[proofplan] We restrict the second-order expansion of $f$ at $a$ to the affine line $t \mapsto a + tv$. The linear term becomes $tDf_a(v)$, while bilinearity turns the quadratic term into $t^2D^2f_a(v,v)$. The only point requiring care is the little-$o$ remainder: when $v \neq 0$, $o(|tv|^2)$ is also $o(t^2)$, and when $v=0$ the curve is constant. [/proofplan] [step:Restrict the second-order expansion of $f$ to the line $a + tv$] Since $U$ is open, $a \in U$, and $f:U \to \mathbb{R}^n$ is twice differentiable at $a$, [citetheorem:8707] gives the second-order expansion \begin{align*} f(a+h)=f(a)+Df_a(h)+\frac{1}{2}D^2f_a(h,h)+R(h) \end{align*} for $h \to 0$ with $a+h \in U$, where the remainder map $R$ is defined on \begin{align*} W:=\{h \in \mathbb{R}^m:a+h \in U\} \end{align*} by \begin{align*} R(h):=f(a+h)-f(a)-Df_a(h)-\frac{1}{2}D^2f_a(h,h) \end{align*} and satisfies $R(h)=o(|h|^2)$ as $h \to 0$ in $W$. For $t \in I$, put $h=tv$. Then $tv \in W$, and the definition of $g$ gives \begin{align*} g(t)=f(a)+Df_a(tv)+\frac{1}{2}D^2f_a(tv,tv)+R(tv). \end{align*} Since $Df_a:\mathbb{R}^m \to \mathbb{R}^n$ is linear and $D^2f_a:\mathbb{R}^m \times \mathbb{R}^m \to \mathbb{R}^n$ is bilinear, this becomes \begin{align*} g(t)=g(0)+tDf_a(v)+\frac{1}{2}t^2D^2f_a(v,v)+R(tv). \end{align*} [guided] The input theorem [citetheorem:8707] applies because the present hypotheses match its hypotheses: $U \subset \mathbb{R}^m$ is open, $a \in U$, and $f:U \to \mathbb{R}^n$ is twice differentiable at $a$. Therefore there is a second-order Peano expansion of $f$ at $a$: \begin{align*} f(a+h)=f(a)+Df_a(h)+\frac{1}{2}D^2f_a(h,h)+R(h) \end{align*} as $h \to 0$ with $a+h \in U$. Here the remainder map is the function \begin{align*} R:W &\to \mathbb{R}^n \end{align*} defined on \begin{align*} W:=\{h \in \mathbb{R}^m:a+h \in U\} \end{align*} by \begin{align*} R(h):=f(a+h)-f(a)-Df_a(h)-\frac{1}{2}D^2f_a(h,h), \end{align*} and the assertion $R(h)=o(|h|^2)$ means \begin{align*} \frac{|R(h)|}{|h|^2}\to 0 \end{align*} as $h \to 0$ through nonzero vectors in $W$. We now follow the line specified in the theorem. For each $t \in I$, the hypothesis $a+tv \in U$ implies $tv \in W$, so we may substitute $h=tv$ into the expansion. This gives \begin{align*} g(t)=f(a+tv)=f(a)+Df_a(tv)+\frac{1}{2}D^2f_a(tv,tv)+R(tv). \end{align*} The point of this substitution is that the line parameter $t$ can be pulled out of the first and [second derivative](/page/Second%20Derivative) terms. Since $Df_a$ is linear, \begin{align*} Df_a(tv)=tDf_a(v). \end{align*} Since $D^2f_a$ is bilinear, \begin{align*} D^2f_a(tv,tv)=t^2D^2f_a(v,v). \end{align*} Also $g(0)=f(a)$ because $a+0v=a$. Hence the restricted expansion is \begin{align*} g(t)=g(0)+tDf_a(v)+\frac{1}{2}t^2D^2f_a(v,v)+R(tv). \end{align*} [/guided] [/step] [step:Convert the restricted remainder into a one-variable $o(t^2)$ term] If $v=0$, then $g(t)=f(a)$ for every $t \in I$, so $g$ is constant near $0$ and $g''(0)=0=D^2f_a(0,0)$ by bilinearity. Assume now that $v \neq 0$. Define the one-variable remainder map \begin{align*} \rho:I &\to \mathbb{R}^n, \qquad t \mapsto R(tv). \end{align*} For $t \neq 0$ sufficiently close to $0$, we have \begin{align*} \frac{|\rho(t)|}{t^2}=|v|^2\frac{|R(tv)|}{|tv|^2}. \end{align*} Since $tv \to 0$ in $W$ as $t \to 0$ and $R(h)=o(|h|^2)$, the right-hand side tends to $0$. Therefore $\rho(t)=o(t^2)$ as $t \to 0$. Thus, in both cases, \begin{align*} g(t)=g(0)+tDf_a(v)+\frac{1}{2}t^2D^2f_a(v,v)+o(t^2) \end{align*} as $t \to 0$ in $I$. [/step] [step:Read off the second derivative of the one-dimensional map] The preceding expansion has the defining form of second-order Fréchet differentiability for the map $g:I \to \mathbb{R}^n$ at $0$: a constant term, a [linear map](/page/Linear%20Map) $\ell:\mathbb{R}\to\mathbb{R}^n$ given by $\ell(s)=sDf_a(v)$, a bilinear map $Q:\mathbb{R}\times\mathbb{R}\to\mathbb{R}^n$ given by $Q(s_1,s_2)=s_1s_2D^2f_a(v,v)$, and a remainder $o(t^2)$. Therefore $g$ is twice differentiable at $0$ in the one-dimensional Fréchet sense, with \begin{align*} Dg_0(1)=Df_a(v) \end{align*} and \begin{align*} D^2g_0(1,1)=D^2f_a(v,v). \end{align*} By the standard notation $g''(0):=D^2g_0(1,1)$ for maps of one real variable, we obtain \begin{align*} g''(0)=D^2f_a(v,v). \end{align*} This is the desired identity. [/step]