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. Let $I=\{\tau\in\mathbb{R}:x+\tau h\in W\}$. Then $I$ is an open subset of $\mathbb{R}$ containing $[0,1]$: it contains $[0,1]$ because the segment lies in $W$, and for $\tau_0\in I$ the point $x+\tau_0h$ has an open ball of some radius $r>0$ contained in $W$, so every $\tau$ with $|\tau-\tau_0|<r/|h|$ lies in $I$. We next fix an open interval $J$ with $[0,1]\subseteq J\subseteq I$. If no $r>0$ satisfied $(-r,1+r)\subseteq I$, then for every natural number $n$ there would be $\tau_n\notin I$ with $-\tfrac1n<\tau_n<1+\tfrac1n$; the bounded sequence $(\tau_n)$ has, by the Bolzano-Weierstrass theorem, a subsequence converging to some $\tau^*$, and the two-sided bounds force $0\le\tau^*\le1$ (if $\tau^*>1$, taking $\varepsilon=\tau^*-1$ in the definition of convergence produces $\tau_{n_k}>1+\varepsilon/2\ge1+\tfrac{1}{n_k}$ for large $k$, a contradiction; symmetrically for $\tau^*<0$), so $\tau^*\in[0,1]\subseteq I$; since $I$ is open, an open interval around $\tau^*$ lies inside $I$, and the subsequence eventually enters it, contradicting $\tau_{n_k}\notin I$. So fix $r>0$ with $J=(-r,1+r)\subseteq I$, an open interval containing $[0,1]$.

Consider the affine map $a:J\to\mathbb{R}^n$, $a(\tau)=x+\tau h$. Each component $a_k(\tau)=x_k+\tau h_k$ has partial derivative with respect to the single coordinate equal to the constant $h_k$ (the difference quotients are constant), and constants are continuous, so $a$ is a $C^1$ map on $J$; for one-variable functions the partial derivative with respect to the only coordinate coincides with the derivative, both being the same limit of difference quotients.

Set $F=f\circ a:J\to\mathbb{R}$. By the chain rule for $C^1$ maps, $F$ is $C^1$ on the open set $J$ 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). $F$ is defined on the interval $J\supseteq[0,1]$, continuous on $[0,1]$ and differentiable on $(0,1)$, so by the mean value theorem there is $\xi\in(0,1)$ 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 each $\partial_i f$ is again a $C^1$ map on $W$. By the chain rule each $\tau\mapsto\partial_i f(a(\tau))$ is $C^1$ on $J$ with derivative $\sum_{j=1}^n\partial_j\partial_i f(a(\tau))\,h_j$, and by the one-dimensional sum and product rules on the interval $J$ the finite sum $F'$ is differentiable on $J$ with

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 continuous on $[0,1]$ and differentiable on $J$ with $H'(\tau)=F'(\tau)-F'(0)-2c\,\tau$, so Rolle's theorem gives $\xi_1\in(0,1)$ with $H'(\xi_1)=0$. Also $H'(0)=0$, and $H'$ is continuous on $[0,\xi_1]$ and differentiable on $(0,\xi_1)$ with $H''(\tau)=F''(\tau)-2c$, so 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 $2c=\varphi''(\xi_2)$) 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).