{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-convex-function-01",
  "mko_id": "mko-convex-function",
  "title": {
    "zh-Hant": "凸函數：二次函數 Jensen gap 的完整證書與單一 Jensen 反例",
    "en": "Convex function: a complete certificate via the Jensen gap of a quadratic, and a single Jensen counterexample"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-convex-function-01",
    "source": {
      "mko_id": "mko-convex-function",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "d5af460bbe41c8e967949f9398bde368d76471cd2b4a24e4f49f14ad90a05fed",
      "english_sha256": "147a00f6af20512702925d6854df7d6be01eeefa9aca4e6b2e1b268412020da4"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "order_basics",
      "square_nonneg",
      "distributive",
      "power_laws",
      "forall_intro",
      "quantifier_negation"
    ],
    "goal": {
      "zh-Hant": "（一）證明 f(x)=2x²−3x+1 在區間 I=[−1,4] 上是凸函數：先確認 I 是凸集合，再對所有 x,y∈I 與所有 λ∈[0,1] 證明 f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y)，方法是證明二次函數的 Jensen gap 等於 aλ(1−λ)(x−y)²。（二）由同一恆等式得仿射函數 ℓ(x)=3x−5 在 I 上取等號。（三）用一個精確的 Jensen 違反否證 g(x)=−x²+x 在 J=[0,2] 上的凸性，並說明幾個通過的樣本為何不能證明凸性。",
      "en": "(1) Prove that f(x)=2x²−3x+1 is convex on the interval I=[−1,4]: first confirm that I is convex, then prove f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y) for all x,y∈I and all λ∈[0,1], by showing that the Jensen gap of a quadratic equals aλ(1−λ)(x−y)². (2) From the same identity, the affine function ℓ(x)=3x−5 gives equality on I. (3) Refute the convexity of g(x)=−x²+x on J=[0,2] with one exact Jensen violation, and explain why a few passing samples cannot prove convexity."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "I=[−1,4]；f:I→ℝ，f(x)=2x²−3x+1（a=2、b=−3、c=1）。",
          "en": "I=[−1,4]; f:I→ℝ, f(x)=2x²−3x+1 (a=2, b=−3, c=1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "ℓ:I→ℝ，ℓ(x)=3x−5（a=0、b=3、c=−5，仿射）。",
          "en": "ℓ:I→ℝ, ℓ(x)=3x−5 (a=0, b=3, c=−5, affine)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "J=[0,2]；g:J→ℝ，g(x)=−x²+x（a=−1、b=1、c=0）。",
          "en": "J=[0,2]; g:J→ℝ, g(x)=−x²+x (a=−1, b=1, c=0)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_cvxf"
        ],
        "before_tex": "f\\ \\text{convex on}\\ I",
        "after_tex": "I\\ \\text{convex}\\ \\land\\ \\forall x,y\\in I\\ \\forall\\lambda\\in[0,1]:\\ f(\\lambda x+(1-\\lambda)y)\\le\\lambda f(x)+(1-\\lambda)f(y)",
        "relation": "equivalent",
        "rule": "mko_cvxf",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 的定義展開。定義有兩部分：定義域必須是凸集合（否則 λx+(1−λ)y 可能不在定義域，f 在那裡沒有值），以及對所有 x、y、λ 的 Jensen 不等式。兩部分都要證。",
          "en": "Unfold the MKO's definition. It has two parts: the domain must be convex (otherwise λx+(1−λ)y may leave the domain, where f has no value), and the Jensen inequality must hold for all x, y, λ. Both parts must be proved."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_cvxf",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "展開與收回定義互逆，雙向成立。",
            "en": "Unfolding and folding the definition undo each other, so the step holds both ways."
          }
        },
        "leaf_basis": "mko_cvxf + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "L_interval",
          "mko_cvxset"
        ],
        "before_tex": "x,y\\in[-1,4],\\ \\lambda\\in[0,1]",
        "after_tex": "z=\\lambda x+(1-\\lambda)y\\in[-1,4];\\ I\\ \\text{convex}",
        "relation": "implies",
        "rule": "L_interval",
        "conditions": [
          "\\lambda\\in[0,1]"
        ],
        "explanation": {
          "zh-Hant": "用引理 L_interval（α=−1、β=4、s=x、t=y）：z 仍在 [−1,4] 中。x、y、λ 任意，所以 I 符合凸集合的定義；f(z) 因此有定義，Jensen 不等式才有意義。引理在 S02.1、S02.2 中證明。",
          "en": "Apply Lemma L_interval (α=−1, β=4, s=x, t=y): z stays in [−1,4]. Since x, y, λ are arbitrary, I satisfies the definition of a convex set; therefore f(z) is defined and the Jensen inequality makes sense. The lemma is proved in S02.1 and S02.2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_interval",
            "mko_cvxset"
          ],
          "explanation": {
            "zh-Hant": "需要 x、y 在區間內與 λ∈[0,1]；不宣稱反向。",
            "en": "It needs x, y in the interval and λ∈[0,1]; no converse is claimed."
          }
        },
        "leaf_basis": "L_interval + forall_intro"
      },
      {
        "id": "S02.1",
        "uses": [
          "mono_nonneg",
          "ineq_add",
          "distributive"
        ],
        "before_tex": "\\alpha\\le s,\\ \\alpha\\le t,\\ \\lambda\\ge 0,\\ 1-\\lambda\\ge 0",
        "after_tex": "\\alpha=\\lambda\\alpha+(1-\\lambda)\\alpha\\le\\lambda s+(1-\\lambda)t",
        "relation": "implies",
        "rule": "mono_nonneg",
        "conditions": [
          "0\\le\\lambda\\le 1"
        ],
        "explanation": {
          "zh-Hant": "由 0≤λ≤1 得 λ≥0 與 1−λ≥0。α≤s 乘以 λ 得 λα≤λs，α≤t 乘以 1−λ 得 (1−λ)α≤(1−λ)t；相加得 λα+(1−λ)α≤λs+(1−λ)t，而依分配律左邊等於 (λ+1−λ)α=α。",
          "en": "From 0≤λ≤1 we get λ≥0 and 1−λ≥0. Multiplying α≤s by λ gives λα≤λs, and α≤t by 1−λ gives (1−λ)α≤(1−λ)t; adding gives λα+(1−λ)α≤λs+(1−λ)t, and by distributivity the left side is (λ+1−λ)α=α."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mono_nonneg",
            "ineq_add",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "需要兩個權重非負；證明 L_interval 的下界，沒有使用 L_interval 本身。",
            "en": "It needs both weights nonnegative; it proves the lower bound of L_interval without using L_interval itself."
          }
        },
        "leaf_basis": "mono_nonneg + ineq_add + distributive + order_basics"
      },
      {
        "id": "S02.2",
        "uses": [
          "S02.1",
          "mono_nonneg",
          "ineq_add",
          "distributive"
        ],
        "before_tex": "s\\le\\beta,\\ t\\le\\beta",
        "after_tex": "\\lambda s+(1-\\lambda)t\\le\\lambda\\beta+(1-\\lambda)\\beta=\\beta",
        "relation": "implies",
        "rule": "mono_nonneg",
        "conditions": [
          "0\\le\\lambda\\le 1"
        ],
        "explanation": {
          "zh-Hant": "上界同理：s≤β 乘以 λ，t≤β 乘以 1−λ，相加，再用分配律化簡右邊為 β。S02.1 與 S02.2 合起來證得引理 L_interval。",
          "en": "The upper bound follows similarly: multiply s≤β by λ and t≤β by 1−λ, add, and simplify the right side to β by distributivity. S02.1 and S02.2 together prove Lemma L_interval."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mono_nonneg",
            "ineq_add",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "與 S02.1 對稱；不使用 L_interval 本身。",
            "en": "Symmetric to S02.1; L_interval itself is not used."
          }
        },
        "leaf_basis": "mono_nonneg + ineq_add + distributive"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "S02"
        ],
        "before_tex": "f(z)\\le\\lambda f(x)+(1-\\lambda)f(y)",
        "after_tex": "G_f(x,y,\\lambda)=\\lambda f(x)+(1-\\lambda)f(y)-f(z)\\ge 0",
        "relation": "equivalent",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "定義 Jensen gap G_f=λf(x)+(1−λ)f(y)−f(z)。不等式兩邊同減 f(z)，方向不變，所以 Jensen 不等式等價於 G_f≥0。之後只需研究 G_f 的符號。",
          "en": "Define the Jensen gap G_f=λf(x)+(1−λ)f(y)−f(z). Subtracting f(z) from both sides keeps the direction, so the Jensen inequality is equivalent to G_f≥0. From now on only the sign of G_f matters."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "order_basics",
            "def_intro"
          ],
          "explanation": {
            "zh-Hant": "同減 f(z) 與同加 f(z) 互逆，兩者都保持不等號方向，雙向成立。",
            "en": "Subtracting f(z) and adding f(z) undo each other and both keep the direction, so the step holds both ways."
          }
        },
        "leaf_basis": "order_basics + def_intro"
      },
      {
        "id": "S04",
        "uses": [
          "L_gap"
        ],
        "before_tex": "G_q(x,y,\\lambda)\\ \\text{for } q(x)=ax^2+bx+c",
        "after_tex": "G_q(x,y,\\lambda)=a\\lambda(1-\\lambda)(x-y)^2",
        "relation": "equivalent",
        "rule": "L_gap",
        "conditions": [],
        "explanation": {
          "zh-Hant": "引理 L_gap：任何一元二次（含 a=0 的仿射）函數的 Jensen gap 等於 aλ(1−λ)(x−y)²，對所有實數 x、y、λ 成立。證明分三個子步：常數與一次部分抵消（S04.1），展開平方（S04.2），二次部分化成完全平方（S04.3）。N1 是整個恆等式的多項式核對。",
          "en": "Lemma L_gap: the Jensen gap of any univariate quadratic (including the affine case a=0) equals aλ(1−λ)(x−y)², for all reals x, y, λ. The proof has three sub-steps: the constant and linear parts cancel (S04.1), the square is expanded (S04.2), and the quadratic part becomes a perfect square (S04.3). N1 checks the whole identity as a polynomial identity."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_gap"
          ],
          "explanation": {
            "zh-Hant": "這是多項式恆等式，兩邊是同一個多項式，雙向成立，沒有附加條件。",
            "en": "This is a polynomial identity; both sides are the same polynomial, so it holds both ways with no side condition."
          }
        },
        "leaf_basis": "distributive + power_laws + add_inverse"
      },
      {
        "id": "S04.1",
        "uses": [
          "distributive"
        ],
        "before_tex": "\\lambda(bx+c)+(1-\\lambda)(by+c)-\\big(b(\\lambda x+(1-\\lambda)y)+c\\big)",
        "after_tex": "0",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "一次部分：λbx+(1−λ)by−b(λx+(1−λ)y)=0（分配律）。常數部分：λc+(1−λ)c−c=(λ+1−λ)c−c=0。所以 gap 只剩二次部分。",
          "en": "Linear part: λbx+(1−λ)by−b(λx+(1−λ)y)=0 by distributivity. Constant part: λc+(1−λ)c−c=(λ+1−λ)c−c=0. So only the quadratic part of the gap remains."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "多項式恆等式，雙向成立。",
            "en": "A polynomial identity, valid both ways."
          }
        },
        "leaf_basis": "distributive + add_inverse + mul_one"
      },
      {
        "id": "S04.2",
        "uses": [
          "distributive",
          "power_laws"
        ],
        "before_tex": "(\\lambda x+(1-\\lambda)y)^2",
        "after_tex": "\\lambda^2x^2+2\\lambda(1-\\lambda)xy+(1-\\lambda)^2y^2",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "用 (p+q)²=p²+2pq+q²（分配律展開），p=λx、q=(1−λ)y。",
          "en": "Use (p+q)²=p²+2pq+q² (expansion by distributivity) with p=λx and q=(1−λ)y."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "power_laws"
          ],
          "explanation": {
            "zh-Hant": "展開與因式分解互逆，雙向成立。",
            "en": "Expansion and factoring undo each other, so it holds both ways."
          }
        },
        "leaf_basis": "distributive + power_laws + mul_comm"
      },
      {
        "id": "S04.3",
        "uses": [
          "S04.1",
          "S04.2",
          "distributive"
        ],
        "before_tex": "a\\big(\\lambda x^2+(1-\\lambda)y^2-(\\lambda x+(1-\\lambda)y)^2\\big)",
        "after_tex": "a\\lambda(1-\\lambda)(x-y)^2",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入 S04.2：x² 的係數是 λ−λ²=λ(1−λ)；y² 的係數是 (1−λ)−(1−λ)²=(1−λ)(1−(1−λ))=λ(1−λ)；xy 的係數是 −2λ(1−λ)。提出 λ(1−λ)：λ(1−λ)(x²−2xy+y²)=λ(1−λ)(x−y)²。乘上 a，再加上 S04.1 的 0，得 G_q=aλ(1−λ)(x−y)²。S04.1 至 S04.3 證得引理 L_gap。",
          "en": "Substitute S04.2: the coefficient of x² is λ−λ²=λ(1−λ); that of y² is (1−λ)−(1−λ)²=(1−λ)(1−(1−λ))=λ(1−λ); that of xy is −2λ(1−λ). Factor out λ(1−λ): λ(1−λ)(x²−2xy+y²)=λ(1−λ)(x−y)². Multiply by a and add the 0 of S04.1 to get G_q=aλ(1−λ)(x−y)². S04.1 to S04.3 prove Lemma L_gap."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "power_laws",
            "S04.2"
          ],
          "explanation": {
            "zh-Hant": "展開 aλ(1−λ)(x−y)² 即回到左邊；不使用 L_gap 本身。",
            "en": "Expanding aλ(1−λ)(x−y)² returns the left side; L_gap itself is not used."
          }
        },
        "leaf_basis": "distributive + power_laws + mul_comm + add_comm"
      },
      {
        "id": "S05",
        "uses": [
          "S04",
          "G1"
        ],
        "before_tex": "G_f(x,y,\\lambda)",
        "after_tex": "G_f(x,y,\\lambda)=2\\lambda(1-\\lambda)(x-y)^2",
        "relation": "equivalent",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 L_gap 用在 a=2、b=−3、c=1 的 f 上（全稱命題代入特例）。b、c 不出現在結果中，只有 a=2 決定 gap。",
          "en": "Apply L_gap to f with a=2, b=−3, c=1 (instantiating a universal statement). b and c do not appear in the result; only a=2 determines the gap."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S04",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "代入後仍是恆等式，兩邊相等，雙向成立。",
            "en": "After substitution it is still an identity; both sides are equal, valid both ways."
          }
        },
        "leaf_basis": "forall_elim + eq_replace"
      },
      {
        "id": "S06",
        "uses": [
          "S02.1",
          "mono_nonneg",
          "mul_zero"
        ],
        "before_tex": "0\\le\\lambda\\le 1",
        "after_tex": "\\lambda(1-\\lambda)\\ge 0",
        "relation": "implies",
        "rule": "mono_nonneg",
        "conditions": [
          "\\lambda\\ge 0",
          "1-\\lambda\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "0≤1−λ 兩邊乘以 λ≥0：λ·0≤λ(1−λ)，而 λ·0=0。所以 λ(1−λ)≥0。",
          "en": "Multiply 0≤1−λ by λ≥0: λ·0≤λ(1−λ), and λ·0=0. So λ(1−λ)≥0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mono_nonneg",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "需要 λ∈[0,1]；λ>1 或 λ<0 時 λ(1−λ)<0，這正是定義限制 λ 範圍的原因。",
            "en": "It needs λ∈[0,1]; for λ>1 or λ<0 one has λ(1−λ)<0, which is why the definition restricts λ."
          }
        },
        "leaf_basis": "mono_nonneg + mul_zero"
      },
      {
        "id": "S07",
        "uses": [
          "S05",
          "S06",
          "square_nonneg",
          "mono_nonneg"
        ],
        "before_tex": "2>0,\\ \\lambda(1-\\lambda)\\ge 0,\\ (x-y)^2\\ge 0",
        "after_tex": "G_f(x,y,\\lambda)=2\\lambda(1-\\lambda)(x-y)^2\\ge 0",
        "relation": "implies",
        "rule": "mono_nonneg",
        "conditions": [],
        "explanation": {
          "zh-Hant": "實數平方非負：(x−y)²≥0。0≤(x−y)² 乘以 λ(1−λ)≥0 得 0≤λ(1−λ)(x−y)²；再乘以 2>0 得 0≤2λ(1−λ)(x−y)²。依 S05，G_f≥0。",
          "en": "Squares are nonnegative: (x−y)²≥0. Multiplying 0≤(x−y)² by λ(1−λ)≥0 gives 0≤λ(1−λ)(x−y)²; multiplying by 2>0 gives 0≤2λ(1−λ)(x−y)². By S05, G_f≥0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S06",
            "square_nonneg",
            "mono_nonneg"
          ],
          "explanation": {
            "zh-Hant": "關鍵是 a=2≥0；若 a<0，同一論證給出 gap≤0（見 S14）。不宣稱反向。",
            "en": "The key is a=2≥0; if a<0, the same argument gives gap≤0 (see S14). No converse is claimed."
          }
        },
        "leaf_basis": "square_nonneg + mono_nonneg + mul_zero"
      },
      {
        "id": "S08",
        "uses": [
          "S01",
          "S02",
          "S03",
          "S07",
          "forall_intro"
        ],
        "before_tex": "G_f(x,y,\\lambda)\\ge 0\\ \\text{for arbitrary } x,y\\in I,\\ \\lambda\\in[0,1]",
        "after_tex": "f\\ \\text{convex on}\\ I",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x、y、λ 是任取的，S03 至 S07 沒有對它們加額外假設，依 ∀ 引入，Jensen 不等式對全部 x,y∈I、λ∈[0,1] 成立；加上 S02 的 I 凸，依 S01，f 在 I 上是凸函數。這是覆蓋全部情況的正證書（MKO a2）。",
          "en": "x, y, λ were arbitrary and S03 to S07 put no extra assumption on them, so by ∀-introduction the Jensen inequality holds for all x,y∈I and λ∈[0,1]; together with the convexity of I from S02 and by S01, f is convex on I. This is a positive certificate covering every case (MKO assumption a2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S02",
            "S03",
            "S07",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要兩部分（定義域凸與 Jensen）都已證，以及 x、y、λ 的任意性。",
            "en": "It needs both parts (convex domain and Jensen) proved, and the arbitrariness of x, y, λ."
          }
        },
        "leaf_basis": "forall_intro + and_intro_elim"
      },
      {
        "id": "S09",
        "uses": [
          "S05",
          "G1"
        ],
        "before_tex": "x=-1,\\ y=4,\\ \\lambda=\\tfrac15",
        "after_tex": "z=3,\\ f(3)=10\\le 18=\\tfrac15 f(-1)+\\tfrac45 f(4),\\ G_f=8=2\\cdot\\tfrac15\\cdot\\tfrac45\\cdot 25",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "一個具體實例作示範：z=(1/5)(−1)+(4/5)·4=3；f(3)=18−9+1=10；f(−1)=6、f(4)=21，(1/5)·6+(4/5)·21=18。gap 18−10=8，與 S05 的公式 2·(1/5)·(4/5)·(−5)²=8 一致。這只是例示，不是證明；證明是 S02 至 S08。",
          "en": "One concrete instance as an illustration: z=(1/5)(−1)+(4/5)·4=3; f(3)=18−9+1=10; f(−1)=6 and f(4)=21, so (1/5)·6+(4/5)·21=18. The gap 18−10=8 agrees with the formula of S05, 2·(1/5)·(4/5)·(−5)²=8. This is an illustration only, not a proof; the proof is S02 to S08."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確的有理數計算；只說明這一組 x、y、λ，不推出任何全稱結論。",
            "en": "An exact rational computation; it concerns this one choice of x, y, λ and implies no universal conclusion."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S04",
          "S02",
          "G2",
          "forall_intro"
        ],
        "before_tex": "G_\\ell(x,y,\\lambda)\\ \\text{for}\\ \\ell(x)=3x-5",
        "after_tex": "G_\\ell=0\\cdot\\lambda(1-\\lambda)(x-y)^2=0;\\ \\ell\\ \\text{convex on}\\ I\\ \\text{with equality}",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "ℓ 是 a=0 的情形，依 L_gap，G_ℓ=0 對所有 x、y、λ 成立，即 ℓ(z)=λℓ(x)+(1−λ)ℓ(y)，Jensen 取等號（MKO para0）。定義域 I 已在 S02 證明是凸的，所以 ℓ 在 I 上是凸函數（而且反向不等式也成立）。",
          "en": "ℓ is the case a=0, so by L_gap G_ℓ=0 for all x, y, λ, that is ℓ(z)=λℓ(x)+(1−λ)ℓ(y): Jensen holds with equality (MKO para0). The domain I was shown convex in S02, so ℓ is convex on I (and the reverse inequality holds too)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S02",
            "forall_elim",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要 L_gap 與定義域凸。等號成立並不表示仿射以外的函數也取等號。",
            "en": "It needs L_gap and the convex domain. Equality here does not mean non-affine functions also give equality."
          }
        },
        "leaf_basis": "forall_elim + mul_zero + forall_intro"
      },
      {
        "id": "S11",
        "uses": [
          "G3",
          "mko_cvxf",
          "L_interval",
          "quantifier_negation"
        ],
        "before_tex": "\\neg(g\\ \\text{convex on}\\ J)",
        "after_tex": "\\exists x,y\\in J\\ \\exists\\lambda\\in[0,1]:\\ g(\\lambda x+(1-\\lambda)y)>\\lambda g(x)+(1-\\lambda)g(y)",
        "relation": "equivalent",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "J=[0,2] 依 L_interval 是凸的，所以定義的第一部分成立；g 不凸等價於 Jensen 部分的否定。¬∀ 等價於 ∃¬，而 ¬(u≤v) 等價於 u>v。所以只要一組 x、y、λ 使 g(z) 嚴格大於右邊。",
          "en": "J=[0,2] is convex by L_interval, so the first part of the definition holds; g is not convex exactly when the Jensen part fails. ¬∀ is equivalent to ∃¬, and ¬(u≤v) is equivalent to u>v. So one choice of x, y, λ with g(z) strictly greater than the right side is enough."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "quantifier_negation",
            "order_basics",
            "L_interval"
          ],
          "explanation": {
            "zh-Hant": "量詞否定與序的三分律都是雙向的；J 凸是此等價的前提。",
            "en": "Quantifier negation and trichotomy of order are both two-way; the convexity of J is a premise of this equivalence."
          }
        },
        "leaf_basis": "quantifier_negation + order_basics + L_interval"
      },
      {
        "id": "S12",
        "uses": [
          "G3"
        ],
        "before_tex": "x=0,\\ y=2,\\ \\lambda=\\tfrac12",
        "after_tex": "z=1\\in J,\\ g(1)=0,\\ \\tfrac12 g(0)+\\tfrac12 g(2)=\\tfrac12\\cdot 0+\\tfrac12\\cdot(-2)=-1",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取 J 的兩端點 x=0、y=2 與 λ=1/2∈[0,1]：z=1∈J。精確求值：g(1)=−1+1=0，g(0)=0，g(2)=−4+2=−2，右邊 (1/2)·0+(1/2)·(−2)=−1。",
          "en": "Take the endpoints x=0, y=2 of J and λ=1/2∈[0,1]: z=1∈J. Exact evaluation: g(1)=−1+1=0, g(0)=0, g(2)=−4+2=−2, and the right side is (1/2)·0+(1/2)·(−2)=−1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確計算；x、y 確實在 J 中，λ 確實在 [0,1] 中。",
            "en": "Exact computation; x and y do lie in J and λ in [0,1]."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S11",
          "S12",
          "exists_intro"
        ],
        "before_tex": "g(1)=0,\\ \\tfrac12 g(0)+\\tfrac12 g(2)=-1",
        "after_tex": "0>-1;\\ g\\ \\text{is not convex on}\\ J",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0>−1，所以這組 x、y、λ 違反 Jensen 不等式。依 ∃ 引入，S11 的存在命題成立，g 在 J 上不是凸函數。一個精確違反就足夠（MKO a2）。",
          "en": "0>−1, so this choice of x, y, λ violates the Jensen inequality. By ∃-introduction the existential statement of S11 holds, and g is not convex on J. One exact violation suffices (MKO assumption a2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "S12",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "需要見證在範圍內且不等式嚴格反向，兩者都已核對。",
            "en": "It needs the witness to lie in range and the inequality to be strictly reversed; both were checked."
          }
        },
        "leaf_basis": "exists_intro + order_basics"
      },
      {
        "id": "S14",
        "uses": [
          "S04",
          "S06",
          "S12",
          "square_nonneg"
        ],
        "before_tex": "G_g(x,y,\\lambda)",
        "after_tex": "G_g=-\\lambda(1-\\lambda)(x-y)^2;\\ G_g(0,2,\\tfrac12)=-1",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "交叉核對：g 的 a=−1，依 L_gap，G_g=−λ(1−λ)(x−y)²。在 (0,2,1/2)：−(1/2)(1/2)·4=−1，正好等於 S12 的右邊減 g(z)：−1−0=−1。更一般地，只要 x≠y 且 0<λ<1，G_g<0：每一個非平凡的 Jensen 實例都失敗。",
          "en": "Cross-check: g has a=−1, so by L_gap G_g=−λ(1−λ)(x−y)². At (0,2,1/2): −(1/2)(1/2)·4=−1, exactly the right side of S12 minus g(z): −1−0=−1. More generally, whenever x≠y and 0<λ<1, G_g<0: every nontrivial Jensen instance fails."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S12",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "這是與 S13 獨立的核對；S13 的結論不依賴本步。「x≠y 且 0<λ<1 時 G_g<0」用到 λ(1−λ)>0 與 (x−y)²>0，兩者在這些條件下成立。",
            "en": "This is a check independent of S13; S13's conclusion does not depend on it. 'G_g<0 when x≠y and 0<λ<1' uses λ(1−λ)>0 and (x−y)²>0, both of which hold under these conditions."
          }
        },
        "leaf_basis": "forall_elim + square_nonneg + order_basics"
      },
      {
        "id": "S15",
        "uses": [
          "S13",
          "S14"
        ],
        "before_tex": "\\lambda\\in\\{0,1\\}\\ \\text{or}\\ x=y",
        "after_tex": "G_g=0\\ \\text{(samples pass)},\\ \\text{yet } g\\ \\text{not convex}",
        "relation": "implies",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "當 λ=0、λ=1 或 x=y 時，λ(1−λ)(x−y)²=0，所以 g 在這些樣本上 Jensen 都以等號通過。但 S13 已證 g 不凸。因此「若干 Jensen 樣本通過」不能推出凸性；正面結論必須像 S02 至 S08 那樣覆蓋全部 x、y、λ。",
          "en": "When λ=0, λ=1 or x=y, λ(1−λ)(x−y)²=0, so g passes Jensen with equality on these samples. Yet S13 proved g is not convex. Hence 'some Jensen samples pass' does not imply convexity; a positive conclusion must cover every x, y, λ, as S02 to S08 do."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "S14",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "只用乘以零與 S13 的反例；結論是「通過的樣本不足以證凸」，由 g 這個具體反例支持。",
            "en": "It uses only multiplication by zero and the counterexample of S13; the conclusion, that passing samples are not enough to prove convexity, is supported by the concrete counterexample g."
          }
        },
        "leaf_basis": "mul_zero + square_nonneg"
      }
    ],
    "result": {
      "zh-Hant": "I=[−1,4] 是凸集合，且對所有 x,y∈I、λ∈[0,1]，f(x)=2x²−3x+1 的 Jensen gap 等於 2λ(1−λ)(x−y)²≥0，所以 f 在 I 上是凸函數。ℓ(x)=3x−5 的 gap 恆為 0，Jensen 取等號。g(x)=−x²+x 在 J=[0,2] 上不凸：x=0、y=2、λ=1/2 時 g(1)=0>−1=(1/2)g(0)+(1/2)g(2)。",
      "en": "I=[−1,4] is convex, and for all x,y∈I and λ∈[0,1] the Jensen gap of f(x)=2x²−3x+1 equals 2λ(1−λ)(x−y)²≥0, so f is convex on I. The gap of ℓ(x)=3x−5 is identically 0, so Jensen holds with equality. g(x)=−x²+x is not convex on J=[0,2]: with x=0, y=2, λ=1/2, g(1)=0>−1=(1/2)g(0)+(1/2)g(2)."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "g 在 λ=0、λ=1 與 x=y 的樣本上都滿足 Jensen，所以 g 在 [0,2] 上凸",
          "en": "g satisfies Jensen on the samples with λ=0, λ=1 and x=y, so g is convex on [0,2]"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x=0、y=2、λ=1/2：g(1)=0>−1。",
          "en": "x=0, y=2, λ=1/2: g(1)=0>−1."
        }
      },
      {
        "claim": {
          "zh-Hant": "f 在 x=−1、y=4、λ=1/5 的 Jensen 實例通過，這就證明 f 在 I 上凸",
          "en": "The Jensen instance of f at x=−1, y=4, λ=1/5 passes, which proves that f is convex on I"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "單一通過實例不能證凸（MKO a2）；結論成立是因為 S04 至 S08 覆蓋全部 x、y、λ。",
          "en": "One passing instance cannot prove convexity (MKO assumption a2); the conclusion holds because S04 to S08 cover every x, y, λ."
        }
      },
      {
        "claim": {
          "zh-Hant": "x² 在 D=[0,1]∪[2,3] 上是 MKO 意義下的凸函數",
          "en": "x² on D=[0,1]∪[2,3] is a convex function in the MKO's sense"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "D 不是凸集合：1、2∈D 但 (1/2)·1+(1/2)·2=3/2∉D；定義要求凸定義域。",
          "en": "D is not convex: 1, 2∈D but (1/2)·1+(1/2)·2=3/2∉D; the definition requires a convex domain."
        }
      },
      {
        "claim": {
          "zh-Hant": "每個二次函數都是凸函數",
          "en": "Every quadratic function is convex"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "g(x)=−x²+x 的 a=−1，在 x=0、y=2、λ=1/2 違反 Jensen。",
          "en": "g(x)=−x²+x has a=−1 and violates Jensen at x=0, y=2, λ=1/2."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "square_nonneg",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。",
          "en": "Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=0."
        }
      },
      {
        "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": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "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": "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": "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": "def_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。",
          "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable."
        }
      },
      {
        "id": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "id": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "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": "forall_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。",
          "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds."
        }
      },
      {
        "id": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "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": "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": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "id": "mko_cvxf",
        "kind": "cited",
        "source": {
          "mko_id": "mko-convex-function",
          "sha256": "d5af460bbe41c8e967949f9398bde368d76471cd2b4a24e4f49f14ad90a05fed",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在凸定義域 C 上，f 為凸若對所有 x,y∈C 及 λ∈[0,1]，f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y)。單一 Jensen 違反足以否證；單一通過實例不能證凸，正證書需覆蓋全部 x、y、λ。",
          "en": "On a convex domain C, f is convex if f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y) for every x,y∈C and every λ∈[0,1]. One Jensen violation suffices to refute convexity, but one passing instance cannot prove it; positive certificates must cover every x, y and λ."
        }
      },
      {
        "id": "mko_cvxset",
        "kind": "cited",
        "source": {
          "mko_id": "mko-convex-set",
          "sha256": "0d52be71dca290f856018d176bbb26dcd877c887031f690bd1d656db017ce02d",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "實向量空間的子集 C 為凸，若對所有 x,y∈C 及所有 λ∈[0,1]，λx+(1−λ)y 仍屬於 C。",
          "en": "A subset C of a real vector space is convex if λx+(1−λ)y belongs to C for every x,y∈C and every λ∈[0,1]."
        }
      },
      {
        "id": "mono_nonneg",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以非負數保持不等號：若 u≤v 且 w≥0，則 wu≤wv（w>0 時由實數的序，w=0 時兩邊都是 0）。",
          "en": "Multiplying by a nonnegative number keeps the inequality: if u≤v and w≥0, then wu≤wv (for w>0 by the order of the reals; for w=0 both sides are 0)."
        }
      },
      {
        "id": "ineq_add",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "不等式相加：若 u₁≤v₁ 且 u₂≤v₂，則 u₁+u₂≤v₁+v₂。",
          "en": "Adding inequalities: if u₁≤v₁ and u₂≤v₂, then u₁+u₂≤v₁+v₂."
        }
      },
      {
        "id": "L_interval",
        "kind": "lemma",
        "proved_in": [
          "S02.1",
          "S02.2"
        ],
        "statement": {
          "zh-Hant": "若 α≤s≤β、α≤t≤β 且 λ∈[0,1]，則 α≤λs+(1−λ)t≤β；因此每個閉區間 [α,β] 都是凸集合。",
          "en": "If α≤s≤β, α≤t≤β and λ∈[0,1], then α≤λs+(1−λ)t≤β; hence every closed interval [α,β] is convex."
        }
      },
      {
        "id": "L_gap",
        "kind": "lemma",
        "proved_in": [
          "S04.1",
          "S04.2",
          "S04.3"
        ],
        "statement": {
          "zh-Hant": "對 q(x)=ax²+bx+c 與任意實數 x、y、λ：λq(x)+(1−λ)q(y)−q(λx+(1−λ)y)=aλ(1−λ)(x−y)²。",
          "en": "For q(x)=ax²+bx+c and any reals x, y, λ: λq(x)+(1−λ)q(y)−q(λx+(1−λ)y)=aλ(1−λ)(x−y)²."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S04",
        "vars": [
          "a",
          "b",
          "c",
          "l",
          "x",
          "y"
        ],
        "degree": 2,
        "lhs": "l*(a*x**2+b*x+c)+(1-l)*(a*y**2+b*y+c)-(a*(l*x+(1-l)*y)**2+b*(l*x+(1-l)*y)+c)",
        "rhs": "a*l*(1-l)*(x-y)**2"
      },
      {
        "id": "N2",
        "step": "S04.1",
        "vars": [
          "b",
          "c",
          "l",
          "x",
          "y"
        ],
        "degree": 2,
        "lhs": "l*(b*x+c)+(1-l)*(b*y+c)-(b*(l*x+(1-l)*y)+c)",
        "rhs": "0"
      },
      {
        "id": "N3",
        "step": "S04.2",
        "vars": [
          "l",
          "x",
          "y"
        ],
        "degree": 2,
        "lhs": "(l*x+(1-l)*y)**2",
        "rhs": "l**2*x**2+2*l*(1-l)*x*y+(1-l)**2*y**2"
      },
      {
        "id": "N4",
        "step": "S04.3",
        "vars": [
          "l",
          "x",
          "y"
        ],
        "degree": 2,
        "lhs": "l*x**2+(1-l)*y**2-(l*x+(1-l)*y)**2",
        "rhs": "l*(1-l)*(x-y)**2"
      },
      {
        "id": "N5",
        "step": "S02.1",
        "vars": [
          "l"
        ],
        "degree": 1,
        "lhs": "-1*l+(-1)*(1-l)",
        "rhs": "-1"
      },
      {
        "id": "N6",
        "step": "S02.2",
        "vars": [
          "l"
        ],
        "degree": 1,
        "lhs": "4*l+4*(1-l)",
        "rhs": "4"
      },
      {
        "id": "N7",
        "step": "S09",
        "lhs": "(1/5)*(-1)+(4/5)*4",
        "rhs": "3"
      },
      {
        "id": "N8",
        "step": "S09",
        "lhs": "2*3**2-3*3+1",
        "rhs": "10"
      },
      {
        "id": "N9",
        "step": "S09",
        "lhs": "(1/5)*(2*(-1)**2-3*(-1)+1)+(4/5)*(2*4**2-3*4+1)",
        "rhs": "18"
      },
      {
        "id": "N10",
        "step": "S09",
        "lhs": "18-10",
        "rhs": "2*(1/5)*(4/5)*(-1-4)**2"
      },
      {
        "id": "N11",
        "step": "S10",
        "vars": [
          "l",
          "x",
          "y"
        ],
        "degree": 2,
        "lhs": "l*(3*x-5)+(1-l)*(3*y-5)-(3*(l*x+(1-l)*y)-5)",
        "rhs": "0"
      },
      {
        "id": "N12",
        "step": "S12",
        "lhs": "(1/2)*0+(1/2)*2",
        "rhs": "1"
      },
      {
        "id": "N13",
        "step": "S12",
        "lhs": "-(1)**2+1",
        "rhs": "0"
      },
      {
        "id": "N14",
        "step": "S12",
        "lhs": "(1/2)*(-(0)**2+0)+(1/2)*(-(2)**2+2)",
        "rhs": "-1"
      },
      {
        "id": "N15",
        "step": "S13",
        "lhs": "-(1)**2+1",
        "relation": ">",
        "rhs": "(1/2)*(-(0)**2+0)+(1/2)*(-(2)**2+2)"
      },
      {
        "id": "N16",
        "step": "S14",
        "lhs": "-1*(1/2)*(1/2)*(0-2)**2",
        "rhs": "-1"
      },
      {
        "id": "N17",
        "lhs": "(1/2)*1+(1/2)*2",
        "rhs": "3/2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "停在 MKO 的切片內：明示的非空實區間作定義域，一元仿射與二次函數，精確係數。正面結論用覆蓋全部 x、y、λ 的代數證書；反面結論用單一精確違反。不處理嚴格凸、次梯度、二階導數判別或高維凸函數；不改動 canonical MKO 或英文 edition。",
      "en": "Stays inside the MKO's slice: explicitly declared nonempty real intervals as domains, univariate affine and quadratic functions, exact coefficients. Positive conclusions use algebraic certificates covering every x, y, λ; negative ones use a single exact violation. Strict convexity, subgradients, second-derivative tests and multivariate convex functions are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "I=[-1,4],\\ J=[0,2]",
        "明示的定義域區間",
        "the declared domain intervals"
      ],
      [
        "\\lambda\\in[0,1]",
        "Jensen 組合的權重",
        "the weight of the Jensen combination"
      ],
      [
        "z=\\lambda x+(1-\\lambda)y",
        "x 與 y 的凸組合",
        "the convex combination of x and y"
      ],
      [
        "G_q(x,y,\\lambda)",
        "函數 q 的 Jensen gap：λq(x)+(1−λ)q(y)−q(z)",
        "the Jensen gap of q: λq(x)+(1−λ)q(y)−q(z)"
      ],
      [
        "a,\\ b,\\ c",
        "二次函數 ax²+bx+c 的係數",
        "coefficients of the quadratic ax²+bx+c"
      ]
    ],
    "md_error_controls": [
      {
        "title": "幾個 Jensen 樣本通過不證凸 / A few passing Jensen samples do not prove convexity",
        "zh": "MKO 指出的誤解是由幾個 Jensen 樣本通過就宣稱全域凸。g 在 λ=0、λ=1 與 x=y 的所有樣本上都通過（S15），卻不是凸函數。S09 對 f 的實例同樣只是例示；f 的凸性來自 S04 至 S08 對全部 x、y、λ 的恆等式與符號論證。",
        "en": "The mistake named by the MKO is to declare global convexity because a few Jensen samples pass. g passes on every sample with λ=0, λ=1 or x=y (S15), yet it is not convex. The instance S09 for f is likewise only an illustration; the convexity of f comes from the identity and sign argument of S04 to S08 covering all x, y, λ."
      },
      {
        "title": "定義域必須凸 / The domain must be convex",
        "zh": "MKO 的另一個誤解是在非凸定義域上談一般凸函數。若定義域是 D=[0,1]∪[2,3]，x=1、y=2、λ=1/2 給出 3/2∉D，函數在 3/2 沒有值，Jensen 不等式根本無法陳述。所以 S02 先證 I 是凸的，S11 也先確認 J 是凸的。",
        "en": "Another misconception identified in the MKO is to speak of a convex function in the general sense on a non-convex domain. If the domain is D=[0,1]∪[2,3], then x=1, y=2, λ=1/2 gives 3/2∉D, where the function has no value, so the Jensen inequality cannot even be stated. That is why S02 first proves that I is convex, and S11 first confirms that J is."
      },
      {
        "title": "a 的正負決定 gap 的正負 / The sign of a decides the sign of the gap",
        "zh": "依 L_gap，二次函數的 Jensen gap 是 aλ(1−λ)(x−y)²。a=2 時 gap≥0（凸），a=0 時 gap=0（仿射，取等號），a=−1 時 gap≤0，而且 x≠y、0<λ<1 時嚴格小於 0（不凸）。「二次函數都是凸的」是錯的。",
        "en": "By L_gap, the Jensen gap of a quadratic is aλ(1−λ)(x−y)². For a=2 the gap is ≥0 (convex), for a=0 it is 0 (affine, equality), and for a=−1 it is ≤0, strictly negative when x≠y and 0<λ<1 (not convex). 'Every quadratic is convex' is false."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "mono_nonneg 與 ineq_add 是本稿的前提，不在本稿中由實數公理推出（mono_nonneg 在凸集合教學稿中以引理證明）。",
        "en": "mono_nonneg and ineq_add are premises of this guide and are not derived here from the axioms of the reals (mono_nonneg is proved as a lemma in the convex-set guide)."
      },
      {
        "zh-Hant": "N7 至 N10 只核 S09 的一個實例；它們是例示，證明不依賴它們。",
        "en": "N7 to N10 check only the one instance of S09; they are illustrations and the proof does not depend on them."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "3f396010579d87300f52e4a41245fcd8d02d372b3712f863225c958c4632db0c",
    "original_source_sha256": "3f396010579d87300f52e4a41245fcd8d02d372b3712f863225c958c4632db0c",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}