Proof of A Scaled Squared Distance to a Point is of Class $C^2$, with Gradient and Hessian

Throughout, $\lVert\,\cdot\,\rVert$ is the Euclidean norm on $\mathbb{R}^n$, $|\,\cdot\,|$ is the absolute value on $\mathbb{R}$, and finite sums are the finite sums in the field of real numbers. Claims are cited by number from Elementary Order Arithmetic in an Ordered Field (below, the order arithmetic lemma; claim 5 products of positives, claim 7 inverses of positives, claim 10 multiplication by a positive element), from Elementary Arithmetic in an Ordered Field (claim 1 $0\le1$, claim 2 sums, claim 5 multiplication by a nonnegative element), and from Properties of the Absolute Value in an Ordered Field (below, the absolute-value lemma).

Preliminaries. By claim 2 and then claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, for every $x=(x_1,\dots,x_n)\in\mathbb{R}^n$,

the $k$th coordinate of $x-a$ being $x_k-a_k$ by Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$. Hence, by claim 3 (homogeneity) of Properties of Finite Sums,

Square comparison. We record an elementary fact used twice below: if $r,s\in\mathbb{R}$ satisfy $0\le r$ and $0<s$, then $r<s$ if and only if $r^2<s^2$. Indeed, if $r<s$ then $r\cdot r\le r\cdot s$ by multiplication by the nonnegative $r$ and $r\cdot s<s\cdot s$ by strict compatibility with multiplication by the positive $s$, so $r^2<s^2$ by mixed transitivity. Conversely, if $s\le r$ then $s\cdot s\le s\cdot r$ and $s\cdot r\le r\cdot r$ by multiplication by the nonnegative elements $s$ and $r$, so $s^2\le r^2$ and $r^2<s^2$ fails; hence $r^2<s^2$ forces $r<s$.

Translation of Euclidean continuity. Let $E\subseteq\mathbb{R}^n$, let $f:E\to\mathbb{R}$ and let $b\in E$. We claim that $f$ is continuous at $b$ if and only if for every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ there is $\delta\in\mathbb{R}$ with $0<\delta$ such that every $x\in E$ with $d_E(x,b)<\delta$ satisfies $|f(x)-f(b)|<\varepsilon$. This is only a rewriting of that definition, taken with $m=1$ and with $f$ as the single coordinate function. In it, the hypothesis is $\sum_{k=1}^{n}(x_k-b_k)^2<\delta^2$, which by the displayed identity above equals $d_E(x,b)^2<\delta^2$ and hence, by the square comparison applied with $r=d_E(x,b)$ and $s=\delta$, is equivalent to $d_E(x,b)<\delta$; the conclusion is $(f(x)-f(b))^2<\varepsilon^2$, and since $t^2=|t|\cdot|t|$ for every real $t$ (by claim 1 of the absolute-value lemma, $|t|$ is $t$ or $-t$, and in either case $|t|\cdot|t|=t\cdot t$), the square comparison applied with $r=|f(x)-f(b)|$ and $s=\varepsilon$ shows that this is equivalent to $|f(x)-f(b)|<\varepsilon$. Below, continuous for real-valued functions on subsets of $\mathbb{R}^n$ always refers to this criterion.

Step 1: proof of claim 1. Fix $b=(b_1,\dots,b_n)\in U$ and $j\in\{1,\dots,n\}$, and apply Slice Function and the Partial Derivative to $q$ at $b$ in the $j$th variable: let $\rho\in\mathbb{R}$ with $0<\rho$ be as in claim 1 of that lemma, and let $I$ and the slice function $g:I\to\mathbb{R}$, $g(s)=q(b[s])$, be as in claim 2, where $b[s]$ is the point of $\mathbb{R}^n$ whose $j$th coordinate is $s$ and whose $k$th coordinate is $b_k$ for $k\ne j$, so that $b[b_j]=b$.

Let $h\in\mathbb{R}$ satisfy $0<|h|$ and $b_j+h\in I$. Using the displayed formula for $q$ and claims 2 and 3 of Properties of Finite Sums (additivity, and homogeneity with the scalar $-1$, which together give that a difference of two finite sums is the finite sum of the differences),

For $k\ne j$ we have $(b[b_j+h])_k=b_k$, so the $k$th summand is $0$; hence by claim 7 (a single possibly nonzero summand) of Properties of Finite Sums,

Write $t=b_j-a_j$, so that $b_j+h-a_j=t+h$. Elementary field arithmetic gives $(t+h)^2-t^2=h\cdot(2t+h)$, so

the division being legitimate because $h\ne0$. By claim 4 of the absolute-value lemma,

Now let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. The element $1+|c|$ satisfies $0<1+|c|$, since $0\le 1$, $0\le|c|$ and $1+|c|\ne 0$ (were $1+|c|=0$ we would get $1=-|c|\le 0$, contradicting $0\le 1$ and $1\ne 0$); hence $\delta=\varepsilon\cdot(1+|c|)^{-1}$ satisfies $0<\delta$. If $0<|h|<\delta$, then $|c|\,|h|\le(1+|c|)\,|h|$ by multiplication of $|c|\le 1+|c|$ by the nonnegative $|h|$, and $(1+|c|)\,|h|<(1+|c|)\,\delta=\varepsilon$ by strict compatibility with multiplication by the positive $1+|c|$; so

By Derivative at an Interior Point, $g$ is differentiable at $b_j$ with $g'(b_j)=2c\,t=2c\,(b_j-a_j)$, and by claim 2 of Slice Function and the Partial Derivative the partial derivative of $q$ with respect to the $j$th variable exists at $b$ and equals $2c\,(b_j-a_j)$. Since $b\in U$ and $j$ were arbitrary, claim 1 holds.

Step 2: continuity of $q$ and of its first partial derivatives. For $j\in\{1,\dots,n\}$ let $p_j:U\to\mathbb{R}$ be the function $p_j(x)=2c\,(x_j-a_j)$; by claim 1, $p_j$ is the function $\partial q/\partial x_j$ on $U$.

(i) Each $p_j$ is continuous at every point of $U$. Let $b\in U$ and $0<\varepsilon$. For $x\in U$ we have $p_j(x)-p_j(b)=2c\,(x_j-b_j)$, so by claim 4 of the absolute-value lemma and then claim 4 (coordinate bound) and claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$,

With $\delta=\varepsilon\cdot(1+|2c|)^{-1}$, which is positive for the reason given in Step 1, $d_E(x,b)<\delta$ gives $|2c|\,d_E(x,b)\le(1+|2c|)\,d_E(x,b)<(1+|2c|)\,\delta=\varepsilon$, whence $|p_j(x)-p_j(b)|<\varepsilon$.

(ii) $q$ is continuous at every point of $U$. Let $b\in U$ and $0<\varepsilon$, and set $r=\lVert x-a\rVert$ for a given $x\in U$ and $s=\lVert b-a\rVert$. By claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, $q(x)=c\,r^2$ and $q(b)=c\,s^2$, and $0\le r$, $0\le s$. Since $\mathbb{R}^n$ is a real vector space by Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, we have $x-a=(x-b)+(b-a)$, so claim 6 (triangle inequality) of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ gives $r\le\lVert x-b\rVert+s$, that is, $r-s\le\lVert x-b\rVert$. Exchanging the roles of $x$ and $b$ and using claim 5 of that lemma with $\lambda=-1$ together with $b-x=-(x-b)$ gives $s-r\le\lVert b-x\rVert=\lVert x-b\rVert$. By claim 6 (two-sided bound) of the absolute-value lemma, $|r-s|\le\lVert x-b\rVert=d_E(x,b)$. Also $r+s\le d_E(x,b)+2s$ and $0\le r+s$.

By elementary field arithmetic $r^2-s^2=(r-s)(r+s)$, so by claim 4 of the absolute-value lemma and $|r+s|=r+s$,

the last inequality being obtained by multiplying $|r-s|\le d_E(x,b)$ and $r+s\le d_E(x,b)+2s$ by nonnegative elements and using mixed transitivity. Let $\delta$ be the minimum of $1$ and $\varepsilon\cdot\bigl(1+|c|\,(1+2s)\bigr)^{-1}$; by claim 2 of Elementary Properties of the Minimum of Two Elements the value $\delta$ is one of these two positive numbers, so $0<\delta$, and by claim 1 of that lemma $\delta\le1$ and $\delta\le\varepsilon\cdot(1+|c|\,(1+2s))^{-1}$. If $x\in U$ satisfies $d_E(x,b)<\delta$, then $d_E(x,b)+2s<1+2s$, hence

where the middle inequality multiplies $d_E(x,b)+2s\le 1+2s$ by the nonnegative $|c|\,d_E(x,b)$, and the strict inequality multiplies $d_E(x,b)<\delta$ by the positive $1+|c|\,(1+2s)$ after bounding $|c|\,(1+2s)\le 1+|c|\,(1+2s)$. Thus $q$ is continuous at $b$.

Step 3: second partial derivatives. Fix $b=(b_1,\dots,b_n)\in U$ and $i,j\in\{1,\dots,n\}$, and apply Slice Function and the Partial Derivative to $p_j$ at $b$ in the $i$th variable, obtaining $\rho$, $I$ and the slice function $g_2(s)=p_j(b\langle s\rangle)$, where $b\langle s\rangle$ has $i$th coordinate $s$ and $k$th coordinate $b_k$ for $k\ne i$.

If $i=j$, then $g_2(s)=2c\,(s-a_j)$, so for $h\ne0$ with $b_i+h\in I$ we get $g_2(b_i+h)-g_2(b_i)=2c\,h$ and hence the difference quotient equals $2c$ exactly; so for every $\varepsilon$ with $0<\varepsilon$ any positive $\delta$ witnesses Derivative at an Interior Point, and $g_2'(b_i)=2c$.

If $i\ne j$, then $(b\langle s\rangle)_j=b_j$ for every $s\in I$, so $g_2$ is the constant function with value $2c\,(b_j-a_j)$, which by claim 2 of Sum and Product Rules for One-Dimensional Derivatives and Continuity is differentiable at $b_i$ with $g_2'(b_i)=0$.

By claim 2 of Slice Function and the Partial Derivative, the partial derivative of $p_j$ with respect to the $i$th variable exists at every point of $U$, and with the notation of C^2 Real-Valued Map on an Open Subset of Euclidean Space,

These values do not depend on $b$, so each function $\partial^2q/\partial x_i\,\partial x_j$ is constant on $U$ and hence continuous at every point of $U$ (for any $\varepsilon$ with $0<\varepsilon$, any positive $\delta$ works, since the difference of values is $0$ and $|0|=0<\varepsilon$).

Step 4: proof of claim 2. By Step 2(ii) the function $q$ is continuous at every point of $U$, by claim 1 its first partial derivatives exist at every point of $U$, and by Step 2(i) each function $\partial q/\partial x_j=p_j$ is continuous at every point of $U$; so $q$ is of class $C^1$ on $U$, that definition being applied with $m=1$ and with $q$ as the single coordinate function. Fix $j\in\{1,\dots,n\}$. The function $p_j$ is continuous at every point of $U$ by Step 2(i), and by Step 3 its first partial derivatives exist at every point of $U$ and are continuous there; so $p_j$ is of class $C^1$ on $U$. By C^2 Real-Valued Map on an Open Subset of Euclidean Space, $q$ is of class $C^2$ on $U$.

Step 5: proof of claim 3. Let $x=(x_1,\dots,x_n)\in U$. By Gradient of a Real-Valued Function on a Euclidean Open Set and claim 1,

and since the $k$th coordinate of $x-a$ is $x_k-a_k$ and the $k$th coordinate of the scalar multiple $(2c)(x-a)$ is $2c\,(x_k-a_k)$ by Scalar Multiple of a Point of $\mathbb{R}^n$, this is exactly $(2c)(x-a)$.

By Hessian Matrix of a C^2 Function and Step 3, the entry of $D^2q(x)$ in row $i$ and column $j$ is $2c$ if $i=j$ and $0$ if $i\ne j$. By Identity Matrix the corresponding entry of $I_n$ is $1$ if $i=j$ and $0$ otherwise, and by Scalar Multiple of a Real Matrix the entry of $(2c)I_n$ in row $i$ and column $j$ is $2c$ times that, that is, $2c$ if $i=j$ and $0$ otherwise. Since two real $n\times n$ matrices with the same entries are equal, $D^2q(x)=(2c)I_n$.