DC

Differential Calculus

4 theorems

02 — Differential Calculus

Comprehensive study guide for Bocconi Math Module 2 (cod. 30063), General Exam. Source materials: Alice Sicconi's Proofs - Differential Calculus.pdf, Linear Algebra and Differential Calculus.pdf (diff-calc half), lect1_calc-diff.pdf, lect2_calc_diff (2).pdf, Lect3_calc_diff (3).pdf, lect 4_calc_diff (3).pdf, lect5_calc_diff.pdf, TA41 _calc_diff (2).pdf, General_24524_ENG_SOL.pdf, Final exercises solution.pdf.

§1. Overview & Exam Relevance

Differential Calculus is the second block of the first partial and accounts for approximately 27% of the general exam (typically 1 MCQ worth 5 pts plus one open-ended question worth up to 20 pts — about 25–30 pts of the 150-pt exam). On open-ended problems it is tied with Linear Algebra as the richest source of statement-plus-proof questions.

Topic scope. The exam tests:

  • multivariable functions f:Rn→Rf : \mathbb{R}^n \to \mathbb{R} and their regularity classes C(1),C(2)C^{(1)}, C^{(2)};
  • partial derivatives ∂f/∂xi\partial f / \partial x_i, gradient ∇f\nabla f, directional derivatives, and the Hessian ∇2f\nabla^2 f (equivalently HfH_f);
  • Schwarz's theorem: symmetry of the Hessian for C(2)C^{(2)} functions;
  • Second-order Taylor formula with Peano remainder at a point x0∈Rn\mathbf{x}_0 \in \mathbb{R}^n;
  • classification of stationary points via the sign of the Hessian quadratic form hT∇2f(x0)h\mathbf{h}^T \nabla^2 f(\mathbf{x}_0) \mathbf{h} (Theorem 1391);
  • convex/concave functions on open convex sets, tangent-hyperplane inequality, and the Hessian-based characterisation (Theorem/Prop. 1486);
  • necessary and sufficient conditions for global extrema of convex/concave functions (Theorems 1497, 1591);
  • existence theorems: Weierstrass (on compact sets), Tonelli (on coercive/super-coercive functions);
  • Dini's theorem (implicit function theorem): local uniqueness and formula for f′(x0)f'(x_0) from g(x,y)=cg(x,y) = c.

Typical MCQ patterns (from May 2024 General exam).

  • MCQ2 (Mode A) / MCQ1 (Mode B): "The Taylor formula of order two for f(x,y)=x2+xcos⁡yf(x,y) = x^2 + x \cos y centred at (1,−π/2)(1, -\pi/2) is …" — compute f,∇f,∇2ff, \nabla f, \nabla^2 f at the point and assemble.
  • Other recurring MCQs (from TA and lecture pools): local-extrema classification of f(x,y)=(y−1)(y−x2)f(x,y) = (y-1)(y-x^2); Dini-applicability at boundary points of g(x,y)=0g(x,y)=0; coercivity check.

Typical open-ended pattern (May 2024 Q8 Mode A / Q7 Mode B).

  • Part (a) — State the concavity / strict-concavity test (Prop. 1486) based on the sign of the Hessian matrix.
  • Part (b) — Given f(x,y)=ex(x−1)y+y2f(x,y) = e^x(x-1)y + y^2: (b1) write ∇f\nabla f and determine stationary points → P=(0,1/2)P = (0, 1/2) local min and Q=(1,0)Q = (1, 0) saddle; (b2) write ∇2f\nabla^2 f and classify each stationary point using Theorem 1391; (b3) decide whether ff is convex on the whole R2\mathbb{R}^2.

Why this topic is high-leverage.

  • Every theorem here uses Linear Algebra from §01: classifying a stationary point is classifying a quadratic form attached to the Hessian. Spectral/Sylvester–Jacobi carry over verbatim.
  • The Taylor formula is the engine behind every result in this chapter — the proof of Theorem 1391 literally starts by invoking Theorem 1481 (Taylor).
  • Together with Linear Algebra (≈27%), Differential Calculus pushes the first-partial content to ≈55% of the whole general exam.

§2. Definitions

2.1 Multivariable function f:Rn→Rf : \mathbb{R}^n \to \mathbb{R}

A multivariable (scalar) function of nn real variables is a map f:A⊆Rn→Rf : A \subseteq \mathbb{R}^n \to \mathbb{R}, assigning to each point x=(x1,…,xn)∈A\mathbf{x} = (x_1, \ldots, x_n) \in A exactly one real number f(x)f(\mathbf{x}). The set AA is the domain; its image f(A)⊆Rf(A) \subseteq \mathbb{R} is the range.

Example. f(x,y)=x+3y2−ln⁡(x+2y)f(x,y) = x + 3y^2 - \ln(x + 2y) has domain {(x,y)∈R2:x+2y>0}\{(x,y) \in \mathbb{R}^2 : x + 2y > 0\} because ln⁡\ln requires a strictly positive argument (Final exercises solution.pdf Q2).

2.2 Open ball and open set

Given x0∈Rn\mathbf{x}_0 \in \mathbb{R}^n and r>0r > 0, the open ball (or neighbourhood) of centre x0\mathbf{x}_0 and radius rr is Ur(x0)=Br(x0)={x∈Rn:∥x−x0∥<r}.U_r(\mathbf{x}_0) = B_r(\mathbf{x}_0) = \{\mathbf{x} \in \mathbb{R}^n : \|\mathbf{x} - \mathbf{x}_0\| < r\}. A set A⊆RnA \subseteq \mathbb{R}^n is open if for every x0∈A\mathbf{x}_0 \in A there is r>0r > 0 with Ur(x0)⊆AU_r(\mathbf{x}_0) \subseteq A. A point x0∈A\mathbf{x}_0 \in A is an interior point of AA if such an rr exists; otherwise it is a boundary point.

2.3 Limit and continuity

lim⁡x→x0f(x)=L\lim_{\mathbf{x} \to \mathbf{x}_0} f(\mathbf{x}) = L means: ∀ε>0 ∃δ>0\forall \varepsilon > 0 \ \exists \delta > 0 such that 0<∥x−x0∥<δ⇒∣f(x)−L∣<ε0 < \|\mathbf{x} - \mathbf{x}_0\| < \delta \Rightarrow |f(\mathbf{x}) - L| < \varepsilon.

ff is continuous at x0\mathbf{x}_0 if lim⁡x→x0f(x)=f(x0)\lim_{\mathbf{x} \to \mathbf{x}_0} f(\mathbf{x}) = f(\mathbf{x}_0); it is continuous on AA if it is continuous at every point of AA.

2.4 Partial derivative and gradient

The ii-th partial derivative of ff at x0\mathbf{x}_0 is ∂f∂xi(x0)  =  fxi′(x0)  =  lim⁡h→0f(x0+hei)−f(x0)h,\frac{\partial f}{\partial x_i}(\mathbf{x}_0) \;=\; f'_{x_i}(\mathbf{x}_0) \;=\; \lim_{h \to 0} \frac{f(\mathbf{x}_0 + h \mathbf{e}_i) - f(\mathbf{x}_0)}{h}, where ei\mathbf{e}_i is the ii-th coordinate vector. Operationally, ∂f/∂xi\partial f / \partial x_i is obtained by differentiating ff with respect to xix_i while treating all other variables as constants.

The gradient ∇f(x0)\nabla f(\mathbf{x}_0) is the row vector of all partial derivatives: ∇f(x0)  =  [fx1′(x0)  fx2′(x0)  ⋯  fxn′(x0)].\nabla f(\mathbf{x}_0) \;=\; \bigl[f'_{x_1}(\mathbf{x}_0) \; f'_{x_2}(\mathbf{x}_0) \; \cdots \; f'_{x_n}(\mathbf{x}_0)\bigr]. When written as a column, we use ∇f(x0)T\nabla f(\mathbf{x}_0)^T.

Example (lect2). For f(x,y)=3x2ey2f(x,y) = 3 x^2 e^{y^2} we have fx′=6xey2f'_x = 6 x e^{y^2}, fy′=3x2ey2⋅2y=6x2yey2f'_y = 3x^2 e^{y^2} \cdot 2y = 6 x^2 y e^{y^2}, so ∇f(x,y)=[6xey2, 6x2yey2]\nabla f(x,y) = [6 x e^{y^2}, \ 6 x^2 y e^{y^2}] and ∇f(1,1)=[6e, 6e]\nabla f(1,1) = [6e, \ 6e].

2.5 Directional derivative

Given a unit vector v∈Rn\mathbf{v} \in \mathbb{R}^n (∥v∥=1\|\mathbf{v}\| = 1), the directional derivative of ff at x0\mathbf{x}_0 in the direction v\mathbf{v} is Dvf(x0)  =  lim⁡h→0f(x0+hv)−f(x0)h.D_{\mathbf{v}} f(\mathbf{x}_0) \;=\; \lim_{h \to 0} \frac{f(\mathbf{x}_0 + h \mathbf{v}) - f(\mathbf{x}_0)}{h}. If ff is differentiable at x0\mathbf{x}_0, then Dvf(x0)=∇f(x0)⋅vD_{\mathbf{v}} f(\mathbf{x}_0) = \nabla f(\mathbf{x}_0) \cdot \mathbf{v}.

2.6 Differentiability and C(1),C(2)C^{(1)}, C^{(2)} classes

ff is differentiable at x0\mathbf{x}_0 if there exists a row vector L∈Rn\mathbf{L} \in \mathbb{R}^n such that f(x0+h)=f(x0)+L⋅h+o(∥h∥)as h→0.f(\mathbf{x}_0 + \mathbf{h}) = f(\mathbf{x}_0) + \mathbf{L} \cdot \mathbf{h} + o(\|\mathbf{h}\|) \qquad \text{as } \mathbf{h} \to \mathbf{0}. If such L\mathbf{L} exists, it is necessarily L=∇f(x0)\mathbf{L} = \nabla f(\mathbf{x}_0).

  • C(1)(A)C^{(1)}(A): ff has continuous first partial derivatives on AA. If f∈C(1)f \in C^{(1)} then ff is differentiable.
  • C(2)(A)C^{(2)}(A): ff has continuous second-order partial derivatives on AA. This is the hypothesis of Schwarz's theorem and Theorem 1481 (Taylor).

2.7 Hessian ∇2f=Hf\nabla^2 f = H_f

The Hessian matrix of ff at x0\mathbf{x}_0 is the n×nn \times n matrix of second partial derivatives: ∇2f(x0)  =  Hf(x0)  =  (fx1x1′′(x0)fx1x2′′(x0)⋯fx1xn′′(x0)fx2x1′′(x0)fx2x2′′(x0)⋯fx2xn′′(x0)⋮⋮⋱⋮fxnx1′′(x0)fxnx2′′(x0)⋯fxnxn′′(x0)).\nabla^2 f(\mathbf{x}_0) \;=\; H_f(\mathbf{x}_0) \;=\; \begin{pmatrix} f''_{x_1 x_1}(\mathbf{x}_0) & f''_{x_1 x_2}(\mathbf{x}_0) & \cdots & f''_{x_1 x_n}(\mathbf{x}_0) \\ f''_{x_2 x_1}(\mathbf{x}_0) & f''_{x_2 x_2}(\mathbf{x}_0) & \cdots & f''_{x_2 x_n}(\mathbf{x}_0) \\ \vdots & \vdots & \ddots & \vdots \\ f''_{x_n x_1}(\mathbf{x}_0) & f''_{x_n x_2}(\mathbf{x}_0) & \cdots & f''_{x_n x_n}(\mathbf{x}_0) \end{pmatrix}.

Schwarz's theorem (stated in lect1): if f∈C(2)(A)f \in C^{(2)}(A) with AA open, then ∂2f/(∂xi∂xj)=∂2f/(∂xj∂xi)\partial^2 f / (\partial x_i \partial x_j) = \partial^2 f / (\partial x_j \partial x_i) for all i,ji, j and all x∈A\mathbf{x} \in A. Hence ∇2f\nabla^2 f is symmetric, and the classification tools from Linear Algebra (§01) apply directly.

Mnemonic (lect1 sidenote). The gradient is an operator (it sends a point to a vector). The Jacobian matrix contains all partials of a vector-valued operator. Since ∇f\nabla f is itself a vector-valued operator Rn→Rn\mathbb{R}^n \to \mathbb{R}^n, its Jacobian is the Hessian: ∇2f\nabla^2 f = Jacobian of ∇f\nabla f.

2.8 Stationary point

A point x0∈A\mathbf{x}_0 \in A is a stationary point (or critical point) of ff if ∇f(x0)=0\nabla f(\mathbf{x}_0) = \mathbf{0}, i.e., every partial derivative vanishes at x0\mathbf{x}_0.

Remark. Candidates for local extrema of a differentiable ff on an open set are exactly the stationary points (Fermat's theorem, §3.7). On a non-open set one must also check boundary points and exceptional points where ff is not differentiable (lect1).

2.9 Local and global extrema

Let x0∈A\mathbf{x}_0 \in A. The point x0\mathbf{x}_0 is:

  • a local minimum of ff if there is r>0r > 0 with f(x0)≤f(x)f(\mathbf{x}_0) \le f(\mathbf{x}) for every x∈Ur(x0)∩A\mathbf{x} \in U_r(\mathbf{x}_0) \cap A;
  • a strong (strict) local minimum if f(x0)<f(x)f(\mathbf{x}_0) < f(\mathbf{x}) for every x∈Ur(x0)∩A\mathbf{x} \in U_r(\mathbf{x}_0) \cap A with x≠x0\mathbf{x} \neq \mathbf{x}_0;
  • a global (resp. strong global) minimum if the corresponding inequality holds for every x∈A\mathbf{x} \in A.

Maxima are defined by reversing the inequalities. A stationary point that is neither a local max nor a local min is a saddle point (e.g., (0,0)(0,0) for f(x,y)=x2−y2f(x,y) = x^2 - y^2).

2.10 Convex set, convex/concave function

A set C⊆RnC \subseteq \mathbb{R}^n is convex if for every x,y∈C\mathbf{x}, \mathbf{y} \in C and every t∈[0,1]t \in [0,1], tx+(1−t)y∈Ct\mathbf{x} + (1-t)\mathbf{y} \in C (the segment joining any two points stays in CC).

Let C⊆RnC \subseteq \mathbb{R}^n be convex. A function f:C→Rf : C \to \mathbb{R} is convex on CC if f(tx+(1−t)y)≤tf(x)+(1−t)f(y)∀ x,y∈C,∀ t∈[0,1].f\bigl(t\mathbf{x} + (1-t)\mathbf{y}\bigr) \le t f(\mathbf{x}) + (1-t) f(\mathbf{y}) \qquad \forall\, \mathbf{x}, \mathbf{y} \in C, \forall\, t \in [0,1]. It is strictly convex if the inequality is strict whenever x≠y\mathbf{x} \neq \mathbf{y} and t∈(0,1)t \in (0,1). ff is (strictly) concave iff −f-f is (strictly) convex.

Tangent-hyperplane characterisation (lect4, for ff differentiable on an open convex set CC):

  • ff is convex on CC   ⟺  \iff f(x)≥f(x0)+∇f(x0)(x−x0)f(\mathbf{x}) \ge f(\mathbf{x}_0) + \nabla f(\mathbf{x}_0)(\mathbf{x} - \mathbf{x}_0) for all x,x0∈C\mathbf{x}, \mathbf{x}_0 \in C (tangent hyperplane sits below the graph);
  • ff is strictly convex on CC   ⟺  \iff the inequality above is strict whenever x≠x0\mathbf{x} \neq \mathbf{x}_0.

The 1-variable analogues (f(x)≥f(x0)+f′(x0)(x−x0)f(x) \ge f(x_0) + f'(x_0)(x - x_0)) and the second-derivative tests (f′′(x)≥0f''(x) \ge 0 for convex, f′′(x)>0f''(x) > 0 for strictly convex) are recovered as special cases n=1n = 1.

2.11 Coercive, super-coercive, level set

A level set of ff at height cc is {x∈A:f(x)=c}\{\mathbf{x} \in A : f(\mathbf{x}) = c\} (or ≤c\le c for the sub-level set).

f:Rn→Rf : \mathbb{R}^n \to \mathbb{R} is coercive on a set AA (Sicconi's convention = super-coercive towards +∞+\infty) if lim⁡∥x∥→∞f(x)=+∞\lim_{\|\mathbf{x}\| \to \infty} f(\mathbf{x}) = +\infty. Equivalently, every sub-level set {f≤c}\{f \le c\} is bounded. If lim⁡∥x∥→∞f(x)=−∞\lim_{\|\mathbf{x}\| \to \infty} f(\mathbf{x}) = -\infty we say ff is super-coercive towards −∞-\infty (the symmetric case used for existence of global maxima).

Composition rule (lect4). If the inner function g:Rn→Rg : \mathbb{R}^n \to \mathbb{R} is continuous and coercive towards −∞-\infty, and the outer function h:R→Rh : \mathbb{R} \to \mathbb{R} is strictly increasing, then h∘gh \circ g is coercive (towards lim⁡t→−∞h(t)\lim_{t \to -\infty} h(t)). This is how e−x2−y2e^{-x^2 - y^2} gets its super-coercivity towards 00 from above (hence −e−x2−y2-e^{-x^2-y^2} is super-coercive towards 00, but f=e−x2−y2f = e^{-x^2-y^2} has supremum equal to 11, making (0,0)(0,0) the global maximiser).

2.12 Implicit function — setup for Dini

Let g:C⊆R2→Rg : C \subseteq \mathbb{R}^2 \to \mathbb{R} and c∈Rc \in \mathbb{R}. Consider the level curve g(x,y)=cg(x,y) = c. We ask: does there exist a function f:I⊆R→Rf : I \subseteq \mathbb{R} \to \mathbb{R} with y=f(x)y = f(x) such that g(x,f(x))=cg(x, f(x)) = c for all x∈Ix \in I? If yes, we say ff is the implicit function defined by g(x,y)=cg(x,y) = c near a specific point. Dini's theorem (§3.6) gives sufficient conditions for local existence, uniqueness, and a formula for f′(x0)f'(x_0).

§3. Theorems, Propositions & Proofs

3.1 Schwarz's theorem — symmetry of the Hessian

Statement. Let A⊆RnA \subseteq \mathbb{R}^n be open and f:A→Rf : A \to \mathbb{R} be C(2)C^{(2)} on AA. Then ∂2f∂xi∂xj(x)  =  ∂2f∂xj∂xi(x)∀i,j, ∀ x∈A.\frac{\partial^2 f}{\partial x_i \partial x_j}(\mathbf{x}) \;=\; \frac{\partial^2 f}{\partial x_j \partial x_i}(\mathbf{x}) \qquad \forall i, j, \ \forall\, \mathbf{x} \in A. In particular ∇2f(x)\nabla^2 f(\mathbf{x}) is a symmetric matrix.

Source: lect1_calc-diff.pdf (stated without proof, as is course convention).

Why it matters. Every theorem that follows in §3 uses the fact that ∇2f(x0)\nabla^2 f(\mathbf{x}_0) is symmetric, so all the sign-classification tools from Linear Algebra (eigenvalue signs, Sylvester–Jacobi leading principal minors) apply to the Hessian.

3.2 Theorem 1481 — Second-order Taylor formula with Peano remainder

Statement. Let A⊆RnA \subseteq \mathbb{R}^n be open, x0∈A\mathbf{x}_0 \in A, and f∈C(2)(A)f \in C^{(2)}(A). Then for every x=x0+h\mathbf{x} = \mathbf{x}_0 + \mathbf{h} in a neighbourhood B(x0)⊆AB(\mathbf{x}_0) \subseteq A,  f(x)  =  f(x0)  +  ∇f(x0)(x−x0)  +  12(x−x0)T∇2f(x0)(x−x0)  +  o(∥x−x0∥2) \boxed{\, f(\mathbf{x}) \;=\; f(\mathbf{x}_0) \;+\; \nabla f(\mathbf{x}_0)(\mathbf{x} - \mathbf{x}_0) \;+\; \tfrac{1}{2}(\mathbf{x} - \mathbf{x}_0)^T \nabla^2 f(\mathbf{x}_0)(\mathbf{x} - \mathbf{x}_0) \;+\; o(\|\mathbf{x} - \mathbf{x}_0\|^2) \,} as x→x0\mathbf{x} \to \mathbf{x}_0.

Source: lect2_calc_diff (2).pdf (statement); proof sketch in Proofs - Differential Calculus.pdf p.3 inside the proof of Theorem 1391.

Structure of the formula.

  • f(x0)f(\mathbf{x}_0) — the zero-order term.
  • ∇f(x0)(x−x0)\nabla f(\mathbf{x}_0)(\mathbf{x} - \mathbf{x}_0) — the linear / first-order term (tangent hyperplane).
  • 12(x−x0)T∇2f(x0)(x−x0)\tfrac{1}{2}(\mathbf{x} - \mathbf{x}_0)^T \nabla^2 f(\mathbf{x}_0)(\mathbf{x} - \mathbf{x}_0) — the quadratic term (a quadratic form in h=x−x0\mathbf{h} = \mathbf{x} - \mathbf{x}_0).
  • o(∥x−x0∥2)o(\|\mathbf{x} - \mathbf{x}_0\|^2) — the Peano remainder: a function R(h)R(\mathbf{h}) with R(h)/∥h∥2→0R(\mathbf{h}) / \|\mathbf{h}\|^2 \to 0 as h→0\mathbf{h} \to \mathbf{0}.

The n=1n=1 special case. f(x)=f(x0)+f′(x0)(x−x0)+12f′′(x0)(x−x0)2+o((x−x0)2).f(x) = f(x_0) + f'(x_0)(x - x_0) + \tfrac{1}{2} f''(x_0)(x - x_0)^2 + o\bigl((x - x_0)^2\bigr). Maclaurin is Taylor with x0=0\mathbf{x}_0 = \mathbf{0}.

Remark on notation. The factor 1/21/2 in the quadratic term comes from 1/2!1/2!. For general order-kk expansion the coefficient would be 1/k!1/k!, but the course stops at second order.

3.3 Theorem 1391 — Second-order conditions for unconstrained local max/min (with proof)

Statement. Let A⊆RnA \subseteq \mathbb{R}^n be open, f∈C(2)(A)f \in C^{(2)}(A), and x0∈A\mathbf{x}_0 \in A be a stationary point (i.e., ∇f(x0)=0\nabla f(\mathbf{x}_0) = \mathbf{0}). Then:

(1) Necessary condition. If x0\mathbf{x}_0 is a local maximiser (resp. minimiser) of ff, then the quadratic form hT∇2f(x0)h\mathbf{h}^T \nabla^2 f(\mathbf{x}_0) \mathbf{h} is negative semi-definite (resp. positive semi-definite).

(2) Sufficient condition. If the quadratic form hT∇2f(x0)h\mathbf{h}^T \nabla^2 f(\mathbf{x}_0) \mathbf{h} is negative definite (resp. positive definite), then x0\mathbf{x}_0 is a strong local maximiser (resp. strong local minimiser) of ff.

Consequently, if ∇2f(x0)\nabla^2 f(\mathbf{x}_0) is indefinite, x0\mathbf{x}_0 is a saddle point.

Source: Proofs - Differential Calculus.pdf p.3–4 (handwritten theorem 1391, "only (S)" — only the sufficient half is proved in the compendium, following Sicconi's syllabus).

Proof of (2) — sufficient condition, maximiser case.

Assume ∇2f(x0)\nabla^2 f(\mathbf{x}_0) defines a negative-definite quadratic form. We show x0\mathbf{x}_0 is a strong local maximiser.

Step 1 — Apply Taylor (Theorem 1481). Since f∈C(2)f \in C^{(2)} on a neighbourhood U(x0)U(\mathbf{x}_0), for every x∈U(x0)\mathbf{x} \in U(\mathbf{x}_0), f(x)  =  f(x0)+∇f(x0)(x−x0)+12(x−x0)T∇2f(x0)(x−x0)+o(∥x−x0∥2),x→x0.f(\mathbf{x}) \;=\; f(\mathbf{x}_0) + \nabla f(\mathbf{x}_0)(\mathbf{x} - \mathbf{x}_0) + \tfrac{1}{2}(\mathbf{x} - \mathbf{x}_0)^T \nabla^2 f(\mathbf{x}_0)(\mathbf{x} - \mathbf{x}_0) + o(\|\mathbf{x} - \mathbf{x}_0\|^2), \quad \mathbf{x} \to \mathbf{x}_0. (With ∇2f(x0)\nabla^2 f(\mathbf{x}_0) symmetric, by Schwarz.) Because x0\mathbf{x}_0 is stationary, ∇f(x0)=0\nabla f(\mathbf{x}_0) = \mathbf{0}, so the linear term vanishes: f(x)−f(x0)  =  12(x−x0)T∇2f(x0)(x−x0)+o(∥x−x0∥2).f(\mathbf{x}) - f(\mathbf{x}_0) \;=\; \tfrac{1}{2}(\mathbf{x} - \mathbf{x}_0)^T \nabla^2 f(\mathbf{x}_0)(\mathbf{x} - \mathbf{x}_0) + o(\|\mathbf{x} - \mathbf{x}_0\|^2).

Step 2 — Substitute h=x−x0\mathbf{h} = \mathbf{x} - \mathbf{x}_0. f(x)−f(x0)  =  12 hT∇2f(x0) h  +  o(∥h∥2)h→0.f(\mathbf{x}) - f(\mathbf{x}_0) \;=\; \tfrac{1}{2}\,\mathbf{h}^T \nabla^2 f(\mathbf{x}_0)\, \mathbf{h} \;+\; o(\|\mathbf{h}\|^2) \qquad \mathbf{h} \to \mathbf{0}.

Step 3 — Renormalise. Assume x≠x0\mathbf{x} \neq \mathbf{x}_0, so h≠0\mathbf{h} \neq \mathbf{0}, and multiply–divide the right-hand side by ∥h∥2\|\mathbf{h}\|^2: f(x)−f(x0)  =  ∥h∥2⋅[ 12 hT∇2f(x0) h∥h∥2  +  o(∥h∥2)∥h∥2 ]  =  ∥h∥2⋅[ 12 vT∇2f(x0)v+o(1) ],f(\mathbf{x}) - f(\mathbf{x}_0) \;=\; \|\mathbf{h}\|^2 \cdot \left[\, \tfrac{1}{2}\,\frac{\mathbf{h}^T \nabla^2 f(\mathbf{x}_0)\, \mathbf{h}}{\|\mathbf{h}\|^2} \;+\; \frac{o(\|\mathbf{h}\|^2)}{\|\mathbf{h}\|^2}\, \right] \;=\; \|\mathbf{h}\|^2 \cdot \left[\, \tfrac{1}{2}\, \mathbf{v}^T \nabla^2 f(\mathbf{x}_0) \mathbf{v} + o(1)\, \right], where v=h/∥h∥\mathbf{v} = \mathbf{h} / \|\mathbf{h}\| is a unit vector, and we used the definition of little-oo to write o(∥h∥2)/∥h∥2=o(1)o(\|\mathbf{h}\|^2) / \|\mathbf{h}\|^2 = o(1) as h→0\mathbf{h} \to \mathbf{0}.

Step 4 — Maximise the quadratic form over the unit sphere. Consider the continuous function Q(v)=vT∇2f(x0)vonS={v∈Rn:∥v∥=1},Q(\mathbf{v}) = \mathbf{v}^T \nabla^2 f(\mathbf{x}_0) \mathbf{v} \qquad \text{on} \qquad S = \{\mathbf{v} \in \mathbb{R}^n : \|\mathbf{v}\| = 1\}, which is the unit sphere — a compact set (closed and bounded). By Weierstrass, QQ attains a maximum q∗q^{\ast} on SS. Because ∇2f(x0)\nabla^2 f(\mathbf{x}_0) is negative definite, Q(v)<0Q(\mathbf{v}) < 0 for every v≠0\mathbf{v} \neq \mathbf{0}; in particular Q(v)<0Q(\mathbf{v}) < 0 for every v∈S\mathbf{v} \in S (note 0∉S\mathbf{0} \notin S), so q∗  =  max⁡v∈SQ(v)  <  0.q^{\ast} \;=\; \max_{\mathbf{v} \in S} Q(\mathbf{v}) \;<\; 0.

Step 5 — Bound f(x)−f(x0)f(\mathbf{x}) - f(\mathbf{x}_0) strictly below zero. Substituting into Step 3: f(x)−f(x0)  ≤  ∥h∥2[12q∗+o(1)].f(\mathbf{x}) - f(\mathbf{x}_0) \;\le\; \|\mathbf{h}\|^2 \Bigl[\tfrac{1}{2} q^{\ast} + o(1)\Bigr]. We want this bracket to be negative. Pick any ε>0\varepsilon > 0; by the definition of o(1)o(1), there is δ>0\delta > 0 such that 0<∥h∥<δ0 < \|\mathbf{h}\| < \delta implies ∣o(1)∣<ε|o(1)| < \varepsilon, i.e., −ε<o(1)<ε-\varepsilon < o(1) < \varepsilon. Choosing the concrete value ε‾=−14q∗\overline{\varepsilon} = -\tfrac{1}{4} q^{\ast} (which is strictly positive because q∗<0q^{\ast} < 0), we get for ∥h∥<δ‾\|\mathbf{h}\| < \overline{\delta}: 12q∗+o(1)  <  12q∗−14q∗  =  14q∗  <  0.\tfrac{1}{2} q^{\ast} + o(1) \;<\; \tfrac{1}{2} q^{\ast} - \tfrac{1}{4} q^{\ast} \;=\; \tfrac{1}{4} q^{\ast} \;<\; 0. Therefore f(x)−f(x0)  ≤  ∥h∥2⏟>0⋅(14q∗)⏟<0  <  0,f(\mathbf{x}) - f(\mathbf{x}_0) \;\le\; \underbrace{\|\mathbf{h}\|^2}_{>0} \cdot \underbrace{\bigl(\tfrac{1}{4} q^{\ast}\bigr)}_{<0} \;<\; 0, i.e., f(x)<f(x0)f(\mathbf{x}) < f(\mathbf{x}_0) for every x≠x0\mathbf{x} \neq \mathbf{x}_0 in a neighbourhood of x0\mathbf{x}_0. Hence x0\mathbf{x}_0 is a strong local maximiser. ■\blacksquare

Minimiser case. Replace "negative definite" by "positive definite"; then q∗>0q^{\ast} > 0 (in fact q⋆=min⁡Q>0q_\star = \min Q > 0 by the same Weierstrass argument applied to −Q-Q), and the inequality flips to f(x)>f(x0)f(\mathbf{x}) > f(\mathbf{x}_0).

Saddle case. If ∇2f(x0)\nabla^2 f(\mathbf{x}_0) is indefinite there exist unit vectors v+,v−\mathbf{v}_+, \mathbf{v}_- with Q(v+)>0Q(\mathbf{v}_+) > 0 and Q(v−)<0Q(\mathbf{v}_-) < 0. Taylor along each direction shows ff increases along v+\mathbf{v}_+ and decreases along v−\mathbf{v}_-, so x0\mathbf{x}_0 is a saddle point.

Necessary condition (part 1), sketch. If x0\mathbf{x}_0 is a local maximiser, then f(x)≤f(x0)f(\mathbf{x}) \le f(\mathbf{x}_0) nearby, so the Taylor remainder 12hT∇2f(x0)h+o(∥h∥2)≤0\tfrac{1}{2}\mathbf{h}^T \nabla^2 f(\mathbf{x}_0)\mathbf{h} + o(\|\mathbf{h}\|^2) \le 0. Dividing by ∥h∥2\|\mathbf{h}\|^2 and taking h→0\mathbf{h} \to \mathbf{0} along any direction v\mathbf{v} gives vT∇2f(x0)v≤0\mathbf{v}^T \nabla^2 f(\mathbf{x}_0) \mathbf{v} \le 0, i.e., negative semi-definite.

Summary table.

| Sign of ∇2f(x0)\nabla^2 f(\mathbf{x}_0) | Conclusion | Role | |---|---|---| | positive definite | strong local min | sufficient | | negative definite | strong local max | sufficient | | indefinite | saddle | sufficient | | positive semi-definite (not def.) | local min is possible, not guaranteed | necessary only | | negative semi-definite (not def.) | local max is possible, not guaranteed | necessary only |

Critical pitfall. Semi-definite ⇒\Rightarrow cannot conclude. Need a direct inspection (as in Example 4.3 below, f(x,y)=x2yf(x,y) = x^2 y at (0,y)(0,y)).

3.4 Theorem/Proposition 1486 — Concavity/strict-concavity test via the Hessian

Statement. Let C⊆RnC \subseteq \mathbb{R}^n be an open convex set and f∈C(2)(C)f \in C^{(2)}(C). Then: (a) ff is convex on CC   ⟺  \iff ∇2f(x)\nabla^2 f(\mathbf{x}) is positive semi-definite for every x∈C\mathbf{x} \in C. (b) ff is strictly convex on CC   ⟺  \iff ∇2f(x)\nabla^2 f(\mathbf{x}) is positive definite for every x∈C\mathbf{x} \in C.

Symmetric statements for concave / strictly concave by flipping all signs.

Source: Proofs - Differential Calculus.pdf (ref to prop 1486); lect 4_calc_diff (3).pdf (statement); General_24524_ENG_SOL.pdf Q8(a) Mode A / Q7(a) Mode B cites "see the textbook, proposition 1486" for the concavity test.

Proof (convex case ⇐, the direction most used on exams).

Assume ∇2f(x)\nabla^2 f(\mathbf{x}) is positive semi-definite for every x∈C\mathbf{x} \in C. Pick x,x0∈C\mathbf{x}, \mathbf{x}_0 \in C. By the 2nd-order Taylor formula with Lagrange remainder (the integral / mean-value form, course-level), there exists ξ\xi on the segment [x0,x][\mathbf{x}_0, \mathbf{x}] (which stays in CC because CC is convex) such that f(x)  =  f(x0)+∇f(x0)(x−x0)+12(x−x0)T∇2f(ξ)(x−x0).f(\mathbf{x}) \;=\; f(\mathbf{x}_0) + \nabla f(\mathbf{x}_0)(\mathbf{x} - \mathbf{x}_0) + \tfrac{1}{2}(\mathbf{x} - \mathbf{x}_0)^T \nabla^2 f(\xi)(\mathbf{x} - \mathbf{x}_0). Since ∇2f(ξ)\nabla^2 f(\xi) is PSD, the last term is ≥0\ge 0, so f(x)  ≥  f(x0)+∇f(x0)(x−x0).f(\mathbf{x}) \;\ge\; f(\mathbf{x}_0) + \nabla f(\mathbf{x}_0)(\mathbf{x} - \mathbf{x}_0). This is the tangent-hyperplane inequality for convex functions, which (lect4) is equivalent to convexity of ff on CC.

Strict case (b), ⇐. If ∇2f(x)\nabla^2 f(\mathbf{x}) is positive definite everywhere, then for x≠x0\mathbf{x} \neq \mathbf{x}_0 the last term is strictly positive, giving the strict tangent-hyperplane inequality — equivalent to strict convexity.

Converse (⇒), sketch. Convexity implies f(x)≥f(x0)+∇f(x0)(x−x0)f(\mathbf{x}) \ge f(\mathbf{x}_0) + \nabla f(\mathbf{x}_0)(\mathbf{x} - \mathbf{x}_0). Apply Taylor, subtract the linear part, divide by ∥x−x0∥2\|\mathbf{x} - \mathbf{x}_0\|^2: the lower bound ≥0\ge 0 transfers to the quadratic form vT∇2f(x0)v≥0\mathbf{v}^T \nabla^2 f(\mathbf{x}_0) \mathbf{v} \ge 0 for every unit direction v\mathbf{v}, i.e., PSD. ■\blacksquare

One-variable shadow. The standard "convex   ⟺  \iff f′′(x)≥0f''(x) \ge 0" is exactly the n=1n = 1 case.

Lecture 4 sidenote. For a strictly convex function the Hessian can fail to be positive definite at isolated points (classic example f(x)=x4f(x) = x^4 has f′′(0)=0f''(0) = 0 but is strictly convex on R\mathbb{R}). The textbook equivalence (b) as stated holds under the C(2)C^{(2)} assumption; in practice the PSD-everywhere plus "PD somewhere" condition is what is checked.

3.5 Theorem 1497 — Global extrema of differentiable convex/concave functions

Statement. Let C⊆RnC \subseteq \mathbb{R}^n be an open convex set, f:C→Rf : C \to \mathbb{R} differentiable, and x0∈C\mathbf{x}_0 \in C. (1) If ff is concave on CC, then x0 is a global maximiser for f  ⟺  ∇f(x0)=0.\mathbf{x}_0 \text{ is a global maximiser for } f \iff \nabla f(\mathbf{x}_0) = \mathbf{0}. (2) If ff is convex on CC, then x0 is a global minimiser for f  ⟺  ∇f(x0)=0.\mathbf{x}_0 \text{ is a global minimiser for } f \iff \nabla f(\mathbf{x}_0) = \mathbf{0}.

Source: Proofs - Differential Calculus.pdf p.5 (handwritten theorem 1497).

Proof of (1), ⇒\Rightarrow (global max ⇒∇f(x0)=0\Rightarrow \nabla f(\mathbf{x}_0) = \mathbf{0}).

Since x0\mathbf{x}_0 is a global maximiser, it is also a local maximiser. By Fermat's theorem (§3.7), and because CC is open (so x0\mathbf{x}_0 is an interior point), ∇f(x0)=0\nabla f(\mathbf{x}_0) = \mathbf{0}.

Proof of (1), ⇐\Leftarrow (∇f(x0)=0⇒\nabla f(\mathbf{x}_0) = \mathbf{0} \Rightarrow global max).

Since ff is concave and differentiable on the open convex set CC, the tangent-hyperplane characterisation of concavity gives f(x)  ≤  f(x0)+∇f(x0)(x−x0)∀ x∈C.f(\mathbf{x}) \;\le\; f(\mathbf{x}_0) + \nabla f(\mathbf{x}_0)(\mathbf{x} - \mathbf{x}_0) \qquad \forall\, \mathbf{x} \in C. (Concave: the tangent hyperplane sits above the graph.) Because ∇f(x0)=0\nabla f(\mathbf{x}_0) = \mathbf{0}, the linear term vanishes and f(x)≤f(x0)∀ x∈C.f(\mathbf{x}) \le f(\mathbf{x}_0) \qquad \forall\, \mathbf{x} \in C. Hence x0\mathbf{x}_0 is a global maximiser.

Proof of (2). Identical, with ≤\le replaced by ≥\ge, concave replaced by convex, maximiser by minimiser. ■\blacksquare

Why this is the single most useful theorem for open-ended exams. If you can prove the Hessian is PSD/PD everywhere on CC, then any stationary point is automatically a global minimiser (not just local) — no neighbourhood argument required. This is exactly the punchline of Final exercises Q2(b): the Hessian of x+3y2−ln⁡(x+2y)x + 3y^2 - \ln(x+2y) is PD throughout the domain, so (1/3,1/3)(1/3, 1/3) is a global min.

3.6 Theorem 1591 — Global maximum of a concave function (extension to continuous case)

Statement. Let C⊆RnC \subseteq \mathbb{R}^n be a convex set (not necessarily open), f:C→Rf : C \to \mathbb{R} be continuous on CC, concave on CC, and differentiable on int(C)\text{int}(C). Let x^∈int(C)\hat{\mathbf{x}} \in \text{int}(C). Then x^ is a global maximiser of f on C  ⟺  ∇f(x^)=0.\hat{\mathbf{x}} \text{ is a global maximiser of } f \text{ on } C \iff \nabla f(\hat{\mathbf{x}}) = \mathbf{0}. Symmetric statement for convex functions and global minimisers.

Source: Sicconi's course reference (textbook theorem 1591). Not proved in the compendium; follows from 1497 plus the standard continuous-extension argument.

Remark. This upgrades Theorem 1497 from open convex sets to arbitrary convex sets, provided the candidate stationary point lies in the interior. Points on the boundary still need separate treatment.

3.7 Fermat's theorem — necessary condition for interior extrema

Statement. Let A⊆RnA \subseteq \mathbb{R}^n and f:A→Rf : A \to \mathbb{R}. If x0\mathbf{x}_0 is an interior point of AA, ff is differentiable at x0\mathbf{x}_0, and x0\mathbf{x}_0 is a local extremum (min or max) of ff, then ∇f(x0)=0.\nabla f(\mathbf{x}_0) = \mathbf{0}.

Source: lect 4_calc_diff (3).pdf (applied inside the proof of Theorem 1497).

Proof. Without loss of generality, assume x0\mathbf{x}_0 is a local maximiser. Fix any direction v∈Rn\mathbf{v} \in \mathbb{R}^n with ∥v∥=1\|\mathbf{v}\| = 1, and consider the one-variable function φ(t):=f(x0+tv)\varphi(t) := f(\mathbf{x}_0 + t \mathbf{v}), defined on a small interval (−δ,δ)(-\delta, \delta) (possible because x0\mathbf{x}_0 is interior). Then t=0t = 0 is a local maximum of φ\varphi, so by the one-variable Fermat theorem φ′(0)=0\varphi'(0) = 0. By the chain rule, φ′(0)=∇f(x0)⋅v\varphi'(0) = \nabla f(\mathbf{x}_0) \cdot \mathbf{v}. Since this must equal 00 for every direction v\mathbf{v}, we get ∇f(x0)=0\nabla f(\mathbf{x}_0) = \mathbf{0}. ■\blacksquare

Converse is false. Not every stationary point is an extremum — saddles are also stationary. Fermat is a necessary, not sufficient, condition.

3.8 Dini's theorem (Implicit function theorem, local 2D version)

Statement. Let g:A⊆R2→Rg : A \subseteq \mathbb{R}^2 \to \mathbb{R} and let P=(x0,y0)∈AP = (x_0, y_0) \in A satisfy g(x0,y0)=cg(x_0, y_0) = c. Suppose: (i) g∈C(1)g \in C^{(1)} in a neighbourhood B(x0,y0)⊆AB(x_0, y_0) \subseteq A; (ii) ∂g∂y(x0,y0)≠0\dfrac{\partial g}{\partial y}(x_0, y_0) \neq 0.

Then there exist neighbourhoods U(x0)⊆RU(x_0) \subseteq \mathbb{R} and V(y0)⊆RV(y_0) \subseteq \mathbb{R} (with U×V⊆B(x0,y0)U \times V \subseteq B(x_0, y_0)) and a unique function f:U(x0)→V(y0)f : U(x_0) \to V(y_0) with f(x0)=y0f(x_0) = y_0 such that g(x,f(x))=c∀x∈U(x0).g(x, f(x)) = c \qquad \forall x \in U(x_0). Furthermore ff is differentiable at x0x_0 and  f′(x0)  =  − ∂g/∂x (x0,y0)∂g/∂y (x0,y0) .\boxed{\, f'(x_0) \;=\; -\, \frac{\partial g / \partial x \,(x_0, y_0)}{\partial g / \partial y \,(x_0, y_0)} \,}.

Source: lect5_calc_diff.pdf "Dini's Theorem: Implicit Theorem, local version"; Final exercises solution.pdf Q3 applies the formula with the minus sign.

Idea. The non-vanishing of ∂g/∂y\partial g / \partial y at PP says the level curve g=cg = c is non-vertical at PP — no vertical tangent — so locally we can invert the yy-coordinate and express yy as a function of xx. The derivative formula comes from differentiating g(x,f(x))=cg(x, f(x)) = c implicitly and using the chain rule: ∂g∂x(x,f(x))+∂g∂y(x,f(x))⋅f′(x)  =  0  ⟹  f′(x)  =  − gxgy.\frac{\partial g}{\partial x}(x, f(x)) + \frac{\partial g}{\partial y}(x, f(x)) \cdot f'(x) \;=\; 0 \;\Longrightarrow\; f'(x) \;=\; -\, \frac{g_x}{g_y}.

Global existence/uniqueness (lect5). If instead of just PP we work on a rectangular domain A×B⊆R2A \times B \subseteq \mathbb{R}^2 (with BB an interval) and we have (1) inf⁡y∈Bg(x,y)<c<sup⁡y∈Bg(x,y)\inf_{y \in B} g(x, y) < c < \sup_{y \in B} g(x, y) for every x∈Ax \in A; (2) gg continuous in yy (for each fixed xx); (3) gg strictly monotonic in yy (for each fixed xx) — equivalently ∂g/∂y\partial g / \partial y has constant sign,

then there exists a unique global f:A→Bf : A \to B with g(x,f(x))=cg(x, f(x)) = c for every x∈Ax \in A.

Caveat (lect5 MCQ). If ∂g/∂y(x0,y0)=0\partial g / \partial y (x_0, y_0) = 0 (vertical tangent), Dini fails. Example: an ellipse at its leftmost / rightmost points fails the vertical-line test: a small move to the right leaves the ellipse entirely, so no function y=f(x)y = f(x) is definable there.

3.9 Weierstrass theorem and coercive-function existence (Tonelli)

Weierstrass (stated in lect4). If K⊆RnK \subseteq \mathbb{R}^n is compact (closed and bounded) and f:K→Rf : K \to \mathbb{R} is continuous, then ff attains both its maximum and its minimum on KK: ∃ xmax⁡,xmin⁡∈K\exists\, \mathbf{x}_{\max}, \mathbf{x}_{\min} \in K with f(xmax⁡)=max⁡Kff(\mathbf{x}_{\max}) = \max_K f, f(xmin⁡)=min⁡Kff(\mathbf{x}_{\min}) = \min_K f.

Tonelli / coercive theorem (stated in lect4). If f:Rn→Rf : \mathbb{R}^n \to \mathbb{R} is continuous and coercive (lim⁡∥x∥→∞f(x)=+∞\lim_{\|\mathbf{x}\|\to\infty} f(\mathbf{x}) = +\infty), then ff attains a global minimum. If ff is continuous and super-coercive towards −∞-\infty (lim⁡∥x∥→∞f(x)=−∞\lim_{\|\mathbf{x}\|\to\infty} f(\mathbf{x}) = -\infty), then ff attains a global maximum.

Why coercivity alone gives existence. A continuous coercive ff has bounded sub-level sets. Pick any c>inf⁡fc > \inf f; the sub-level set K={f≤c}K = \{f \le c\} is non-empty, closed (continuity), and bounded (coercivity) — hence compact. By Weierstrass, f∣Kf|_K attains its minimum; that minimum is automatically the global minimum of ff on Rn\mathbb{R}^n (points outside KK have ff-values >c≥min⁡Kf> c \ge \min_K f).

Composition rule (lect4). If g:Rn→Rg : \mathbb{R}^n \to \mathbb{R} is continuous with g(x)→−∞g(\mathbf{x}) \to -\infty as ∥x∥→∞\|\mathbf{x}\| \to \infty, and h:R→Rh : \mathbb{R} \to \mathbb{R} is continuous and strictly increasing, then h∘gh \circ g inherits the coercivity behaviour of gg filtered through hh. Example: f(x,y)=e−x2−y2f(x,y) = e^{-x^2 - y^2}: inner −x2−y2→−∞-x^2 - y^2 \to -\infty (super-coercive towards −∞-\infty), outer ete^t strictly increasing, so f→e−∞=0+f \to e^{-\infty} = 0^+ as ∥(x,y)∥→∞\|(x,y)\| \to \infty; hence the supremum 11 must be attained at a stationary point, which is (0,0)(0,0) — a global maximum.

Relaxation of constraints (lect4). When the admissible set is not compact and ff not coercive, one can sometimes relax the constraint and solve the unconstrained problem, then check whether the unconstrained optimum lies in the original set. If yes, it is the constrained optimum too; if no, the method fails and Lagrange multipliers or boundary analysis are needed.

§4. Worked Examples

Example 4.1 — Gradient, Hessian, and stationary points of f(x,y)=ex(x−1) y+y2f(x,y) = e^x(x-1)\,y + y^2

Source: May 2024 Q8 (Mode A) / Q7 (Mode B), General_24524_ENG_SOL.pdf.

Gradient. Treat yy as a constant: fx′=∂∂x[y ex(x−1)]=y[ex(x−1)+ex]=y⋅ex⋅x=xyex.f'_x = \frac{\partial}{\partial x}\bigl[y\, e^x(x-1)\bigr] = y\bigl[e^x(x-1) + e^x\bigr] = y \cdot e^x \cdot x = x y e^x. Treat xx as a constant: fy′=∂∂y[y ex(x−1)+y2]=ex(x−1)+2y.f'_y = \frac{\partial}{\partial y}\bigl[y\, e^x(x-1) + y^2\bigr] = e^x(x-1) + 2y. ∇f(x,y)  =  [ xyex,2y+ex(x−1) ].\boxed{\nabla f(x,y) \;=\; \bigl[\,x y e^x, \quad 2y + e^x(x-1)\,\bigr].}

Stationary points. Solve ∇f(x,y)=0\nabla f(x,y) = \mathbf{0}: {xyex=02y+ex(x−1)=0\begin{cases} x y e^x = 0 \\ 2y + e^x(x-1) = 0 \end{cases} The first equation gives x=0x = 0 or y=0y = 0 (since ex>0e^x > 0 always).

  • Case x=0x = 0. The second equation becomes 2y+e0⋅(0−1)=2y−1=02y + e^0 \cdot (0 - 1) = 2y - 1 = 0, so y=1/2y = 1/2. Stationary point P=(0,1/2)P = (0, 1/2).
  • Case y=0y = 0. The second equation becomes 0+ex(x−1)=0⇒x−1=0⇒x=10 + e^x(x - 1) = 0 \Rightarrow x - 1 = 0 \Rightarrow x = 1 (since ex>0e^x > 0). Stationary point Q=(1,0)Q = (1, 0).

Hessian. Compute second derivatives: fxx′′=∂∂x(xyex)=yex+xyex=yex(x+1),f''_{xx} = \frac{\partial}{\partial x}(x y e^x) = y e^x + x y e^x = y e^x(x + 1), fxy′′=∂∂y(xyex)=xex(=fyx′′ by Schwarz),f''_{xy} = \frac{\partial}{\partial y}(x y e^x) = x e^x \quad (= f''_{yx} \text{ by Schwarz}), fyy′′=∂∂y(2y+ex(x−1))=2.f''_{yy} = \frac{\partial}{\partial y}\bigl(2y + e^x(x-1)\bigr) = 2. ∇2f(x,y)  =  (yex(x+1)xexxex2).\nabla^2 f(x,y) \;=\; \begin{pmatrix} y e^x (x+1) & x e^x \\ x e^x & 2 \end{pmatrix}.

Classify P=(0,1/2)P = (0, 1/2). ∇2f(0,1/2)=((1/2)e0(0+1)002)=(1/2002).\nabla^2 f(0, 1/2) = \begin{pmatrix} (1/2) e^0 (0+1) & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1/2 & 0 \\ 0 & 2 \end{pmatrix}. Leading principal minors: det⁡A1=1/2>0\det A_1 = 1/2 > 0 and det⁡A2=1>0\det A_2 = 1 > 0. Positive definite. By Theorem 1391, P=(0,1/2)P = (0, 1/2) is a strong local minimiser.

Classify Q=(1,0)Q = (1, 0). ∇2f(1,0)=(0⋅e⋅21⋅ee2)=(0ee2).\nabla^2 f(1, 0) = \begin{pmatrix} 0 \cdot e \cdot 2 & 1 \cdot e \\ e & 2 \end{pmatrix} = \begin{pmatrix} 0 & e \\ e & 2 \end{pmatrix}. det⁡A1=0\det A_1 = 0, det⁡A2=0⋅2−e2=−e2<0\det A_2 = 0 \cdot 2 - e^2 = -e^2 < 0. Since det⁡A2≠0\det A_2 \neq 0, we can finalise: one eigenvalue is positive, the other negative (product =det⁡A2<0= \det A_2 < 0), so the Hessian is indefinite. By Theorem 1391, Q=(1,0)Q = (1, 0) is a saddle point — not a local optimiser.

Global convexity of ff on R2\mathbb{R}^2? By Prop. 1486 (§3.4), ff is convex on R2\mathbb{R}^2 iff ∇2f(x,y)\nabla^2 f(x,y) is PSD at every point. We found ∇2f(1,0)\nabla^2 f(1, 0) is indefinite — so ff is not convex on R2\mathbb{R}^2. (Official May 2024 solution: "The function ff is not convex on all R2\mathbb{R}^2, because ∇2f(x,y)\nabla^2 f(x,y) is not positive semidefinite on all (x,y)∈R2(x,y) \in \mathbb{R}^2. For example, this does not happen with point (1,0)(1, 0).")

Example 4.2 — Second-order Taylor of f(x,y)=x2+xcos⁡yf(x,y) = x^2 + x \cos y at (1,−π/2)(1, -\pi/2)

Source: May 2024 MCQ2 (Mode A) / MCQ1 (Mode B).

Step 1 — function value. f(1,−π/2)=12+1⋅cos⁡(−π/2)=1+0=1f(1, -\pi/2) = 1^2 + 1 \cdot \cos(-\pi/2) = 1 + 0 = 1.

Step 2 — gradient. fx′=2x+cos⁡yf'_x = 2x + \cos y, fy′=−xsin⁡yf'_y = -x \sin y. At (1,−π/2)(1, -\pi/2):

  • fx′(1,−π/2)=2+cos⁡(−π/2)=2+0=2f'_x(1, -\pi/2) = 2 + \cos(-\pi/2) = 2 + 0 = 2;
  • fy′(1,−π/2)=−1⋅sin⁡(−π/2)=−1⋅(−1)=1f'_y(1, -\pi/2) = -1 \cdot \sin(-\pi/2) = -1 \cdot (-1) = 1.

Thus ∇f(1,−π/2)=[2  1]\nabla f(1, -\pi/2) = [2 \; 1].

Step 3 — Hessian. fxx′′=2f''_{xx} = 2, fxy′′=−sin⁡yf''_{xy} = -\sin y, fyy′′=−xcos⁡yf''_{yy} = -x \cos y. At (1,−π/2)(1, -\pi/2):

  • fxx′′=2f''_{xx} = 2; fxy′′=−sin⁡(−π/2)=−(−1)=1f''_{xy} = -\sin(-\pi/2) = -(-1) = 1; fyy′′=−1⋅cos⁡(−π/2)=−1⋅0=0f''_{yy} = -1 \cdot \cos(-\pi/2) = -1 \cdot 0 = 0.

Thus ∇2f(1,−π/2)=(2110)\nabla^2 f(1, -\pi/2) = \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}.

Step 4 — assemble Taylor formula. With h=(x−1,  y+π/2)\mathbf{h} = (x - 1, \; y + \pi/2): f(x,y)  =  1+[2  1](x−1y+π/2)+12[x−1    y+π/2](2110)(x−1y+π/2)+o(∥(x−1,y+π/2)∥2).f(x,y) \;=\; 1 + [2 \; 1]\begin{pmatrix} x-1 \\ y + \pi/2 \end{pmatrix} + \tfrac{1}{2}[x-1 \; \; y + \pi/2]\begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} x-1 \\ y + \pi/2 \end{pmatrix} + o\bigl(\|(x-1, y+\pi/2)\|^2\bigr).

Step 5 — expand. Linear part: 2(x−1)+(y+π/2)2(x-1) + (y + \pi/2). Quadratic form: 12[2(x−1)2+2(x−1)(y+π/2)]=(x−1)2+(x−1)(y+π/2)\tfrac{1}{2}[2(x-1)^2 + 2(x-1)(y+\pi/2)] = (x-1)^2 + (x-1)(y+\pi/2). After simplification the official answer is: f(x,y)  =  π2x+x2+xy+o(∥(x−1,y+π/2)∥2).f(x,y) \;=\; \tfrac{\pi}{2} x + x^2 + x y + o\bigl(\|(x-1, y+\pi/2)\|^2\bigr).

Answer. None of the listed options matches exactly ⇒ the correct MCQ choice is "none of the others" (Mode A option D, Mode B option C).

Example 4.3 — A case where the Hessian is semi-definite (lect 4_calc_diff (3).pdf)

f(x,y)=x2yf(x,y) = x^2 y. Find all extrema.

Stationary points. ∇f=[2xy,  x2]=[0  0]\nabla f = [2xy, \; x^2] = [0 \; 0] gives x2=0⇒x=0x^2 = 0 \Rightarrow x = 0, and then 2⋅0⋅y=02 \cdot 0 \cdot y = 0 for every yy — so the entire yy-axis is stationary: (0,y)(0, y) for every y∈Ry \in \mathbb{R}.

Hessian. ∇2f=(2y2x2x0)\nabla^2 f = \begin{pmatrix} 2y & 2x \\ 2x & 0 \end{pmatrix}, so ∇2f(0,y)=(2y000)\nabla^2 f(0, y) = \begin{pmatrix} 2y & 0 \\ 0 & 0 \end{pmatrix}.

  • If y>0y > 0: Hessian is positive semi-definite (eigenvalues 2y,02y, 0), not definite. Theorem 1391 inconclusive.
  • If y<0y < 0: Hessian is negative semi-definite, not definite. Theorem 1391 inconclusive.
  • If y=0y = 0: Hessian is the zero matrix, the most degenerate case.

Direct inspection. Write f(x,y)=x2yf(x, y) = x^2 y near (0,y0)(0, y_0).

  • If y0>0y_0 > 0: in a small nbhd, yy stays positive, so f(x,y)=x2y≥0=f(0,y0)f(x, y) = x^2 y \ge 0 = f(0, y_0), with equality only on the yy-axis. Hence (0,y0)(0, y_0) is a weak (non-strict) local minimum.
  • If y0<0y_0 < 0: symmetrically, (0,y0)(0, y_0) is a weak local maximum.
  • If y0=0y_0 = 0: the origin. For any neighbourhood, pick y>0y > 0: f(x,y)>0f(x,y) > 0; pick y<0y < 0: f(x,y)<0f(x,y) < 0. So (0,0)(0,0) is a saddle point — neither max nor min.

Lesson. Semi-definite Hessian ⇒ you must argue by direct inspection or higher-order expansion. Theorem 1391 is silent.

Example 4.4 — f(x,y)=x+3y2−ln⁡(x+2y)f(x,y) = x + 3 y^2 - \ln(x + 2y) (domain, stationary point, convexity, Taylor)

Source: Final exercises solution.pdf Q2.

Domain. The logarithm requires x+2y>0x + 2y > 0, so Dom(f)={(x,y)∈R2:x+2y>0}.\text{Dom}(f) = \{(x, y) \in \mathbb{R}^2 : x + 2y > 0\}. This is the open half-plane above the line x+2y=0x + 2y = 0; it is open and convex.

Gradient. fx′=1−1x+2y,fy′=6y−2x+2y.f'_x = 1 - \frac{1}{x + 2y}, \qquad f'_y = 6y - \frac{2}{x + 2y}. ∇f(x,y)=[ 1−1x+2y,  6y−2x+2y ].\nabla f(x,y) = \Bigl[\,1 - \tfrac{1}{x+2y},\ \ 6y - \tfrac{2}{x+2y}\,\Bigr].

Stationary points. Set ∇f=0\nabla f = \mathbf{0}: {1−1x+2y=06y−2x+2y=0  ⟺  {x+2y=16y=2x+2y=2  ⟺  {x+2y=1y=1/3  ⟺  x=1−2/3=1/3, y=1/3.\begin{cases} 1 - \frac{1}{x+2y} = 0 \\ 6y - \frac{2}{x+2y} = 0 \end{cases} \iff \begin{cases} x + 2y = 1 \\ 6y = \frac{2}{x+2y} = 2 \end{cases} \iff \begin{cases} x + 2y = 1 \\ y = 1/3 \end{cases} \iff x = 1 - 2/3 = 1/3,\ y = 1/3. Unique stationary point: (1/3,1/3)(1/3, 1/3). Check domain: 1/3+2/3=1>01/3 + 2/3 = 1 > 0. ✓\checkmark

Hessian. Let u=x+2yu = x + 2y. Then 1u\frac{1}{u} has derivatives ∂x=−1/u2\partial_x = -1/u^2 and ∂y=−2/u2\partial_y = -2/u^2. fxx′′=∂x[1−u−1]=u−2=(x+2y)−2,f''_{xx} = \partial_x\Bigl[1 - u^{-1}\Bigr] = u^{-2} = (x + 2y)^{-2}, fxy′′=∂y[1−u−1]=2u−2=2(x+2y)−2,f''_{xy} = \partial_y\Bigl[1 - u^{-1}\Bigr] = 2 u^{-2} = 2(x+2y)^{-2}, fyy′′=∂y[6y−2u−1]=6+4u−2=6+4(x+2y)−2.f''_{yy} = \partial_y\bigl[6y - 2 u^{-1}\bigr] = 6 + 4 u^{-2} = 6 + 4(x+2y)^{-2}. ∇2f(x,y)=((x+2y)−22(x+2y)−22(x+2y)−26+4(x+2y)−2).\nabla^2 f(x,y) = \begin{pmatrix} (x+2y)^{-2} & 2(x+2y)^{-2} \\ 2(x+2y)^{-2} & 6 + 4(x+2y)^{-2} \end{pmatrix}.

At (1/3,1/3)(1/3, 1/3) where x+2y=1x + 2y = 1: ∇2f(1/3,1/3)=(12210).\nabla^2 f(1/3, 1/3) = \begin{pmatrix} 1 & 2 \\ 2 & 10 \end{pmatrix}. Leading principal minors: det⁡A1=1>0\det A_1 = 1 > 0, det⁡A2=10−4=6>0\det A_2 = 10 - 4 = 6 > 0. Positive definite ⇒ by Theorem 1391, (1/3,1/3)(1/3, 1/3) is a strong local minimiser.

Is it also a global minimiser? Check PSD-ness of ∇2f(x,y)\nabla^2 f(x,y) everywhere on the domain:

  • fxx′′=(x+2y)−2>0f''_{xx} = (x+2y)^{-2} > 0 since the denominator is positive (domain condition).
  • det⁡∇2f(x,y)=(x+2y)−2⋅[6+4(x+2y)−2]−4(x+2y)−4=6(x+2y)−2+4(x+2y)−4−4(x+2y)−4=6(x+2y)−2>0\det \nabla^2 f(x,y) = (x+2y)^{-2} \cdot [6 + 4(x+2y)^{-2}] - 4(x+2y)^{-4} = 6(x+2y)^{-2} + 4(x+2y)^{-4} - 4(x+2y)^{-4} = 6(x+2y)^{-2} > 0.

Both leading principal minors positive everywhere — Hessian is positive definite on the whole (convex, open) domain. By Prop. 1486, ff is strictly convex on Dom(f)\text{Dom}(f). By Theorem 1497, the unique stationary point (1/3,1/3)(1/3, 1/3) is the global minimiser.

Value at the minimum. f(1/3,1/3)=1/3+3⋅(1/9)−ln⁡(1)=1/3+1/3−0=2/3f(1/3, 1/3) = 1/3 + 3 \cdot (1/9) - \ln(1) = 1/3 + 1/3 - 0 = 2/3.

Second-order Taylor at (1/3,1/3)(1/3, 1/3). With ∇f(1/3,1/3)=(0,0)\nabla f(1/3, 1/3) = (0, 0) (stationary point) and ∇2f(1/3,1/3)=(12210)\nabla^2 f(1/3, 1/3) = \bigl(\begin{smallmatrix}1 & 2\\ 2 & 10\end{smallmatrix}\bigr): f(x,y)  =  23+12[(x−13)2+10(y−13)2+4(x−13)(y−13)]+o(∥(x−1/3,y−1/3)∥2).f(x, y) \;=\; \tfrac{2}{3} + \tfrac{1}{2}\Bigl[(x - \tfrac{1}{3})^2 + 10(y - \tfrac{1}{3})^2 + 4(x - \tfrac{1}{3})(y - \tfrac{1}{3})\Bigr] + o\bigl(\|(x - 1/3, y - 1/3)\|^2\bigr).

Example 4.5 — Dini's theorem applied to g(x,y)=2yex+xey−2g(x,y) = 2 y e^x + x e^y - 2 at g=−2g = -2

Source: Final exercises solution.pdf Q3.

Find f′(0)f'(0) where y=f(x)y = f(x) is defined implicitly by g(x,y)=−2g(x, y) = -2.

Step 1 — locate the base point (x0,y0)(x_0, y_0). We need x0=0x_0 = 0 and y0y_0 such that g(0,y0)=−2g(0, y_0) = -2. Compute g(0,y)=2ye0+0⋅ey−2=2y−2g(0, y) = 2y e^0 + 0 \cdot e^y - 2 = 2y - 2. Setting 2y−2=−22y - 2 = -2 gives y=0y = 0. So (x0,y0)=(0,0)(x_0, y_0) = (0, 0).

Step 2 — verify Dini's hypothesis. Compute ∂g/∂y\partial g / \partial y: ∂g∂y(x,y)=2ex+xey.\frac{\partial g}{\partial y}(x, y) = 2 e^x + x e^y. At (0,0)(0, 0): gy(0,0)=2e0+0⋅e0=2≠0g_y(0, 0) = 2 e^0 + 0 \cdot e^0 = 2 \neq 0. ✓\checkmark Dini applies.

Step 3 — compute ∂g/∂x\partial g / \partial x. ∂g∂x(x,y)=2yex+ey.\frac{\partial g}{\partial x}(x, y) = 2 y e^x + e^y. At (0,0)(0, 0): gx(0,0)=2⋅0⋅e0+e0=1g_x(0, 0) = 2 \cdot 0 \cdot e^0 + e^0 = 1.

Step 4 — apply Dini's formula. f′(0)  =  − gx(0,0)gy(0,0)  =  − 12.f'(0) \;=\; -\,\frac{g_x(0, 0)}{g_y(0, 0)} \;=\; -\,\frac{1}{2}.

 f′(0)  =  −12. \boxed{\, f'(0) \;=\; -\tfrac{1}{2}. \,}

Critical: do not drop the minus sign.

Example 4.6 — Existence of a global maximiser via composition-coercivity (lect4)

Find the global maximiser of f(x,y)=e−x2−y2f(x,y) = e^{-x^2 - y^2} on R2\mathbb{R}^2.

Step 1 — existence. The inner function g(x,y)=−x2−y2g(x,y) = -x^2 - y^2 is continuous with g(x,y)→−∞g(x,y) \to -\infty as ∥(x,y)∥→∞\|(x,y)\| \to \infty. The outer function h(t)=eth(t) = e^t is strictly increasing and continuous. By the composition rule (§3.9), ff is continuous and f(x,y)→e−∞=0f(x,y) \to e^{-\infty} = 0 at infinity, i.e., ff is super-coercive towards 00 from above. In particular, on any sub-level set {f≥c}\{f \ge c\} for small c>0c > 0, ff is bounded — and f≤1=f(0,0)f \le 1 = f(0,0) so the supremum 11 is attained. Tonelli-type argument: ff has a global maximum.

Step 2 — candidates. Compute ∇f\nabla f: ∇f(x,y)=[ −2x e−x2−y2,  −2y e−x2−y2 ].\nabla f(x,y) = \bigl[\,-2x\, e^{-x^2 - y^2},\ \ -2y\, e^{-x^2 - y^2}\,\bigr]. Setting ∇f=0\nabla f = \mathbf{0}: the exponential is never 00, so −2x=0⇒x=0-2x = 0 \Rightarrow x = 0 and −2y=0⇒y=0-2y = 0 \Rightarrow y = 0. Unique stationary point (0,0)(0, 0).

Step 3 — conclusion. Since the global maximum exists (Tonelli) and there is only one candidate, (0,0)(0, 0) is the global maximum of ff on R2\mathbb{R}^2, with f(0,0)=1f(0, 0) = 1.

Example 4.7 — Stationary-point classification with three cases (lect3 TA)

f(x,y)=(y−1)(y−x2)f(x, y) = (y - 1)(y - x^2). Identify all extrema. (MCQ trap: the function expands to y2−yx2−y+x2y^2 - y x^2 - y + x^2.)

Gradient. fx′=−2xy+2xf'_x = -2xy + 2x, fy′=2y−x2−1f'_y = 2y - x^2 - 1.

Stationary points. Solve −2xy+2x=0-2xy + 2x = 0 and 2y−x2−1=02y - x^2 - 1 = 0.

  • From first: 2x(1−y)=0⇒x=02x(1 - y) = 0 \Rightarrow x = 0 or y=1y = 1.
  • Substitute into second: 2y=x2+12y = x^2 + 1, so y=(x2+1)/2y = (x^2 + 1)/2.
    • Case x=0x = 0: y=1/2y = 1/2. Point (0,1/2)(0, 1/2).
    • Case y=1y = 1: 1=(x2+1)/2⇒x2=1⇒x=±11 = (x^2 + 1)/2 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1. Points (1,1)(1, 1) and (−1,1)(-1, 1).

Three stationary points: (0,1/2),(1,1),(−1,1)(0, 1/2), (1, 1), (-1, 1).

Hessian. fxx′′=−2y+2f''_{xx} = -2y + 2, fxy′′=−2xf''_{xy} = -2x, fyy′′=2f''_{yy} = 2. ∇2f(x,y)=(−2y+2−2x−2x2).\nabla^2 f(x, y) = \begin{pmatrix} -2y + 2 & -2x \\ -2x & 2 \end{pmatrix}.

At (0,1/2)(0, 1/2): ∇2f(0,1/2)=(1002)\nabla^2 f(0, 1/2) = \bigl(\begin{smallmatrix} 1 & 0 \\ 0 & 2 \end{smallmatrix}\bigr). det⁡A1=1>0\det A_1 = 1 > 0, det⁡A2=2>0\det A_2 = 2 > 0. Positive definite ⇒ strong local minimiser.

At (1,1)(1, 1): ∇2f(1,1)=(0−2−22)\nabla^2 f(1, 1) = \bigl(\begin{smallmatrix} 0 & -2 \\ -2 & 2 \end{smallmatrix}\bigr). det⁡A1=0\det A_1 = 0, det⁡A2=0⋅2−4=−4<0\det A_2 = 0 \cdot 2 - 4 = -4 < 0. Indefinite (eigenvalue product = −4<0-4 < 0) ⇒ saddle point.

At (−1,1)(-1, 1): ∇2f(−1,1)=(0222)\nabla^2 f(-1, 1) = \bigl(\begin{smallmatrix} 0 & 2 \\ 2 & 2 \end{smallmatrix}\bigr). det⁡A2=−4<0\det A_2 = -4 < 0. Indefinite ⇒ saddle point.

Answer. The function has a unique local minimiser and two saddle points — matches MCQ choice (C) "a unique local minimiser".

§5. Solution Methods

Each method is a named algorithm with input → steps → output → pitfalls. Cross-references indicate May 2024 exam use.

M-DC-1 — Compute the gradient ∇f\nabla f and Hessian ∇2f\nabla^2 f

Used on: May 2024 MCQ2 / MCQ1 (Taylor), Q8 / Q7 (stationary points), Final exercises Q2.

Input. A function f:Rn→Rf : \mathbb{R}^n \to \mathbb{R} of class at least C(2)C^{(2)}.

Steps.

  1. Compute each fxi′f'_{x_i}: differentiate ff with respect to xix_i, treating the other variables as constants. Assemble the row vector ∇f=[fx1′⋯fxn′]\nabla f = [f'_{x_1} \cdots f'_{x_n}].
  2. For each pair (i,j)(i, j), compute fxixj′′=∂/∂xj(fxi′)f''_{x_i x_j} = \partial / \partial x_j \bigl(f'_{x_i}\bigr).
  3. Assemble ∇2f\nabla^2 f as the n×nn \times n matrix with entry (i,j)(i, j) equal to fxixj′′f''_{x_i x_j}.
  4. Schwarz sanity check: verify fxixj′′=fxjxi′′f''_{x_i x_j} = f''_{x_j x_i} (symmetry). If not, you have a calculation error.

Output. ∇f(x)\nabla f(\mathbf{x}) and ∇2f(x)\nabla^2 f(\mathbf{x}) as functions of x\mathbf{x}.

Pitfalls.

  • Forgetting the product and chain rules in expressions like ex(x−1)e^x(x-1) or ln⁡(x+2y)\ln(x + 2y).
  • Miscomputing fyy′′f''_{yy} by forgetting that the remaining xx-dependence is held constant.
  • Not using Schwarz: if fxy′′≠fyx′′f''_{xy} \neq f''_{yx} symbolically (for a C(2)C^{(2)} function), you have almost certainly made a sign error.
  • Writing the gradient as a column when the convention is a row (affects matrix products in Taylor).

M-DC-2 — Find all stationary points (solve ∇f=0\nabla f = \mathbf{0})

Used on: May 2024 Q8 / Q7, Final exercises Q2, lecture examples.

Input. The gradient ∇f(x)\nabla f(\mathbf{x}) as an nn-tuple of expressions, and the domain.

Steps.

  1. Write the system ∇f=0\nabla f = \mathbf{0} as nn scalar equations fxi′(x)=0f'_{x_i}(\mathbf{x}) = 0.
  2. Factor carefully. If an equation factors as A(x)⋅B(x)=0A(\mathbf{x}) \cdot B(\mathbf{x}) = 0 with A,BA, B not identically zero on the domain, split into cases A=0A = 0 and B=0B = 0.
  3. Within each case, substitute into the remaining equations and solve.
  4. Check the domain. Discard any algebraic solutions that lie outside Dom(f)\text{Dom}(f) (e.g., require x+2y>0x + 2y > 0 for a log term).
  5. List all surviving stationary points.

Output. A finite list of points x(1),…,x(k)\mathbf{x}^{(1)}, \ldots, \mathbf{x}^{(k)} where ∇f=0\nabla f = \mathbf{0}.

Common case splits.

  • xyex=0x y e^x = 0 (May 2024 Q8): ex>0e^x > 0 always, so xy=0⇒x=0xy = 0 \Rightarrow x = 0 or y=0y = 0.
  • ex+2y=2λx−2λe^{x+2y} = 2\lambda x - 2\lambda (lect4 constrained): rule out eanything=0e^{\text{anything}} = 0 (impossible), keep the algebraic case.
  • Curves of stationary points (Example 4.3): if one factor is always zero (e.g., x2=0⇒x=0x^2 = 0 \Rightarrow x = 0 but then 2xy=02xy = 0 automatically for every yy), the stationary set is a curve, not a discrete set. Each point on the curve must be analysed separately (often by direct inspection, since the Hessian is semi-definite).

Pitfalls.

  • Forgetting domain constraints. Final exercises Q2 yields the candidate (1/3,1/3)(1/3, 1/3) which happens to lie in the domain; had it lain on x+2y=0x + 2y = 0, we would have rejected it (boundary of domain, ln⁡\ln undefined).
  • Dropping solutions from y=0y = 0 or x=0x = 0 branches. Always enumerate both.
  • Dividing by λ\lambda / by a factor that might be zero. Either confirm the factor is non-zero (cross-check) or track the λ=0\lambda = 0 case separately.

M-DC-3 — Classify a stationary point via the Hessian

Used on: May 2024 Q8 / Q7 (b2), Final exercises Q2 (a), lecture TA problems.

Input. A stationary point x0\mathbf{x}_0 and the Hessian ∇2f(x0)\nabla^2 f(\mathbf{x}_0) (a symmetric n×nn \times n matrix).

Steps.

  1. Compute the leading principal minors det⁡A1,det⁡A2,…,det⁡An\det A_1, \det A_2, \ldots, \det A_n of ∇2f(x0)\nabla^2 f(\mathbf{x}_0).
  2. Apply Sylvester–Jacobi (§01 M-LA-5):
    • All >0> 0 ⇒ positive definite ⇒ strong local min.
    • Alternating starting <0<0 (−,+,−,…-, +, -, \ldots) ⇒ negative definite ⇒ strong local max.
    • Some minor =0= 0: test inconclusive; fall through to step 3.
    • All nonzero but neither pattern ⇒ indefinite ⇒ saddle.
  3. If Sylvester is inconclusive, compute all principal minors (not just leading), or compute eigenvalues:
    • All eigenvalues ≥0\ge 0, not all >0>0 ⇒ positive semi-definite ⇒ Theorem 1391 is inconclusive, fall through to step 4.
    • All ≤0\le 0, some =0= 0 ⇒ negative semi-definite ⇒ inconclusive.
    • Mixed signs ⇒ indefinite ⇒ saddle.
  4. Direct inspection fallback. Compute f(x)−f(x0)f(\mathbf{x}) - f(\mathbf{x}_0) near x0\mathbf{x}_0 and determine its sign by factoring or by a Taylor expansion of higher order.

Output. One of: strong local min, strong local max, saddle, inconclusive (semi-definite case).

Pitfalls.

  • Confusing semi-definite with definite. Semi-definite is only a necessary condition for a local extremum — not sufficient.
  • Sign of alternating pattern. The negative-definite pattern starts negative (det⁡A1<0\det A_1 < 0, det⁡A2>0\det A_2 > 0, det⁡A3<0\det A_3 < 0, …). All-negative is not negative-definite for n≥2n \ge 2.
  • Forgetting the case det⁡A2<0\det A_2 < 0. In 2D this means the two eigenvalues have opposite signs — automatically indefinite, regardless of det⁡A1\det A_1. (Example 4.1, (1,0)(1,0): det⁡A1=0\det A_1 = 0 but det⁡A2=−e2<0\det A_2 = -e^2 < 0 ⇒ indefinite ⇒ saddle.)

M-DC-4 — Check global convexity/concavity via the Hessian

Used on: May 2024 Q8 / Q7 (b3), Final exercises Q2 (b).

Input. A C(2)C^{(2)} function ff defined on an open convex set CC, and its Hessian ∇2f(x)\nabla^2 f(\mathbf{x}).

Steps.

  1. Study the sign of ∇2f(x)\nabla^2 f(\mathbf{x}) as a function of x\mathbf{x} ranging over CC, not just at one point.
  2. If ∇2f(x)\nabla^2 f(\mathbf{x}) is positive definite (resp. PSD) for every x∈C\mathbf{x} \in C, then ff is strictly convex (resp. convex) on CC by Prop. 1486.
  3. If there is even one point x∗∈C\mathbf{x}^\ast \in C where ∇2f(x∗)\nabla^2 f(\mathbf{x}^\ast) fails to be PSD (i.e., it is indefinite, or negative (semi-)definite, or has a negative eigenvalue), then ff is not convex on CC.

Output. One of: ff strictly convex on CC, ff convex on CC, ff not convex on CC (with a witness point). Symmetric statements for concave.

Consequence for global extrema.

  • If ff is strictly convex on an open convex domain and has a stationary point x0\mathbf{x}_0, then by Theorem 1497 x0\mathbf{x}_0 is the unique global minimiser.
  • If ff is strictly concave, the unique global maximiser is any stationary point.

Pitfalls.

  • Testing the Hessian only at the stationary point. That tells you local behaviour, not global. Global (strict) convexity requires PSD/PD everywhere on the domain.
  • Ignoring the convexity of the domain. Prop. 1486 requires CC to be a convex set. For {x+2y>0}\{x + 2y > 0\}, CC is a half-plane — convex ✓\checkmark.
  • Witness point. To disprove convexity on all of R2\mathbb{R}^2, cite one specific point where the Hessian fails PSD (May 2024 Q8 b3: "(1,0)(1,0)").

M-DC-5 — Write the second-order Taylor expansion at (a,b)(a, b)

Used on: May 2024 MCQ2 / MCQ1, Final exercises Q2(c).

Input. A C(2)C^{(2)} function ff and an expansion point x0=(a,b)\mathbf{x}_0 = (a, b).

Steps.

  1. Compute f(a,b)f(a, b) — zero-order term.
  2. Compute ∇f(a,b)=[fx′(a,b),  fy′(a,b)]\nabla f(a, b) = [f'_x(a,b), \; f'_y(a,b)] — linear coefficient.
  3. Compute ∇2f(a,b)=(fxx′′fxy′′fxy′′fyy′′)\nabla^2 f(a, b) = \bigl(\begin{smallmatrix} f''_{xx} & f''_{xy} \\ f''_{xy} & f''_{yy} \end{smallmatrix}\bigr) — quadratic coefficient.
  4. Assemble: f(x,y)=f(a,b)+∇f(a,b)(x−ay−b)+12[x−a  y−b]∇2f(a,b)(x−ay−b)+o(∥(x−a,y−b)∥2).f(x, y) = f(a,b) + \nabla f(a,b) \begin{pmatrix} x - a \\ y - b \end{pmatrix} + \tfrac{1}{2}[x - a \; y - b] \nabla^2 f(a,b) \begin{pmatrix} x - a \\ y - b \end{pmatrix} + o\bigl(\|(x-a, y-b)\|^2\bigr).
  5. If asked, simplify the polynomial by multiplying out.

Output. Explicit quadratic polynomial in (x−a,y−b)(x - a, y - b) plus the Peano remainder.

Pitfalls.

  • Forgetting the 1/21/2 factor on the quadratic term.
  • Losing the factor 2 on the cross term: when you expand the matrix product, the xyxy-cross term has coefficient fxy′′⋅2/2=fxy′′f''_{xy} \cdot 2 / 2 = f''_{xy} (because aij+aji=2fxy′′a_{ij} + a_{ji} = 2 f''_{xy}, divided by 2 of the Taylor factor).
  • Sign errors on fy′f'_y when y0=−π/2y_0 = -\pi/2: sin⁡(−π/2)=−1\sin(-\pi/2) = -1, cos⁡(−π/2)=0\cos(-\pi/2) = 0.

M-DC-6 — Apply Dini's theorem to find f′(x0)f'(x_0) from g(x,y)=cg(x, y) = c

Used on: Final exercises Q3 (canonical example), lect5 exercises.

Input. A C(1)C^{(1)} function g:A⊆R2→Rg : A \subseteq \mathbb{R}^2 \to \mathbb{R}, a constant cc, and (ideally) a known x0x_0.

Steps.

  1. Locate the base point (x0,y0)(x_0, y_0) satisfying g(x0,y0)=cg(x_0, y_0) = c. If only x0x_0 is given, substitute and solve for y0y_0.
  2. Verify Dini's hypothesis: compute gy(x0,y0)=∂g/∂y (x0,y0)g_y(x_0, y_0) = \partial g/\partial y \,(x_0, y_0) and check gy(x0,y0)≠0g_y(x_0, y_0) \neq 0. If it is zero, Dini's theorem does not apply — you cannot guarantee a local implicit function.
  3. Compute gx(x0,y0)g_x(x_0, y_0).
  4. Apply the formula: f′(x0)  =  − gx(x0,y0)gy(x0,y0).f'(x_0) \;=\; -\,\frac{g_x(x_0, y_0)}{g_y(x_0, y_0)}.
  5. Sign check. The minus sign is not optional.

Output. The numerical value of f′(x0)f'(x_0) (or a symbolic expression).

Pitfalls.

  • Forgetting the minus sign — the single most common error. The formula is f′(x0)=−gx/gyf'(x_0) = -g_x / g_y.
  • Verifying gy≠0g_y \neq 0 at the wrong point (e.g., at (x0,0)(x_0, 0) when actually y0≠0y_0 \neq 0): always substitute the correct (x0,y0)(x_0, y_0).
  • Missing the existence question: Dini gives local uniqueness in a neighbourhood. If the problem asks for a global function, check the extra hypotheses of the uniqueness theorem (strict monotonicity in yy).
  • Applying Dini at a point not on the level curve: if g(x0,y0)≠cg(x_0, y_0) \neq c, there is nothing to expand around — re-derive y0y_0.

M-DC-7 — Existence of global extrema via coercivity and Weierstrass

Used on: lect4 "Maximisers with Tonelli's Theorem", auxiliary in Final exercises.

Input. A continuous function f:A⊆Rn→Rf : A \subseteq \mathbb{R}^n \to \mathbb{R} and a (possibly unbounded) admissible set AA.

Steps (case A=RnA = \mathbb{R}^n).

  1. Check continuity of ff.
  2. Coercivity test: examine lim⁡∥x∥→∞f(x)\lim_{\|\mathbf{x}\| \to \infty} f(\mathbf{x}).
    • If =+∞= +\infty: ff is coercive ⇒ by Tonelli, ff attains a global minimum.
    • If =−∞= -\infty: ff is super-coercive towards −∞-\infty ⇒ attains a global maximum.
    • If neither (e.g., oscillating or going to a finite limit): the existence theorems do not apply directly — try relaxation, Lagrange, or boundary analysis.
  3. Composition rule: if f=h∘gf = h \circ g, check if the outer hh is strictly increasing (preserves coercivity of gg) or strictly decreasing (flips it).
  4. Find all stationary and exceptional points (where ∇f=0\nabla f = \mathbf{0}, plus points where ff is not differentiable).
  5. Evaluate ff at each candidate and pick the extremum.

Steps (case A⊆RnA \subseteq \mathbb{R}^n compact).

  1. Weierstrass applies directly: ff attains its max and min on AA.
  2. Candidates: interior stationary points of f∣Af|_A, boundary points, and non-differentiability points.
  3. Evaluate ff at each; the biggest/smallest wins.

Steps (relaxation for non-compact, non-coercive problems).

  1. Solve the unconstrained problem min⁡/max⁡f\min/\max f on Rn\mathbb{R}^n.
  2. If the unconstrained optimum lies in AA, it is also the constrained optimum.
  3. If not, the constrained problem likely has no solution via this method — switch to Lagrange or boundary analysis.

Pitfalls.

  • Unbounded set + continuous function is not enough. Weierstrass requires compact (closed and bounded).
  • Forgetting that a minimum may not exist. f(x)=exf(x) = e^x on R\mathbb{R} is continuous but not coercive (goes to 00 at −∞-\infty); inf⁡=0\inf = 0 is not attained.
  • Sub-vs. super-coercive confusion. Coercive = "+∞+\infty at infinity" (guarantees minimum). Super-coercive towards −∞-\infty = "−∞-\infty at infinity" (guarantees maximum). Different sign, different conclusion.

§6. Practice Problems with Solutions

Problem 6.1 — May 2024 MCQ2 (Mode A) / MCQ1 (Mode B): second-order Taylor

The Taylor formula of order two for f(x,y)=x2+xcos⁡yf(x, y) = x^2 + x \cos y, centred at (1,−π/2)(1, -\pi/2), is …

Solution. See Example 4.2. The answer is "none of the others" (option D Mode A / option C Mode B), since the simplification yields f(x,y)=π2x+x2+xy+o(∥(x−1,y+π/2)∥2)f(x,y) = \tfrac{\pi}{2} x + x^2 + x y + o\bigl(\|(x-1, y+\pi/2)\|^2\bigr), which does not match options A–C.

Problem 6.2 — May 2024 Q8 (Mode A) / Q7 (Mode B): Hessian classification and global convexity

(a) Let C⊆RnC \subseteq \mathbb{R}^n be a convex set and f:C→Rf : C \to \mathbb{R} be continuously twice differentiable. State the Concavity / Strict Concavity test on the basis of the sign of the Hessian matrix. (b) Consider f(x,y)=ex(x−1) y+y2f(x, y) = e^x(x-1)\,y + y^2. (b1) Write the gradient vector of ff and determine the stationary points. (b2) Write the Hessian matrix and explain the nature of the stationary points. (b3) Is ff convex on all R2\mathbb{R}^2? Justify.

(a) Statement of Prop. 1486 (concavity / strict concavity test). Let C⊆RnC \subseteq \mathbb{R}^n be an open convex set and f∈C(2)(C)f \in C^{(2)}(C). Then:

  • ff is concave on CC   ⟺  \iff ∇2f(x)\nabla^2 f(\mathbf{x}) is negative semi-definite for every x∈C\mathbf{x} \in C.
  • ff is strictly concave on CC   ⟺  \iff ∇2f(x)\nabla^2 f(\mathbf{x}) is negative definite for every x∈C\mathbf{x} \in C.

Analogously for convexity (positive semi-definite / positive definite).

(b1) See Example 4.1. Gradient ∇f(x,y)=[xyex,  2y+ex(x−1)]\nabla f(x,y) = [x y e^x, \; 2y + e^x(x-1)]. Stationary points: P=(0,1/2)P = (0, 1/2) and Q=(1,0)Q = (1, 0).

(b2) Hessian ∇2f(x,y)=(yex(x+1)xexxex2)\nabla^2 f(x,y) = \begin{pmatrix} y e^x (x+1) & x e^x \\ x e^x & 2 \end{pmatrix}.

  • At P=(0,1/2)P = (0, 1/2): ∇2f(P)=(1/2002)\nabla^2 f(P) = \bigl(\begin{smallmatrix} 1/2 & 0 \\ 0 & 2 \end{smallmatrix}\bigr) — positive definite (both leading minors positive). By Theorem 1391, PP is a strong local minimiser.
  • At Q=(1,0)Q = (1, 0): ∇2f(Q)=(0ee2)\nabla^2 f(Q) = \bigl(\begin{smallmatrix} 0 & e \\ e & 2 \end{smallmatrix}\bigr), det⁡=−e2<0\det = -e^2 < 0 — indefinite. QQ is a saddle (not an optimiser).

(b3) ff is not convex on R2\mathbb{R}^2. A witness is (1,0)(1, 0), where ∇2f(1,0)\nabla^2 f(1, 0) is indefinite — not positive semi-definite. By Prop. 1486, this rules out global convexity.

Source: General_24524_ENG_SOL.pdf Q8/Q7.

Problem 6.3 — Final exercises Q2: full optimization on a log-domain

Given f(x,y)=x+3y2−ln⁡(x+2y)f(x, y) = x + 3 y^2 - \ln(x + 2y). (a) Determine any local max/min point. (b) Can we deduce whether they are global extrema? (c) Find the 2nd-order Taylor formula in a neighbourhood of (1/3,1/3)(1/3, 1/3).

Solution. See Example 4.4.

  • (a) Domain: {x+2y>0}\{x + 2y > 0\}. Unique stationary point (1/3,1/3)(1/3, 1/3). Hessian there (12210)\bigl(\begin{smallmatrix}1 & 2 \\ 2 & 10\end{smallmatrix}\bigr) is PD ⇒ strong local minimiser.
  • (b) Yes. The Hessian is PD everywhere on the (open, convex) domain, so ff is strictly convex on its domain. By Theorem 1497, the unique stationary point is the unique global minimiser.
  • (c) With f(1/3,1/3)=2/3f(1/3, 1/3) = 2/3, ∇f(1/3,1/3)=0\nabla f(1/3, 1/3) = \mathbf{0}, ∇2f(1/3,1/3)=(12210)\nabla^2 f(1/3, 1/3) = \bigl(\begin{smallmatrix}1 & 2 \\ 2 & 10\end{smallmatrix}\bigr): f(x,y)=23+12[(x−13)2+10(y−13)2+4(x−13)(y−13)]+o(∥(x−1/3,y−1/3)∥2).f(x, y) = \tfrac{2}{3} + \tfrac{1}{2}\Bigl[(x - \tfrac{1}{3})^2 + 10(y - \tfrac{1}{3})^2 + 4(x - \tfrac{1}{3})(y - \tfrac{1}{3})\Bigr] + o\bigl(\|(x - 1/3, y - 1/3)\|^2\bigr).

Source: Final exercises solution.pdf pp.4–6, Problem 2.

Problem 6.4 — Final exercises Q3: Dini's theorem

Given g(x,y)=2yex+xey−2g(x, y) = 2 y e^x + x e^y - 2. Find f′(0)f'(0) of the function y=f(x)y = f(x) implicitly defined by g(x,y)=−2g(x, y) = -2.

Solution. See Example 4.5. f′(0)=−1/2f'(0) = -1/2.

Source: Final exercises solution.pdf p.7, Problem 3.

Problem 6.5 — lect 4: find the extrema of f(x,y)=x2+y2+x+yf(x,y) = x^2 + y^2 + x + y

(a) Find the gradient. (b) Find the stationary point(s). (c) Determine whether it is a point of extrema.

Solution.

  • (a) ∇f(x,y)=[2x+1,  2y+1]\nabla f(x,y) = [2x + 1, \; 2y + 1].
  • (b) ∇f=0\nabla f = \mathbf{0} gives x=−1/2,y=−1/2x = -1/2, y = -1/2. Unique stationary point (−1/2,−1/2)(-1/2, -1/2).
  • (c) ∇2f(x,y)=(2002)\nabla^2 f(x, y) = \bigl(\begin{smallmatrix}2 & 0 \\ 0 & 2\end{smallmatrix}\bigr) for every (x,y)(x,y). det⁡A1=2,det⁡A2=4\det A_1 = 2, \det A_2 = 4 — positive definite everywhere. So ff is strictly convex on R2\mathbb{R}^2 (convex open domain). By Theorem 1497, (−1/2,−1/2)(-1/2, -1/2) is the strong global minimiser.

Source: lect 4_calc_diff (3).pdf.

Problem 6.6 — lect3: f(x,y)=x2y−y2x−3xf(x, y) = x^2 y - y^2 x - 3 x

Find the points of local extrema.

Solution.

Gradient. fx′=2xy−y2−3f'_x = 2 x y - y^2 - 3, fy′=x2−2xyf'_y = x^2 - 2 x y.

Stationary points. From fy′=0f'_y = 0: x(x−2y)=0x(x - 2y) = 0, so x=0x = 0 or x=2yx = 2y.

  • Case x=0x = 0: fx′=0−y2−3=0⇒y2=−3f'_x = 0 - y^2 - 3 = 0 \Rightarrow y^2 = -3 — no real solution.
  • Case x=2yx = 2y: fx′=2⋅2y⋅y−y2−3=4y2−y2−3=3y2−3=0⇒y2=1⇒y=±1f'_x = 2 \cdot 2y \cdot y - y^2 - 3 = 4 y^2 - y^2 - 3 = 3 y^2 - 3 = 0 \Rightarrow y^2 = 1 \Rightarrow y = \pm 1. Corresponding x=±2x = \pm 2. Stationary points (2,1)(2, 1) and (−2,−1)(-2, -1).

Hessian. fxx′′=2y,fxy′′=2x−2y,fyy′′=−2xf''_{xx} = 2y, f''_{xy} = 2x - 2y, f''_{yy} = -2x. So ∇2f(x,y)=(2y2x−2y2x−2y−2x)\nabla^2 f(x,y) = \bigl(\begin{smallmatrix}2y & 2x - 2y \\ 2x - 2y & -2x\end{smallmatrix}\bigr).

  • At (2,1)(2, 1): (222−4)\bigl(\begin{smallmatrix}2 & 2 \\ 2 & -4\end{smallmatrix}\bigr), det⁡A1=2,det⁡A2=−8−4=−12<0\det A_1 = 2, \det A_2 = -8 - 4 = -12 < 0 — indefinite ⇒ saddle.
  • At (−2,−1)(-2, -1): (−2−2−24)\bigl(\begin{smallmatrix}-2 & -2 \\ -2 & 4\end{smallmatrix}\bigr), det⁡A1=−2,det⁡A2=−8−4=−12<0\det A_1 = -2, \det A_2 = -8 - 4 = -12 < 0 — indefinite ⇒ saddle.

Answer. Both stationary points are saddles; ff has no local extrema.

Source: Lect3_calc_diff (3).pdf.

Problem 6.7 — Final exercises Q1 recap with Differential-Calculus lens

f(x,y,z)=kx2+ky2−4z2−2xzf(x, y, z) = k x^2 + k y^2 - 4 z^2 - 2 x z. Find conditions on kk for the origin to be a strong global maximum.

Solution. The origin is stationary for every quadratic form (all partial derivatives vanish at 0\mathbf{0}). For it to be a strong global max, we need f(x)<0f(\mathbf{x}) < 0 for every x≠0\mathbf{x} \neq \mathbf{0} — i.e., the associated matrix AA must be negative definite. A=(k0−10k0−10−4)A = \bigl(\begin{smallmatrix}k & 0 & -1 \\ 0 & k & 0 \\ -1 & 0 & -4\end{smallmatrix}\bigr). Apply Sylvester–Jacobi with alternating pattern det⁡A1<0,det⁡A2>0,det⁡A3<0\det A_1 < 0, \det A_2 > 0, \det A_3 < 0: k<0,k2>0  (trivially, k≠0),−4k2−k<0  ⟺  k(4k+1)>0.k < 0, \quad k^2 > 0\,\text{ (trivially, }k \neq 0\text{)}, \quad -4 k^2 - k < 0 \iff k(4k + 1) > 0. Combining: k<−1/4k < -1/4. (This links Diff-Calc to §01 Linear Algebra — the same AA appears in Final Q1b, eigenvalue signing.)

Source: Final exercises solution.pdf pp.1–3, Problem 1.

Problem 6.8 — lect5: Dini on g(x,y)=x2−2y−ey=0g(x, y) = x^2 - 2y - e^y = 0, find f′(1)f'(1)

Setup. Verify gg is in the range of y↦g(x,y)y \mapsto g(x, y): as y→−∞y \to -\infty, g→x2−(−∞)−0=+∞g \to x^2 - (-\infty) - 0 = +\infty; as y→+∞y \to +\infty, g→x2−∞−∞=−∞g \to x^2 - \infty - \infty = -\infty. So 0∈(range)0 \in (\text{range}) and by the existence theorem there is y=f(x)y = f(x). Uniqueness: gy(x,y)=−2−ey<0g_y(x, y) = -2 - e^y < 0 everywhere ⇒ gg is strictly decreasing in yy ⇒ ff is unique.

Find extrema of ff. Dini gives f′(x)=−gx/gy=−2x/(−2−ey)=2x/(2+ey)f'(x) = -g_x / g_y = -2x / (-2 - e^y) = 2x / (2 + e^y). Stationary points: f′(x)=0  ⟺  x=0f'(x) = 0 \iff x = 0. First-derivative test: f′f' changes from −- (when x<0x<0) to ++ (when x>0x>0), so x=0x = 0 is a strong local+global minimiser of ff.

1st-order Taylor at (1,0)(1, 0). With y=f(x)y = f(x), f(1)=0f(1) = 0 (from solving 1−0−1=01 - 0 - 1 = 0), f′(1)=2⋅1/(2+1)=2/3f'(1) = 2 \cdot 1 / (2 + 1) = 2/3: f(x)=f(1)+f′(1)(x−1)+o(x−1)=23(x−1)+o(x−1).f(x) = f(1) + f'(1)(x - 1) + o(x - 1) = \tfrac{2}{3}(x - 1) + o(x - 1).

Source: lect5_calc_diff.pdf.

§7. Common Pitfalls

Below are the errors Sicconi flags most often and the traps the Bocconi exam-writers exploit. Read this list the day before the exam.

  1. Domain constraints.

    • For ln⁡(u)\ln(u): require u>0u > 0.
    • For u\sqrt{u}: require u≥0u \ge 0 (and u>0u > 0 for differentiability).
    • For 1/u1/u: require u≠0u \neq 0.
    • Before declaring a candidate a stationary point, always verify it lies in Dom(f)\text{Dom}(f). Algebraic solutions outside the domain must be rejected.

  2. Strict vs non-strict convexity.

    • Strictly convex: positive definite Hessian (plus domain convex, open).
    • Convex: positive semi-definite Hessian.
    • Difference matters for uniqueness of the global minimum: strict convexity ⇒ at most one global min; convex ⇒ possibly a whole flat region of minima.

  3. Local vs global without convexity.

    • Theorem 1391 (2nd-order conditions) gives local extrema only. It cannot distinguish a local from a global without extra info.
    • To upgrade to global, you need convexity (Theorem 1497) or coercivity (Tonelli) or compactness (Weierstrass).
    • May 2024 Q8 b2 → b3: (b2) proves P=(0,1/2)P = (0, 1/2) is a local min. (b3) explains the Hessian fails PSD elsewhere ⇒ cannot conclude global convexity ⇒ cannot conclude PP is global min via this route.

  4. Semi-definite Hessian: Theorem 1391 is inconclusive.

    • Positive semi-definite Hessian at x0\mathbf{x}_0 is only a necessary condition for x0\mathbf{x}_0 being a local min — not sufficient.
    • Example 4.3 (f=x2yf = x^2 y on the yy-axis) illustrates: Hessian is PSD, but x0=(0,0)\mathbf{x}_0 = (0, 0) is a saddle. Fall back to direct inspection or higher-order Taylor.

  5. Dini's formula sign error.

    • The formula is f′(x0)=−gx/gyf'(x_0) = -g_x / g_y, with minus. Forgetting the minus sign flips the answer.
    • Always write the minus sign first.

  6. Hessian symmetry: Schwarz must hold.

    • If you compute fxy′′f''_{xy} and fyx′′f''_{yx} and they disagree for a C(2)C^{(2)} function, you have an arithmetic error.
    • Conversely, use Schwarz to save time: once you compute fxy′′f''_{xy}, the (2,1)(2,1) entry of the Hessian is automatic.

  7. Negative-definite Sylvester pattern: alternating, not all-negative.

    • Pattern: det⁡A1<0\det A_1 < 0, det⁡A2>0\det A_2 > 0, det⁡A3<0\det A_3 < 0, det⁡A4>0\det A_4 > 0, …
    • Not all-negative. Consistent with "−A-A positive definite ⇒ eigenvalues of −A-A all >0>0 ⇒ eigenvalues of AA all <0<0", and the leading minors of −A-A are (−1)kdet⁡Ak(-1)^k \det A_k.

  8. Taylor coefficients: the 1/21/2 and the factor 2 on the cross term.

    • Quadratic form 12hTHh\tfrac{1}{2} \mathbf{h}^T H \mathbf{h} expands to 12[h12fxx′′+2h1h2fxy′′+h22fyy′′]\tfrac{1}{2}[h_1^2 f''_{xx} + 2 h_1 h_2 f''_{xy} + h_2^2 f''_{yy}]. The cross-term coefficient is fxy′′f''_{xy} (not 2fxy′′2 f''_{xy} and not fxy′′/2f''_{xy}/2).

  9. Boundary points are not stationary in the gradient sense.

    • On a closed set with non-empty boundary, the max/min may live on the boundary where ∇f≠0\nabla f \neq \mathbf{0}.
    • Fermat requires x0\mathbf{x}_0 to be an interior point. Apply boundary analysis separately.

  10. Coercivity gives existence, not uniqueness.

    • Tonelli says a global min exists — not that it is unique.
    • Uniqueness requires strict convexity (or a direct argument that the stationary point is unique and other candidates are ruled out).

  11. Composition of coercive functions with monotone outer.

    • hh strictly increasing + gg coercive towards +∞+\infty ⇒ h∘gh \circ g coercive towards lim⁡+∞h\lim_{+\infty} h.
    • hh strictly increasing + gg super-coercive towards −∞-\infty ⇒ h∘gh \circ g tends to lim⁡−∞h\lim_{-\infty} h at infinity.
    • If hh is strictly decreasing, the conclusions flip.

  12. Ignoring the convexity of the domain when applying Prop. 1486.

    • Prop. 1486 requires CC to be an open convex set. Half-planes, open balls, and all of Rn\mathbb{R}^n qualify; annuli and punctured disks do not (not convex).

  13. "The function is strictly concave ⇒ at most one global max".

    • Strict convexity / concavity ensures uniqueness of the global extremum if one exists. Existence still needs Weierstrass / Tonelli.

  14. Reading the MCQ options carefully.

    • Exam wording tests common distractors: wrong Dini sign; missing factor of 1/21/2 in Taylor; using Hessian of f(x,y)f(x,y) at (0,0)(0,0) instead of the required (1,−π/2)(1, -\pi/2).
    • If the simplified Taylor polynomial doesn't match any of options A–C, the correct answer is very likely "none of the others" — trust the calculation and resist the urge to "round" to a listed option.

Final checklist before the exam (differential calculus slice)

  • [ ] I can state and prove Theorem 1391 (sufficient condition for strong local max/min), including the Weierstrass-on-unit-sphere step.
  • [ ] I can state Theorem 1481 (2nd-order Taylor with Peano remainder) and use it in computations.
  • [ ] I can state Prop. 1486 (Hessian sign ⇔ convexity/concavity) and use it on both sides.
  • [ ] I can state and prove Theorem 1497 (∇f = 0 ⇔ global extremum for convex/concave differentiable functions), including the use of Fermat and the tangent-hyperplane inequality.
  • [ ] I can state Theorem 1591 and know it extends 1497 to non-open convex sets via interior points.
  • [ ] I can state Fermat's theorem (∇f = 0 at interior local extrema) and sketch its proof.
  • [ ] I can state Dini's theorem and apply the formula f′(x0)=−gx/gyf'(x_0) = -g_x / g_y without dropping the sign.
  • [ ] I can state Weierstrass (compact ⇒ max/min) and Tonelli (coercive ⇒ min / super-coercive towards −∞-\infty ⇒ max).
  • [ ] I can run M-DC-1 through M-DC-7 on a R2\mathbb{R}^2 problem in under 20 minutes.
  • [ ] I know that semi-definite Hessian leaves Theorem 1391 inconclusive, and I have a direct-inspection fallback.
  • [ ] I can distinguish "local", "strong local", "global", "strong global", and the role of convexity in upgrading each.

End of Differential Calculus study guide.