[proofplan]
We use the definition of twice differentiability at a point: the derivative maps $x \mapsto Df_x$ and $x \mapsto Dg_x$ are defined near $a$ and differentiable at $a$. First we prove directly from the first-order expansion that $\alpha f+\beta g$ is differentiable near $a$ and that its derivative is the corresponding linear combination of $Df_x$ and $Dg_x$. Then we differentiate this derivative map at $a$ and translate the resulting [linear map](/page/Linear%20Map) into the associated bilinear [second derivative](/page/Second%20Derivative).
[/proofplan]
[step:Choose a neighbourhood where the derivative maps are defined]
Since $f$ and $g$ are twice differentiable at $a$, there exists $r>0$ such that $B(a,r) \subset U$, both $f$ and $g$ are differentiable at every point of $B(a,r)$, and the derivative maps
\begin{align*}
F: B(a,r) \to \operatorname{Lin}(\mathbb{R}^m,\mathbb{R}^n)
\end{align*}
are defined as follows. Here $B(a,r):=\{x \in \mathbb{R}^m: |x-a|<r\}$ is the open Euclidean ball centered at $a$ with radius $r$, and $\operatorname{Lin}(\mathbb{R}^m,\mathbb{R}^n)$ denotes the real [vector space](/page/Vector%20Space) of linear maps from $\mathbb{R}^m$ to $\mathbb{R}^n$. The first derivative map of $f$ is
and the first derivative map of $g$ is
\begin{align*}
G: B(a,r) \to \operatorname{Lin}(\mathbb{R}^m,\mathbb{R}^n).
\end{align*}
They are defined by $F(x)=Df_x$ and $G(x)=Dg_x$, and they are differentiable at $a$. Let
\begin{align*}
H: U \to \mathbb{R}^n
\end{align*}
be the map $H=\alpha f+\beta g$.
[/step]
[step:Differentiate the linear combination once]
Fix $x \in B(a,r)$. Since $f$ and $g$ are differentiable at $x$, for $q \in \mathbb{R}^m$ with $x+q \in U$ and $q \to 0$ there are remainder functions $\rho_f$ and $\rho_g$ satisfying
\begin{align*}
f(x+q)=f(x)+Df_x(q)+\rho_f(q)
\end{align*}
with $\rho_f(q)=o(|q|)$, and
\begin{align*}
g(x+q)=g(x)+Dg_x(q)+\rho_g(q)
\end{align*}
with $\rho_g(q)=o(|q|)$. Therefore
\begin{align*}
H(x+q)=H(x)+\bigl(\alpha Df_x+\beta Dg_x\bigr)(q)+\alpha\rho_f(q)+\beta\rho_g(q).
\end{align*}
Because $\alpha\rho_f(q)+\beta\rho_g(q)=o(|q|)$, the map $H$ is differentiable at $x$ and
\begin{align*}
DH_x=\alpha Df_x+\beta Dg_x.
\end{align*}
Since $x \in B(a,r)$ was arbitrary, $H$ is differentiable throughout $B(a,r)$.
[guided]
Fix a point $x \in B(a,r)$. The goal is to compute the derivative of $H=\alpha f+\beta g$ at $x$ directly from the defining first-order expansions. Because $f$ is differentiable at $x$, there is a remainder function $\rho_f$ such that
\begin{align*}
f(x+q)=f(x)+Df_x(q)+\rho_f(q)
\end{align*}
and $\rho_f(q)=o(|q|)$ as $q \to 0$. Likewise, because $g$ is differentiable at $x$, there is a remainder function $\rho_g$ such that
\begin{align*}
g(x+q)=g(x)+Dg_x(q)+\rho_g(q)
\end{align*}
and $\rho_g(q)=o(|q|)$.
Now multiply the first expansion by $\alpha$, multiply the second expansion by $\beta$, and add them. This gives
\begin{align*}
H(x+q)=H(x)+\alpha Df_x(q)+\beta Dg_x(q)+\alpha\rho_f(q)+\beta\rho_g(q).
\end{align*}
The map $\alpha Df_x+\beta Dg_x$ is linear from $\mathbb{R}^m$ to $\mathbb{R}^n$, since it is a linear combination of linear maps. The remaining error term is still negligible compared with $|q|$, because
\begin{align*}
\frac{|\alpha\rho_f(q)+\beta\rho_g(q)|}{|q|}\le |\alpha|\frac{|\rho_f(q)|}{|q|}+|\beta|\frac{|\rho_g(q)|}{|q|}
\end{align*}
for $q \ne 0$, and the right-hand side tends to $0$ as $q \to 0$. Therefore $\alpha\rho_f(q)+\beta\rho_g(q)=o(|q|)$, so the defining expansion for differentiability gives
\begin{align*}
DH_x=\alpha Df_x+\beta Dg_x.
\end{align*}
Since this argument applies to every $x \in B(a,r)$, the derivative map of $H$ is defined on $B(a,r)$.
[/guided]
[/step]
[step:Differentiate the derivative map at $a$]
Let
\begin{align*}
A: \mathbb{R}^m \to \operatorname{Lin}(\mathbb{R}^m,\mathbb{R}^n)
\end{align*}
denote the derivative of $F$ at $a$, and let
\begin{align*}
B: \mathbb{R}^m \to \operatorname{Lin}(\mathbb{R}^m,\mathbb{R}^n)
\end{align*}
denote the derivative of $G$ at $a$. Equip $\operatorname{Lin}(\mathbb{R}^m,\mathbb{R}^n)$ with the operator norm
\begin{align*}
\|T\|_{\mathrm{op}}:=\sup\{|T(v)|: v \in \mathbb{R}^m, |v|\le 1\}
\end{align*}
for each $T \in \operatorname{Lin}(\mathbb{R}^m,\mathbb{R}^n)$. Thus, as $h \to 0$ in $\mathbb{R}^m$,
\begin{align*}
F(a+h)=F(a)+A(h)+R_f(h)
\end{align*}
where $\|R_f(h)\|_{\mathrm{op}}=o(|h|)$, and
\begin{align*}
G(a+h)=G(a)+B(h)+R_g(h)
\end{align*}
where $\|R_g(h)\|_{\mathrm{op}}=o(|h|)$. From the first-derivative formula in the previous step, the derivative map of $H$ on $B(a,r)$ is $\alpha F+\beta G$. Hence
\begin{align*}
DH_{a+h}=DH_a+\bigl(\alpha A+\beta B\bigr)(h)+\alpha R_f(h)+\beta R_g(h).
\end{align*}
Moreover,
\begin{align*}
\|\alpha R_f(h)+\beta R_g(h)\|_{\mathrm{op}}\le |\alpha|\|R_f(h)\|_{\mathrm{op}}+|\beta|\|R_g(h)\|_{\mathrm{op}},
\end{align*}
so $\|\alpha R_f(h)+\beta R_g(h)\|_{\mathrm{op}}=o(|h|)$. Therefore the derivative map $x \mapsto DH_x$ is differentiable at $a$, with derivative
\begin{align*}
D(DH)_a=\alpha A+\beta B.
\end{align*}
Thus $H=\alpha f+\beta g$ is twice differentiable at $a$.
[/step]
[step:Identify the resulting second derivative as a bilinear map]
By the definition of the bilinear second derivative as the derivative of the first derivative map evaluated on a second direction, for $h,k \in \mathbb{R}^m$ we have
\begin{align*}
D^2H_a(h,k)=\bigl(D(DH)_a(h)\bigr)(k).
\end{align*}
Using $D(DH)_a=\alpha A+\beta B$, we obtain
\begin{align*}
D^2H_a(h,k)=\bigl(\alpha A(h)+\beta B(h)\bigr)(k).
\end{align*}
Since scalar multiplication and addition in $\operatorname{Lin}(\mathbb{R}^m,\mathbb{R}^n)$ are pointwise,
\begin{align*}
D^2H_a(h,k)=\alpha A(h)(k)+\beta B(h)(k).
\end{align*}
Again using the same identification, $A(h)(k)=D^2f_a(h,k)$ and $B(h)(k)=D^2g_a(h,k)$. Therefore
\begin{align*}
D^2(\alpha f+\beta g)_a(h,k)=\alpha D^2f_a(h,k)+\beta D^2g_a(h,k)
\end{align*}
for all $h,k \in \mathbb{R}^m$, which is the desired formula.
[/step]
