[proofplan]
We use the definition of twice Fréchet differentiability as differentiability of the derivative map $x \mapsto Df_x$ at $a$. Openness of $U$ lets us restrict to small vectors $h$ for which the segment from $a$ to $a+h$ lies in $U$. Along this segment, we write the increment of $f$ as the one-dimensional integral of its [directional derivative](/page/Directional%20Derivative), then substitute the first-order expansion of $Df$ at $a$ and integrate the linear term in $t$.
[/proofplan]
[step:Choose a ball on which the line segment remains inside the differentiability neighbourhood]
For $c \in \mathbb{R}^m$ and $r>0$, write $B(c,r):=\{x \in \mathbb{R}^m: |x-c|<r\}$ for the open Euclidean ball. By the theorem statement, there is an open neighbourhood $V \subset U$ of $a$ on which $f$ is continuously differentiable and the derivative map $x \mapsto Df_x$ is defined as a map into $\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$. Since $V$ is open and $a \in V$, there exists $\rho>0$ such that $B(a,\rho) \subset V$. In the sequel we consider $h \in \mathbb{R}^m$ with $0<|h|<\rho$. Define the line-segment map
\begin{align*}
\gamma_h: [0,1] &\to V
\end{align*}
by $\gamma_h(t)=a+th$. For every $t \in [0,1]$,
\begin{align*}
|\gamma_h(t)-a|=t|h|<\rho,
\end{align*}
so $\gamma_h(t) \in B(a,\rho) \subset V \subset U$.
[/step]
[step:Express the increment of $f$ as an integral along the segment]
Define
\begin{align*}
\varphi_h: [0,1] &\to \mathbb{R}^n
\end{align*}
by $\varphi_h(t)=f(a+th)$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Since $f$ is continuously differentiable on $B(a,\rho)$, the one-dimensional chain rule applied to $\varphi_h=f\circ\gamma_h$ gives
\begin{align*}
\varphi_h'(t)=Df_{a+th}(h)
\end{align*}
for $t \in [0,1]$. The map $t \mapsto Df_{a+th}(h)$ is continuous, so $\varphi_h$ is continuously differentiable and hence absolutely continuous on $[0,1]$. The vector-valued [fundamental theorem of calculus](/theorems/632) applies to this absolutely continuous map and gives
\begin{align*}
f(a+h)-f(a)=\int_0^1 Df_{a+th}(h)\,d\mathcal{L}^1(t).
\end{align*}
[guided]
The purpose of introducing $\varphi_h$ is to reduce the multivariable increment of $f$ to a one-variable increment. We define
\begin{align*}
\varphi_h: [0,1] &\to \mathbb{R}^n
\end{align*}
by $\varphi_h(t)=f(a+th)$. The previous step shows that $a+th \in U$ for all $t \in [0,1]$, so this map is well-defined. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$.
Because $f$ is continuously differentiable on $B(a,\rho)$ and the map $t \mapsto a+th$ has derivative $h$, the one-dimensional chain rule applied to $\varphi_h=f\circ\gamma_h$ gives
\begin{align*}
\varphi_h'(t)=Df_{a+th}(h).
\end{align*}
The derivative $\varphi_h'$ is continuous because $x \mapsto Df_x$ is continuous on $B(a,\rho)$ and $t \mapsto a+th$ is continuous. Therefore $\varphi_h$ is absolutely continuous on $[0,1]$, which is the regularity required to use the vector-valued fundamental theorem of calculus. Applying that theorem to the absolutely [continuous function](/page/Continuous%20Function) $\varphi_h$ yields
\begin{align*}
\varphi_h(1)-\varphi_h(0)=\int_0^1 \varphi_h'(t)\,d\mathcal{L}^1(t).
\end{align*}
Substituting $\varphi_h(1)=f(a+h)$, $\varphi_h(0)=f(a)$, and $\varphi_h'(t)=Df_{a+th}(h)$ gives
\begin{align*}
f(a+h)-f(a)=\int_0^1 Df_{a+th}(h)\,d\mathcal{L}^1(t).
\end{align*}
This identity is the bridge from differentiability of the derivative map to a Taylor expansion for $f$ itself.
[/guided]
[/step]
[step:Expand the derivative map at $a$]
Let $\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$ denote the space of linear maps from $\mathbb{R}^m$ to $\mathbb{R}^n$, equipped with the operator norm $\|\cdot\|_{\mathrm{op}}$. The theorem statement defines $D^2f_a$ from the Fréchet derivative of $x \mapsto Df_x$ by the rule $D^2f_a(k,v)=DF_a(k)(v)$ for $k,v \in \mathbb{R}^m$; hence $D^2f_a(k,\cdot)$ is the [linear map](/page/Linear%20Map) $v \mapsto D^2f_a(k,v)$. Let
\begin{align*}
R: B(0,\rho) &\to \mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)
\end{align*}
be the remainder map defined by
\begin{align*}
R(k)=Df_{a+k}-Df_a-D^2f_a(k,\cdot).
\end{align*}
Differentiability of the derivative map $x \mapsto Df_x$ at $a$ means
\begin{align*}
\frac{\|R(k)\|_{\mathrm{op}}}{|k|}\to 0
\end{align*}
as $k \to 0$ in $\mathbb{R}^m$. Hence, for $t \in [0,1]$,
\begin{align*}
Df_{a+th}=Df_a+D^2f_a(th,\cdot)+R(th).
\end{align*}
Evaluating this linear map at $h$ gives
\begin{align*}
Df_{a+th}(h)=Df_a(h)+tD^2f_a(h,h)+R(th)(h).
\end{align*}
[/step]
[step:Integrate the expansion and identify the remainder]
Substituting the expansion of $Df_{a+th}(h)$ into the segment integral gives
\begin{align*}
f(a+h)-f(a)=\int_0^1 Df_a(h)\,d\mathcal{L}^1(t)+\int_0^1 tD^2f_a(h,h)\,d\mathcal{L}^1(t)+\int_0^1 R(th)(h)\,d\mathcal{L}^1(t).
\end{align*}
Since $Df_a(h)$ and $D^2f_a(h,h)$ are independent of $t$,
\begin{align*}
\int_0^1 Df_a(h)\,d\mathcal{L}^1(t)=Df_a(h)
\end{align*}
and
\begin{align*}
\int_0^1 tD^2f_a(h,h)\,d\mathcal{L}^1(t)=\frac{1}{2}D^2f_a(h,h).
\end{align*}
Define the remainder vector
\begin{align*}
E(h)=\int_0^1 R(th)(h)\,d\mathcal{L}^1(t).
\end{align*}
Then
\begin{align*}
f(a+h)=f(a)+Df_a(h)+\frac{1}{2}D^2f_a(h,h)+E(h).
\end{align*}
[/step]
[step:Show the integrated remainder is $o(|h|^2)$]
Fix $\varepsilon>0$. Since $\|R(k)\|_{\mathrm{op}}/|k|\to 0$ as $k \to 0$, there exists $\delta>0$ such that $0<|k|<\delta$ implies
\begin{align*}
\|R(k)\|_{\mathrm{op}}\leq \varepsilon |k|.
\end{align*}
For $0<|h|<\delta$ and $t \in [0,1]$, this gives
\begin{align*}
\|R(th)\|_{\mathrm{op}}\leq \varepsilon t|h|
\end{align*}
when $t>0$, while the value at $t=0$ does not affect the integral. Therefore
\begin{align*}
|E(h)|\leq \int_0^1 \|R(th)\|_{\mathrm{op}}|h|\,d\mathcal{L}^1(t)\leq \varepsilon |h|^2\int_0^1 t\,d\mathcal{L}^1(t)=\frac{\varepsilon}{2}|h|^2.
\end{align*}
Thus
\begin{align*}
\frac{|E(h)|}{|h|^2}\to 0
\end{align*}
as $h \to 0$. Hence $E(h)=o(|h|^2)$, and the desired expansion follows.
[/step]
