{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-bounded-linear-operator-01",
  "mko_id": "mko-bounded-linear-operator",
  "title": {
    "zh-Hant": "有界線性算子：矩陣在 ℓ^∞ 與 ℓ¹ 下的精確誘導算子範數",
    "en": "Bounded linear operator: exact induced operator norms of a matrix under ℓ^∞ and ℓ¹"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-bounded-linear-operator-01",
    "source": {
      "mko_id": "mko-bounded-linear-operator",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "c673fba2cd8ab7e707c55dc85252e9d84fa9bed89d6cfd51ac95bcc4cee707f9",
      "english_sha256": "3ac597900945ab4e80b91b57863d6afe3de538143013f1cd21732bdc4e4906a2"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "abs_rules",
      "order_basics",
      "matrix_product_def",
      "forall_intro",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "設 A 為第一行 (1,−2)、第二行 (3,4) 的整數矩陣，T(x)=Ax:ℝ²→ℝ²。（一）在兩端都用 ℓ^∞ 範數時，證明對所有 x 有 ‖Ax‖_∞≤7‖x‖_∞，且沒有更小的常數可用，所以 ‖T‖_{∞→∞}=7（最大行絕對值和）。（二）兩端都用 ℓ¹ 範數時，證明 ‖T‖_{1→1}=6（最大列絕對值和）。（三）說明非最佳常數、有限輸入與值域無界這三個陷阱。",
      "en": "Let A be the integer matrix with first row (1,−2) and second row (3,4), and T(x)=Ax:ℝ²→ℝ². (1) With the ℓ^∞ norm at both ends, prove that ‖Ax‖_∞≤7‖x‖_∞ for all x and that no smaller constant works, so ‖T‖_{∞→∞}=7 (the maximum absolute row sum). (2) With the ℓ¹ norm at both ends, prove ‖T‖_{1→1}=6 (the maximum absolute column sum). (3) Explain three traps: non-optimal constants, finitely many inputs, and unbounded range."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "ℝ² 上的兩個命名範數：‖x‖_∞=max(|x₁|,|x₂|)，‖x‖₁=|x₁|+|x₂|；每個任務在來源與目標兩端使用同一個指定名稱。",
          "en": "Two named norms on ℝ²: ‖x‖_∞=max(|x₁|,|x₂|) and ‖x‖₁=|x₁|+|x₂|; each task uses one specified name at both the source and the target."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "A 的列為 (1,−2) 與 (3,4)；T(x)=Ax。",
          "en": "A has rows (1,−2) and (3,4); T(x)=Ax."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "測試向量：x*=(1,1)、e₁=(1,0)、e₂=(0,1)。",
          "en": "Test vectors: x*=(1,1), e₁=(1,0), e₂=(0,1)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_bounded_def",
          "op_norm_def"
        ],
        "before_tex": "T:(\\mathbb{R}^2,\\|\\cdot\\|_\\infty)\\to(\\mathbb{R}^2,\\|\\cdot\\|_\\infty)\\ \\text{bounded}",
        "after_tex": "\\exists C\\ge0\\ \\forall x:\\ \\|Tx\\|_\\infty\\le C\\|x\\|_\\infty;\\ \\ \\|T\\|_{\\infty\\to\\infty}=\\text{least admissible}\\ C",
        "relation": "definition",
        "rule": "mko_bounded_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，有界表示存在一個常數 C，同一個 C 控制所有輸入 x（量詞順序是「存在 C，對所有 x」）。算子範數是最小的可用常數。兩端的範數名稱先固定為 ℓ^∞。",
          "en": "By the MKO, bounded means there is one constant C that controls every input x (the quantifier order is 'there exists C such that for all x'). The operator norm is the least admissible constant. The norm names at both ends are first fixed as ℓ^∞."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_bounded_def",
            "op_norm_def"
          ],
          "explanation": {
            "zh-Hant": "只展開定義；若把量詞順序寫成「對每個 x 有某個 C」，條件就會對每個線性映射平凡成立，失去意義。",
            "en": "It only unfolds the definition; with the quantifiers reversed ('for each x some C') the condition would hold trivially for every linear map and say nothing."
          }
        },
        "leaf_basis": "mko_bounded_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G2",
          "mko_linear_map_def",
          "matrix_product_def"
        ],
        "before_tex": "T(x)=Ax",
        "after_tex": "Ax=(x_1-2x_2,\\ 3x_1+4x_2),\\ \\ T\\ \\text{linear}",
        "relation": "equivalent",
        "rule": "matrix_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依矩陣乘積，Ax 的兩個坐標是兩行與 x 的內積。矩陣映射是線性的（引用線性映射節點的全稱證書），所以 T 是線性算子，只剩有界性要證。",
          "en": "By the matrix product, the two coordinates of Ax are the rows paired with x. A matrix map is linear (citing the universal certificate of the linear-map node), so T is a linear operator and only boundedness remains."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "matrix_product_def"
          ],
          "explanation": {
            "zh-Hant": "坐標公式是乘積的展開，兩邊同一向量；線性性質是引用，不在本步重證。",
            "en": "The coordinate formula expands the product, so both sides are the same vector; linearity is cited, not reproved in this step."
          }
        },
        "leaf_basis": "matrix_product_def + mko_linear_map_def"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G1"
        ],
        "before_tex": "\\|Ax\\|_\\infty",
        "after_tex": "\\max(|x_1-2x_2|,\\ |3x_1+4x_2|)",
        "relation": "equivalent",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "展開目標端的 ℓ^∞ 範數：取兩個坐標絕對值的最大者。",
          "en": "Unfold the ℓ^∞ norm at the target: take the larger absolute value of the two coordinates."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "定義展開，兩向成立。",
            "en": "Unfolding a definition holds both ways."
          }
        },
        "leaf_basis": "def_unfold"
      },
      {
        "id": "S04",
        "uses": [
          "S03"
        ],
        "before_tex": "|x_1-2x_2|",
        "after_tex": "|x_1-2x_2|\\le(|1|+|-2|)\\|x\\|_\\infty=3\\|x\\|_\\infty",
        "relation": "implies",
        "rule": "L_row_bound",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第一行 (1,−2)：依引理 L_row_bound（c₁=1、c₂=−2），坐標絕對值至多是行絕對值和 3 乘 ‖x‖_∞。引理在子步 S04.1、S04.2 證明，對所有 x 成立。",
          "en": "First row (1,−2): by Lemma L_row_bound (c₁=1, c₂=−2), the absolute value of the coordinate is at most the absolute row sum 3 times ‖x‖_∞. The lemma is proved in sub-steps S04.1 and S04.2 for all x."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_row_bound"
          ],
          "explanation": {
            "zh-Hant": "這是上界，不宣稱等號；等號何時達成由 S07 另行處理。",
            "en": "This is an upper bound and claims no equality; attainment is handled separately in S07."
          }
        },
        "leaf_basis": "abs_rules + order_basics + max_rules"
      },
      {
        "id": "S04.1",
        "uses": [
          "abs_rules"
        ],
        "before_tex": "|c_1x_1+c_2x_2|",
        "after_tex": "|c_1x_1+c_2x_2|\\le|c_1||x_1|+|c_2||x_2|",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三角不等式 |u+v|≤|u|+|v|，再用 |c_kx_k|=|c_k||x_k|。",
          "en": "The triangle inequality |u+v|≤|u|+|v|, then |c_kx_k|=|c_k||x_k|."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "單向不等式，不宣稱等號。",
            "en": "A one-way inequality; no equality is claimed."
          }
        },
        "leaf_basis": "abs_rules"
      },
      {
        "id": "S04.2",
        "uses": [
          "S04.1",
          "max_rules",
          "order_basics"
        ],
        "before_tex": "|c_1||x_1|+|c_2||x_2|",
        "after_tex": "\\le|c_1|\\|x\\|_\\infty+|c_2|\\|x\\|_\\infty=(|c_1|+|c_2|)\\|x\\|_\\infty",
        "relation": "implies",
        "rule": "max_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 max 的性質 |x_k|≤‖x‖_∞；乘上非負數 |c_k| 保持不等號（順序公理），兩項相加後提出 ‖x‖_∞。S04.1 與 S04.2 合起來證得引理 L_row_bound。",
          "en": "By the property of max, |x_k|≤‖x‖_∞; multiplying by the nonnegative |c_k| keeps the inequality (order axioms), and adding the two terms and factoring out ‖x‖_∞ finishes. S04.1 and S04.2 together prove Lemma L_row_bound."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "max_rules",
            "order_basics",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "需要 |c_k|≥0；若乘以負數，不等號會翻轉。",
            "en": "It needs |c_k|≥0; multiplying by a negative number would flip the inequality."
          }
        },
        "leaf_basis": "max_rules + order_basics + distributive"
      },
      {
        "id": "S05",
        "uses": [
          "S03",
          "L_row_bound"
        ],
        "before_tex": "|3x_1+4x_2|",
        "after_tex": "|3x_1+4x_2|\\le(|3|+|4|)\\|x\\|_\\infty=7\\|x\\|_\\infty",
        "relation": "implies",
        "rule": "L_row_bound",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第二行 (3,4)：同一引理（c₁=3、c₂=4）給出上界 7‖x‖_∞。",
          "en": "Second row (3,4): the same lemma (c₁=3, c₂=4) gives the bound 7‖x‖_∞."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_row_bound"
          ],
          "explanation": {
            "zh-Hant": "上界，不宣稱等號。",
            "en": "An upper bound; no equality is claimed."
          }
        },
        "leaf_basis": "abs_rules + order_basics + max_rules"
      },
      {
        "id": "S06",
        "uses": [
          "S03",
          "S04",
          "S05",
          "max_rules"
        ],
        "before_tex": "\\max(|x_1-2x_2|,|3x_1+4x_2|)",
        "after_tex": "\\le\\max(3,7)\\|x\\|_\\infty=7\\|x\\|_\\infty\\ \\ \\forall x\\ \\Rightarrow\\ T\\ \\text{bounded with}\\ C=7",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個坐標都 ≤7‖x‖_∞（因 3‖x‖_∞≤7‖x‖_∞），所以最大者也 ≤7‖x‖_∞。x 是任意的，依全稱引入，同一個 C=7 控制所有輸入：T 有界。7 正是最大行絕對值和 max(1+2, 3+4)。",
          "en": "Both coordinates are ≤7‖x‖_∞ (since 3‖x‖_∞≤7‖x‖_∞), so their maximum is ≤7‖x‖_∞ as well. x was arbitrary, so by generalization the single constant C=7 controls every input: T is bounded. 7 is exactly the maximum absolute row sum max(1+2, 3+4)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S05",
            "max_rules",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "證明沒有代入特定 x；N3 在整數格上作有限核對，只是輔助（N2 核行和）。",
            "en": "No particular x was substituted; N3 is a finite check on an integer grid and only auxiliary (N2 checks the row sums)."
          }
        },
        "leaf_basis": "max_rules + forall_intro + order_basics"
      },
      {
        "id": "S07",
        "uses": [
          "S02",
          "G3"
        ],
        "before_tex": "x^{*}=(1,1),\\ \\|x^{*}\\|_\\infty=1",
        "after_tex": "Ax^{*}=(-1,7),\\ \\|Ax^{*}\\|_\\infty=7=7\\|x^{*}\\|_\\infty",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取 x* 的坐標為第二行係數的符號（3、4 都為正，所以取 1、1）：Ax*=(1−2, 3+4)=(−1,7)，‖Ax*‖_∞=7，而 ‖x*‖_∞=1。上界 7 被達成。",
          "en": "Take x* with coordinates equal to the signs of the second-row coefficients (3 and 4 are positive, so 1 and 1): Ax*=(1−2, 3+4)=(−1,7), so ‖Ax*‖_∞=7 while ‖x*‖_∞=1. The bound 7 is attained."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確計算（N1）；這一個點只用來證明下界，不用來證明上界。",
            "en": "Exact computation (N1); this single point is used only for the lower bound, never for the upper bound."
          }
        },
        "leaf_basis": "numeric_evaluation + matrix_product_def"
      },
      {
        "id": "S08",
        "uses": [
          "S06",
          "S07",
          "op_norm_def"
        ],
        "before_tex": "C\\ \\text{admissible}\\ \\Rightarrow\\ 7=\\|Ax^{*}\\|_\\infty\\le C\\|x^{*}\\|_\\infty=C",
        "after_tex": "C\\ge7\\ \\text{for every admissible}\\ C;\\ \\ \\|T\\|_{\\infty\\to\\infty}=7",
        "relation": "implies",
        "rule": "op_norm_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "任何可用常數 C 都必須對 x* 成立，所以 7≤C。C=7 本身可用（S06），而且不大於任何可用常數，依定義 ‖T‖_{∞→∞}=7，與 MKO 的最大行絕對值和公式一致。",
          "en": "Any admissible constant C must work at x*, so 7≤C. C=7 itself is admissible (S06) and is at most every admissible constant, so by definition ‖T‖_{∞→∞}=7, in agreement with the MKO's maximum-absolute-row-sum formula."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S06",
            "S07",
            "op_norm_def"
          ],
          "explanation": {
            "zh-Hant": "目標：‖T‖=7。足夠條件：7 可用（S06）且每個可用 C≥7（S07）；兩者都已正向證得。",
            "en": "Goal: ‖T‖=7. Sufficient condition: 7 is admissible (S06) and every admissible C is ≥7 (S07); both are proved forwards."
          }
        },
        "leaf_basis": "op_norm_def + order_basics"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "G1",
          "abs_rules"
        ],
        "before_tex": "\\|Ax\\|_1=|x_1-2x_2|+|3x_1+4x_2|",
        "after_tex": "\\|Ax\\|_1\\le 6\\|x\\|_1\\ \\ \\forall x",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "改用兩端 ℓ¹ 範數。子步 S09.1 依列重組得 ‖Ax‖₁≤4|x₁|+6|x₂|，S09.2 再放大成 6‖x‖₁。6 是最大列絕對值和 max(|1|+|3|, |−2|+|4|)。",
          "en": "Now use ℓ¹ at both ends. Sub-step S09.1 regroups by columns to get ‖Ax‖₁≤4|x₁|+6|x₂|, and S09.2 enlarges this to 6‖x‖₁. 6 is the maximum absolute column sum max(|1|+|3|, |−2|+|4|)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "order_basics",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "全稱上界；N5 是有限格點輔助核對。",
            "en": "A universal bound; N5 is an auxiliary finite grid check."
          }
        },
        "leaf_basis": "abs_rules + order_basics + forall_intro"
      },
      {
        "id": "S09.1",
        "uses": [
          "abs_rules"
        ],
        "before_tex": "|x_1-2x_2|+|3x_1+4x_2|",
        "after_tex": "\\le(|x_1|+2|x_2|)+(3|x_1|+4|x_2|)=4|x_1|+6|x_2|",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "每一行用三角不等式與 |cx|=|c||x|，再把同一個 |x_k| 的係數合併：|x₁| 的係數是第一列的絕對值和 1+3=4，|x₂| 的是第二列的 2+4=6。",
          "en": "Apply the triangle inequality and |cx|=|c||x| in each row, then collect the coefficients of each |x_k|: the coefficient of |x₁| is the first column's absolute sum 1+3=4, and that of |x₂| is the second column's 2+4=6."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "add_comm",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "單向不等式加上重組；ℓ¹ 自然地按列合併，ℓ^∞ 則按行取最大。",
            "en": "A one-way inequality plus regrouping; ℓ¹ naturally collects by columns, whereas ℓ^∞ takes the maximum over rows."
          }
        },
        "leaf_basis": "abs_rules + add_comm + distributive"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "order_basics"
        ],
        "before_tex": "4|x_1|+6|x_2|",
        "after_tex": "\\le6|x_1|+6|x_2|=6\\|x\\|_1",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "因 |x₁|≥0 且 4≤6，4|x₁|≤6|x₁|；加上 6|x₂| 並提出 6。",
          "en": "Since |x₁|≥0 and 4≤6, 4|x₁|≤6|x₁|; add 6|x₂| and factor out 6."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "需要 |x₁|≥0；不宣稱等號。",
            "en": "It needs |x₁|≥0; no equality is claimed."
          }
        },
        "leaf_basis": "order_basics + distributive + abs_rules"
      },
      {
        "id": "S10",
        "uses": [
          "S02",
          "S09",
          "G3",
          "op_norm_def"
        ],
        "before_tex": "e_2=(0,1),\\ \\|e_2\\|_1=1",
        "after_tex": "Ae_2=(-2,4),\\ \\|Ae_2\\|_1=6\\ \\Rightarrow\\ \\|T\\|_{1\\to1}=6",
        "relation": "implies",
        "rule": "op_norm_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "Ae₂ 是 A 的第二列 (−2,4)，ℓ¹ 範數 2+4=6，而 ‖e₂‖₁=1。任何可用常數 C 都必須 ≥6；6 本身可用（S09），所以 ‖T‖_{1→1}=6。",
          "en": "Ae₂ is the second column of A, (−2,4), with ℓ¹ norm 2+4=6, while ‖e₂‖₁=1. Every admissible constant C must be ≥6; 6 itself is admissible (S09), so ‖T‖_{1→1}=6."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S09",
            "op_norm_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "目標：‖T‖_{1→1}=6。足夠條件：6 可用（S09）且 e₂ 迫使 C≥6（N4）。",
            "en": "Goal: ‖T‖_{1→1}=6. Sufficient condition: 6 is admissible (S09) and e₂ forces C≥6 (N4)."
          }
        },
        "leaf_basis": "op_norm_def + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S06",
          "S08",
          "S09",
          "S10"
        ],
        "before_tex": "\\|T\\|_{\\infty\\to\\infty}=7,\\ \\|T\\|_{1\\to1}=6",
        "after_tex": "C=7\\ \\text{admissible for}\\ \\ell^1\\ \\text{but}\\ \\neq\\|T\\|_{1\\to1};\\ C=10\\ \\text{admissible for}\\ \\ell^\\infty\\ \\text{but}\\ \\neq\\|T\\|_{\\infty\\to\\infty}",
        "relation": "implies",
        "rule": "op_norm_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "算子範數依兩端範數名稱而定：同一個 A 在 ℓ^∞ 下是 7、在 ℓ¹ 下是 6。非最佳常數仍可證有界：由 S09，‖Ax‖₁≤6‖x‖₁≤7‖x‖₁，所以 C=7 對 ℓ¹ 可用；由 S06，C=10 對 ℓ^∞ 可用。但它們不是算子範數，MKO 禁止把任意可用上界冒充算子範數。",
          "en": "The operator norm depends on the norm names at both ends: the same A gives 7 under ℓ^∞ and 6 under ℓ¹. Non-optimal constants still prove boundedness: by S09, ‖Ax‖₁≤6‖x‖₁≤7‖x‖₁, so C=7 is admissible for ℓ¹; by S06, C=10 is admissible for ℓ^∞. But they are not operator norms, and the MKO forbids passing off an arbitrary admissible bound as the operator norm."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S08",
            "S09",
            "S10",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "可用常數的集合向上封閉（大於可用常數者仍可用），算子範數是其最小者（N8）。",
            "en": "The set of admissible constants is closed upwards (anything larger than an admissible constant is admissible), and the operator norm is its least element (N8)."
          }
        },
        "leaf_basis": "op_norm_def + order_basics"
      },
      {
        "id": "S12",
        "uses": [
          "S02",
          "S07",
          "G3"
        ],
        "before_tex": "C=5:\\ \\|Ae_1\\|_\\infty=\\|(1,3)\\|_\\infty=3\\le5,\\ \\|Ae_2\\|_\\infty=\\|(-2,4)\\|_\\infty=4\\le5",
        "after_tex": "\\|Ax^{*}\\|_\\infty=7>5=5\\|x^{*}\\|_\\infty\\ \\Rightarrow\\ C=5\\ \\text{not admissible}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "只看基向量 e₁、e₂，C=5 都通過；但 x*=(1,1) 給出 7>5，所以 C=5 不可用。有限個輸入不能證明一個界，同一個 C 必須控制所有輸入；MKO 指出 instance 通過不能取代全稱證書。",
          "en": "Looking only at the basis vectors e₁ and e₂, C=5 passes both; but x*=(1,1) gives 7>5, so C=5 is not admissible. Finitely many inputs cannot prove a bound, since one C must control every input; the MKO says an instance pass does not replace a universal certificate."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "quantifier_negation",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一個反例 x* 足以否定 C=5（N7）。",
            "en": "One counterexample, x*, suffices to refute C=5 (N7)."
          }
        },
        "leaf_basis": "quantifier_negation + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S07",
          "mko_linear_map_def"
        ],
        "before_tex": "T(tx^{*})=tAx^{*}=(-t,7t),\\ t\\in\\mathbb{R}",
        "after_tex": "\\|T(tx^{*})\\|_\\infty=7|t|\\ \\text{unbounded in}\\ t,\\ \\text{yet}\\ T\\ \\text{is bounded}",
        "relation": "implies",
        "rule": "archimedean",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依線性，T(tx*)=tAx*，其範數 7|t|。對任何正數 K，依阿基米德性質取 t>K/7，則 7|t|>K，所以 T 的值域無界。但 T 是有界算子：有界指對所有 x≠0 的比值 ‖Tx‖/‖x‖ 有共同上界，等價於對所有 x 有 ‖Tx‖≤C‖x‖；x=0 時用 T0=0，不取 0/0。這不是值域有界。MKO 把「值域看似有界」列為誤解。",
          "en": "By linearity T(tx*)=tAx*, with norm 7|t|. For any positive K, the Archimedean property gives t>K/7, and then 7|t|>K, so the range of T is unbounded. Yet T is a bounded operator: boundedness gives a common bound on ‖Tx‖/‖x‖ for x≠0, equivalently ‖Tx‖≤C‖x‖ for every x. At x=0 use T0=0 rather than 0/0. This does not bound the range. The MKO lists 'the range looks bounded' as a misconception."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "archimedean",
            "mko_linear_map_def"
          ],
          "explanation": {
            "zh-Hant": "需要 T 線性與 Ax*≠0；只有零算子的值域有界。",
            "en": "It needs T linear and Ax*≠0; only the zero operator has a bounded range."
          }
        },
        "leaf_basis": "archimedean + mko_linear_map_def + abs_rules"
      },
      {
        "id": "S14",
        "uses": [
          "S08",
          "S10",
          "S11",
          "S12",
          "S13"
        ],
        "before_tex": "\\text{S08},\\ \\text{S10},\\ \\text{S11},\\ \\text{S12},\\ \\text{S13}",
        "after_tex": "\\|T\\|_{\\infty\\to\\infty}=7,\\ \\|T\\|_{1\\to1}=6;\\ \\text{traps refuted}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合併：在 ℓ^∞ 下 T 有界，算子範數 7（最大行絕對值和，於 (1,1) 達成）；在 ℓ¹ 下算子範數 6（最大列絕對值和，於 e₂ 達成）。非最佳常數只證有界；有限輸入不能證界；值域無界不妨礙有界性。",
          "en": "Combine: under ℓ^∞, T is bounded with operator norm 7 (the maximum absolute row sum, attained at (1,1)); under ℓ¹ the operator norm is 6 (the maximum absolute column sum, attained at e₂). Non-optimal constants only prove boundedness; finitely many inputs cannot prove a bound; an unbounded range does not prevent boundedness."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S08",
            "S10",
            "S11",
            "S12",
            "S13",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標是三個任務的合取；各部分由所行步驟正向證得。",
            "en": "The goal is the conjunction of the three tasks; each part is proved forwards by the listed steps."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "對 A（行 (1,−2),(3,4)）：所有 x 都有 ‖Ax‖_∞≤7‖x‖_∞，且 x*=(1,1) 給 ‖Ax*‖_∞=7，所以 ‖T‖_{∞→∞}=7（最大行絕對值和）；所有 x 都有 ‖Ax‖₁≤6‖x‖₁，且 e₂ 給 ‖Ae₂‖₁=6，所以 ‖T‖_{1→1}=6（最大列絕對值和）。C=10 或對 ℓ¹ 的 C=7 只證有界、不是算子範數；C=5 通過 e₁、e₂ 卻在 x* 失敗；T 的值域無界，但 T 是有界算子。",
      "en": "For A (rows (1,−2),(3,4)): every x has ‖Ax‖_∞≤7‖x‖_∞, and x*=(1,1) gives ‖Ax*‖_∞=7, so ‖T‖_{∞→∞}=7 (the maximum absolute row sum); every x has ‖Ax‖₁≤6‖x‖₁, and e₂ gives ‖Ae₂‖₁=6, so ‖T‖_{1→1}=6 (the maximum absolute column sum). C=10, or C=7 for ℓ¹, proves boundedness but is not the operator norm; C=5 passes e₁ and e₂ but fails at x*; the range of T is unbounded, yet T is a bounded operator."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "‖Ae₁‖_∞=3≤5 且 ‖Ae₂‖_∞=4≤5，所以對所有 x 有 ‖Ax‖_∞≤5‖x‖_∞",
          "en": "‖Ae₁‖_∞=3≤5 and ‖Ae₂‖_∞=4≤5, so ‖Ax‖_∞≤5‖x‖_∞ for all x"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x*=(1,1)：‖Ax*‖_∞=7>5。",
          "en": "x*=(1,1): ‖Ax*‖_∞=7>5."
        }
      },
      {
        "claim": {
          "zh-Hant": "因為 ‖Ax‖_∞≤10‖x‖_∞ 對所有 x 成立，所以 ‖T‖_{∞→∞}=10",
          "en": "Since ‖Ax‖_∞≤10‖x‖_∞ for all x, ‖T‖_{∞→∞}=10"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "10 只是可用常數；最小可用常數是 7。",
          "en": "10 is only an admissible constant; the least admissible constant is 7."
        }
      },
      {
        "claim": {
          "zh-Hant": "ℓ^∞ 誘導算子範數是最大列絕對值和 max(4,6)=6",
          "en": "The ℓ^∞ induced operator norm is the maximum absolute column sum max(4,6)=6"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "列和屬於 ℓ¹；ℓ^∞ 用行和，得 7，x* 給出 7>6。",
          "en": "Column sums belong to ℓ¹; ℓ^∞ uses row sums and gives 7, and x* gives 7>6."
        }
      },
      {
        "claim": {
          "zh-Hant": "T(tx*)=(−t,7t) 的範數可任意大，所以 T 不是有界算子",
          "en": "T(tx*)=(−t,7t) has arbitrarily large norm, so T is not a bounded operator"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有界指 ‖Tx‖≤7‖x‖ 對所有 x 成立，與值域是否有界無關。",
          "en": "Boundedness means ‖Tx‖≤7‖x‖ for all x, which has nothing to do with whether the range is bounded."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "abs_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "matrix_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "矩陣乘法：若 A 為 m×n、B 為 n×p，則 (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}。矩陣乘法一般不可交換。",
          "en": "Matrix product: for A of size m×n and B of size n×p, (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}. Matrix multiplication is in general not commutative."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "archimedean",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "阿基米德性質：對任意實數 x，存在自然數 n>x；因此對任意 ε>0，存在 n 使 1/n<ε。",
          "en": "Archimedean property: for every real x there is a natural number n>x; hence for every ε>0 there is n with 1/n<ε."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_bounded_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-bounded-linear-operator",
          "sha256": "c673fba2cd8ab7e707c55dc85252e9d84fa9bed89d6cfd51ac95bcc4cee707f9",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "獨立來源追加式勘誤（使用者已採用於本地修訂；未合併或發布）：先要求 T:V→W 是線性映射，再要求存在 C≥0 使所有 x 有 ‖Tx‖_W≤C‖x‖_V，兩項共同構成有界線性算子資格。單有此增長界不推出線性，例如 x↦|x|。本輪整矩陣在 ℓ¹ 下採最大列絕對值和，在 ℓ^∞ 下採最大行絕對值和作精確誘導範數證書；instance 通過不代替全稱證書。原始引文與 canonical MKO 保留。",
          "en": "Independent append-only source erratum (owner-adopted in this local revision; not merged or published): first require T:V→W to be linear, and separately require C≥0 with ‖Tx‖_W≤C‖x‖_V for every x. Together these give a bounded linear operator. The growth bound alone does not imply linearity, as x↦|x| shows. For the integer matrices here, exact induced-norm certificates use the maximum absolute column sum for ℓ¹ and row sum for ℓ^∞; an instance pass does not replace a universal certificate. The original citation and canonical MKO are retained."
        }
      },
      {
        "id": "mko_linear_map_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-general-linear-map",
          "sha256": "184a885b199aff8dc05ec273c0fe5b2dcdd38a9ef98c8ec790ce2668da9a07c7",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "實線性映射滿足 T(αu+βv)=αT(u)+βT(v)；有限維矩陣公式提供全稱座標證書。",
          "en": "A real linear map satisfies T(αu+βv)=αT(u)+βT(v); a finite-dimensional matrix formula supplies a universal coordinate certificate."
        }
      },
      {
        "id": "op_norm_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有界算子的誘導算子範數 ‖T‖ 是所有可用常數 C（對所有 x 有 ‖Tx‖≤C‖x‖）的下確界；若某個可用常數 C₀ 使每個可用常數都 ≥C₀，則 ‖T‖=C₀。",
          "en": "The induced operator norm ‖T‖ of a bounded operator is the infimum of all admissible constants C (those with ‖Tx‖≤C‖x‖ for all x); if some admissible C₀ is at most every admissible constant, then ‖T‖=C₀."
        }
      },
      {
        "id": "max_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "max(s,t)≥s 且 max(s,t)≥t；max(s,t)≤K 當且僅當 s≤K 且 t≤K。",
          "en": "max(s,t)≥s and max(s,t)≥t; max(s,t)≤K iff s≤K and t≤K."
        }
      },
      {
        "id": "L_row_bound",
        "kind": "lemma",
        "proved_in": [
          "S04.1",
          "S04.2"
        ],
        "statement": {
          "zh-Hant": "對實數 c₁,c₂ 與 x∈ℝ²：|c₁x₁+c₂x₂|≤(|c₁|+|c₂|)‖x‖_∞。",
          "en": "For reals c₁,c₂ and x∈ℝ²: |c₁x₁+c₂x₂|≤(|c₁|+|c₂|)‖x‖_∞."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S07",
        "lhs": "max(abs(1-2*1),abs(3*1+4*1))",
        "rhs": "7"
      },
      {
        "id": "N2",
        "step": "S06",
        "lhs": "max(abs(1)+abs(-2),abs(3)+abs(4))",
        "rhs": "7"
      },
      {
        "id": "N3",
        "step": "S06",
        "vars": [
          "x1",
          "x2"
        ],
        "range": [
          -4,
          4
        ],
        "lhs": "max(abs(x1-2*x2),abs(3*x1+4*x2))",
        "relation": "<=",
        "rhs": "7*max(abs(x1),abs(x2))"
      },
      {
        "id": "N4",
        "step": "S10",
        "lhs": "abs(-2)+abs(4)",
        "rhs": "6"
      },
      {
        "id": "N5",
        "step": "S09",
        "vars": [
          "x1",
          "x2"
        ],
        "range": [
          -4,
          4
        ],
        "lhs": "abs(x1-2*x2)+abs(3*x1+4*x2)",
        "relation": "<=",
        "rhs": "6*(abs(x1)+abs(x2))"
      },
      {
        "id": "N6",
        "step": "S10",
        "lhs": "max(abs(1)+abs(3),abs(-2)+abs(4))",
        "rhs": "6"
      },
      {
        "id": "N7",
        "step": "S12",
        "lhs": "max(abs(1-2*1),abs(3*1+4*1))",
        "relation": ">",
        "rhs": "5*max(abs(1),abs(1))"
      },
      {
        "id": "N8",
        "step": "S11",
        "lhs": "7",
        "relation": "!=",
        "rhs": "6"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：來源與目標都是明示的 ℝ²，兩端範數名稱固定（假設 a1）；ℓ¹ 用最大列絕對值和、ℓ^∞ 用最大行絕對值和，並對這個矩陣完整證明上界與達成（假設 a2）。一般 m×n 矩陣的公式只由 MKO 敘述，不另證。不處理連續性、對偶或譜理論，不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: source and target are the explicit ℝ² with fixed norm names at both ends (assumption a1); ℓ¹ uses the maximum absolute column sum and ℓ^∞ the maximum absolute row sum, and for this matrix both the bound and its attainment are proved completely (assumption a2). The formula for general m-by-n matrices is taken from the MKO's statement and not proved separately. Continuity, duality and spectral theory are not treated, and the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "A=\\begin{pmatrix}1&-2\\\\3&4\\end{pmatrix}",
        "給定的整數矩陣",
        "the given integer matrix"
      ],
      [
        "\\|x\\|_\\infty=\\max(|x_1|,|x_2|)",
        "ℓ^∞ 範數",
        "the ℓ^∞ norm"
      ],
      [
        "\\|x\\|_1=|x_1|+|x_2|",
        "ℓ¹ 範數",
        "the ℓ¹ norm"
      ],
      [
        "\\|T\\|_{\\infty\\to\\infty},\\ \\|T\\|_{1\\to1}",
        "兩種誘導算子範數",
        "the two induced operator norms"
      ],
      [
        "C",
        "可用的界常數",
        "an admissible bounding constant"
      ],
      [
        "x^{*}=(1,1)",
        "達成 ℓ^∞ 上界的向量",
        "the vector attaining the ℓ^∞ bound"
      ]
    ],
    "md_error_controls": [
      {
        "title": "有限輸入不證有界 / Finitely many inputs do not prove a bound",
        "zh": "MKO 指出的誤解之一是只看有限輸入。C=5 對 e₁、e₂ 都成立，但 x*=(1,1) 給 7>5。同一個 C 必須控制所有輸入，所以上界要用 S04 到 S06 的全稱估計證明。",
        "en": "One misconception named by the MKO is to look only at finitely many inputs. C=5 holds at e₁ and e₂, but x*=(1,1) gives 7>5. One C must control every input, so the bound must be proved by the universal estimates of S04 to S06."
      },
      {
        "title": "可用上界不一定是算子範數 / An admissible bound need not equal the operator norm",
        "zh": "C=10 證明 T 有界，但算子範數是最小的可用常數 7，需要 S07 的達成向量才能確定。同樣在 ℓ¹ 下 7 可用，但算子範數是 6。",
        "en": "C=10 proves that T is bounded, but the operator norm is the least admissible constant 7, which is pinned down only by the attaining vector of S07. Likewise under ℓ¹, 7 is admissible but the operator norm is 6."
      },
      {
        "title": "有界不是值域有界 / Bounded does not mean bounded range",
        "zh": "非零線性算子的值域永遠無界（S13）。有界算子的意思是 ‖Tx‖ 被 C‖x‖ 控制，等價於 x≠0 時的比值有共同上界；x=0 的不等式由 T0=0 處理，不取 0/0；把「值域看似有界」當定義是 MKO 指出的誤解。",
        "en": "The range of a nonzero linear operator is always unbounded (S13). A bounded operator is one with ‖Tx‖ controlled by C‖x‖, equivalently a common ratio bound for x≠0; at x=0 use T0=0, not 0/0; taking 'the range looks bounded' as the definition is a misconception named by the MKO."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "一般 m×n 矩陣的最大行和／最大列和公式由 MKO 敘述；本稿只對這個 2×2 矩陣完整證明上界與達成。",
        "en": "The maximum-row-sum and maximum-column-sum formulas for general m-by-n matrices are taken from the MKO's statement; this guide proves the bound and its attainment completely only for this 2-by-2 matrix."
      },
      {
        "zh-Hant": "N3、N5 是整數格點上的有限核對，只是輔助；全稱上界由 S04 到 S06 與 S09 證明。",
        "en": "N3 and N5 are finite checks on integer grids and only auxiliary; the universal bounds are proved by S04 to S06 and S09."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "62f768caa7fd6bfdba8a4015f71bafa6a99724583bfdb2b12927754626981870",
    "original_source_sha256": "63dd897c1a76886df54bea7e1ee19386d14e12f7b203715b1bb3773f979351a1",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M082",
      "BR-M083"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}