Proof of The Quartic Bump: Making a Local Maximum Strict without Changing the Test Data

Conventions. The order $\le$ and the arithmetic of $\mathbb{R}$ are those of the ordered field of real numbers; among its axioms is the compatibility of $\le$ with addition, which is what we use for non-strict inequalities, claim 1 of Elementary Order Arithmetic in an Ordered Field being its strict counterpart. Multiplication by a nonnegative real number preserves $\le$: if $a\le b$ and $0\le\lambda$ then either $a=b$, and the products are equal, or $a<b$, and then claim 10 of Elementary Order Arithmetic in an Ordered Field applies when $0<\lambda$, while $\lambda=0$ makes both products $0$ by claim 1 of Zero Products and Elementary Identities in a Field. Partial derivatives are those of that definition, the gradient is as in that definition and the Hessian as in that definition, so that the $i$th coordinate of $D f(x)$ is $\partial_{i}f(x)$ and the entry of $D^{2}f(x)$ in row $i$ and column $j$ is $\partial_{i}\partial_{j}f(x)$. Being of class $C^{2}$ on an open set means, by clauses 2 and 3 there, being of class $C^{1}$ with every first partial derivative of class $C^{1}$.

Step 1: the squared norm as a quadratic. By Euclidean Space and Real Matrices: Standing Notation (iii) the matrix $2I_{n}$ lies in $\mathcal{S}(n)$, so Quadratic and Affine Functions of Class $C^2$, Translation, and Quadratic Perturbation of Semiconvexity (i) applies to the function $Q:\mathbb{R}^{n}\to\mathbb{R}$ given by

$$Q(w)=\tfrac{1}{2}\,w\cdot\bigl((2I_{n})w\bigr)+0_{\mathbb{R}^{n}}\cdot w+0 :$$

$Q$ is of class $C^{2}$ on $\mathbb{R}^{n}$, its restriction to any open subset of $\mathbb{R}^{n}$ is of class $C^{2}$ there, and at every $w$ the gradient is $(2I_{n})w$ and the Hessian is $2I_{n}$.

Moreover $Q(w)=\lVert w\rVert^{2}$ for every $w\in\mathbb{R}^{n}$. Indeed $(2I_{n})w=2(I_{n}w)=2w$ by claims 1 and 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum; then $w\cdot(2w)=2(w\cdot w)$ by claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite and $w\cdot w=\lVert w\rVert^{2}$ by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, so the first summand is $\tfrac{1}{2}\bigl(2\lVert w\rVert^{2}\bigr)=\lVert w\rVert^{2}$; and $0_{\mathbb{R}^{n}}\cdot w=0$ by the coordinate formula for the dot product, every summand there being a product with $0$, which vanishes by claim 1 of Zero Products and Elementary Identities in a Field. Consequently $DQ(w)=2w$ and $D^{2}Q(w)=2I_{n}$ for every $w$.

Step 2: the squared distance to $z$. Applying Quadratic and Affine Functions of Class $C^2$, Translation, and Quadratic Perturbation of Semiconvexity (ii) to the open set $V$ with $b=z$ shows that $V-z=\{w\in\mathbb{R}^{n}:w+z\in V\}$ is open. By Step 1 the restriction $Q|_{V-z}$ is of class $C^{2}$ on $V-z$, with gradient $2w$ and Hessian $2I_{n}$ at each $w\in V-z$. Applying Quadratic and Affine Functions of Class $C^2$, Translation, and Quadratic Perturbation of Semiconvexity (ii) again, now to the open set $V-z$, the function $Q|_{V-z}$ and $b=-z$, and noting that

$$(V-z)-(-z)=\{y\in\mathbb{R}^{n}:y-z\in V-z\}=\{y\in\mathbb{R}^{n}:(y-z)+z\in V\}=V,$$

we conclude that the function $g:V\to\mathbb{R}$ given by $g(y)=Q(y-z)=\lVert y-z\rVert^{2}$ is of class $C^{2}$ on $V$, with

$$Dg(y)=2(y-z),\qquad D^{2}g(y)=2I_{n}\qquad(y\in V).$$

In particular $z-z=0_{\mathbb{R}^{n}}$, so $g(z)=\lVert 0_{\mathbb{R}^{n}}\rVert^{2}=0$ by claim 3 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, and $Dg(z)=2\cdot0_{\mathbb{R}^{n}}=0_{\mathbb{R}^{n}}$, every coordinate of the scalar multiple being a product with $0$ and so vanishing by claim 1 of Zero Products and Elementary Identities in a Field; whence $\partial_{i}g(z)=0$ for every $i\in\{1,\dots,n\}$, the coordinates of the gradient being the first partial derivatives.

Step 3: proof of claim 1. For $x\in V$ we have $\beta(x)=g(x)\,g(x)$, so $\beta$ is the pointwise product of $g$ with itself and is of class $C^{2}$ on $V$ by claim 3 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions.

By claim 1 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions, for every $j\in\{1,\dots,n\}$ and every $x\in V$,

$$\partial_{j}\beta(x)=\partial_{j}g(x)\,g(x)+g(x)\,\partial_{j}g(x).$$

Since $g$ is of class $C^{2}$ on $V$, each $\partial_{j}g$ is of class $C^{1}$ on $V$, so claim 1 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions may be applied once more to each of the two products on the right: for all $i,j\in\{1,\dots,n\}$ and every $x\in V$,

$$\partial_{i}\partial_{j}\beta(x)=\partial_{i}\partial_{j}g(x)\,g(x)+\partial_{j}g(x)\,\partial_{i}g(x)+\partial_{i}g(x)\,\partial_{j}g(x)+g(x)\,\partial_{i}\partial_{j}g(x).$$

Evaluating at $x=z$ and using $g(z)=0$ and $\partial_{i}g(z)=\partial_{j}g(z)=0$ from Step 2, together with claim 1 of Zero Products and Elementary Identities in a Field, every summand in both displays vanishes. Hence $\partial_{j}\beta(z)=0$ for every $j$, so $D\beta(z)=0_{\mathbb{R}^{n}}$, and $\partial_{i}\partial_{j}\beta(z)=0$ for all $i,j$, so every entry of $D^{2}\beta(z)$ is $0$, that is, $D^{2}\beta(z)=0_{n}$. This proves claim 1.

Step 4: proof of claim 2. Let $x\in V$. By claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square the number $\lVert x-z\rVert$ is nonnegative, hence so is $\lVert x-z\rVert^{2}$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, which applied with $0$ and $\lVert x-z\rVert$ gives $0\cdot0\le\lVert x-z\rVert^{2}$, and $0\cdot0=0$ by claim 1 of Zero Products and Elementary Identities in a Field; multiplying the inequality $0\le\lVert x-z\rVert^{2}$ by the nonnegative number $\lVert x-z\rVert^{2}$ and using claim 1 of Zero Products and Elementary Identities in a Field gives $0\le\beta(x)$.

If $\beta(x)=0$, that is $\lVert x-z\rVert^{2}\lVert x-z\rVert^{2}=0$, then $\lVert x-z\rVert^{2}=0$ by claim 3 of Zero Products and Elementary Identities in a Field, and then $\lVert x-z\rVert=0$ by that same claim, so $x-z=0_{\mathbb{R}^{n}}$ by claim 3 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and hence $x=z$. Conversely $\beta(z)=\lVert 0_{\mathbb{R}^{n}}\rVert^{2}\lVert 0_{\mathbb{R}^{n}}\rVert^{2}=0$ by that same claim 3 and claim 1 of Zero Products and Elementary Identities in a Field. This proves claim 2.

Step 5: proof of claim 3. The function $\psi=\varphi+\beta$ is the pointwise sum of two functions of class $C^{2}$ on $V$, hence is of class $C^{2}$ on $V$ by claim 3 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions. By claim 1 there, $\partial_{j}\psi(z)=\partial_{j}\varphi(z)+\partial_{j}\beta(z)=\partial_{j}\varphi(z)$ for every $j$, using the theorem's claim 1, proved in Step 3; and applying claim 1 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions to the sum $\partial_{j}\psi=\partial_{j}\varphi+\partial_{j}\beta$ gives $\partial_{i}\partial_{j}\psi(z)=\partial_{i}\partial_{j}\varphi(z)+\partial_{i}\partial_{j}\beta(z)=\partial_{i}\partial_{j}\varphi(z)$ for all $i,j$. Since the coordinates of a gradient are the first partial derivatives and the entries of a Hessian the second, $D\psi(z)=D\varphi(z)$ and $D^{2}\psi(z)=D^{2}\varphi(z)$.

By the definition of a local maximum there is a positive $\delta\in\mathbb{R}$ such that every $x\in S$ with $d_{E}(z,x)<\delta$ satisfies

$$h(x)-\varphi(x)\le h(z)-\varphi(z).$$

Let $x\in S$ satisfy $d_{E}(z,x)<\delta$ and $x\ne z$. By claim 2 we have $0\le\beta(x)$ and $\beta(x)\ne0$, so $0<\beta(x)$ and hence $-\beta(x)<0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field. Adding $-\beta(x)$ to both sides of the displayed inequality, which the compatibility of $\le$ with addition permits, gives

$$h(x)-\psi(x)=\bigl(h(x)-\varphi(x)\bigr)-\beta(x)\le\bigl(h(z)-\varphi(z)\bigr)-\beta(x),$$

while adding $h(z)-\varphi(z)$ to $-\beta(x)<0$ gives, by claim 1 of Elementary Order Arithmetic in an Ordered Field,

$$\bigl(h(z)-\varphi(z)\bigr)-\beta(x)<h(z)-\varphi(z)=h(z)-\psi(z),$$

the last equality because $\beta(z)=0$ by claim 2. Claim 2 of Elementary Order Arithmetic in an Ordered Field combines the two comparisons into $h(x)-\psi(x)<h(z)-\psi(z)$. The same $\delta$ therefore witnesses, by the definition of a strict local maximum, that $x\mapsto h(x)-\psi(x)$ has a strict local maximum at $z$ relative to $S$. This proves claim 3.