Proof of Second-Order Taylor Expansion with Peano Remainder

Throughout we use the following elementary consequence of Elementary Order Arithmetic in an Ordered Field: for real $s,c$ with $0\le s$ and $0<c$, we have $s<c$ if and only if $s^2<c^2$. Indeed, if $s<c$ then either $s=0$, in which case $s^2=0<c^2$, or $0<s$, in which case multiplying $s<c$ by $s$ and then by $c$ (claim 10 there) gives $s^2<sc$ and $sc<c^2$, whence $s^2<c^2$ by claim 2 there; and conversely, if $c\le s$ then the same argument with the roles reversed gives $c^2\le s^2$.

Fix $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$, and set $\bar\varepsilon=2\varepsilon/n$, which is a positive real number because $n\ge1$.

Step 1: a ball around $x$ inside $U$. Since $U$ is open and $x\in U$, there is a real $r>0$ such that every $y=(y_1,\dots,y_n)\in\mathbb{R}^n$ with $\sum_{i=1}^n(y_i-x_i)^2<r^2$ belongs to $U$. By claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ the left-hand side equals $\lVert y-x\rVert^2$, so by the remark above,

Step 2: the second partial derivatives are continuous. By hypothesis $f$ is of class $C^2$ on $U$. Unwinding clause 2 of that definition with $k=1$ (and clause 3, since $f$ is real-valued): $f$ is of class $C^1$ on $U$ and, for every $i\in\{1,\dots,n\}$, the function $\partial_i f:U\to\mathbb{R}$ is of class $C^1$ on $U$. Applying clause 1 to $\partial_i f$ shows that for every $j\in\{1,\dots,n\}$ the partial derivative of $\partial_i f$ with respect to the $j$-th coordinate exists at every point of $U$ — so the function $\partial_j\partial_i f:U\to\mathbb{R}$ of clause 4 is defined on all of $U$ — and that it is continuous at every point of $U$.

Step 3: choice of $\delta$. Fix $i,j\in\{1,\dots,n\}$. Continuity of $\partial_j\partial_i f$ at $x$, applied with the positive number $\bar\varepsilon$, yields a real $\delta_{ij}>0$ such that every $z=(z_1,\dots,z_n)\in U$ with $\sum_{k=1}^n(z_k-x_k)^2<\delta_{ij}^2$ satisfies

By claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ the hypothesis here reads $\lVert z-x\rVert^2<\delta_{ij}^2$, which by the opening remark is equivalent to $\lVert z-x\rVert<\delta_{ij}$; and, again by that remark applied to $s=|\partial_j\partial_i f(z)-\partial_j\partial_i f(x)|$ and $c=\bar\varepsilon$, using claim 1 of Properties of the Absolute Value in an Ordered Field to identify $s^2$ with the squared difference, the conclusion reads $|\partial_j\partial_i f(z)-\partial_j\partial_i f(x)|<\bar\varepsilon$.

Let $\delta$ be the least element of the finite list consisting of $r$ and the numbers $\delta_{ij}$ for $i,j\in\{1,\dots,n\}$; by repeated use of claim 9 of Elementary Order Arithmetic in an Ordered Field, $\delta$ is one of these numbers and $\delta\le r$ and $\delta\le\delta_{ij}$ for all $i,j$, so $0<\delta$.

Step 4: the segment from $x$ to $x+h$. Let $h=(h_1,\dots,h_n)\in\mathbb{R}^n$ with $\lVert h\rVert<\delta$, and put $y=x+h$. For $\tau\in\mathbb{R}$ with $0\le\tau\le1$, claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ gives $\lVert\tau h\rVert=|\tau|\,\lVert h\rVert=\tau\,\lVert h\rVert$, and multiplying $\tau\le1$ by the nonnegative number $\lVert h\rVert$ gives $\tau\,\lVert h\rVert\le\lVert h\rVert$. Hence

so $x+\tau h\in U$ by Step 1. Taking $\tau=1$ gives $x+h\in U$, as asserted; and since $y-x=h$, the whole segment $\{x+\tau(y-x):\tau\in[0,1]\}$ is contained in $U$. Moreover every point $z=x+\tau h$ of that segment satisfies $\lVert z-x\rVert<\delta\le\delta_{ij}$, so by Step 3,

Step 5: conclusion. The hypotheses of part (iii) of Multivariate Taylor Expansion with Uniform Second-Order Remainder are now met, with $U$ in the role of the open set there, the points $x$ and $y=x+h$, and the constant $\bar\varepsilon$: $f$ is of class $C^2$ on $U$, the segment from $x$ to $y$ lies in $U$ by Step 4, and the oscillation bound just displayed holds on that segment. The quantity written $|h|$ there is the Euclidean distance between $x$ and $y$, which equals $\lVert h\rVert$ by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$. Part (iii) therefore gives

the final equality because $\tfrac{1}{2}n\cdot(2\varepsilon/n)=\varepsilon$. Since $\varepsilon>0$ was arbitrary and $\delta>0$ was produced from it, this is the assertion. $\blacksquare$