Proof of Multivariate Taylor Expansion with Uniform Second-Order Remainder

If $h=0$ then $y=x$ and every asserted inequality reads $0\le0$, so assume $h\neq0$.

Setup. For $\tau\in\mathbb{R}$ write $a(\tau)=x+\tau h$. By claim 1 of Derivatives Along a Segment for C^1 Functions on a Euclidean Open Set (the segment hypothesis giving $a(\tau)\in W$ for $0\le\tau\le1$), there is a real $r>0$ with $a(\tau)\in W$ for every $\tau$ with $-r<\tau<1+r$. Set $J=(-r,1+r)$, an open interval containing $[0,1]$; by An Open Interval is an Interval All of Whose Points Are Interior, $J$ is an interval all of whose points are interior points.

Set $F:J\to\mathbb{R}$, $F(\tau)=f(a(\tau))$. By claims 2 and 3 of Derivatives Along a Segment for C^1 Functions on a Euclidean Open Set (every point of $J$ being interior), $F$ is continuous at every point of $J$ relative to $J$ and is differentiable at every $\tau\in J$ with

An elementary inequality. $\big(\sum_{i=1}^n|h_i|\big)^2=\sum_{i,j}|h_i||h_j|\le\sum_{i,j}\tfrac{1}{2}(h_i^2+h_j^2)=n\,|h|^2$, using $2|h_i||h_j|\le h_i^2+h_j^2$; hence $\sum_i|h_i|\le\sqrt{n}\,|h|$.

Part (i). By claims 1 and 2 of Restriction Stability of Continuity and of the Derivative, the restriction of $F$ to $[0,1]$ is continuous relative to $[0,1]$ and differentiable at every point of $(0,1)$ with the same derivative, so the mean value theorem gives $\xi\in(0,1)$ with $F(1)-F(0)=F'(\xi)$. Since $a(\xi)$ lies on the segment, $|F'(\xi)|\le\sum_i|\partial_i f(a(\xi))||h_i|\le M_1\sum_i|h_i|\le\sqrt{n}\,M_1|h|$, and $F(1)-F(0)=f(y)-f(x)$. This proves (i), which used only that $f$ is $C^1$.

Second-derivative setup for (ii) and (iii). Assume now that $f$ is of class $C^2$ on $W$. By claim 4 of Derivatives Along a Segment for C^1 Functions on a Euclidean Open Set (every point of $J$ being interior), the function $\tau\mapsto\sum_{i=1}^n\partial_i f(a(\tau))\,h_i$, which equals $F'$ on $J$, is differentiable at every $\tau\in J$. Thus $F'$ is differentiable on $J$; writing $F''$ for its derivative there,

Part (ii). Define $\psi(\tau)=F(\tau)-F(0)-F'(0)\,\tau$ on $J$, set $c=\psi(1)$, and put $H(\tau)=\psi(\tau)-c\,\tau^2$. Then $H(0)=0$ and $H(1)=\psi(1)-c=0$, and $H$ is differentiable on $J$ with $H'(\tau)=F'(\tau)-F'(0)-2c\,\tau$ — by claims 1 and 2 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives, together with the constant difference quotients giving $(\tau\mapsto\tau)'=1$ and claim 3 there giving $(\tau\mapsto\tau^2)'=2\tau$ — hence continuous at every point of $J$, with restriction to $[0,1]$ continuous relative to $[0,1]$ and differentiable on $(0,1)$ by claims 1 and 2 of Restriction Stability of Continuity and of the Derivative; so Rolle's theorem gives $\xi_1\in(0,1)$ with $H'(\xi_1)=0$. Also $H'(0)=0$, and $H'$ is differentiable on $J$ with $H''(\tau)=F''(\tau)-2c$ (claims 1 and 2 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives, since $F'$ is differentiable on $J$), hence continuous at every point of $J$ by Differentiability at an Interior Point Implies Continuity There; restricting to $[0,\xi_1]$ by claims 1 and 2 of Restriction Stability of Continuity and of the Derivative, Rolle's theorem applied to $H'$ on $[0,\xi_1]$ gives $\xi_2\in(0,\xi_1)$ with $H''(\xi_2)=0$, that is, $2c=F''(\xi_2)$. Since $a(\xi_2)$ lies on the segment,

and $\psi(1)=f(y)-f(x)-\sum_i\partial_i f(x)h_i$, which is claim (ii).

Part (iii). Define $\varphi(\tau)=F(\tau)-F(0)-F'(0)\,\tau-\tfrac{1}{2}F''(0)\,\tau^2$ on $J$, where $F''(0)=\sum_{i,j}\partial_j\partial_i f(x)h_ih_j$. Then $\varphi(0)=0$, $\varphi'(\tau)=F'(\tau)-F'(0)-F''(0)\tau$ satisfies $\varphi'(0)=0$, and $\varphi''(\tau)=F''(\tau)-F''(0)=\sum_{i,j}\big(\partial_j\partial_i f(a(\tau))-\partial_j\partial_i f(x)\big)h_ih_j$, so for $\tau\in(0,1)$ the point $a(\tau)$ lies on the segment and $|\varphi''(\tau)|\le\bar{\varepsilon}\big(\sum_i|h_i|\big)^2\le n\,\bar{\varepsilon}\,|h|^2$. Repeating the double application of Rolle's theorem from Part (ii) with $\varphi$ in place of $\psi$ (set $c=\varphi(1)$, $H(\tau)=\varphi(\tau)-c\tau^2$; then $H(0)=H(1)=0$ yields $\xi_1$, and $H'(0)=H'(\xi_1)=0$ yields $\xi_2$ with $2c=\varphi''(\xi_2)$; the citations of Part (ii) apply verbatim, $\varphi$, $\varphi'$ and $\varphi''$ existing on $J$ by the second-derivative setup) gives

Since $\varphi(1)=f(y)-f(x)-\sum_i\partial_i f(x)h_i-\tfrac{1}{2}\sum_{i,j}\partial_j\partial_i f(x)h_ih_j$, this is claim (iii).