Proof of First- and Second-Order Conditions at a Local Extremum

Throughout, the arithmetic and order of $\mathbb{R}$ are used through The Structure of ℝ as a Field and an Ordered Ring (i), The Structure of ℝ as a Field and an Ordered Ring (ii), Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (v), Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (ix), Rules of Arithmetic and Order in an Ordered Field (iii) and The Quotient 1/x, Order Rules for Sums, and Nesting of Closed Intervals in ℝ (i); $\mathbb{R}^{n}$ and $\mathbb{R}$ carry the metrics $d_{E}$ and $d_{\mathbb{R}}$ of $\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence (i) and The Absolute Value Metric on the Real Numbers (i); and $d_{E}(a,a+h)=\lVert h\rVert$ for $h\in\mathbb{R}^{n}$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (ii)). When $f$ has a local maximum or a local minimum at $a$ relative to $U$, we write $\delta_{0}>0$ for a radius as in Local Maximum of a Function Relative to a Subset of a Metric Space (i) or Local Minimum of a Function Relative to a Subset of a Metric Space (i).

We begin with (i). Suppose that $f$ has a local maximum at $a$; the case of a local minimum is the same with the inequalities reversed. Fix $i\in[n]$. Since $U$ is open, we may choose $\rho\in(0,\delta_{0}]$ with $B(a,\rho)\subseteq U$. On the interval $J=(-\rho,\rho)$ consider the curve

$$\gamma(t)=a+te_{i},\qquad t\in J,$$

whose values lie in $B(a,\rho)$, since $d_{E}(a,\gamma(t))=|t|\,\lVert e_{i}\rVert=|t|$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v), Standard Basis Vectors of ℝ^q: Coordinates, Orthonormality and Unit Length (iii)). The difference quotients of $\gamma$ are all equal to $e_{i}$, so $\gamma'(t)=e_{i}$ by The Derivative of a Vector-Valued Function as a Limit of Difference Quotients in Norm (i). By the chain rule Chain Rule for Differentiable Functions Between Euclidean Spaces (iv), the function $\psi=f\circ\gamma$ is differentiable at $0$, and since the $i$-th coordinate of a vector is its dot product with $e_{i}$ (Standard Basis Vectors of ℝ^q: Coordinates, Orthonormality and Unit Length (i), The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite (i)) and the $i$-th coordinate of $\nabla f(a)$ is $\partial_{i}f(a)$ (Gradient of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$ (i)),

$$\psi'(0)=\nabla f(a)\cdot e_{i}=\partial_{i}f(a).$$

For $t\in J$ we have $d_{E}(a,\gamma(t))=|t|<\delta_{0}$, hence $\psi(t)\le\psi(0)$; so $\psi$ attains a maximum over $J$ at $0$. The derivative vanishes at such an interior maximum, so $\psi'(0)=0$ (Vanishing of the Derivative at a Local Extremum Strictly Inside an Interval (i) applies since $-\rho/2<0<\rho/2$ in $J$, with $\delta=\rho$; Vanishing of the Derivative at a Local Extremum Strictly Inside an Interval (ii) in the case of a local minimum). Thus every component $\partial_{i}f(a)$ of $\nabla f(a)$ is $0$, and $\nabla f(a)=0$ by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iii).

We turn to (ii), and first to a local minimum. A function of class $C^{2}$ on $U$ is of class $C^{1}$, hence differentiable at $a$ (Basic Properties of Cᵏ Functions (ii), Basic Properties of Cᵏ Functions (v)), so $\nabla f(a)=0$ by the first part. Here $D^{2}f(a)\in\mathcal{S}(n)$ by Equality of Mixed Partial Derivatives, and Symmetry and Continuity of the Hessian (ii), and $-D^{2}f(a)\in\mathcal{S}(n)$ by Rules for Symmetric and Positive Semidefinite Matrices: Closure, Self-Adjointness, Polarization, Congruence and Cones (i). Suppose, for a contradiction, that some $v\in\mathbb{R}^{n}$ satisfies

$$c=v\cdot\bigl(D^{2}f(a)\,v\bigr)<0;$$

then $v\neq0$. The choices are made in this order: first $v$ and $c$; then

$$\varepsilon=\frac{-c}{4\lVert v\rVert^{2}}>0;$$

then $\delta>0$ for this $\varepsilon$ by the Peano form of the remainder, Second-Order Taylor Expansion (ii); and finally a real number $t>0$ with $t\lVert v\rVert<\min(\delta,\delta_{0})$. Put $h=tv$. By bilinearity of the dot product and linearity of the matrix-vector product (The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite (ii), The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite (iii), Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v)), $h\cdot\bigl(D^{2}f(a)\,h\bigr)=t^{2}c$, and $\lVert h\rVert=t\lVert v\rVert<\delta$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v)). Since $\nabla f(a)=0$, the Peano bound gives $a+h\in U$ and

$$f(a+h)-f(a)\le\tfrac{1}{2}t^{2}c+\varepsilon t^{2}\lVert v\rVert^{2}=\tfrac{1}{4}t^{2}c<0.$$

On the other hand $d_{E}(a,a+h)=\lVert h\rVert<\delta_{0}$, so $f(a)\le f(a+h)$ by the choice of $\delta_{0}$, a contradiction. Hence $v\cdot\bigl(D^{2}f(a)\,v\bigr)\ge0$ for every $v\in\mathbb{R}^{n}$; that is, $D^{2}f(a)$ is positive semidefinite (Positive Semidefinite and Positive Definite Real Matrices (i)).

For a local maximum the argument is the same. If $-D^{2}f(a)$ were not positive semidefinite, there would be $v$ with $c=v\cdot\bigl((-D^{2}f(a))\,v\bigr)<0$, that is, $v\cdot\bigl(D^{2}f(a)\,v\bigr)=-c>0$. Making the same choices of $\varepsilon$, $\delta$ and $t$, the Peano bound now gives

$$f(a+h)-f(a)\ge-\tfrac{1}{4}t^{2}c>0,$$

which contradicts $f(a+h)\le f(a)$.

Finally, (iii). Suppose first that $A=D^{2}f(a)$ is positive definite. Here $A\in\mathcal{S}(n)$ by Equality of Mixed Partial Derivatives, and Symmetry and Continuity of the Hessian (ii). We need a lower bound for the quadratic form of $A$. Let

$$S=\{u\in\mathbb{R}^{n}:\lVert u\rVert=1\}$$

be the unit sphere. It is compact by The Heine-Borel Theorem in $\mathbb{R}^n$, and Compactness of Closed Balls and the Unit Sphere (iii), for the topology of open subsets of $(\mathbb{R}^{n},d_{E})$ (The Open Sets of a Metric Space Form a Set, Given by a Predicative Formula (ii)), and nonempty, since $e_{1}\in S$. The function $q(u)=u\cdot(Au)$ is continuous on $S$ (Continuous Map Between Metric Spaces (ii); given $\varepsilon>0$ take $\delta=\varepsilon/(2\lVert A\rVert+1)$), because by The Operator Norm of a Symmetric Matrix: Quadratic Forms, Eigenvalues and the Semidefinite Ordering (x) and Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vi)

$$|q(u)-q(w)|\le2\lVert A\rVert\,\lVert u-w\rVert\qquad(u,w\in S).$$

By the extreme value theorem Extreme Value Theorem on a Compact Subset of a Metric Space (i) there is $u_{0}\in S$ with $q(u_{0})\le q(u)$ for all $u\in S$. Put $m=q(u_{0})$; it is positive because $u_{0}\neq0$ and $A$ is positive definite (Positive Semidefinite and Positive Definite Real Matrices (ii)). For $h\neq0$, writing $h=\lVert h\rVert u$ with $u=h/\lVert h\rVert\in S$, we obtain (The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite (ii), The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite (iii), Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v), Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v))

$$h\cdot(Ah)=\lVert h\rVert^{2}q(u)\ge m\lVert h\rVert^{2}.$$

Now the strict minimum. The choices are made in this order: first $m$, which depends only on $A$; then $\varepsilon=m/4$; then $\delta>0$ for this $\varepsilon$ by Second-Order Taylor Expansion (ii). Let $y\in U$ with $d_{E}(a,y)<\delta$ and $y\neq a$, and put $h=y-a$, so that $0<\lVert h\rVert<\delta$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (iii)). Since $\nabla f(a)=0$, the Peano bound and the lower bound for the quadratic form give

$$f(y)-f(a)\ge\tfrac{1}{2}m\lVert h\rVert^{2}-\tfrac{1}{4}m\lVert h\rVert^{2}=\tfrac{1}{4}m\lVert h\rVert^{2}>0.$$

So $f$ has a strict local minimum at $a$ relative to $U$, with radius $\delta$ (Local Minimum of a Function Relative to a Subset of a Metric Space (ii)).

If instead $-A$ is positive definite, where $-A\in\mathcal{S}(n)$ by Rules for Symmetric and Positive Semidefinite Matrices: Closure, Self-Adjointness, Polarization, Congruence and Cones (i), the lower bound applied to $-A$ gives $m>0$ with $h\cdot(Ah)\le-m\lVert h\rVert^{2}$ for $h\neq0$, and the same choices of $\varepsilon$ and $\delta$ give

$$f(y)-f(a)\le-\tfrac{1}{4}m\lVert h\rVert^{2}<0$$

for every $y\in U$ with $d_{E}(a,y)<\delta$ and $y\neq a$: a strict local maximum at $a$ relative to $U$ (Local Maximum of a Function Relative to a Subset of a Metric Space (ii)).