge1}$, is an $F_{\\sigma}$–set of first (Baire) category, completing part (C).\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.492402",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensional setting: the problem is lifted from ℝ² to arbitrary\n   ℝᵏ, forcing the solver to manage k–dimensional dyadic lattices and\n   multilinear interpolation.\n\n2. Stronger approximation: the task is no longer merely to obtain\n   continuous approximants but smooth (C^∞) ones with an explicit global\n   Lipschitz control that worsens at a prescribed rate.\n\n3. Exact interpolation constraint: the approximants must coincide with f\n   on an exponentially dense lattice – this severely restricts the\n   admissible smoothing procedures and forces a delicate combination of\n   Whitney–type extension and mollification.\n\n4. Uniform convergence on compacta away from D(f) must be demonstrated,\n   introducing an interplay between pointwise estimates, moduli of\n   continuity, and compactness arguments.\n\n5. Additional conclusions about the structure of the discontinuity set\n   (F_σ, first category, measure 0) require both Baire-category theory\n   and measure-theoretic covering arguments, far beyond the scope of the\n   original exercise.\n\nCollectively these enhancements demand mastery of interpolation theory,\nmollifiers, Lipschitz estimates, Baire category, and measure theory,\nmaking the variant significantly harder than both the original problem\nand the earlier kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}