{
  "index": "1961-B-7",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "7. Given a sequence \\( \\left\\{a_{n}\\right\\} \\) of non-negative real numbers such that \\( a_{n+m} \\leq \\) \\( a_{n} a_{m} \\) for all pairs of positive integers, \\( m \\) and \\( n \\), prove that the sequence \\( \\left\\{\\sqrt[n]{a_{n}}\\right\\} \\) has a limit as \\( n \\rightarrow \\infty \\).",
  "solution": "Solution. If \\( a_{p}=0 \\) for some \\( p \\), then from \\( a_{p+m} \\leq 0 \\cdot a_{m} \\) we have \\( a_{p+1} \\) \\( =0, a_{p+2}=0, \\ldots \\), and indeed \\( a_{n}=0 \\) for all \\( n \\geq p \\). In this event \\( \\lim a_{n}{ }^{1 / n}=0 \\).\n\nIf no \\( a_{p}=0 \\), then \\( a_{n}>0 \\) for all \\( n \\), and we can put\n\\[\nb_{n}=\\log a_{n} .\n\\]\n\nThen the given inequality becomes\n\\[\nb_{m+n} \\leq b_{m}+b_{n}\n\\]\nand it follows that\n\\[\nb_{k m+t} \\leq k b_{m}+b_{t}\n\\]\nfor any non-negative integers \\( k, m, t \\).\nWe fix the positive integer \\( m \\), and for any \\( n \\), let \\( n=k(n) m+t(n) \\) where \\( 0 \\leq t(n)<m \\). Let \\( c=\\max \\left(b_{0}, b_{1}, \\ldots, b_{m-1}\\right) \\)\n\\[\n\\begin{aligned}\nb_{n}=b_{k(n) m+t(n)} & \\leq k(n) b_{m}+b_{t(n)} \\\\\n& \\leq k(n) b_{m}+c\n\\end{aligned}\n\\]\nand hence\n\\[\n\\frac{b_{n}}{n}<\\frac{k(n)}{n} b_{m}+\\frac{c}{n}\n\\]\n\nAs \\( n \\rightarrow \\infty, \\frac{k(n)}{n} \\rightarrow \\frac{1}{m} \\), hence\n\\[\n\\lim \\sup \\frac{1}{n} b_{n} \\leq \\frac{1}{m} b_{m}\n\\]\n\nIf \\( \\alpha \\) is a number exceeding \\( \\lim \\inf b_{n} / n \\), we can choose \\( m \\) so that \\( b_{m} / m \\) \\( <\\alpha \\). Then (1) shows that\n\\[\n\\lim \\sup \\frac{1}{n} b_{n} \\leq \\alpha\n\\]\n\nHence we conclude\n\\[\n\\lim \\sup \\frac{1}{n} b_{n} \\leq \\lim \\inf \\frac{1}{n} b_{n} .\n\\]\n\nThus, \\( \\lim b_{n} / n \\) exists, but possibly improperly. It is less than \\( b_{1} \\), for we may take \\( m=1 \\) in (1), but it may be \\( -\\infty \\). In any case\n\\[\n\\lim _{n \\rightarrow \\infty} a_{n}^{1 / n}=\\lim _{n \\rightarrow \\infty} \\exp \\left(\\frac{1}{n} b_{n}\\right)\n\\]\nexists.",
  "vars": [
    "a_n",
    "a_m",
    "a_p",
    "b_n",
    "b_m",
    "b_t",
    "b_0",
    "b_1",
    "n",
    "m",
    "p",
    "t",
    "k"
  ],
  "params": [
    "c",
    "\\\\alpha"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a_n": "termgeneral",
        "a_m": "termindexm",
        "a_p": "termindexp",
        "b_n": "loggeneral",
        "b_m": "logindexm",
        "b_t": "logindext",
        "b_0": "logzero",
        "b_1": "logone",
        "n": "countern",
        "m": "counterm",
        "p": "counterp",
        "t": "countert",
        "k": "counterk",
        "c": "maxbound",
        "\\alpha": "alphaval"
      },
      "question": "7. Given a sequence \\( \\left\\{termgeneral\\right\\} \\) of non-negative real numbers such that \\( a_{countern+counterm} \\leq termgeneral\\,termindexm \\) for all pairs of positive integers, \\( counterm \\) and \\( countern \\), prove that the sequence \\( \\left\\{\\sqrt[countern]{termgeneral}\\right\\} \\) has a limit as \\( countern \\rightarrow \\infty \\).",
      "solution": "Solution. If \\( termindexp =0 \\) for some \\( counterp \\), then from \\( a_{counterp+counterm} \\leq 0 \\cdot termindexm \\) we have \\( a_{counterp+1}=0, a_{counterp+2}=0, \\ldots \\), and indeed \\( termgeneral =0 \\) for all \\( countern \\geq counterp \\). In this event \\( \\lim termgeneral^{1 / countern}=0 \\).\n\nIf no \\( termindexp =0 \\), then \\( termgeneral >0 \\) for all \\( countern \\), and we can put\n\\[\nloggeneral = \\log termgeneral .\n\\]\nThen the given inequality becomes\n\\[\nb_{counterm+countern} \\leq logindexm + loggeneral\n\\]\nand it follows that\n\\[\nb_{counterk\\,counterm+countert} \\leq counterk\\,logindexm + logindext\n\\]\nfor any non-negative integers \\( counterk, counterm, countert \\).\n\nWe fix the positive integer \\( counterm \\), and for any \\( countern \\), let \\( countern = counterk(countern)\\,counterm + countert(countern) \\) where \\( 0 \\leq countert(countern) < counterm \\). Let\n\\[\nmaxbound = \\max \\left(logzero,\\,logone,\\,\\ldots,\\,b_{counterm-1}\\right).\n\\]\nThen\n\\[\n\\begin{aligned}\nloggeneral = b_{counterk(countern)\\,counterm + countert(countern)} &\\leq counterk(countern)\\,logindexm + b_{countert(countern)}\\\\[2pt]\n&\\leq counterk(countern)\\,logindexm + maxbound,\n\\end{aligned}\n\\]\nand hence\n\\[\n\\frac{loggeneral}{countern} < \\frac{counterk(countern)}{countern}\\,logindexm + \\frac{maxbound}{countern}.\n\\]\nAs \\( countern \\rightarrow \\infty, \\; \\dfrac{counterk(countern)}{countern} \\rightarrow \\dfrac{1}{counterm} \\); hence\n\\[\n\\limsup \\frac{1}{countern} \\, loggeneral \\leq \\frac{1}{counterm}\\,logindexm.\n\\]\nIf \\( alphaval \\) is a number exceeding \\( \\liminf \\, loggeneral / countern \\), we can choose \\( counterm \\) so that \\( logindexm / counterm < alphaval \\). Then the previous inequality shows that\n\\[\n\\limsup \\frac{1}{countern} \\, loggeneral \\leq alphaval.\n\\]\nHence we conclude\n\\[\n\\limsup \\frac{1}{countern} \\, loggeneral \\leq \\liminf \\frac{1}{countern} \\, loggeneral.\n\\]\nThus, \\( \\lim loggeneral / countern \\) exists, though possibly improperly. It is less than \\( logone \\) (take \\( counterm = 1 \\) in the earlier bound), but it may be \\( -\\infty \\). In any case\n\\[\n\\lim_{countern \\rightarrow \\infty} termgeneral^{1 / countern} = \\lim_{countern \\rightarrow \\infty} \\exp\\!\bigl(\\tfrac{1}{countern}\\,loggeneral\\bigr)\n\\]\nexists."
    },
    "descriptive_long_confusing": {
      "map": {
        "a_n": "cantaloupe",
        "a_m": "saxophone",
        "a_p": "butterfly",
        "b_n": "watermelon",
        "b_m": "telescope",
        "b_t": "paintbrush",
        "b_0": "spaceship",
        "b_1": "chocolate",
        "n": "pineapple",
        "m": "strawberry",
        "p": "kangaroo",
        "t": "alligator",
        "k": "porcupine",
        "c": "zeppelin",
        "\\\\alpha": "quartzite"
      },
      "question": "7. Given a sequence \\( \\left\\{cantaloupe_{pineapple}\\right\\} \\) of non-negative real numbers such that \\( a_{n+m} \\leq \\) \\( cantaloupe_{pineapple} saxophone_{strawberry} \\) for all pairs of positive integers, \\( strawberry \\) and \\( pineapple \\), prove that the sequence \\( \\left\\{\\sqrt[pineapple]{cantaloupe_{pineapple}}\\right\\} \\) has a limit as \\( pineapple \\rightarrow \\infty \\).",
      "solution": "Solution. If \\( butterfly_{kangaroo}=0 \\) for some \\( kangaroo \\), then from \\( a_{p+m} \\leq 0 \\cdot saxophone_{strawberry} \\) we have \\( a_{p+1}=0, a_{p+2}=0, \\ldots \\), and indeed \\( cantaloupe_{pineapple}=0 \\) for all \\( pineapple \\geq kangaroo \\). In this event \\( \\lim cantaloupe_{pineapple}^{1 / pineapple}=0 \\).\n\nIf no \\( butterfly_{kangaroo}=0 \\), then \\( cantaloupe_{pineapple}>0 \\) for all \\( pineapple \\), and we can put\n\\[\nwatermelon_{pineapple}=\\log cantaloupe_{pineapple} .\n\\]\n\nThen the given inequality becomes\n\\[\nb_{m+n} \\leq telescope_{strawberry}+watermelon_{pineapple}\n\\]\nand it follows that\n\\[\nb_{k m+t} \\leq porcupine telescope_{strawberry}+paintbrush_{alligator}\n\\]\nfor any non-negative integers \\( porcupine, strawberry, alligator \\).\nWe fix the positive integer \\( strawberry \\), and for any \\( pineapple \\), let \\( pineapple=porcupine(pineapple) strawberry+alligator(pineapple) \\) where \\( 0 \\leq alligator(pineapple)<strawberry \\). Let \\( zeppelin=\\max \\left(spaceship, chocolate, \\ldots, b_{m-1}\\right) \\)\n\\[\n\\begin{aligned}\nwatermelon_{pineapple}=b_{k(n) m+t(n)} & \\leq k(n) telescope_{strawberry}+paintbrush_{alligator(pineapple)} \\\\\n& \\leq k(n) telescope_{strawberry}+zeppelin\n\\end{aligned}\n\\]\nand hence\n\\[\n\\frac{watermelon_{pineapple}}{pineapple}<\\frac{k(n)}{n} telescope_{strawberry}+\\frac{zeppelin}{pineapple}\n\\]\n\nAs \\( pineapple \\rightarrow \\infty, \\frac{k(n)}{n} \\rightarrow \\frac{1}{strawberry} \\), hence\n\\[\n\\lim \\sup \\frac{1}{pineapple} watermelon_{pineapple} \\leq \\frac{1}{strawberry} telescope_{strawberry}\n\\]\n\nIf \\( quartzite \\) is a number exceeding \\( \\lim \\inf watermelon_{pineapple} / pineapple \\), we can choose \\( strawberry \\) so that \\( telescope_{strawberry} / strawberry <quartzite \\). Then (1) shows that\n\\[\n\\lim \\sup \\frac{1}{pineapple} watermelon_{pineapple} \\leq quartzite\n\\]\n\nHence we conclude\n\\[\n\\lim \\sup \\frac{1}{pineapple} watermelon_{pineapple} \\leq \\lim \\inf \\frac{1}{pineapple} watermelon_{pineapple} .\n\\]\n\nThus, \\( \\lim watermelon_{pineapple} / pineapple \\) exists, but possibly improperly. It is less than \\( chocolate \\), for we may take \\( strawberry=1 \\) in (1), but it may be \\( -\\infty \\). In any case\n\\[\n\\lim _{pineapple \\rightarrow \\infty} cantaloupe_{pineapple}^{1 / pineapple}=\\lim _{pineapple \\rightarrow \\infty} \\exp \\left(\\frac{1}{pineapple} watermelon_{pineapple}\\right)\n\\]\nexists."
    },
    "descriptive_long_misleading": {
      "map": {
        "a_n": "barrenvalue",
        "a_m": "infertilevalue",
        "a_p": "voidmetric",
        "b_n": "exponentialdatum",
        "b_m": "exponentialitem",
        "b_t": "exponentialindex",
        "b_0": "exponentialzero",
        "b_1": "exponentialone",
        "n": "outcomevalue",
        "m": "startvalue",
        "p": "continuation",
        "t": "eternity",
        "k": "zerocount",
        "c": "negativity",
        "\\\\alpha": "endingpoint"
      },
      "question": "Given a sequence \\( \\left\\{barrenvalue\\right\\} \\) of non-negative real numbers such that \\( a_{outcomevalue+startvalue} \\leq \\) \\( barrenvalue infertilevalue \\) for all pairs of positive integers, \\( startvalue \\) and \\( outcomevalue \\), prove that the sequence \\( \\left\\{\\sqrt[outcomevalue]{barrenvalue}\\right\\} \\) has a limit as \\( outcomevalue \\rightarrow \\infty \\).",
      "solution": "Solution. If \\( voidmetric =0 \\) for some \\( continuation \\), then from \\( a_{continuation+startvalue} \\leq 0 \\cdot infertilevalue \\) we have \\( a_{continuation+1}=0, a_{continuation+2}=0, \\ldots \\), and indeed \\( a_{outcomevalue}=0 \\) for all \\( outcomevalue \\geq continuation \\). In this event \\( \\lim barrenvalue^{1 / outcomevalue}=0 \\).\n\nIf no \\( voidmetric =0 \\), then \\( a_{outcomevalue}>0 \\) for all \\( outcomevalue \\), and we can put\n\\[\nexponentialdatum =\\log barrenvalue .\n\\]\n\nThen the given inequality becomes\n\\[\nb_{startvalue+outcomevalue} \\leq exponentialitem + exponentialdatum\n\\]\nand it follows that\n\\[\nb_{zerocount\\, startvalue+eternity} \\leq zerocount\\,exponentialitem + exponentialindex\n\\]\nfor any non-negative integers \\( zerocount, startvalue, eternity \\).\nWe fix the positive integer \\( startvalue \\), and for any \\( outcomevalue \\), let \\( outcomevalue=zerocount(outcomevalue)\\,startvalue+eternity(outcomevalue) \\) where \\( 0 \\leq eternity(outcomevalue)<startvalue \\). Let \\( negativity=\\max \\left(exponentialzero , exponentialone , \\ldots , b_{startvalue-1}\\right) \\)\n\\[\n\\begin{aligned}\n b_{outcomevalue}&=b_{zerocount(outcomevalue)\\,startvalue+eternity(outcomevalue)} \\\\\n & \\leq zerocount(outcomevalue)\\,exponentialitem + b_{eternity(outcomevalue)} \\\\\n & \\leq zerocount(outcomevalue)\\,exponentialitem + negativity\n\\end{aligned}\n\\]\nand hence\n\\[\n\\frac{b_{outcomevalue}}{outcomevalue}<\\frac{zerocount(outcomevalue)}{outcomevalue}\\,exponentialitem+\\frac{negativity}{outcomevalue}\n\\]\n\nAs \\( outcomevalue \\rightarrow \\infty, \\frac{zerocount(outcomevalue)}{outcomevalue} \\rightarrow \\frac{1}{startvalue} \\), hence\n\\[\n\\lim \\sup \\frac{1}{outcomevalue}\\,b_{outcomevalue} \\leq \\frac{1}{startvalue}\\,exponentialitem\n\\]\n\nIf \\( endingpoint \\) is a number exceeding \\( \\lim \\inf b_{outcomevalue}/outcomevalue \\), we can choose \\( startvalue \\) so that \\( exponentialitem/startvalue < endingpoint \\). Then the preceding inequality shows that\n\\[\n\\lim \\sup \\frac{1}{outcomevalue}\\,b_{outcomevalue} \\leq endingpoint\n\\]\n\nHence we conclude\n\\[\n\\lim \\sup \\frac{1}{outcomevalue}\\,b_{outcomevalue} \\leq \\lim \\inf \\frac{1}{outcomevalue}\\,b_{outcomevalue} .\n\\]\n\nThus, \\( \\lim b_{outcomevalue}/outcomevalue \\) exists, but possibly improperly. It is less than \\( exponentialitem \\), for we may take \\( startvalue=1 \\) in the above estimate, but it may be \\( -\\infty \\). In any case\n\\[\n\\lim _{outcomevalue \\rightarrow \\infty} barrenvalue^{1 / outcomevalue}=\\lim _{outcomevalue \\rightarrow \\infty} \\exp \\left(\\frac{1}{outcomevalue}\\,exponentialdatum\\right)\n\\]\nexists."
    },
    "garbled_string": {
      "map": {
        "a_n": "qzxwvtnp",
        "a_m": "hjgrksla",
        "a_p": "xbejqotn",
        "b_n": "farksqud",
        "b_m": "glimvzhy",
        "b_t": "wcnporad",
        "b_0": "yuqslfok",
        "b_1": "inpxghat",
        "n": "rmucodve",
        "m": "kpjazlwh",
        "p": "tsbneqor",
        "t": "odvhwaer",
        "k": "lzjskrwp",
        "c": "vdmraqhe",
        "\\\\alpha": "qvnbjser"
      },
      "question": "7. Given a sequence \\( \\left\\{qzxwvtnp\\right\\} \\) of non-negative real numbers such that \\( a_{rmucodve+kpjazlwh} \\leq qzxwvtnp hjgrksla \\) for all pairs of positive integers, \\( kpjazlwh \\) and \\( rmucodve \\), prove that the sequence \\( \\left\\{\\sqrt[rmucodve]{qzxwvtnp}\\right\\} \\) has a limit as \\( rmucodve \\rightarrow \\infty \\).",
      "solution": "Solution. If \\( xbejqotn=0 \\) for some \\( tsbneqor \\), then from \\( a_{tsbneqor+kpjazlwh} \\leq 0 \\cdot hjgrksla \\) we have \\( a_{tsbneqor+1}=0, a_{tsbneqor+2}=0, \\ldots \\), and indeed \\( qzxwvtnp=0 \\) for all \\( rmucodve \\geq tsbneqor \\). In this event \\( \\lim qzxwvtnp^{1 / rmucodve}=0 \\).\n\nIf no \\( xbejqotn=0 \\), then \\( qzxwvtnp>0 \\) for all \\( rmucodve \\), and we can put\n\\[\nfarksqud=\\log qzxwvtnp .\n\\]\n\nThen the given inequality becomes\n\\[\nb_{kpjazlwh+rmucodve} \\leq glimvzhy+farksqud\n\\]\nand it follows that\n\\[\nb_{lzjskrwp kpjazlwh+odvhwaer} \\leq lzjskrwp \\, glimvzhy+wcnporad\n\\]\nfor any non-negative integers \\( lzjskrwp, kpjazlwh, odvhwaer \\).\nWe fix the positive integer \\( kpjazlwh \\), and for any \\( rmucodve \\), let \\( rmucodve=lzjskrwp(rmucodve) kpjazlwh+odvhwaer(rmucodve) \\) where \\( 0 \\leq odvhwaer(rmucodve)<kpjazlwh \\). Let \\( vdmraqhe=\\max \\left(yuqslfok, inpxghat, \\ldots, b_{kpjazlwh-1}\\right) \\)\n\\[\n\\begin{aligned}\nfarksqud=b_{lzjskrwp(rmucodve) kpjazlwh+odvhwaer(rmucodve)} & \\leq lzjskrwp(rmucodve) \\, glimvzhy+wcnporad(rmucodve) \\\\& \\leq lzjskrwp(rmucodve) \\, glimvzhy+vdmraqhe\n\\end{aligned}\n\\]\nand hence\n\\[\n\\frac{farksqud}{rmucodve}<\\frac{lzjskrwp(rmucodve)}{rmucodve} \\, glimvzhy+\\frac{vdmraqhe}{rmucodve}\n\\]\n\nAs \\( rmucodve \\rightarrow \\infty, \\frac{lzjskrwp(rmucodve)}{rmucodve} \\rightarrow \\frac{1}{kpjazlwh} \\), hence\n\\[\n\\lim \\sup \\frac{1}{rmucodve} farksqud \\leq \\frac{1}{kpjazlwh} \\, glimvzhy\n\\]\n\nIf \\( qvnbjser \\) is a number exceeding \\( \\lim \\inf farksqud / rmucodve \\), we can choose \\( kpjazlwh \\) so that \\( glimvzhy / kpjazlwh <qvnbjser \\). Then (1) shows that\n\\[\n\\lim \\sup \\frac{1}{rmucodve} farksqud \\leq qvnbjser\n\\]\n\nHence we conclude\n\\[\n\\lim \\sup \\frac{1}{rmucodve} farksqud \\leq \\lim \\inf \\frac{1}{rmucodve} farksqud .\n\\]\n\nThus, \\( \\lim farksqud / rmucodve \\) exists, but possibly improperly. It is less than \\( inpxghat \\), for we may take \\( kpjazlwh=1 \\) in (1), but it may be \\( -\\infty \\). In any case\n\\[\n\\lim _{rmucodve \\rightarrow \\infty} qzxwvtnp^{1 / rmucodve}=\\lim _{rmucodve \\rightarrow \\infty} \\exp \\left(\\frac{1}{rmucodve} farksqud\\right)\n\\]\nexists."
    },
    "kernel_variant": {
      "question": "Let $d\\ge 2$ and denote by  \n\\[\n\\mathbb{N}_{0}^{d}:=\\bigl\\{(n_{1},\\dots ,n_{d})\\; ;\\; n_{i}\\in\\mathbb{N}_{0}\\bigr\\}.\n\\]  \nFix a family of non–negative real numbers  \n\\[\n\\bigl(x_{n}\\bigr)_{\\,n\\in\\mathbb{N}_{0}^{d}}\n\\qquad\\bigl(\\text{with the convention }x_{\\mathbf{0}}:=1\\bigr)\n\\]  \nsatisfying\n\n(1)  (sub–multiplicativity)  \n\\[\nx_{m+n}\\le x_{m}\\,x_{n}\\qquad\\forall\\,m,n\\in\\mathbb{N}_{0}^{d};\n\\]\n\n(2)  (non–degeneracy on the coordinate axes)  \nfor every unit vector $e_{i}$, $1\\le i\\le d$, the sequence $\\bigl(x_{k e_{i}}\\bigr)_{k\\ge 1}$ is **not** eventually $0$;\n\n(3)  (mild exponential upper bound)  \nthere exists a constant $C>0$ such that  \n\\[\nx_{n}\\le C^{\\,\\lVert n\\rVert_{1}}\\qquad\\forall\\,n\\in\\mathbb{N}_{0}^{d},\n\\qquad\\hbox{where }\\lVert n\\rVert_{1}:=n_{1}+\\dots +n_{d}.\n\\]\n\nFor a rational direction $\\theta=(\\theta_{1},\\dots ,\\theta_{d})$ with $\\theta_{i}\\ge 0$ and\n$\\theta_{1}+\\dots +\\theta_{d}=1$, let $s$ be any common denominator of the $\\theta_{i}$.  \nWhenever $k\\in s\\mathbb{N}$ we put $k\\theta:=(k\\theta_{1},\\dots ,k\\theta_{d})\\in\\mathbb{N}_{0}^{d}$.\n\n(a)  (directional limits)  \nShow that the limit  \n\\[\nL(\\theta)\\ :=\\ \\lim_{k\\to\\infty}x_{k\\theta}^{\\,1/k}\n\\]\nexists (it may be $0$) for every rational direction $\\theta$.  \nDefine  \n\\[\nL_{i}\\ :=\\ \\lim_{k\\to\\infty}x_{k e_{i}}^{\\,1/k}\\qquad(1\\le i\\le d).\n\\]\n\n(b)  (asymptotic upper envelope)  \nProve that  \n\\[\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}x_{n}^{\\,1/\\lVert n\\rVert_{1}}\n\\;=\\;\n\\max_{1\\le i\\le d}L_{i}.\n\\tag{$\\star$}\n\\]\n\n(c)  (a necessary condition for a global limit)  \nShow that if the global limit  \n\\[\n\\lim_{\\lVert n\\rVert_{1}\\to\\infty}x_{n}^{\\,1/\\lVert n\\rVert_{1}}\n\\]\nexists, then necessarily $L_{1}=L_{2}=\\dots =L_{d}$.\n\n(d)  (failure of sufficiency)  \nFor $d=2$ construct an explicit family $\\bigl(x_{n}\\bigr)$ satisfying (1)–(3) with\n$L_{1}=L_{2}$ but for which the global limit in (c) does **not** exist.\nExplain why this shows that the condition found in (c) is not sufficient.\n\nThroughout you may set $\\log 0:=-\\infty$ and work in the extended real line\n$\\mathbb{R}\\cup\\{-\\infty\\}$.\n\n───────────────────────────────────────────",
      "solution": "Throughout we write  \n\\[\nb_{n}:=\\log x_{n}\\in\\mathbb{R}\\cup\\{-\\infty\\},\n\\qquad\\bigl(\\log 0:=-\\infty,\\; \\log 1 =0\\bigr).\n\\]  \nWith the normalisation $x_{\\mathbf{0}}=1$ we have $b_{\\mathbf{0}}=0$.  \nAssumptions (1) and (3) become\n\n(S)\\; $b_{m+n}\\le b_{m}+b_{n}$  for all $m,n\\in\\mathbb{N}_{0}^{d}$ (sub–additivity),  \n\n(B)\\; $b_{n}\\le (\\log C)\\lVert n\\rVert_{1}$ for all $n$ (linear upper bound).\n\nStep 1.  Directional limits – part (a).  \n\nFix any non–zero $v\\in\\mathbb{N}_{0}^{d}$.  \nThe sequence $k\\longmapsto b_{k v}$ is sub–additive, so by Fekete’s lemma the limit  \n\\[\n\\beta(v):=\\lim_{k\\to\\infty}\\frac{b_{k v}}{k}\\in[-\\infty,\\infty)\n\\]  \nexists.  \nGiven a rational $\\theta$ choose the **primitive** integer vector $v$ with\n$\\theta=v/\\lVert v\\rVert_{1}$.  \nIf $k\\in s\\mathbb{N}$ then $k\\theta=(k/s)v$, whence\n\\[\n\\frac{b_{k\\theta}}{k}\\;=\\;\\frac{b_{(k/s)v}}{(k/s)}\\cdot\\frac{1}{\\lVert v\\rVert_{1}}\n\\;\\xrightarrow{k\\to\\infty}\\;\n\\frac{\\beta(v)}{\\lVert v\\rVert_{1}}.\n\\]\nExponentiating yields\n\\[\nL(\\theta)=\\exp\\!\\Bigl(\\beta(v)/\\lVert v\\rVert_{1}\\Bigr).\n\\tag{1}\n\\]\nIn particular, for $v=e_{i}$ we get $L_{i}=e^{\\,\\beta(e_{i})}$.  \nThus the limits in (a) exist.  \\hfill$\\square$ (a)\n\nStep 2.  Coordinate–wise decomposition.  \n\nClaim.  For every $n=(n_{1},\\dots ,n_{d})\\in\\mathbb{N}_{0}^{d}$ one has  \n\\[\nb_{n}\\le b_{n_{1}e_{1}}+\\dots +b_{n_{d}e_{d}}.\n\\tag{2}\n\\]\n\nProof by induction on $m:=\\lVert n\\rVert_{1}$.  \nFor $m=0$ both sides equal $0$.  \nAssume (2) is known for all vectors of 1–norm $<m$ and take\n$n$ with $\\lVert n\\rVert_{1}=m\\ge 1$.  \nChoose an index $j$ with $n_{j}\\ge 1$ and write\n\\[\nn=(n_{1},\\dots ,n_{j}-1,\\dots ,n_{d})+e_{j}.\n\\]\nBoth addends are in $\\mathbb{N}_{0}^{d}$, the first one having strictly smaller\n1–norm.  Sub–additivity (S) gives\n\\[\nb_{n}\\le b_{n-e_{j}}+b_{e_{j}}\n        \\le\\bigl[b_{(n_{1}e_{1})}+\\dots +b_{(n_{j}-1)e_{j}}+\\dots +b_{n_{d}e_{d}}\\bigr]+b_{e_{j}}\n        = b_{n_{1}e_{1}}+\\dots +b_{n_{d}e_{d}},\n\\]\nwhere the middle inequality uses the inductive hypothesis for $n-e_{j}$.  \nThus (2) holds for all $n$. \\hfill$\\square$\n\nStep 3.  Proof of the asymptotic upper envelope – part (b).  \n\nLet\n\\[\nM_{\\beta}:=\\max_{1\\le i\\le d}\\beta(e_{i})\\quad\\Bigl(=\\log\\!\\bigl[\\max_{i}L_{i}\\bigr]\\Bigr).\n\\]\n\n(A)  The **finite** case $\\;M_{\\beta}>-\\infty$.  \n\nFix $\\varepsilon>0$.  \nFor each $i$ the limit $\\beta(e_{i})$ exists, hence there are infinitely many\n$k$ with $x_{k e_{i}}>0$ and\n\\[\nb_{k e_{i}}\\le k\\bigl(M_{\\beta}+\\varepsilon/2\\bigr).\n\\tag{3}\n\\]\nPick one such integer and denote it by $K_{i}\\ge 1$.  \nDefine\n\\[\nK_{\\max}:=\\max_{i}K_{i},\n\\qquad\nM:=\\max_{1\\le i\\le d}\\;\\max_{0\\le r<K_{i},\\;x_{r e_{i}}>0} b_{r e_{i}},\n\\]\nwhich is finite by (B).  \n\nNow let $n\\in\\mathbb{N}_{0}^{d}$ with $\\lVert n\\rVert_{1}\\gg 1$.\nWrite $n_{i}=q_{i}K_{i}+r_{i}$ with $0\\le r_{i}<K_{i}$.  \nUsing (2) and (3),\n\\[\n\\begin{aligned}\nb_{n}\\ &\\le \\sum_{i=1}^{d} b_{n_{i}e_{i}}\n      \\le \\sum_{i=1}^{d}\\bigl[q_{i}\\,b_{K_{i}e_{i}}+b_{r_{i}e_{i}}\\bigr]  \\\\\n      &\\le \\sum_{i=1}^{d}q_{i}K_{i}\\bigl(M_{\\beta}+\\varepsilon/2\\bigr)+dM\n      \\le \\lVert n\\rVert_{1}\\bigl(M_{\\beta}+\\varepsilon/2\\bigr)+dM.\n\\end{aligned}\n\\]\nDividing by $\\lVert n\\rVert_{1}$ and letting $\\lVert n\\rVert_{1}\\to\\infty$ gives\n\\[\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}\\frac{b_{n}}{\\lVert n\\rVert_{1}}\n\\le M_{\\beta}+\\varepsilon.\n\\]\nBecause $\\varepsilon>0$ is arbitrary,\n\\[\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}\\frac{b_{n}}{\\lVert n\\rVert_{1}}\n\\le M_{\\beta}.\n\\tag{4}\n\\]\n\n(B)  The **degenerate** case $\\;M_{\\beta}=-\\infty$ (equivalently $L_{i}=0$ for all $i$).  \n\nChoose any positive number $M_{0}>0$.  \nFor every $i$ there exist infinitely many $k$ with $x_{k e_{i}}>0$ and\n$b_{k e_{i}}\\le -kM_{0}$.  Fix one such $k$ and call it $K_{i}$.  \nRepeating the preceding argument with these $K_{i}$’s yields, for all large $n$,\n\\[\n\\frac{b_{n}}{\\lVert n\\rVert_{1}}\\le -M_{0}+\\frac{dM}{\\lVert n\\rVert_{1}}\n\\;<\\;-M_{0}/2.\n\\]\nSince $M_{0}>0$ was arbitrary we deduce\n\\[\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}\\frac{b_{n}}{\\lVert n\\rVert_{1}}\n=-\\infty.\n\\tag{5}\n\\]\n\n(C)  Lower bound in the finite case.  \nTake $n=k e_{j}$ for any index $j$ with $\\beta(e_{j})=M_{\\beta}$.  \nThen $\\displaystyle \\lim_{k\\to\\infty}\\tfrac{b_{k e_{j}}}{k}=M_{\\beta}$, hence\n\\[\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}\\frac{b_{n}}{\\lVert n\\rVert_{1}}\n\\ge M_{\\beta}.\n\\tag{6}\n\\]\n\nCombining (4)–(6) and exponentiating we obtain in **all** cases  \n\\[\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}x_{n}^{\\,1/\\lVert n\\rVert_{1}}\n=\\max_{1\\le i\\le d}L_{i},\n\\]\nwhich is exactly $(\\star)$. \\hfill$\\square$ (b)\n\nStep 4.  Necessity of equal one–dimensional limits – part (c).  \n\nAssume the global limit \n$L:=\\displaystyle\\lim_{\\lVert n\\rVert_{1}\\to\\infty}x_{n}^{\\,1/\\lVert n\\rVert_{1}}$\nexists.  \nEvaluating it along the $i$–th coordinate axis ($n=k e_{i}$) gives\n\\[\n\\lim_{k\\to\\infty}x_{k e_{i}}^{\\,1/k}=L,\n\\quad\\text{i.e. }L_{i}=L\\qquad(1\\le i\\le d).\n\\]\nThus $L_{1}=L_{2}=\\dots =L_{d}$. \\hfill$\\square$ (c)\n\nStep 5.  A counter–example in dimension $2$ – part (d).  \n\nFor $n=(n_{1},n_{2})\\ne\\mathbf{0}$ set  \n\\[\nx_{n}:=\\begin{cases}\n\\mathrm{e}^{\\,n_{1}+n_{2}} & \\text{if }n_{1}=0\\ \\text{or}\\ n_{2}=0,\\\\[4pt]\n1                          & \\text{if }n_{1},n_{2}>0,\n\\end{cases}\n\\qquad\\text{and }x_{\\mathbf{0}}:=1.\n\\tag{7}\n\\]\n\nVerification of the assumptions.  \n\n•  (1)  A direct case distinction using (7) shows  \n$x_{m+n}\\le x_{m}\\,x_{n}$ for all $m,n\\in\\mathbb{N}_{0}^{2}$.  \n\n•  (2)  Along either axis $x_{(k,0)}=x_{(0,k)}=\\mathrm{e}^{\\,k}>0$, hence the axis\nsequences are not eventually $0$.  \n\n•  (3)  For every $n$ one has $x_{n}\\le\\mathrm{e}^{\\,\\lVert n\\rVert_{1}}$.  \n\nDirectional limits.  On the axes $x_{k e_{i}}^{\\,1/k}=\\mathrm{e}$, so\n$L_{1}=L_{2}=\\mathrm{e}$.  \n\nGlobal behaviour.  Along the diagonal $n=(k,k)$ we have $x_{n}=1$, hence\n$x_{n}^{\\,1/\\lVert n\\rVert_{1}}=1$, while along the axis $n=(k,0)$ the same\nquantity equals $\\mathrm{e}$.  Therefore\n\\[\n\\liminf_{\\lVert n\\rVert_{1}\\to\\infty}x_{n}^{\\,1/\\lVert n\\rVert_{1}}\n=1\n\\;<\\;\n\\mathrm{e}\n=\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}x_{n}^{\\,1/\\lVert n\\rVert_{1}}.\n\\]\nThe global limit fails to exist although $L_{1}=L_{2}$, so the condition found\nin (c) is **not** sufficient.  \\hfill$\\square$ (d)\n\nAll four parts (a)–(d) are now complete. \\qed\n\n───────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.533355",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the unknowns now form a d-parameter array on ℕ⁽ᵈ⁾ rather than a single sequence, forcing a multi-variable treatment.  \n• Additional constraints: we must control behaviour in every lattice direction simultaneously and prove independence from the chosen path to infinity.  \n• More sophisticated structures: the proof uses the semigroup ℕ⁽ᵈ⁾, multi-variable subadditivity, primitive lattice directions, and norms on ℝᵈ.  \n• Deeper theory: two separate invocations of Fekete’s lemma (one along rays, one on the minima m_k) are coupled with a delicate sandwich argument to match upper and lower asymptotic slopes.  Handling the quotient b_n/‖n‖₁ demands an “ε–approximation” by thick slabs, borrowing ideas from the multidimensional version of the Subadditive Ergodic Theorem (but staying purely deterministic).  \n• Multiple interacting concepts: lattice geometry, norms, directional limits, homogenisation and subadditivity all interplay.  None of these appear in the original one-dimensional setting, and simple pattern matching is hopeless—one must coordinate several advanced tools to finish the proof."
      }
    },
    "original_kernel_variant": {
      "question": "Let $d\\ge 2$ and denote by  \n\\[\n\\mathbb{N}_{0}^{d}:=\\bigl\\{(n_{1},\\dots ,n_{d})\\; ;\\; n_{i}\\in\\mathbb{N}_{0}\\bigr\\}.\n\\]  \nFix a family of non–negative real numbers  \n\\[\n\\bigl(x_{n}\\bigr)_{\\,n\\in\\mathbb{N}_{0}^{d}}\n\\qquad\\bigl(\\text{with the convention }x_{\\mathbf{0}}:=1\\bigr)\n\\]  \nsatisfying\n\n(1)  (sub–multiplicativity)  \n\\[\nx_{m+n}\\le x_{m}\\,x_{n}\\qquad\\forall\\,m,n\\in\\mathbb{N}_{0}^{d};\n\\]\n\n(2)  (non–degeneracy on the coordinate axes)  \nfor every unit vector $e_{i}$, $1\\le i\\le d$, the sequence $\\bigl(x_{k e_{i}}\\bigr)_{k\\ge 1}$ is **not** eventually $0$;\n\n(3)  (mild exponential upper bound)  \nthere exists a constant $C>0$ such that  \n\\[\nx_{n}\\le C^{\\,\\lVert n\\rVert_{1}}\\qquad\\forall\\,n\\in\\mathbb{N}_{0}^{d},\n\\qquad\\hbox{where }\\lVert n\\rVert_{1}:=n_{1}+\\dots +n_{d}.\n\\]\n\nFor a rational direction $\\theta=(\\theta_{1},\\dots ,\\theta_{d})$ with $\\theta_{i}\\ge 0$ and\n$\\theta_{1}+\\dots +\\theta_{d}=1$, let $s$ be any common denominator of the $\\theta_{i}$.  \nWhenever $k\\in s\\mathbb{N}$ we put $k\\theta:=(k\\theta_{1},\\dots ,k\\theta_{d})\\in\\mathbb{N}_{0}^{d}$.\n\n(a)  (directional limits)  \nShow that the limit  \n\\[\nL(\\theta)\\ :=\\ \\lim_{k\\to\\infty}x_{k\\theta}^{\\,1/k}\n\\]\nexists (it may be $0$) for every rational direction $\\theta$.  \nDefine  \n\\[\nL_{i}\\ :=\\ \\lim_{k\\to\\infty}x_{k e_{i}}^{\\,1/k}\\qquad(1\\le i\\le d).\n\\]\n\n(b)  (asymptotic upper envelope)  \nProve that  \n\\[\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}x_{n}^{\\,1/\\lVert n\\rVert_{1}}\n\\;=\\;\n\\max_{1\\le i\\le d}L_{i}.\n\\tag{$\\star$}\n\\]\n\n(c)  (a necessary condition for a global limit)  \nShow that if the global limit  \n\\[\n\\lim_{\\lVert n\\rVert_{1}\\to\\infty}x_{n}^{\\,1/\\lVert n\\rVert_{1}}\n\\]\nexists, then necessarily $L_{1}=L_{2}=\\dots =L_{d}$.\n\n(d)  (failure of sufficiency)  \nFor $d=2$ construct an explicit family $\\bigl(x_{n}\\bigr)$ satisfying (1)–(3) with\n$L_{1}=L_{2}$ but for which the global limit in (c) does **not** exist.\nExplain why this shows that the condition found in (c) is not sufficient.\n\nThroughout you may set $\\log 0:=-\\infty$ and work in the extended real line\n$\\mathbb{R}\\cup\\{-\\infty\\}$.\n\n───────────────────────────────────────────",
      "solution": "Throughout we write  \n\\[\nb_{n}:=\\log x_{n}\\in\\mathbb{R}\\cup\\{-\\infty\\},\n\\qquad\\bigl(\\log 0:=-\\infty,\\; \\log 1 =0\\bigr).\n\\]  \nWith the normalisation $x_{\\mathbf{0}}=1$ we have $b_{\\mathbf{0}}=0$.  \nAssumptions (1) and (3) become\n\n(S)\\; $b_{m+n}\\le b_{m}+b_{n}$  for all $m,n\\in\\mathbb{N}_{0}^{d}$ (sub–additivity),  \n\n(B)\\; $b_{n}\\le (\\log C)\\lVert n\\rVert_{1}$ for all $n$ (linear upper bound).\n\nStep 1.  Directional limits – part (a).  \n\nFix any non–zero $v\\in\\mathbb{N}_{0}^{d}$.  \nThe sequence $k\\longmapsto b_{k v}$ is sub–additive, so by Fekete’s lemma the limit  \n\\[\n\\beta(v):=\\lim_{k\\to\\infty}\\frac{b_{k v}}{k}\\in[-\\infty,\\infty)\n\\]  \nexists.  \nGiven a rational $\\theta$ choose the **primitive** integer vector $v$ with\n$\\theta=v/\\lVert v\\rVert_{1}$.  \nIf $k\\in s\\mathbb{N}$ then $k\\theta=(k/s)v$, whence\n\\[\n\\frac{b_{k\\theta}}{k}\\;=\\;\\frac{b_{(k/s)v}}{(k/s)}\\cdot\\frac{1}{\\lVert v\\rVert_{1}}\n\\;\\xrightarrow{k\\to\\infty}\\;\n\\frac{\\beta(v)}{\\lVert v\\rVert_{1}}.\n\\]\nExponentiating yields\n\\[\nL(\\theta)=\\exp\\!\\Bigl(\\beta(v)/\\lVert v\\rVert_{1}\\Bigr).\n\\tag{1}\n\\]\nIn particular, for $v=e_{i}$ we get $L_{i}=e^{\\,\\beta(e_{i})}$.  \nThus the limits in (a) exist.  \\hfill$\\square$ (a)\n\nStep 2.  Coordinate–wise decomposition.  \n\nClaim.  For every $n=(n_{1},\\dots ,n_{d})\\in\\mathbb{N}_{0}^{d}$ one has  \n\\[\nb_{n}\\le b_{n_{1}e_{1}}+\\dots +b_{n_{d}e_{d}}.\n\\tag{2}\n\\]\n\nProof by induction on $m:=\\lVert n\\rVert_{1}$.  \nFor $m=0$ both sides equal $0$.  \nAssume (2) is known for all vectors of 1–norm $<m$ and take\n$n$ with $\\lVert n\\rVert_{1}=m\\ge 1$.  \nChoose an index $j$ with $n_{j}\\ge 1$ and write\n\\[\nn=(n_{1},\\dots ,n_{j}-1,\\dots ,n_{d})+e_{j}.\n\\]\nBoth addends are in $\\mathbb{N}_{0}^{d}$, the first one having strictly smaller\n1–norm.  Sub–additivity (S) gives\n\\[\nb_{n}\\le b_{n-e_{j}}+b_{e_{j}}\n        \\le\\bigl[b_{(n_{1}e_{1})}+\\dots +b_{(n_{j}-1)e_{j}}+\\dots +b_{n_{d}e_{d}}\\bigr]+b_{e_{j}}\n        = b_{n_{1}e_{1}}+\\dots +b_{n_{d}e_{d}},\n\\]\nwhere the middle inequality uses the inductive hypothesis for $n-e_{j}$.  \nThus (2) holds for all $n$. \\hfill$\\square$\n\nStep 3.  Proof of the asymptotic upper envelope – part (b).  \n\nLet\n\\[\nM_{\\beta}:=\\max_{1\\le i\\le d}\\beta(e_{i})\\quad\\Bigl(=\\log\\!\\bigl[\\max_{i}L_{i}\\bigr]\\Bigr).\n\\]\n\n(A)  The **finite** case $\\;M_{\\beta}>-\\infty$.  \n\nFix $\\varepsilon>0$.  \nFor each $i$ the limit $\\beta(e_{i})$ exists, hence there are infinitely many\n$k$ with $x_{k e_{i}}>0$ and\n\\[\nb_{k e_{i}}\\le k\\bigl(M_{\\beta}+\\varepsilon/2\\bigr).\n\\tag{3}\n\\]\nPick one such integer and denote it by $K_{i}\\ge 1$.  \nDefine\n\\[\nK_{\\max}:=\\max_{i}K_{i},\n\\qquad\nM:=\\max_{1\\le i\\le d}\\;\\max_{0\\le r<K_{i},\\;x_{r e_{i}}>0} b_{r e_{i}},\n\\]\nwhich is finite by (B).  \n\nNow let $n\\in\\mathbb{N}_{0}^{d}$ with $\\lVert n\\rVert_{1}\\gg 1$.\nWrite $n_{i}=q_{i}K_{i}+r_{i}$ with $0\\le r_{i}<K_{i}$.  \nUsing (2) and (3),\n\\[\n\\begin{aligned}\nb_{n}\\ &\\le \\sum_{i=1}^{d} b_{n_{i}e_{i}}\n      \\le \\sum_{i=1}^{d}\\bigl[q_{i}\\,b_{K_{i}e_{i}}+b_{r_{i}e_{i}}\\bigr]  \\\\\n      &\\le \\sum_{i=1}^{d}q_{i}K_{i}\\bigl(M_{\\beta}+\\varepsilon/2\\bigr)+dM\n      \\le \\lVert n\\rVert_{1}\\bigl(M_{\\beta}+\\varepsilon/2\\bigr)+dM.\n\\end{aligned}\n\\]\nDividing by $\\lVert n\\rVert_{1}$ and letting $\\lVert n\\rVert_{1}\\to\\infty$ gives\n\\[\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}\\frac{b_{n}}{\\lVert n\\rVert_{1}}\n\\le M_{\\beta}+\\varepsilon.\n\\]\nBecause $\\varepsilon>0$ is arbitrary,\n\\[\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}\\frac{b_{n}}{\\lVert n\\rVert_{1}}\n\\le M_{\\beta}.\n\\tag{4}\n\\]\n\n(B)  The **degenerate** case $\\;M_{\\beta}=-\\infty$ (equivalently $L_{i}=0$ for all $i$).  \n\nChoose any positive number $M_{0}>0$.  \nFor every $i$ there exist infinitely many $k$ with $x_{k e_{i}}>0$ and\n$b_{k e_{i}}\\le -kM_{0}$.  Fix one such $k$ and call it $K_{i}$.  \nRepeating the preceding argument with these $K_{i}$’s yields, for all large $n$,\n\\[\n\\frac{b_{n}}{\\lVert n\\rVert_{1}}\\le -M_{0}+\\frac{dM}{\\lVert n\\rVert_{1}}\n\\;<\\;-M_{0}/2.\n\\]\nSince $M_{0}>0$ was arbitrary we deduce\n\\[\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}\\frac{b_{n}}{\\lVert n\\rVert_{1}}\n=-\\infty.\n\\tag{5}\n\\]\n\n(C)  Lower bound in the finite case.  \nTake $n=k e_{j}$ for any index $j$ with $\\beta(e_{j})=M_{\\beta}$.  \nThen $\\displaystyle \\lim_{k\\to\\infty}\\tfrac{b_{k e_{j}}}{k}=M_{\\beta}$, hence\n\\[\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}\\frac{b_{n}}{\\lVert n\\rVert_{1}}\n\\ge M_{\\beta}.\n\\tag{6}\n\\]\n\nCombining (4)–(6) and exponentiating we obtain in **all** cases  \n\\[\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}x_{n}^{\\,1/\\lVert n\\rVert_{1}}\n=\\max_{1\\le i\\le d}L_{i},\n\\]\nwhich is exactly $(\\star)$. \\hfill$\\square$ (b)\n\nStep 4.  Necessity of equal one–dimensional limits – part (c).  \n\nAssume the global limit \n$L:=\\displaystyle\\lim_{\\lVert n\\rVert_{1}\\to\\infty}x_{n}^{\\,1/\\lVert n\\rVert_{1}}$\nexists.  \nEvaluating it along the $i$–th coordinate axis ($n=k e_{i}$) gives\n\\[\n\\lim_{k\\to\\infty}x_{k e_{i}}^{\\,1/k}=L,\n\\quad\\text{i.e. }L_{i}=L\\qquad(1\\le i\\le d).\n\\]\nThus $L_{1}=L_{2}=\\dots =L_{d}$. \\hfill$\\square$ (c)\n\nStep 5.  A counter–example in dimension $2$ – part (d).  \n\nFor $n=(n_{1},n_{2})\\ne\\mathbf{0}$ set  \n\\[\nx_{n}:=\\begin{cases}\n\\mathrm{e}^{\\,n_{1}+n_{2}} & \\text{if }n_{1}=0\\ \\text{or}\\ n_{2}=0,\\\\[4pt]\n1                          & \\text{if }n_{1},n_{2}>0,\n\\end{cases}\n\\qquad\\text{and }x_{\\mathbf{0}}:=1.\n\\tag{7}\n\\]\n\nVerification of the assumptions.  \n\n•  (1)  A direct case distinction using (7) shows  \n$x_{m+n}\\le x_{m}\\,x_{n}$ for all $m,n\\in\\mathbb{N}_{0}^{2}$.  \n\n•  (2)  Along either axis $x_{(k,0)}=x_{(0,k)}=\\mathrm{e}^{\\,k}>0$, hence the axis\nsequences are not eventually $0$.  \n\n•  (3)  For every $n$ one has $x_{n}\\le\\mathrm{e}^{\\,\\lVert n\\rVert_{1}}$.  \n\nDirectional limits.  On the axes $x_{k e_{i}}^{\\,1/k}=\\mathrm{e}$, so\n$L_{1}=L_{2}=\\mathrm{e}$.  \n\nGlobal behaviour.  Along the diagonal $n=(k,k)$ we have $x_{n}=1$, hence\n$x_{n}^{\\,1/\\lVert n\\rVert_{1}}=1$, while along the axis $n=(k,0)$ the same\nquantity equals $\\mathrm{e}$.  Therefore\n\\[\n\\liminf_{\\lVert n\\rVert_{1}\\to\\infty}x_{n}^{\\,1/\\lVert n\\rVert_{1}}\n=1\n\\;<\\;\n\\mathrm{e}\n=\n\\limsup_{\\lVert n\\rVert_{1}\\to\\infty}x_{n}^{\\,1/\\lVert n\\rVert_{1}}.\n\\]\nThe global limit fails to exist although $L_{1}=L_{2}$, so the condition found\nin (c) is **not** sufficient.  \\hfill$\\square$ (d)\n\nAll four parts (a)–(d) are now complete. \\qed\n\n───────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.444442",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the unknowns now form a d-parameter array on ℕ⁽ᵈ⁾ rather than a single sequence, forcing a multi-variable treatment.  \n• Additional constraints: we must control behaviour in every lattice direction simultaneously and prove independence from the chosen path to infinity.  \n• More sophisticated structures: the proof uses the semigroup ℕ⁽ᵈ⁾, multi-variable subadditivity, primitive lattice directions, and norms on ℝᵈ.  \n• Deeper theory: two separate invocations of Fekete’s lemma (one along rays, one on the minima m_k) are coupled with a delicate sandwich argument to match upper and lower asymptotic slopes.  Handling the quotient b_n/‖n‖₁ demands an “ε–approximation” by thick slabs, borrowing ideas from the multidimensional version of the Subadditive Ergodic Theorem (but staying purely deterministic).  \n• Multiple interacting concepts: lattice geometry, norms, directional limits, homogenisation and subadditivity all interplay.  None of these appear in the original one-dimensional setting, and simple pattern matching is hopeless—one must coordinate several advanced tools to finish the proof."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}