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Let $M$ be a compact, finite-dimensional manifold and $\pi : \mathrm{B} \rightarrow M$ a vector bundle over $M$ whose typical fiber is $\mathbb{R}^n$. Denote by $\mathcal{C}^{\infty}(M, \mathrm{B})$ the vector space of all smooth sections of $\mathrm{B}$.

If we choose Riemannian metrics on (the fibers of) $\mathrm{B}$ and on $M$, and then introduce a metric connection $D$ on $B$, we may define a family of seminorms on $\mathcal{C}^{\infty}(M, \mathrm{B})$ by $$\| s \|_n = \sum_{i = 0}^n~\sup_{x \in M} |D^js(x)|,$$ where $|D^0s(x)|$ is just $|s(x)|$, and for $j \geq 1$, $$|D^js(x)| = \sup |(D_{v_1} \circ \dots \circ D_{v_j}s)(x)|,$$ the supremum being taken over all $(v_1, \ldots, v_j) \in (T_xM)^j$ with $|v_k| = 1$, for $k = 1, \ldots, j$. It's not hard to prove that if $\{s_n\}_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathcal{C}^{\infty}(M, \mathrm{B})$ with respect to this family of seminorms, then it converges to a continuous section $s : M \rightarrow \mathrm{B}$. I'd like to prove that $s$ is actually smooth.

QUESTION: What would be the best ($\sim$ least messy, shortest) way to prove this?

Thanks.

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    Isn't that obvious if you locally pick an orthonormal frame $E_i$ for $B$ and a geodesic chart, giving an orthonormal basis $\partial_j$ for $TM$?2012-03-25

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