[proofplan]
We restrict $f$ to the line through $a$ in direction $v$, obtaining a one-variable map $g(t)=f(a+tv)$. Since $f$ is twice continuously Frechet differentiable, this restriction is twice continuously differentiable, and its first two derivatives at $0$ are $D_vf(a)$ and $D_v^2f(a)$. We then prove the one-variable second-order Taylor expansion with little-o remainder for this vector-valued map, componentwise or equivalently in the Euclidean norm. Substituting the computed derivatives gives the asserted expansion.
[/proofplan]
[step:Restrict $f$ to the line through $a$ in direction $v$]
Choose $\varepsilon \in (0,\varepsilon_0]$. Define the affine line segment map
\begin{align*}
\gamma: (-\varepsilon,\varepsilon) \to U
\end{align*}
by $\gamma(t)=a+tv$. Define
\begin{align*}
g: (-\varepsilon,\varepsilon) \to \mathbb{R}^n
\end{align*}
by $g(t)=f(\gamma(t))=f(a+tv)$.
Since $\gamma$ is a smooth map and $f \in C^2(U;\mathbb{R}^n)$, the composite $g=f\circ \gamma$ belongs to $C^2((-\varepsilon,\varepsilon);\mathbb{R}^n)$.
[/step]
[step:Identify the first and second derivatives of the one-variable restriction]
For every $s \in (-\varepsilon,\varepsilon)$, the chain rule gives
\begin{align*}
g'(s)=Df_{\gamma(s)}(\gamma'(s)).
\end{align*}
Since $\gamma'(s)=v$, this becomes
\begin{align*}
g'(s)=Df_{a+sv}(v)=D_vf(a+sv).
\end{align*}
Evaluating at $s=0$ gives
\begin{align*}
g'(0)=D_vf(a).
\end{align*}
Because $f \in C^2(U;\mathbb{R}^n)$, the map $x \mapsto D_vf(x)$ is differentiable at $a$ in the direction $v$. Differentiating $s \mapsto D_vf(a+sv)$ at $s=0$ yields
\begin{align*}
g''(0)=D_v(D_vf)(a)=D_v^2f(a).
\end{align*}
Equivalently, by [citetheorem:9039],
\begin{align*}
D_v^2f(a)=D^2f_a(v,v).
\end{align*}
[guided]
The purpose of passing to $g$ is to convert a multivariable statement into a one-variable Taylor expansion. We have already defined
\begin{align*}
g: (-\varepsilon,\varepsilon) \to \mathbb{R}^n
\end{align*}
by $g(t)=f(a+tv)$. The map inside $f$ is
\begin{align*}
\gamma: (-\varepsilon,\varepsilon) \to U,
\end{align*}
given by $\gamma(t)=a+tv$, and its derivative is the constant vector $\gamma'(t)=v$.
Since $f$ is $C^2$ and $\gamma$ is smooth, the chain rule applies to $g=f\circ\gamma$. For each $s \in (-\varepsilon,\varepsilon)$,
\begin{align*}
g'(s)=Df_{\gamma(s)}(\gamma'(s)).
\end{align*}
Substituting $\gamma(s)=a+sv$ and $\gamma'(s)=v$ gives
\begin{align*}
g'(s)=Df_{a+sv}(v).
\end{align*}
By the definition of the [directional derivative](/page/Directional%20Derivative) through the Frechet derivative, this is exactly
\begin{align*}
g'(s)=D_vf(a+sv).
\end{align*}
In particular, at $s=0$,
\begin{align*}
g'(0)=D_vf(a).
\end{align*}
Now differentiate this identity once more at $s=0$. The function being differentiated is the one-variable map $s \mapsto D_vf(a+sv)$. Since $f \in C^2(U;\mathbb{R}^n)$, the first derivative $Df$ is differentiable, so this directional derivative exists. Therefore
\begin{align*}
g''(0)=D_v(D_vf)(a).
\end{align*}
By definition, $D_v(D_vf)(a)$ is $D_v^2f(a)$. The same identification can also be expressed using the second Frechet derivative: by [citetheorem:9039],
\begin{align*}
D_v^2f(a)=D^2f_a(v,v).
\end{align*}
Thus the first two one-variable derivatives of the restriction $g$ are exactly the directional derivatives appearing in the theorem.
[/guided]
[/step]
[step:Apply the one-variable second-order expansion to the restriction]
For each component index $i \in \{1,\dots,n\}$, define
\begin{align*}
g_i: (-\varepsilon,\varepsilon) \to \mathbb{R}
\end{align*}
by $g_i(t)=(g(t))_i$. Since $g \in C^2((-\varepsilon,\varepsilon);\mathbb{R}^n)$, each $g_i$ belongs to $C^2((-\varepsilon,\varepsilon);\mathbb{R})$.
The one-variable Taylor expansion with little-o remainder at $0$ gives, for each $i$,
\begin{align*}
g_i(t)=g_i(0)+tg_i'(0)+\frac{t^2}{2}g_i''(0)+r_i(t),
\end{align*}
where $r_i(t)/t^2 \to 0$ as $t \to 0$. Define
\begin{align*}
r: (-\varepsilon,\varepsilon) \to \mathbb{R}^n
\end{align*}
by $r(t)=(r_1(t),\dots,r_n(t))$. Since $n$ is finite and each $r_i(t)/t^2 \to 0$,
\begin{align*}
\frac{|r(t)|}{|t|^2} \leq \sum_{i=1}^n \left|\frac{r_i(t)}{t^2}\right| \to 0.
\end{align*}
Hence $r(t)=o(t^2)$ in $\mathbb{R}^n$, and therefore
\begin{align*}
g(t)=g(0)+tg'(0)+\frac{t^2}{2}g''(0)+o(t^2).
\end{align*}
[/step]
[step:Substitute back the directional derivatives]
Using $g(t)=f(a+tv)$, $g(0)=f(a)$, $g'(0)=D_vf(a)$, and $g''(0)=D_v^2f(a)$ in the expansion for $g$, we obtain
\begin{align*}
f(a+tv)=f(a)+tD_vf(a)+\frac{t^2}{2}D_v^2f(a)+o(t^2)
\end{align*}
as $t \to 0$. This is the claimed one-directional Taylor expansion.
[/step]
