{
  "index": "1970-B-5",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "B-5. Let \\( u_{n} \\) denote the \"ramp\" function\n\\[\nu_{n}(x)=\\left\\{\\begin{aligned}\n-n & \\text { for } \\quad x \\leqq-n \\\\\nx & \\text { for }-n<x \\leqq n, \\\\\nn & \\text { for } \\quad x>n,\n\\end{aligned}\\right.\n\\]\nand let \\( F \\) denote a real function of a real variable. Show that \\( F \\) is continuous if and only if \\( u_{n} \\circ F \\) is continuous for all \\( n \\). (Note: \\( \\left(u_{n} \\circ F\\right)(x)=u_{n}[F(x)] \\).)",
  "solution": "B-5 Clearly \\( u_{n} \\) is continuous. So, if \\( F \\) is continuous, then \\( u_{n} \\circ F \\) is the composition of continuous functions and hence is continuous. Conversely, suppose \\( u_{n} \\circ F \\) is continuous for all \\( n \\). To prove \\( F \\) is continuous it is enough to show \\( F^{-1}[(a, b)] \\) is open for every bounded interval \\( (a, b) \\). Let \\( n>\\max (|a|,|b|) \\). Then \\( u_{n}{ }^{-1}[(a, b)]=(a, b) \\) so\n\\[\nF^{-1}[(a, b)]=F^{-1}\\left[u_{n}^{-1}\\{(a, b)\\}\\right]=\\left(u_{n} \\circ F\\right)^{-1}[(a, b)],\n\\]\nwhich is an open set by the continuity of \\( u_{n} \\circ F \\).",
  "vars": [
    "u_n",
    "F",
    "x"
  ],
  "params": [
    "n",
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "u_n": "rampfun",
        "F": "targetf",
        "x": "realvar",
        "n": "indexval",
        "a": "boundlow",
        "b": "boundhigh"
      },
      "question": "B-5. Let \\( rampfun_{indexval} \\) denote the \"ramp\" function\n\\[\nrampfun_{indexval}(realvar)=\\left\\{\\begin{aligned}\n-indexval & \\text { for } \\quad realvar \\leqq-indexval \\\\\nrealvar & \\text { for }-indexval<realvar \\leqq indexval, \\\\\nindexval & \\text { for } \\quad realvar>indexval,\n\\end{aligned}\\right.\n\\]\nand let \\( targetf \\) denote a real function of a real variable. Show that \\( targetf \\) is continuous if and only if \\( rampfun_{indexval} \\circ targetf \\) is continuous for all \\( indexval \\). (Note: \\( \\left(rampfun_{indexval} \\circ targetf\\right)(realvar)=rampfun_{indexval}[targetf(realvar)] \\).)",
      "solution": "B-5 Clearly \\( rampfun_{indexval} \\) is continuous. So, if \\( targetf \\) is continuous, then \\( rampfun_{indexval} \\circ targetf \\) is the composition of continuous functions and hence is continuous. Conversely, suppose \\( rampfun_{indexval} \\circ targetf \\) is continuous for all \\( indexval \\). To prove \\( targetf \\) is continuous it is enough to show \\( targetf^{-1}[(boundlow, boundhigh)] \\) is open for every bounded interval \\( (boundlow, boundhigh) \\). Let \\( indexval>\\max (|boundlow|,|boundhigh|) \\). Then \\( rampfun_{indexval}{ }^{-1}[(boundlow, boundhigh)]=(boundlow, boundhigh) \\) so\n\\[\ntargetf^{-1}[(boundlow, boundhigh)]=targetf^{-1}\\left[rampfun_{indexval}^{-1}\\{(boundlow, boundhigh)\\}\\right]=\\left(rampfun_{indexval} \\circ targetf\\right)^{-1}[(boundlow, boundhigh)],\n\\]\nwhich is an open set by the continuity of \\( rampfun_{indexval} \\circ targetf \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "u_n": "marigoldd",
        "F": "harvestqq",
        "x": "riverbank",
        "n": "lanterner",
        "a": "orchidale",
        "b": "celestium"
      },
      "question": "B-5. Let \\( marigoldd \\) denote the \"ramp\" function\n\\[\nmarigoldd(riverbank)=\\left\\{\\begin{aligned}\n-lanterner & \\text { for } \\quad riverbank \\leqq-lanterner \\\\\nriverbank & \\text { for }-lanterner<riverbank \\leqq lanterner, \\\\\nlanterner & \\text { for } \\quad riverbank>lanterner,\n\\end{aligned}\\right.\n\\]\nand let \\( harvestqq \\) denote a real function of a real variable. Show that \\( harvestqq \\) is continuous if and only if \\( marigoldd \\circ harvestqq \\) is continuous for all \\( lanterner \\). (Note: \\( \\left(marigoldd \\circ harvestqq\\right)(riverbank)=marigoldd[harvestqq(riverbank)] \\).)",
      "solution": "B-5 Clearly \\( marigoldd \\) is continuous. So, if \\( harvestqq \\) is continuous, then \\( marigoldd \\circ harvestqq \\) is the composition of continuous functions and hence is continuous. Conversely, suppose \\( marigoldd \\circ harvestqq \\) is continuous for all \\( lanterner \\). To prove \\( harvestqq \\) is continuous it is enough to show \\( harvestqq^{-1}[(orchidale, celestium)] \\) is open for every bounded interval \\( (orchidale, celestium) \\). Let \\( lanterner>\\max (|orchidale|,|celestium|) \\). Then \\( marigoldd{ }^{-1}[(orchidale, celestium)]=(orchidale, celestium) \\) so\n\\[\nharvestqq^{-1}[(orchidale, celestium)]=harvestqq^{-1}\\left[marigoldd^{-1}\\{(orchidale, celestium)\\}\\right]=\\left(marigoldd \\circ harvestqq\\right)^{-1}[(orchidale, celestium)],\n\\]\nwhich is an open set by the continuity of \\( marigoldd \\circ harvestqq \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "u_n": "valleyfunction",
        "F": "fixconstant",
        "x": "constantval",
        "n": "smallindex",
        "a": "upperboundary",
        "b": "lowerboundary"
      },
      "question": "B-5. Let \\( valleyfunction_{smallindex} \\) denote the \"ramp\" function\n\\[\nvalleyfunction_{smallindex}(constantval)=\\left\\{\\begin{aligned}\n-smallindex & \\text { for } \\quad constantval \\leqq-smallindex \\\\\nconstantval & \\text { for }-smallindex<constantval \\leqq smallindex, \\\\\nsmallindex & \\text { for } \\quad constantval>smallindex,\n\\end{aligned}\\right.\n\\]\nand let \\( fixconstant \\) denote a real function of a real variable. Show that \\( fixconstant \\) is continuous if and only if \\( valleyfunction_{smallindex} \\circ fixconstant \\) is continuous for all \\( smallindex \\). (Note: \\( \\left(valleyfunction_{smallindex} \\circ fixconstant\\right)(constantval)=valleyfunction_{smallindex}[fixconstant(constantval)] \\).)",
      "solution": "B-5 Clearly valleyfunction_{smallindex} is continuous. So, if fixconstant is continuous, then valleyfunction_{smallindex} \\circ fixconstant is the composition of continuous functions and hence is continuous. Conversely, suppose valleyfunction_{smallindex} \\circ fixconstant is continuous for all smallindex. To prove fixconstant is continuous it is enough to show fixconstant^{-1}[(upperboundary, lowerboundary)] is open for every bounded interval (upperboundary, lowerboundary). Let smallindex>\\max (|upperboundary|,|lowerboundary|). Then valleyfunction_{smallindex}{ }^{-1}[(upperboundary, lowerboundary)]=(upperboundary, lowerboundary) so\n\\[\nfixconstant^{-1}[(upperboundary, lowerboundary)]=fixconstant^{-1}\\left[valleyfunction_{smallindex}^{-1}\\{(upperboundary, lowerboundary)\\}\\right]=\\left(valleyfunction_{smallindex} \\circ fixconstant\\right)^{-1}[(upperboundary, lowerboundary)],\n\\]\nwhich is an open set by the continuity of valleyfunction_{smallindex} \\circ fixconstant."
    },
    "garbled_string": {
      "map": {
        "u_n": "qzxwvtnp",
        "F": "hjgrksla",
        "x": "pvlmbczr",
        "n": "wfkajdhe",
        "a": "zhrmtlqc",
        "b": "sfkjdwnr"
      },
      "question": "B-5. Let \\( qzxwvtnp \\) denote the \"ramp\" function\n\\[\nqzxwvtnp(pvlmbczr)=\\left\\{\\begin{aligned}\n-wfkajdhe & \\text { for } \\quad pvlmbczr \\leqq-wfkajdhe \\\\\npvlmbczr & \\text { for }-wfkajdhe<pvlmbczr \\leqq wfkajdhe, \\\\\nwfkajdhe & \\text { for } \\quad pvlmbczr>wfkajdhe,\n\\end{aligned}\\right.\n\\]\nand let \\( hjgrksla \\) denote a real function of a real variable. Show that \\( hjgrksla \\) is continuous if and only if \\( qzxwvtnp \\circ hjgrksla \\) is continuous for all \\( wfkajdhe \\). (Note: \\( \\left(qzxwvtnp \\circ hjgrksla\\right)(pvlmbczr)=qzxwvtnp[hjgrksla(pvlmbczr)] \\).)",
      "solution": "B-5 Clearly \\( qzxwvtnp \\) is continuous. So, if \\( hjgrksla \\) is continuous, then \\( qzxwvtnp \\circ hjgrksla \\) is the composition of continuous functions and hence is continuous. Conversely, suppose \\( qzxwvtnp \\circ hjgrksla \\) is continuous for all \\( wfkajdhe \\). To prove \\( hjgrksla \\) is continuous it is enough to show \\( hjgrksla^{-1}[(zhrmtlqc, sfkjdwnr)] \\) is open for every bounded interval \\( (zhrmtlqc, sfkjdwnr) \\). Let \\( wfkajdhe>\\max (|zhrmtlqc|,|sfkjdwnr|) \\). Then \\( qzxwvtnp^{-1}[(zhrmtlqc, sfkjdwnr)]=(zhrmtlqc, sfkjdwnr) \\) so\n\\[\nhjgrksla^{-1}[(zhrmtlqc, sfkjdwnr)]=hjgrksla^{-1}\\left[qzxwvtnp^{-1}\\{(zhrmtlqc, sfkjdwnr)\\}\\right]=\\left(qzxwvtnp \\circ hjgrksla\\right)^{-1}[(zhrmtlqc, sfkjdwnr)],\n\\]\nwhich is an open set by the continuity of \\( qzxwvtnp \\circ hjgrksla \\)."
    },
    "kernel_variant": {
      "question": "Let $H$ be a separable real Hilbert space with inner product  \n$\\langle \\!\\cdot\\!,\\!\\cdot\\!\\rangle$ and norm $\\|v\\|=\\sqrt{\\langle v,v\\rangle}$.\n\nFor every ordered pair of positive integers $(k,r)$ define the radial  \n“exponential clamp’’  \n\\[\nA_{k,r}\\colon H\\longrightarrow H,\\qquad \nA_{k,r}(0):=0,\\qquad \nA_{k,r}(v):=\n\\begin{cases}\nv, & \\|v\\|\\le k,\\\\[6pt]\n\\displaystyle\\Bigl(k+(\\|v\\|-k)\\,e^{-r(\\|v\\|-k)}\\Bigr)\\,\n\\dfrac{v}{\\|v\\|}, & \\|v\\|>k .\n\\end{cases}\\tag{1}\n\\]\n\nIntroduce the auxiliary scalar function  \n\\[\n\\rho_{k,r}(t)=\n\\begin{cases}\nt,&0\\le t\\le k,\\\\[4pt]\nk+(t-k)e^{-r(t-k)},&t>k,\n\\end{cases}\\qquad t\\ge0,\\tag{2}\n\\]\nso that $A_{k,r}(v)=\\rho_{k,r}(\\|v\\|)\\,v/\\|v\\|$ whenever $v\\neq0$.\n\n0.  (Calculus preliminaries)  Prove  \n\n(a)  $\\rho_{k,r}\\in C^{1}\\bigl((0,\\infty)\\bigr)$, it extends continuously to $[0,\\infty)$ and $\\rho_{k,r}(0)=0$;\n\n(b)  for all $t>0$ one has\n     $\\bigl|\\rho_{k,r}'(t)\\bigr|\\le 1$,  \n     hence $\\rho_{k,r}$ is $1$–Lipschitz on $[0,\\infty)$;\n\n(c)  for every $t>k$\n     \\[\n        k\\;\\le\\;\\rho_{k,r}(t)\\;\\le\\;k+\\dfrac1{re},\\qquad\n        t-\\rho_{k,r}(t)=(t-k)\\bigl(1-e^{-r(t-k)}\\bigr).\\tag{3}\n     \\]\n\n1.  (Fréchet differentiability)  Prove that $A_{k,r}$ is Fréchet differentiable at every $v\\neq0$.  \n    Compute the derivative\n    \\[\n       D A_{k,r}(v)[h]\n       =\n       \\rho_{k,r}'(\\|v\\|)\\frac{\\langle v,h\\rangle}{\\|v\\|}\\frac{v}{\\|v\\|}\n       +\\rho_{k,r}(\\|v\\|)\\Bigl(\\frac{h}{\\|v\\|}\n       -\\frac{\\langle v,h\\rangle}{\\|v\\|^{3}}\\,v\\Bigr),\\qquad h\\in H,\n    \\]\n    and prove the operator–norm estimate $\\|D A_{k,r}(v)\\|_{\\mathcal L(H)}\\le 1$.\n\n2.  (Global Lipschitz regularity)  Show that  \n    \\[\n       \\|A_{k,r}(v)-A_{k,r}(w)\\|\\le\\|v-w\\|\\qquad\\forall v,w\\in H.\n    \\]\n\n3.  (Approximation of the identity)  Let $\\{(k_{n},r_{n})\\}_{n\\ge1}\\subset\\mathbb N^{2}$ satisfy $k_{n}\\to\\infty$ and $r_{n}\\to\\infty$.  \n    Show that  \n\n    (i) $A_{k_{n},r_{n}}(v)\\to v$ for every $v\\in H$;\n\n    (ii) the convergence is uniform on each bounded subset of $H$.",
      "solution": "Throughout we fix $k,r\\in\\mathbb N$ and abbreviate  \n$\\rho:=\\rho_{k,r}$, $A:=A_{k,r}$.\n\n\\bigskip\n0.  Properties of $\\rho$.\n\n(a)  Smoothness.  \nOn $[0,k]$ we have $\\rho(t)=t$, hence $\\rho$ is $C^{\\infty}$ there and $\\rho'(t)\\equiv1$ for $0<t<k$.  \nOn $(k,\\infty)$,\n\\[\n\\rho(t)=k+(t-k)e^{-r(t-k)},\\qquad\n\\rho'(t)=e^{-r(t-k)}\\bigl(1-r(t-k)\\bigr).\n\\]\nAt $t=k$ the one–sided limits agree:\n\\[\n\\lim_{t\\uparrow k}\\rho(t)=k=\\lim_{t\\downarrow k}\\rho(t),\\qquad\n\\lim_{t\\uparrow k}\\rho'(t)=1=\\lim_{t\\downarrow k}\\rho'(t),\n\\]\nso $\\rho\\in C^{1}\\bigl((0,\\infty)\\bigr)$ and extends continuously to $[0,\\infty)$ with $\\rho(0)=0$.\n\n(b)  Lipschitz constant.  \nFor $0<t\\le k$ we have $|\\rho'(t)|=1$.  \nFor $t>k$ set $u:=r(t-k)\\ge0$.  Then\n\\[\n\\rho'(t)=e^{-u}(1-u),\\qquad\n|\\rho'(t)|=e^{-u}|1-u|.\n\\]\nIf $0\\le u\\le1$, then $|1-u|=1-u\\le1$, hence $|\\rho'(t)|\\le e^{-u}\\le1$.  \nIf $u\\ge1$, write $|1-u|=u-1\\le u$, so $|\\rho'(t)|\\le u e^{-u}\\le1/e<1$.  \nThus $|\\rho'(t)|\\le1$ for all $t>0$, and $\\rho$ is $1$–Lipschitz on $[0,\\infty)$.\n\n(c)  Bounds for $t>k$.  \nLet again $u=r(t-k)>0$.  Then\n\\[\n\\rho(t)=k+\\frac{u}{r}\\,e^{-u}.\n\\]\nSince $u e^{-u}\\le1/e$ for all $u>0$, we obtain\n\\[\nk\\le\\rho(t)\\le k+\\frac1{re},\n\\]\nand the identity in (3) is immediate from the definition of $\\rho$.\n\n\\bigskip\n1.  Fréchet differentiability of $A$ on $H\\setminus\\{0\\}$.\n\nLet $v\\neq0$ and $h\\in H$.  Put $\\alpha:=\\|v\\|$, $\\widehat v:=v/\\alpha$.  \nDecompose\n\\[\nh=a\\widehat v+b,\\qquad a:=\\langle \\widehat v,h\\rangle,\\quad b:=h-a\\widehat v,\\quad \\langle \\widehat v,b\\rangle=0.\n\\]\nThen $\\|h\\|^{2}=a^{2}+\\|b\\|^{2}$.\n\n\\emph{Expansion of $A(v+h)$.}  \nSet $\\beta:=\\|v+h\\|$ and $\\delta:=\\beta-\\alpha$.  \nBecause both $t\\mapsto\\rho(t)$ and $t\\mapsto1/t$ are $C^{1}$ on $(0,\\infty)$, a Taylor expansion yields\n\\[\nA(v+h)=\\rho(\\beta)\\,\\frac{v+h}{\\beta}\n      =\\rho(\\alpha)\\widehat v\n      +\\rho'(\\alpha)\\,\\delta\\,\\widehat v\n      +\\rho(\\alpha)\\Bigl(\\frac{h}{\\alpha}-\\frac{\\langle \\widehat v,h\\rangle}{\\alpha}\\widehat v\\Bigr)\n      +o(\\|h\\|).\n\\]\nSince $\\delta=\\langle \\widehat v,h\\rangle+o(\\|h\\|)=a+o(\\|h\\|)$, we obtain\n\\[\nD A(v)[h]=\\rho'(\\alpha)\\,a\\,\\widehat v+\\frac{\\rho(\\alpha)}{\\alpha}\\,b\n=\\rho'(\\alpha)\\frac{\\langle v,h\\rangle}{\\alpha}\\frac{v}{\\alpha}\n+\\rho(\\alpha)\\Bigl(\\frac{h}{\\alpha}-\\frac{\\langle v,h\\rangle}{\\alpha^{3}}\\,v\\Bigr).\n\\]\n\n\\emph{Operator–norm estimate.}  \nUsing the same decomposition $h=a\\widehat v+b$,\n\\[\nD A(v)[h]=\\rho'(\\alpha)\\,a\\,\\widehat v+\\frac{\\rho(\\alpha)}{\\alpha}\\,b,\n\\]\nwhence\n\\[\n\\|D A(v)[h]\\|^{2}=|\\rho'(\\alpha)|^{2}a^{2}\n                     +\\Bigl(\\frac{\\rho(\\alpha)}{\\alpha}\\Bigr)^{2}\\|b\\|^{2}.\n\\]\nFrom Step 0(b) we have $|\\rho'(\\alpha)|\\le1$ and $\\rho(\\alpha)\\le\\alpha$, so\n\\[\n\\|D A(v)[h]\\|^{2}\\le a^{2}+\\|b\\|^{2}=\\|h\\|^{2},\n\\qquad\\text{that is}\\quad\n\\|D A(v)\\|_{\\mathcal L(H)}\\le1.\n\\]\n\n\\bigskip\n2.  Global $1$–Lipschitz property of $A$.\n\nFix $v,w\\in H$ and consider the segment $\\gamma:[0,1]\\to H$, $\\gamma(s):=w+s(v-w)$.  \nThe map $s\\mapsto A(\\gamma(s))$ is absolutely continuous because $A$ is locally Lipschitz (it is smooth away from the origin and $1$–Lipschitz near $0$ by Step 0(b)).  We have, for almost every $s\\in(0,1)$,\n\\[\n\\frac{\\mathrm d}{\\mathrm ds}A(\\gamma(s))\n      =D A(\\gamma(s))[v-w],\n\\]\nhence by Step&nbsp;1\n\\[\n\\Bigl\\|\\frac{\\mathrm d}{\\mathrm ds}A(\\gamma(s))\\Bigr\\|\n      \\le\\|v-w\\|.\n\\]\nIntegrating from $0$ to $1$ we obtain\n\\[\n\\|A(v)-A(w)\\|\n      =\\Bigl\\|\\int_{0}^{1}\\frac{\\mathrm d}{\\mathrm ds}A(\\gamma(s))\\,\\mathrm ds\\Bigr\\|\n      \\le\\int_{0}^{1}\\Bigl\\|\\frac{\\mathrm d}{\\mathrm ds}A(\\gamma(s))\\Bigr\\|\\mathrm ds\n      \\le\\|v-w\\|.\n\\]\nTherefore $A$ is globally $1$–Lipschitz on $H$.\n\n\\bigskip\n3.  Approximation of the identity.  \n\nFix $M>0$ and denote $B_{M}:=\\{v\\in H:\\|v\\|\\le M\\}$.\n\n(i)  Pointwise convergence.  \nGiven $v\\in H$, choose $n_{0}$ such that $k_{n}\\ge\\|v\\|$ for all $n\\ge n_{0}$.  Then $A_{k_{n},r_{n}}(v)=v$ for $n\\ge n_{0}$, so $A_{k_{n},r_{n}}(v)\\to v$.\n\n(ii)  Uniform convergence on $B_{M}$.  \nBecause $k_{n}\\to\\infty$, there exists $N$ such that $k_{n}\\ge M$ for every $n\\ge N$.  \nFor these indices and every $v\\in B_{M}$ we again have $A_{k_{n},r_{n}}(v)=v$.  Hence\n\\[\n\\sup_{v\\in B_{M}}\\|A_{k_{n},r_{n}}(v)-v\\|=0\\qquad\\text{for }n\\ge N,\n\\]\nwhich proves the uniform convergence.  No estimate involving $r_{n}$ is required.\n\n\\bigskip\n\\emph{Remark.}  One can verify that $A_{k,r}$ is only Gâteaux differentiable at $v=0$, with $D A_{k,r}(0)=\\operatorname{Id}_{H}$; the derivative is not continuous there.  The global Lipschitz bound therefore remains the principal tool at the origin.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.595486",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimensional / functional setting:  The problem is lifted from real-valued functions on \\(\\mathbb R\\) to arbitrary maps into an infinite-dimensional Banach space, requiring familiarity with functional analysis (dual space, separating functionals, Hahn–Banach).  \n• Additional family of parameters:  Instead of a single index \\(n\\) we now have a two–parameter family \\(A_{k,r}\\) with a non-trivial radial profile; analysing its properties is technically more involved.  \n• Multiple equivalent conditions:  The solver must prove a cycle of implications (i)⇔(ii)⇔(iii)⇔(iv), not only the two-way implication of the original task.  \n• Use of advanced tools:  The proof employs separation of points by linear functionals, compactness arguments, and equicontinuity via the Arzelà–Ascoli philosophy.  \n• Uniform approximation statement:  Part 2 adds a non-obvious quantitative result (uniform limit), forcing the solver to control the radial decay parameter \\(r\\) simultaneously with the truncation radius \\(k\\).  \nAll these layers make the enhanced variant substantially harder and substantially more technical than both the original problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $H$ be a separable real Hilbert space with inner product  \n$\\langle \\!\\cdot\\!,\\!\\cdot\\!\\rangle$ and norm $\\|v\\|=\\sqrt{\\langle v,v\\rangle}$.\n\nFor every ordered pair of positive integers $(k,r)$ define the radial  \n“exponential clamp’’  \n\\[\nA_{k,r}\\colon H\\longrightarrow H,\\qquad \nA_{k,r}(0):=0,\\qquad \nA_{k,r}(v):=\n\\begin{cases}\nv, & \\|v\\|\\le k,\\\\[6pt]\n\\displaystyle\\Bigl(k+(\\|v\\|-k)\\,e^{-r(\\|v\\|-k)}\\Bigr)\\,\n\\dfrac{v}{\\|v\\|}, & \\|v\\|>k .\n\\end{cases}\\tag{1}\n\\]\n\nIntroduce the auxiliary scalar function  \n\\[\n\\rho_{k,r}(t)=\n\\begin{cases}\nt,&0\\le t\\le k,\\\\[4pt]\nk+(t-k)e^{-r(t-k)},&t>k,\n\\end{cases}\\qquad t\\ge0,\\tag{2}\n\\]\nso that $A_{k,r}(v)=\\rho_{k,r}(\\|v\\|)\\,v/\\|v\\|$ whenever $v\\neq0$.\n\n0.  (Calculus preliminaries)  Prove  \n\n(a)  $\\rho_{k,r}\\in C^{1}\\bigl((0,\\infty)\\bigr)$, it extends continuously to $[0,\\infty)$ and $\\rho_{k,r}(0)=0$;\n\n(b)  for all $t>0$ one has\n     $\\bigl|\\rho_{k,r}'(t)\\bigr|\\le 1$,  \n     hence $\\rho_{k,r}$ is $1$–Lipschitz on $[0,\\infty)$;\n\n(c)  for every $t>k$\n     \\[\n        k\\;\\le\\;\\rho_{k,r}(t)\\;\\le\\;k+\\dfrac1{re},\\qquad\n        t-\\rho_{k,r}(t)=(t-k)\\bigl(1-e^{-r(t-k)}\\bigr).\\tag{3}\n     \\]\n\n1.  (Fréchet differentiability)  Prove that $A_{k,r}$ is Fréchet differentiable at every $v\\neq0$.  \n    Compute the derivative\n    \\[\n       D A_{k,r}(v)[h]\n       =\n       \\rho_{k,r}'(\\|v\\|)\\frac{\\langle v,h\\rangle}{\\|v\\|}\\frac{v}{\\|v\\|}\n       +\\rho_{k,r}(\\|v\\|)\\Bigl(\\frac{h}{\\|v\\|}\n       -\\frac{\\langle v,h\\rangle}{\\|v\\|^{3}}\\,v\\Bigr),\\qquad h\\in H,\n    \\]\n    and prove the operator–norm estimate $\\|D A_{k,r}(v)\\|_{\\mathcal L(H)}\\le 1$.\n\n2.  (Global Lipschitz regularity)  Show that  \n    \\[\n       \\|A_{k,r}(v)-A_{k,r}(w)\\|\\le\\|v-w\\|\\qquad\\forall v,w\\in H.\n    \\]\n\n3.  (Approximation of the identity)  Let $\\{(k_{n},r_{n})\\}_{n\\ge1}\\subset\\mathbb N^{2}$ satisfy $k_{n}\\to\\infty$ and $r_{n}\\to\\infty$.  \n    Show that  \n\n    (i) $A_{k_{n},r_{n}}(v)\\to v$ for every $v\\in H$;\n\n    (ii) the convergence is uniform on each bounded subset of $H$.",
      "solution": "Throughout we fix $k,r\\in\\mathbb N$ and abbreviate  \n$\\rho:=\\rho_{k,r}$, $A:=A_{k,r}$.\n\n\\bigskip\n0.  Properties of $\\rho$.\n\n(a)  Smoothness.  \nOn $[0,k]$ we have $\\rho(t)=t$, hence $\\rho$ is $C^{\\infty}$ there and $\\rho'(t)\\equiv1$ for $0<t<k$.  \nOn $(k,\\infty)$,\n\\[\n\\rho(t)=k+(t-k)e^{-r(t-k)},\\qquad\n\\rho'(t)=e^{-r(t-k)}\\bigl(1-r(t-k)\\bigr).\n\\]\nAt $t=k$ the one–sided limits agree:\n\\[\n\\lim_{t\\uparrow k}\\rho(t)=k=\\lim_{t\\downarrow k}\\rho(t),\\qquad\n\\lim_{t\\uparrow k}\\rho'(t)=1=\\lim_{t\\downarrow k}\\rho'(t),\n\\]\nso $\\rho\\in C^{1}\\bigl((0,\\infty)\\bigr)$ and extends continuously to $[0,\\infty)$ with $\\rho(0)=0$.\n\n(b)  Lipschitz constant.  \nFor $0<t\\le k$ we have $|\\rho'(t)|=1$.  \nFor $t>k$ set $u:=r(t-k)\\ge0$.  Then\n\\[\n\\rho'(t)=e^{-u}(1-u),\\qquad\n|\\rho'(t)|=e^{-u}|1-u|.\n\\]\nIf $0\\le u\\le1$, then $|1-u|=1-u\\le1$, hence $|\\rho'(t)|\\le e^{-u}\\le1$.  \nIf $u\\ge1$, write $|1-u|=u-1\\le u$, so $|\\rho'(t)|\\le u e^{-u}\\le1/e<1$.  \nThus $|\\rho'(t)|\\le1$ for all $t>0$, and $\\rho$ is $1$–Lipschitz on $[0,\\infty)$.\n\n(c)  Bounds for $t>k$.  \nLet again $u=r(t-k)>0$.  Then\n\\[\n\\rho(t)=k+\\frac{u}{r}\\,e^{-u}.\n\\]\nSince $u e^{-u}\\le1/e$ for all $u>0$, we obtain\n\\[\nk\\le\\rho(t)\\le k+\\frac1{re},\n\\]\nand the identity in (3) is immediate from the definition of $\\rho$.\n\n\\bigskip\n1.  Fréchet differentiability of $A$ on $H\\setminus\\{0\\}$.\n\nLet $v\\neq0$ and $h\\in H$.  Put $\\alpha:=\\|v\\|$, $\\widehat v:=v/\\alpha$.  \nDecompose\n\\[\nh=a\\widehat v+b,\\qquad a:=\\langle \\widehat v,h\\rangle,\\quad b:=h-a\\widehat v,\\quad \\langle \\widehat v,b\\rangle=0.\n\\]\nThen $\\|h\\|^{2}=a^{2}+\\|b\\|^{2}$.\n\n\\emph{Expansion of $A(v+h)$.}  \nSet $\\beta:=\\|v+h\\|$ and $\\delta:=\\beta-\\alpha$.  \nBecause both $t\\mapsto\\rho(t)$ and $t\\mapsto1/t$ are $C^{1}$ on $(0,\\infty)$, a Taylor expansion yields\n\\[\nA(v+h)=\\rho(\\beta)\\,\\frac{v+h}{\\beta}\n      =\\rho(\\alpha)\\widehat v\n      +\\rho'(\\alpha)\\,\\delta\\,\\widehat v\n      +\\rho(\\alpha)\\Bigl(\\frac{h}{\\alpha}-\\frac{\\langle \\widehat v,h\\rangle}{\\alpha}\\widehat v\\Bigr)\n      +o(\\|h\\|).\n\\]\nSince $\\delta=\\langle \\widehat v,h\\rangle+o(\\|h\\|)=a+o(\\|h\\|)$, we obtain\n\\[\nD A(v)[h]=\\rho'(\\alpha)\\,a\\,\\widehat v+\\frac{\\rho(\\alpha)}{\\alpha}\\,b\n=\\rho'(\\alpha)\\frac{\\langle v,h\\rangle}{\\alpha}\\frac{v}{\\alpha}\n+\\rho(\\alpha)\\Bigl(\\frac{h}{\\alpha}-\\frac{\\langle v,h\\rangle}{\\alpha^{3}}\\,v\\Bigr).\n\\]\n\n\\emph{Operator–norm estimate.}  \nUsing the same decomposition $h=a\\widehat v+b$,\n\\[\nD A(v)[h]=\\rho'(\\alpha)\\,a\\,\\widehat v+\\frac{\\rho(\\alpha)}{\\alpha}\\,b,\n\\]\nwhence\n\\[\n\\|D A(v)[h]\\|^{2}=|\\rho'(\\alpha)|^{2}a^{2}\n                     +\\Bigl(\\frac{\\rho(\\alpha)}{\\alpha}\\Bigr)^{2}\\|b\\|^{2}.\n\\]\nFrom Step 0(b) we have $|\\rho'(\\alpha)|\\le1$ and $\\rho(\\alpha)\\le\\alpha$, so\n\\[\n\\|D A(v)[h]\\|^{2}\\le a^{2}+\\|b\\|^{2}=\\|h\\|^{2},\n\\qquad\\text{that is}\\quad\n\\|D A(v)\\|_{\\mathcal L(H)}\\le1.\n\\]\n\n\\bigskip\n2.  Global $1$–Lipschitz property of $A$.\n\nFix $v,w\\in H$ and consider the segment $\\gamma:[0,1]\\to H$, $\\gamma(s):=w+s(v-w)$.  \nThe map $s\\mapsto A(\\gamma(s))$ is absolutely continuous because $A$ is locally Lipschitz (it is smooth away from the origin and $1$–Lipschitz near $0$ by Step 0(b)).  We have, for almost every $s\\in(0,1)$,\n\\[\n\\frac{\\mathrm d}{\\mathrm ds}A(\\gamma(s))\n      =D A(\\gamma(s))[v-w],\n\\]\nhence by Step&nbsp;1\n\\[\n\\Bigl\\|\\frac{\\mathrm d}{\\mathrm ds}A(\\gamma(s))\\Bigr\\|\n      \\le\\|v-w\\|.\n\\]\nIntegrating from $0$ to $1$ we obtain\n\\[\n\\|A(v)-A(w)\\|\n      =\\Bigl\\|\\int_{0}^{1}\\frac{\\mathrm d}{\\mathrm ds}A(\\gamma(s))\\,\\mathrm ds\\Bigr\\|\n      \\le\\int_{0}^{1}\\Bigl\\|\\frac{\\mathrm d}{\\mathrm ds}A(\\gamma(s))\\Bigr\\|\\mathrm ds\n      \\le\\|v-w\\|.\n\\]\nTherefore $A$ is globally $1$–Lipschitz on $H$.\n\n\\bigskip\n3.  Approximation of the identity.  \n\nFix $M>0$ and denote $B_{M}:=\\{v\\in H:\\|v\\|\\le M\\}$.\n\n(i)  Pointwise convergence.  \nGiven $v\\in H$, choose $n_{0}$ such that $k_{n}\\ge\\|v\\|$ for all $n\\ge n_{0}$.  Then $A_{k_{n},r_{n}}(v)=v$ for $n\\ge n_{0}$, so $A_{k_{n},r_{n}}(v)\\to v$.\n\n(ii)  Uniform convergence on $B_{M}$.  \nBecause $k_{n}\\to\\infty$, there exists $N$ such that $k_{n}\\ge M$ for every $n\\ge N$.  \nFor these indices and every $v\\in B_{M}$ we again have $A_{k_{n},r_{n}}(v)=v$.  Hence\n\\[\n\\sup_{v\\in B_{M}}\\|A_{k_{n},r_{n}}(v)-v\\|=0\\qquad\\text{for }n\\ge N,\n\\]\nwhich proves the uniform convergence.  No estimate involving $r_{n}$ is required.\n\n\\bigskip\n\\emph{Remark.}  One can verify that $A_{k,r}$ is only Gâteaux differentiable at $v=0$, with $D A_{k,r}(0)=\\operatorname{Id}_{H}$; the derivative is not continuous there.  The global Lipschitz bound therefore remains the principal tool at the origin.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.476836",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimensional / functional setting:  The problem is lifted from real-valued functions on \\(\\mathbb R\\) to arbitrary maps into an infinite-dimensional Banach space, requiring familiarity with functional analysis (dual space, separating functionals, Hahn–Banach).  \n• Additional family of parameters:  Instead of a single index \\(n\\) we now have a two–parameter family \\(A_{k,r}\\) with a non-trivial radial profile; analysing its properties is technically more involved.  \n• Multiple equivalent conditions:  The solver must prove a cycle of implications (i)⇔(ii)⇔(iii)⇔(iv), not only the two-way implication of the original task.  \n• Use of advanced tools:  The proof employs separation of points by linear functionals, compactness arguments, and equicontinuity via the Arzelà–Ascoli philosophy.  \n• Uniform approximation statement:  Part 2 adds a non-obvious quantitative result (uniform limit), forcing the solver to control the radial decay parameter \\(r\\) simultaneously with the truncation radius \\(k\\).  \nAll these layers make the enhanced variant substantially harder and substantially more technical than both the original problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}