{
  "index": "1940-A-7",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "7. If \\( u_{1}{ }^{2}+u_{2}{ }^{2}+\\cdots \\) and \\( v_{1}{ }^{2}+v_{2}{ }^{2}+\\cdots \\) are convergent series of real constants, prove that\n\\[\n\\left(u_{1}-v_{1}\\right)^{p}+\\left(u_{2}-v_{2}\\right)^{p}+\\cdots, p \\text { an integer } \\geq 2\n\\]\nis convergent.",
  "solution": "Solution. Let \\( A=u_{1}{ }^{2}+u_{2}{ }^{2}+\\cdots \\) and \\( B=v_{1}{ }^{2}+v_{2}{ }^{2}+\\cdots \\). Since\n\\[\n\\left(u_{i}+v_{i}\\right)^{2}+\\left(u_{i}-v_{i}\\right)^{2}=2 u_{i}^{2}+2 v_{i}^{2}\n\\]\nwe have, for any positive integer \\( n \\),\n\\[\n\\sum_{i=1}^{n}\\left(u_{i}-v_{i}\\right)^{2} \\leq 2 \\sum_{i=1}^{n} u_{i}^{2}+2 \\sum_{i=1}^{n} v_{i}^{2} \\leq 2 A+2 B\n\\]\n\nSince the terms are all non-negative, it follows that\n\\[\n\\sum_{i=1}^{\\infty}\\left(u_{i}-v_{i}\\right)^{2} \\text { is convergent. }\n\\]\n\nTherefore, the terms approach zero, so there exists an integer \\( \\boldsymbol{k} \\) such that\n\\[\n\\left(u_{i}-v_{i}\\right)^{2}<1 \\quad \\text { for all } i \\geq k\n\\]\n\nIf \\( p \\) is an integer and \\( p \\geq 2 \\), then \\( \\left|u_{i}-v_{i}\\right|^{p} \\leq\\left(u_{i}-v_{i}\\right)^{2} \\) for all \\( i \\geq k \\), so the series\n\\[\n\\sum_{i=1}^{\\infty}\\left(u_{i}-v_{i}\\right)^{p}\n\\]\nis absolutely convergent, and therefore convergent.",
  "vars": [
    "u_1",
    "u_2",
    "u_i",
    "v_1",
    "v_2",
    "v_i",
    "i",
    "n"
  ],
  "params": [
    "A",
    "B",
    "p",
    "k"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "u_1": "firstuvalue",
        "u_2": "seconduvalue",
        "u_i": "generaluvalue",
        "v_1": "firstvvalue",
        "v_2": "secondvvalue",
        "v_i": "generalvvalue",
        "i": "indexvar",
        "n": "termcount",
        "A": "sumuconst",
        "B": "sumvconst",
        "p": "powerparam",
        "k": "cutoffidx"
      },
      "question": "7. If \\( firstuvalue^{2}+seconduvalue^{2}+\\cdots \\) and \\( firstvvalue^{2}+secondvvalue^{2}+\\cdots \\) are convergent series of real constants, prove that\n\\[\n\\left(firstuvalue-firstvvalue\\right)^{powerparam}+\\left(seconduvalue-secondvvalue\\right)^{powerparam}+\\cdots,\\; powerparam \\text { an integer } \\geq 2\n\\]\nis convergent.",
      "solution": "Solution. Let \\( sumuconst=firstuvalue^{2}+seconduvalue^{2}+\\cdots \\) and \\( sumvconst=firstvvalue^{2}+secondvvalue^{2}+\\cdots \\). Since\n\\[\n\\left(generaluvalue+generalvvalue\\right)^{2}+\\left(generaluvalue-generalvvalue\\right)^{2}=2\\,generaluvalue^{2}+2\\,generalvvalue^{2}\n\\]\nwe have, for any positive integer \\( termcount \\),\n\\[\n\\sum_{indexvar=1}^{termcount}\\left(generaluvalue-generalvvalue\\right)^{2}\\le 2\\sum_{indexvar=1}^{termcount}generaluvalue^{2}+2\\sum_{indexvar=1}^{termcount}generalvvalue^{2}\\le 2\\,sumuconst+2\\,sumvconst\n\\]\n\nSince the terms are all non-negative, it follows that\n\\[\n\\sum_{indexvar=1}^{\\infty}\\left(generaluvalue-generalvvalue\\right)^{2}\\text{ is convergent.}\n\\]\n\nTherefore, the terms approach zero, so there exists an integer \\( \\boldsymbol{cutoffidx} \\) such that\n\\[\n\\left(generaluvalue-generalvvalue\\right)^{2}<1\\quad \\text{for all } indexvar \\ge cutoffidx\n\\]\n\nIf \\( powerparam \\) is an integer and \\( powerparam \\ge 2 \\), then \\( \\lvert generaluvalue-generalvvalue \\rvert^{powerparam}\\le\\left(generaluvalue-generalvvalue\\right)^{2} \\) for all \\( indexvar \\ge cutoffidx \\), so the series\n\\[\n\\sum_{indexvar=1}^{\\infty}\\left(generaluvalue-generalvvalue\\right)^{powerparam}\n\\]\nis absolutely convergent, and therefore convergent."
    },
    "descriptive_long_confusing": {
      "map": {
        "u_1": "sunflower",
        "u_2": "paintbrush",
        "u_i": "chandelier",
        "v_1": "tangerine",
        "v_2": "jellyfish",
        "v_i": "sailboat",
        "i": "waterfall",
        "n": "earthquake",
        "A": "labyrinth",
        "B": "butterfly",
        "p": "dragonfly",
        "k": "lighthouse"
      },
      "question": "7. If \\( sunflower^{2}+paintbrush^{2}+\\cdots \\) and \\( tangerine^{2}+jellyfish^{2}+\\cdots \\) are convergent series of real constants, prove that\n\\[\n\\left(sunflower-tangerine\\right)^{dragonfly}+\\left(paintbrush-jellyfish\\right)^{dragonfly}+\\cdots, dragonfly \\text { an integer } \\geq 2\n\\]\nis convergent.",
      "solution": "Solution. Let \\( labyrinth=sunflower^{2}+paintbrush^{2}+\\cdots \\) and \\( butterfly=tangerine^{2}+jellyfish^{2}+\\cdots \\). Since\n\\[\n\\left(chandelier+sailboat\\right)^{2}+\\left(chandelier-sailboat\\right)^{2}=2 chandelier^{2}+2 sailboat^{2}\n\\]\nwe have, for any positive integer \\( earthquake \\),\n\\[\n\\sum_{waterfall=1}^{earthquake}\\left(chandelier-sailboat\\right)^{2} \\leq 2 \\sum_{waterfall=1}^{earthquake} chandelier^{2}+2 \\sum_{waterfall=1}^{earthquake} sailboat^{2} \\leq 2 labyrinth+2 butterfly\n\\]\n\nSince the terms are all non-negative, it follows that\n\\[\n\\sum_{waterfall=1}^{\\infty}\\left(chandelier-sailboat\\right)^{2} \\text { is convergent. }\n\\]\n\nTherefore, the terms approach zero, so there exists an integer \\( \\boldsymbol{lighthouse} \\) such that\n\\[\n\\left(chandelier-sailboat\\right)^{2}<1 \\quad \\text { for all } waterfall \\geq lighthouse\n\\]\n\nIf \\( dragonfly \\) is an integer and \\( dragonfly \\geq 2 \\), then \\( \\left|chandelier-sailboat\\right|^{dragonfly} \\leq\\left(chandelier-sailboat\\right)^{2} \\) for all \\( waterfall \\geq lighthouse \\), so the series\n\\[\n\\sum_{waterfall=1}^{\\infty}\\left(chandelier-sailboat\\right)^{dragonfly}\n\\]\nis absolutely convergent, and therefore convergent."
    },
    "descriptive_long_misleading": {
      "map": {
        "u_1": "downfirst",
        "u_2": "downsecond",
        "u_i": "downindex",
        "v_1": "scalarfirst",
        "v_2": "scalarsecond",
        "v_i": "scalarindex",
        "i": "contentvalue",
        "n": "unnatural",
        "A": "emptiness",
        "B": "fullness",
        "p": "weakness",
        "k": "finality"
      },
      "question": "7. If \\( downfirst^{2}+downsecond^{2}+\\cdots \\) and \\( scalarfirst^{2}+scalarsecond^{2}+\\cdots \\) are convergent series of real constants, prove that\n\\[\n\\left(downfirst-scalarfirst\\right)^{weakness}+\\left(downsecond-scalarsecond\\right)^{weakness}+\\cdots, weakness \\text { an integer } \\geq 2\n\\]\nis convergent.",
      "solution": "Solution. Let \\( emptiness=downfirst^{2}+downsecond^{2}+\\cdots \\) and \\( fullness=scalarfirst^{2}+scalarsecond^{2}+\\cdots \\). Since\n\\[\n\\left(downindex+scalarindex\\right)^{2}+\\left(downindex-scalarindex\\right)^{2}=2 downindex^{2}+2 scalarindex^{2}\n\\]\nwe have, for any positive integer \\( unnatural \\),\n\\[\n\\sum_{contentvalue=1}^{unnatural}\\left(downindex-scalarindex\\right)^{2} \\leq 2 \\sum_{contentvalue=1}^{unnatural} downindex^{2}+2 \\sum_{contentvalue=1}^{unnatural} scalarindex^{2} \\leq 2 emptiness+2 fullness\n\\]\n\nSince the terms are all non-negative, it follows that\n\\[\n\\sum_{contentvalue=1}^{\\infty}\\left(downindex-scalarindex\\right)^{2} \\text { is convergent. }\n\\]\n\nTherefore, the terms approach zero, so there exists an integer \\( \\boldsymbol{finality} \\) such that\n\\[\n\\left(downindex-scalarindex\\right)^{2}<1 \\quad \\text { for all } contentvalue \\geq finality\n\\]\n\nIf \\( weakness \\) is an integer and \\( weakness \\geq 2 \\), then \\( \\left|downindex-scalarindex\\right|^{weakness} \\leq\\left(downindex-scalarindex\\right)^{2} \\) for all \\( contentvalue \\geq finality \\), so the series\n\\[\n\\sum_{contentvalue=1}^{\\infty}\\left(downindex-scalarindex\\right)^{weakness}\n\\]\nis absolutely convergent, and therefore convergent."
    },
    "garbled_string": {
      "map": {
        "u_1": "qzxwvtnp",
        "u_2": "hjgrksla",
        "u_i": "mnbvcxqe",
        "v_1": "lkjhgfds",
        "v_2": "poiuytre",
        "v_i": "zxcvbnma",
        "n": "asdfghjk",
        "A": "qwertyui",
        "B": "qazwsxed",
        "p": "edcrfvtg",
        "k": "plmoknij"
      },
      "question": "7. If \\( qzxwvtnp^{2}+hjgrksla^{2}+\\cdots \\) and \\( lkjhgfds^{2}+poiuytre^{2}+\\cdots \\) are convergent series of real constants, prove that\n\\[\n\\left(qzxwvtnp-lkjhgfds\\right)^{edcrfvtg}+\\left(hjgrksla-poiuytre\\right)^{edcrfvtg}+\\cdots, edcrfvtg \\text { an integer } \\geq 2\n\\]\nis convergent.",
      "solution": "Solution. Let \\( qwertyui=qzxwvtnp^{2}+hjgrksla^{2}+\\cdots \\) and \\( qazwsxed=lkjhgfds^{2}+poiuytre^{2}+\\cdots \\). Since\n\\[\n\\left(mnbvcxqe+zxcvbnma\\right)^{2}+\\left(mnbvcxqe-zxcvbnma\\right)^{2}=2 mnbvcxqe^{2}+2 zxcvbnma^{2}\n\\]\nwe have, for any positive integer \\( asdfghjk \\),\n\\[\n\\sum_{i=1}^{asdfghjk}\\left(mnbvcxqe-zxcvbnma\\right)^{2} \\leq 2 \\sum_{i=1}^{asdfghjk} mnbvcxqe^{2}+2 \\sum_{i=1}^{asdfghjk} zxcvbnma^{2} \\leq 2 qwertyui+2 qazwsxed\n\\]\n\nSince the terms are all non-negative, it follows that\n\\[\n\\sum_{i=1}^{\\infty}\\left(mnbvcxqe-zxcvbnma\\right)^{2} \\text { is convergent. }\n\\]\n\nTherefore, the terms approach zero, so there exists an integer \\( \\boldsymbol{plmoknij} \\) such that\n\\[\n\\left(mnbvcxqe-zxcvbnma\\right)^{2}<1 \\quad \\text { for all } i \\geq plmoknij\n\\]\n\nIf \\( edcrfvtg \\) is an integer and \\( edcrfvtg \\geq 2 \\), then \\( \\left|mnbvcxqe-zxcvbnma\\right|^{edcrfvtg} \\leq\\left(mnbvcxqe-zxcvbnma\\right)^{2} \\) for all \\( i \\geq plmoknij \\), so the series\n\\[\n\\sum_{i=1}^{\\infty}\\left(mnbvcxqe-zxcvbnma\\right)^{edcrfvtg}\n\\]\nis absolutely convergent, and therefore convergent."
    },
    "kernel_variant": {
      "question": "Let 2 ≤ r < s < ∞ be fixed real numbers.  \nFor every pair (m,n)∈ℕ×ℕ let u_{m,n}, v_{m,n}∈ℂ satisfy  \n\n  ∑_{m=1}^{∞}∑_{n=1}^{∞}|u_{m,n}|^{2}<∞ and ∑_{m=1}^{∞}∑_{n=1}^{∞}|v_{m,n}|^{2}<∞.  \n\nPut  \n\n  S_{r,s}(u,v):=∑_{m=1}^{∞}\\Bigl(∑_{n=1}^{∞}|u_{m,n}-v_{m,n}|^{r}\\Bigr)^{s/r}.  \n\n(a)  Prove that S_{r,s}(u,v) is finite.  \n\n(b)  Show the quantitative estimate  \n\n  S_{r,s}(u,v) ≤ 2^{\\,s/2}\\Bigl( ∑_{m=1}^{∞}∑_{n=1}^{∞}\\bigl(|u_{m,n}|^{2}+|v_{m,n}|^{2}\\bigr)\\Bigr)^{s/2}.     (★)\n\n(c)  Prove that the constant 2^{s/2} in (★) is optimal: for every ε>0 there exist double–arrays (u_{m,n}), (v_{m,n}) that satisfy the square–summability hypotheses and for which  \n\n  S_{r,s}(u,v) > (2^{s/2}-ε)\\Bigl( ∑_{m,n}\\bigl(|u_{m,n}|^{2}+|v_{m,n}|^{2}\\bigr)\\Bigr)^{s/2}.  \n\nIn particular, no smaller universal constant (depending only on r and s) can replace 2^{s/2} in (★).",
      "solution": "Throughout set d_{m,n}:=u_{m,n}-v_{m,n}.  \n\nStep 1.  A uniform ℓ²–bound for the differences  \nFor every complex numbers u,v we have  \n\n  |u|^{2}+|v|^{2}=½\\,(|u+v|^{2}+|u-v|^{2}) ≥ ½\\,|u-v|^{2},  \n\nhence  \n\n  |d_{m,n}|^{2} ≤ 2\\bigl(|u_{m,n}|^{2}+|v_{m,n}|^{2}\\bigr).                                              (1)\n\nBecause the right–hand side is absolutely summable,  \n\n  D:=∑_{m,n}|d_{m,n}|^{2} <∞.                                                                            (2)\n\nStep 2.  Row sums  \nDefine S_{m}:=∑_{n=1}^{∞}|d_{m,n}|^{2}.  \nThen ∑_{m=1}^{∞}S_{m}=D.                                                                                  (3)\n\nStep 3.  Comparing row-wise ℓ^{r} and ℓ^{2} norms  \nBecause r ≥ 2, the ℓ^{p} norms decrease with p, thus  \n\n  (∑_{n}|d_{m,n}|^{r})^{1/r} ≤ (∑_{n}|d_{m,n}|^{2})^{1/2}=S_{m}^{1/2}.                                (4)\n\nRaise (4) to the power s to obtain  \n\n  (∑_{n}|d_{m,n}|^{r})^{s/r} ≤ S_{m}^{s/2}.                                                             (5)\n\nStep 4.  Splitting the index set  \nPartition ℕ into  \n\n  A := {m : S_{m} ≥ 1},  B := {m : 0 ≤ S_{m}<1}.  \n\nBecause each m∈A contributes at least 1 to the sum in (3), A is finite. Therefore  \n\n  ∑_{m∈A}S_{m}^{s/2}<∞                                                                                    (6)\n\n(as s/2>1).  \n\nFor m∈B we have S_{m}<1, hence S_{m}^{s/2} ≤ S_{m}.  Using (3) we get  \n\n  ∑_{m∈B}S_{m}^{s/2} ≤ ∑_{m∈B}S_{m} ≤ D<∞.                                                              (7)\n\nStep 5.  Convergence of S_{r,s}(u,v)  \nCombine (5), (6) and (7):  \n\n  S_{r,s}(u,v)=∑_{m}(∑_{n}|d_{m,n}|^{r})^{s/r} ≤ ∑_{m}S_{m}^{s/2}<∞.                                  (8)\n\nThis settles part (a).\n\nStep 6.  Quantitative estimate  \nWrite T_{m}:=∑_{n}\\bigl(|u_{m,n}|^{2}+|v_{m,n}|^{2}\\bigr).  \nBy (1) we have S_{m} ≤ 2T_{m}.  Hence from (5)  \n\n  (∑_{n}|d_{m,n}|^{r})^{s/r} ≤ S_{m}^{s/2} ≤ (2T_{m})^{s/2}=2^{s/2}T_{m}^{s/2}.                        (9)\n\nSumming over m and using the convexity inequality  \n  ∑_{m}T_{m}^{s/2} ≤ (∑_{m}T_{m})^{s/2} (s/2 > 1),  \nwe obtain  \n\n  S_{r,s}(u,v) ≤ 2^{s/2}∑_{m}T_{m}^{s/2} ≤ 2^{s/2}\\Bigl(∑_{m}T_{m}\\Bigr)^{s/2}.                       (10)\n\nBut ∑_{m}T_{m}=∑_{m,n}\\bigl(|u_{m,n}|^{2}+|v_{m,n}|^{2}\\bigr), so (10) is exactly (★).  \nThis completes part (b).\n\nStep 7.  Optimality of the constant  \nFix ε>0 and choose N large enough so that (1+ε)^{1/s}<1+ε/2.  \nDefine arrays  \n\n  u_{1,1}=1, v_{1,1}=-1, and u_{m,n}=v_{m,n}=0 for all (m,n)≠(1,1).  \n\nThen ∑_{m,n}(|u_{m,n}|^{2}+|v_{m,n}|^{2})=2, while  \n\n  S_{r,s}(u,v)=|u_{1,1}-v_{1,1}|^{s}=2^{s}.  \n\nConsequently  \n\n  S_{r,s}(u,v)=2^{s}=2^{s/2}(2)^{s/2}  \n       > (2^{s/2}-ε)(2)^{s/2} = (2^{s/2}-ε)\\Bigl(∑_{m,n}(|u_{m,n}|^{2}+|v_{m,n}|^{2})\\Bigr)^{s/2}.  \n\nSince ε>0 is arbitrary, the factor 2^{s/2} cannot be replaced by a smaller universal constant.  \nThis proves part (c) and finishes the solution.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.372181",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-Dimensional Structure  \n   • The original task deals with a single sequence; the enhanced variant involves a double-indexed array and a mixed (ℓ^{r},ℓ^{s}) norm, introducing an extra layer of summation and an interaction between two different exponents.\n\n2. Additional Interacting Exponents  \n   • Two distinct exponents (r and s, with r<s) must be handled simultaneously.  The solver must understand how ℓ^{p} norms behave when p varies and how to move from an inner r-norm to an outer s-norm.\n\n3. Sophisticated Techniques  \n   • The proof requires monotonicity of ℓ^{p} norms, a careful row–wise decomposition, and a split into “large’’ and “small’’ rows to control powers exceeding 1.  \n   • A quantitative estimate is demanded, forcing the solver to track constants meticulously instead of merely establishing convergence.\n\n4. Deeper Insight Compared with the Kernel Variant  \n   • In the kernel variant a single Hölder–type observation suffices.  \n   • Here one must recognise that square-summability gives only global control, then devise a two-step strategy (row bounds + size partition) to upgrade it to mixed-norm summability.  \n\n5. Non-trivial Generalisation  \n   • Setting r=2 and s=p recovers the kernel statement after collapsing the second index, but the new problem is strictly stronger: it covers every pair 2≤r<s and any array of data, not just sequences, making it significantly more complex than both the original and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let 2 ≤ r < s < ∞ be fixed real numbers.  \nFor every pair (m,n)∈ℕ×ℕ let u_{m,n}, v_{m,n}∈ℂ satisfy  \n\n  ∑_{m=1}^{∞}∑_{n=1}^{∞}|u_{m,n}|^{2}<∞ and ∑_{m=1}^{∞}∑_{n=1}^{∞}|v_{m,n}|^{2}<∞.  \n\nPut  \n\n  S_{r,s}(u,v):=∑_{m=1}^{∞}\\Bigl(∑_{n=1}^{∞}|u_{m,n}-v_{m,n}|^{r}\\Bigr)^{s/r}.  \n\n(a)  Prove that S_{r,s}(u,v) is finite.  \n\n(b)  Show the quantitative estimate  \n\n  S_{r,s}(u,v) ≤ 2^{\\,s/2}\\Bigl( ∑_{m=1}^{∞}∑_{n=1}^{∞}\\bigl(|u_{m,n}|^{2}+|v_{m,n}|^{2}\\bigr)\\Bigr)^{s/2}.     (★)\n\n(c)  Prove that the constant 2^{s/2} in (★) is optimal: for every ε>0 there exist double–arrays (u_{m,n}), (v_{m,n}) that satisfy the square–summability hypotheses and for which  \n\n  S_{r,s}(u,v) > (2^{s/2}-ε)\\Bigl( ∑_{m,n}\\bigl(|u_{m,n}|^{2}+|v_{m,n}|^{2}\\bigr)\\Bigr)^{s/2}.  \n\nIn particular, no smaller universal constant (depending only on r and s) can replace 2^{s/2} in (★).",
      "solution": "Throughout set d_{m,n}:=u_{m,n}-v_{m,n}.  \n\nStep 1.  A uniform ℓ²–bound for the differences  \nFor every complex numbers u,v we have  \n\n  |u|^{2}+|v|^{2}=½\\,(|u+v|^{2}+|u-v|^{2}) ≥ ½\\,|u-v|^{2},  \n\nhence  \n\n  |d_{m,n}|^{2} ≤ 2\\bigl(|u_{m,n}|^{2}+|v_{m,n}|^{2}\\bigr).                                              (1)\n\nBecause the right–hand side is absolutely summable,  \n\n  D:=∑_{m,n}|d_{m,n}|^{2} <∞.                                                                            (2)\n\nStep 2.  Row sums  \nDefine S_{m}:=∑_{n=1}^{∞}|d_{m,n}|^{2}.  \nThen ∑_{m=1}^{∞}S_{m}=D.                                                                                  (3)\n\nStep 3.  Comparing row-wise ℓ^{r} and ℓ^{2} norms  \nBecause r ≥ 2, the ℓ^{p} norms decrease with p, thus  \n\n  (∑_{n}|d_{m,n}|^{r})^{1/r} ≤ (∑_{n}|d_{m,n}|^{2})^{1/2}=S_{m}^{1/2}.                                (4)\n\nRaise (4) to the power s to obtain  \n\n  (∑_{n}|d_{m,n}|^{r})^{s/r} ≤ S_{m}^{s/2}.                                                             (5)\n\nStep 4.  Splitting the index set  \nPartition ℕ into  \n\n  A := {m : S_{m} ≥ 1},  B := {m : 0 ≤ S_{m}<1}.  \n\nBecause each m∈A contributes at least 1 to the sum in (3), A is finite. Therefore  \n\n  ∑_{m∈A}S_{m}^{s/2}<∞                                                                                    (6)\n\n(as s/2>1).  \n\nFor m∈B we have S_{m}<1, hence S_{m}^{s/2} ≤ S_{m}.  Using (3) we get  \n\n  ∑_{m∈B}S_{m}^{s/2} ≤ ∑_{m∈B}S_{m} ≤ D<∞.                                                              (7)\n\nStep 5.  Convergence of S_{r,s}(u,v)  \nCombine (5), (6) and (7):  \n\n  S_{r,s}(u,v)=∑_{m}(∑_{n}|d_{m,n}|^{r})^{s/r} ≤ ∑_{m}S_{m}^{s/2}<∞.                                  (8)\n\nThis settles part (a).\n\nStep 6.  Quantitative estimate  \nWrite T_{m}:=∑_{n}\\bigl(|u_{m,n}|^{2}+|v_{m,n}|^{2}\\bigr).  \nBy (1) we have S_{m} ≤ 2T_{m}.  Hence from (5)  \n\n  (∑_{n}|d_{m,n}|^{r})^{s/r} ≤ S_{m}^{s/2} ≤ (2T_{m})^{s/2}=2^{s/2}T_{m}^{s/2}.                        (9)\n\nSumming over m and using the convexity inequality  \n  ∑_{m}T_{m}^{s/2} ≤ (∑_{m}T_{m})^{s/2} (s/2 > 1),  \nwe obtain  \n\n  S_{r,s}(u,v) ≤ 2^{s/2}∑_{m}T_{m}^{s/2} ≤ 2^{s/2}\\Bigl(∑_{m}T_{m}\\Bigr)^{s/2}.                       (10)\n\nBut ∑_{m}T_{m}=∑_{m,n}\\bigl(|u_{m,n}|^{2}+|v_{m,n}|^{2}\\bigr), so (10) is exactly (★).  \nThis completes part (b).\n\nStep 7.  Optimality of the constant  \nFix ε>0 and choose N large enough so that (1+ε)^{1/s}<1+ε/2.  \nDefine arrays  \n\n  u_{1,1}=1, v_{1,1}=-1, and u_{m,n}=v_{m,n}=0 for all (m,n)≠(1,1).  \n\nThen ∑_{m,n}(|u_{m,n}|^{2}+|v_{m,n}|^{2})=2, while  \n\n  S_{r,s}(u,v)=|u_{1,1}-v_{1,1}|^{s}=2^{s}.  \n\nConsequently  \n\n  S_{r,s}(u,v)=2^{s}=2^{s/2}(2)^{s/2}  \n       > (2^{s/2}-ε)(2)^{s/2} = (2^{s/2}-ε)\\Bigl(∑_{m,n}(|u_{m,n}|^{2}+|v_{m,n}|^{2})\\Bigr)^{s/2}.  \n\nSince ε>0 is arbitrary, the factor 2^{s/2} cannot be replaced by a smaller universal constant.  \nThis proves part (c) and finishes the solution.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.321460",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-Dimensional Structure  \n   • The original task deals with a single sequence; the enhanced variant involves a double-indexed array and a mixed (ℓ^{r},ℓ^{s}) norm, introducing an extra layer of summation and an interaction between two different exponents.\n\n2. Additional Interacting Exponents  \n   • Two distinct exponents (r and s, with r<s) must be handled simultaneously.  The solver must understand how ℓ^{p} norms behave when p varies and how to move from an inner r-norm to an outer s-norm.\n\n3. Sophisticated Techniques  \n   • The proof requires monotonicity of ℓ^{p} norms, a careful row–wise decomposition, and a split into “large’’ and “small’’ rows to control powers exceeding 1.  \n   • A quantitative estimate is demanded, forcing the solver to track constants meticulously instead of merely establishing convergence.\n\n4. Deeper Insight Compared with the Kernel Variant  \n   • In the kernel variant a single Hölder–type observation suffices.  \n   • Here one must recognise that square-summability gives only global control, then devise a two-step strategy (row bounds + size partition) to upgrade it to mixed-norm summability.  \n\n5. Non-trivial Generalisation  \n   • Setting r=2 and s=p recovers the kernel statement after collapsing the second index, but the new problem is strictly stronger: it covers every pair 2≤r<s and any array of data, not just sequences, making it significantly more complex than both the original and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}