[proofplan] We choose a locally finite smooth trivializing cover of $E$ and use each trivialization to put the standard Hermitian form on the corresponding local bundle. A smooth [partition of unity](/page/Partition%20of%20Unity) subordinate to this cover supplies non-negative weights whose sum is one at every point. The weighted locally finite sum of the local Hermitian forms is then smooth, Hermitian, and positive definite because at least one positive weight is present at each point. [/proofplan] [step:Choose local trivializations and local Hermitian forms] By the definition of a smooth complex vector bundle, there is a smooth open cover $(U_i)_{i \in I}$ of $X$ and smooth bundle trivializations \begin{align*} \Phi_i: E|_{U_i} \to U_i \times \mathbb{C}^r \end{align*} over $U_i$. Since $X$ is paracompact, we may choose the cover to be locally finite after passing to a locally finite smooth trivializing refinement. For each $i \in I$ and each $x \in U_i$, let \begin{align*} \Phi_{i,x}: E_x \to \mathbb{C}^r \end{align*} denote the fiberwise complex-linear isomorphism induced by $\Phi_i$. Define the local Hermitian form \begin{align*} h_{i,x}: E_x \times E_x \to \mathbb{C} \end{align*} by \begin{align*} h_{i,x}(u,v) = \sum_{k=1}^{r} (\Phi_{i,x}(u))_k \overline{(\Phi_{i,x}(v))_k} \end{align*} for $u,v \in E_x$. Each $h_i$ is a smooth Hermitian metric on $E|_{U_i}$ because $\Phi_i$ is a smooth bundle trivialization and the displayed formula is the pullback of the standard Hermitian [inner product](/page/Inner%20Product) on $\mathbb{C}^r$. [/step] [step:Choose a smooth partition of unity subordinate to the trivializing cover] The family $(U_i)_{i \in I}$ is an open cover of the paracompact smooth manifold $X$. By the standard partition-of-unity theorem for paracompact smooth manifolds, there exists a smooth partition of unity $(\rho_i)_{i \in I}$ subordinate to $(U_i)_{i \in I}$. Thus each map \begin{align*} \rho_i: X \to [0,\infty) \end{align*} is smooth. For each $i \in I$, define $\operatorname{supp}(\rho_i)$ to be the support of $\rho_i$, namely the closure in $X$ of the set $\{x \in X : \rho_i(x) \neq 0\}$. The family $(\operatorname{supp}(\rho_i))_{i \in I}$ is locally finite, $\operatorname{supp}(\rho_i) \subset U_i$, and \begin{align*} \sum_{i \in I} \rho_i(x) = 1 \end{align*} for every $x \in X$. [/step] [step:Define the global Hermitian form as a locally finite weighted sum] For each $x \in X$ and $u,v \in E_x$, define \begin{align*} h_x: E_x \times E_x \to \mathbb{C} \end{align*} by \begin{align*} h_x(u,v) = \sum_{i \in I} \rho_i(x) h_{i,x}(u,v), \end{align*} where the summand is interpreted as $0$ whenever $\rho_i(x)=0$. This convention is legitimate because $\operatorname{supp}\rho_i \subset U_i$, so $h_{i,x}$ is needed only at points where $\rho_i(x)$ may be nonzero. The sum is finite for each $x$ because the supports of the functions $\rho_i$ are locally finite. For fixed $x \in X$, each map $h_{i,x}$ is Hermitian and each scalar $\rho_i(x)$ is real and non-negative. Hence $h_x$ is Hermitian: it is complex-linear in the first variable, conjugate-linear in the second variable, and satisfies \begin{align*} h_x(v,u) = \overline{h_x(u,v)} \end{align*} for all $u,v \in E_x$. [guided] The only delicate point in defining the sum is that $h_i$ is defined only over $U_i$. The partition of unity is subordinate to the cover in the strong sense that $\operatorname{supp}\rho_i \subset U_i$. Therefore, if $x \notin U_i$, then in particular $x \notin \operatorname{supp}\rho_i$, so $\rho_i(x)=0$. Thus the expression $\rho_i(x)h_{i,x}(u,v)$ only requires the local form $h_{i,x}$ at points where it is actually defined. Now fix $x \in X$ and $u,v \in E_x$. Since the family $(\operatorname{supp}\rho_i)_{i \in I}$ is locally finite, only finitely many indices $i$ satisfy $\rho_i(x)\neq 0$. Hence \begin{align*} h_x(u,v) = \sum_{i \in I} \rho_i(x) h_{i,x}(u,v) \end{align*} is a finite sum of complex numbers. Each $h_{i,x}$ is Hermitian because it is the pullback of the standard Hermitian form on $\mathbb{C}^r$. Each coefficient $\rho_i(x)$ is a non-negative real number. A finite non-negative real linear combination of Hermitian forms is again Hermitian, so $h_x$ is complex-linear in its first argument, conjugate-linear in its second argument, and satisfies \begin{align*} h_x(v,u) = \overline{h_x(u,v)} \end{align*} for all $u,v \in E_x$. [/guided] [/step] [step:Verify smoothness of the global form] Let $x_0 \in X$. Since the family $(\operatorname{supp}\rho_i)_{i \in I}$ is locally finite, there is an open neighbourhood $W \subset X$ of $x_0$ such that $W$ meets $\operatorname{supp}\rho_i$ for only finitely many indices. Let \begin{align*} I_W = \{i \in I : W \cap \operatorname{supp}\rho_i \neq \varnothing\}. \end{align*} Then $I_W$ is finite, and for every $x \in W$, \begin{align*} h_x = \sum_{i \in I_W} \rho_i(x) h_{i,x}. \end{align*} For each $i \in I_W$, define the weighted local form $\rho_i h_i$ on $E|_W$ by using $\rho_i(x)h_{i,x}$ when $x \in U_i$ and by using $0$ when $x \notin U_i$. This extension is smooth: if $x \notin U_i$, then $x \notin \operatorname{supp}\rho_i$ because $\operatorname{supp}\rho_i \subset U_i$, so, by the definition of support just given, there is an open neighbourhood of $x$ on which $\rho_i$ vanishes identically. Hence each weighted local form is a smooth section of $\operatorname{Herm}(E)|_W$, whose fiber over $x \in W$ is the real [vector space](/page/Vector%20Space) of Hermitian forms on $E_x$. Therefore the displayed finite sum defines a smooth section of $\operatorname{Herm}(E)|_W$. Since $x_0$ was arbitrary, $h$ is smooth on all of $X$. [/step] [step:Prove positive definiteness fiberwise] Fix $x \in X$ and a nonzero vector $u \in E_x$. Since $(\rho_i)_{i \in I}$ is a partition of unity, \begin{align*} \sum_{i \in I} \rho_i(x) = 1. \end{align*} Thus there exists an index $j \in I$ such that $\rho_j(x)>0$. For this index, $x \in U_j$, and since $h_{j,x}$ is positive definite, \begin{align*} h_{j,x}(u,u) > 0. \end{align*} For every $i \in I$ with $\rho_i(x)\neq 0$, the number $h_{i,x}(u,u)$ is non-negative. Therefore \begin{align*} h_x(u,u) = \sum_{i \in I} \rho_i(x) h_{i,x}(u,u) \geq \rho_j(x) h_{j,x}(u,u) > 0. \end{align*} Hence each $h_x$ is positive definite. Together with smoothness and the Hermitian symmetry verified above, this proves that $h$ is a smooth Hermitian metric on $E$. [/step]