[proofplan]
Choose a locally finite smooth trivializing cover of $E$ and a smooth [partition of unity](/page/Partition%20of%20Unity) subordinate to it. Each trivialization transfers the Euclidean [inner product](/page/Inner%20Product) on $\mathbb{R}^k$ to a smooth fibre metric on the corresponding local bundle. Multiplying these local metrics by the partition functions and extending by zero gives globally defined smooth symmetric $2$-tensors on $E$; local finiteness makes their sum smooth. Positivity follows because at each point the coefficients are nonnegative, sum to $1$, and at least one coefficient is positive.
[/proofplan]
[step:Choose local trivializations and a subordinate smooth partition of unity]
By the [smooth partition of unity theorem](/theorems/3917) for paracompact smooth manifolds subordinate to open covers, applied to a smooth trivializing cover of $E$ (citing a result not yet in the wiki: smooth partitions of unity subordinate to trivializing covers on paracompact smooth manifolds), choose a locally finite open cover $(U_i)_{i \in I}$ of $M$, smooth bundle trivializations
\begin{align*}
\Phi_i: \pi^{-1}(U_i) &\to U_i \times \mathbb{R}^k,
\end{align*}
and smooth functions
\begin{align*}
\rho_i: M &\to [0,1]
\end{align*}
such that $\operatorname{supp}\rho_i \subset U_i$, the family $(\operatorname{supp}\rho_i)_{i \in I}$ is locally finite, and
\begin{align*}
\sum_{i \in I} \rho_i(p) = 1
\end{align*}
for every $p \in M$. The sum is pointwise finite because the supports are locally finite.
[guided]
The role of paracompactness is to let us pass from local data to global data without losing smoothness. We use the smooth partition of unity theorem for paracompact smooth manifolds subordinate to open covers, applied to a cover by bundle-trivializing open sets (citing a result not yet in the wiki: smooth partitions of unity subordinate to trivializing covers on paracompact smooth manifolds).
Thus we choose a locally finite open cover $(U_i)_{i \in I}$ of $M$ such that $E$ is trivial over each $U_i$. For each index $i \in I$, fix a smooth bundle trivialization
\begin{align*}
\Phi_i: \pi^{-1}(U_i) &\to U_i \times \mathbb{R}^k.
\end{align*}
We also choose smooth functions
\begin{align*}
\rho_i: M &\to [0,1]
\end{align*}
with $\operatorname{supp}\rho_i \subset U_i$, locally finite supports, and
\begin{align*}
\sum_{i \in I} \rho_i(p) = 1
\end{align*}
for every $p \in M$. The condition $\operatorname{supp}\rho_i \subset U_i$ is what makes extension by zero smooth later: near any point outside $U_i$, either the point lies away from the support or the function has already vanished in a neighbourhood. Local finiteness is equally important because it turns the apparently infinite sum of tensor fields into a finite sum in a neighbourhood of every point.
[/guided]
[/step]
[step:Pull back the Euclidean inner product to each trivialized piece]
For each $i \in I$ and $p \in U_i$, define the fibrewise [linear map](/page/Linear%20Map)
\begin{align*}
\lambda_{i,p}: E_p &\to \mathbb{R}^k
\end{align*}
by the rule that $\Phi_i(v) = (p,\lambda_{i,p}(v))$ for $v \in E_p$. Define a local symmetric [bilinear form](/page/Bilinear%20Form)
\begin{align*}
h_i \in \Gamma(U_i,\operatorname{Sym}^2(E|_{U_i})^*)
\end{align*}
by
\begin{align*}
(h_i)_p(v,w) = \sum_{a=1}^k \lambda_{i,p}(v)_a \lambda_{i,p}(w)_a
\end{align*}
for $p \in U_i$ and $v,w \in E_p$. Since $\Phi_i$ is a smooth bundle trivialization and the Euclidean inner product on $\mathbb{R}^k$ has constant coefficients, $h_i$ is a smooth fibre metric on $E|_{U_i}$.
[/step]
[step:Extend the weighted local metrics by zero]
For each $i \in I$, define a global section
\begin{align*}
s_i \in \Gamma(\operatorname{Sym}^2 E^*)
\end{align*}
by setting $(s_i)_p = \rho_i(p)(h_i)_p$ for $p \in U_i$ and $(s_i)_p = 0$ for $p \in M \setminus U_i$. This is well-defined because $\operatorname{supp}\rho_i \subset U_i$. It is smooth on $U_i$ as the product of the smooth function $\rho_i|_{U_i}$ and the smooth tensor field $h_i$. It is smooth on $M \setminus \operatorname{supp}\rho_i$ because it vanishes there. Since
\begin{align*}
M = U_i \cup (M \setminus \operatorname{supp}\rho_i),
\end{align*}
the section $s_i$ is smooth on all of $M$.
[/step]
[step:Sum the extended tensors to obtain a smooth global symmetric form]
Define
\begin{align*}
g \in \Gamma(\operatorname{Sym}^2 E^*)
\end{align*}
pointwise by
\begin{align*}
g_p = \sum_{i \in I} (s_i)_p.
\end{align*}
For every $p \in M$, local finiteness gives an open neighbourhood $V \subset M$ of $p$ meeting only finitely many supports $\operatorname{supp}\rho_i$. Hence, on $V$, the above sum is a finite sum of smooth sections. Therefore $g$ is a smooth section of $\operatorname{Sym}^2 E^*$. Each $s_i$ is symmetric and bilinear on each fibre, so $g_p$ is symmetric and bilinear for every $p \in M$.
[/step]
[step:Verify positive definiteness on each fibre]
Fix $p \in M$ and a nonzero vector $v \in E_p$. Since $\sum_{i \in I}\rho_i(p)=1$ and every $\rho_i(p) \ge 0$, there exists an index $j \in I$ such that $\rho_j(p)>0$. For every index $i$ with $\rho_i(p)>0$, we have $p \in U_i$, and the local form $(h_i)_p$ is positive definite. By the definitions of $g$ and the sections $s_i$,
\begin{align*}
g_p(v,v) = \sum_{i \in I} \rho_i(p)(h_i)_p(v,v).
\end{align*}
All summands are nonnegative, because $\rho_i(p) \ge 0$ and each $(h_i)_p$ is positive definite whenever $\rho_i(p)>0$. Keeping only the $j$-term gives
\begin{align*}
g_p(v,v) \ge \rho_j(p)(h_j)_p(v,v).
\end{align*}
Since $\rho_j(p)>0$ and $(h_j)_p(v,v)>0$ for $v \ne 0$,
\begin{align*}
\rho_j(p)(h_j)_p(v,v) > 0.
\end{align*}
Thus $g_p(v,v)>0$, so $g_p$ is positive definite for every $p \in M$. Therefore $g$ is a smooth fibre metric on $E$.
[/step]
