How to prove an indirectly defined mapping is smooth I'm trying to understand a proof in Lang's Fundamentals of Differential Geometry. It is Chapter VIII Theorem 4.2. We are given a pseudo Riemannian manifold $(X,g)$ and want to show the existence of a unique spray $F$ on $X$ satisfying a certain condition. One form of the condition states that given a chart
$(U,\phi)$ where $\phi\colon U\to E$ where $E$ is a self-dual Banach space, the
symmetric bilinear map $B_U\colon\phi(U)\to L^2_s(E;E)$ associated with the spray for the open set $U$ satisfies a certain formula (MS 1). This formula defines
$B_U(\phi x)$ for $x\in U$, by giving the values of a complicated expression involving
$B_U(\phi x)$ indirectly. I actually have no problem with the definition (which I'll give below)
of $B_U$. My problem is that Lang gives no
indication of how to prove that $B_U$ is smooth, which I believe is required.
To explain the formula (MS 1), I first need to describe a nasty bit of Lang terminology.
$g\colon X\to (L^2_s)_X(TX)$ is the given pseudo Riemannian metric, but locally, he
reuses the letter $g$ to mean $g\colon\phi(U)\to L(E,E)$, which
"denotes the operator defining the metric relative to the given non-singular form on E."
As I mentioned, part of the definition of a
pseudo Riemannian manifold is that it is modeled on self-dual Banach spaces, so $E$ is self-dual, by means of a non-singular bilinear form
$\langle\cdot,\cdot\rangle_E$. Therefore, any bilinear form $\Omega$ on $E$ has
associated with it an operator $A$ on $E$ such that
$$\Omega(v,w)=\langle Av,w\rangle_E\qquad(v,w\in E).$$
So the quoted statement means that $G(x)(\tau_x^{-1}v,\tau_x^{-1}w)=\langle g(\phi x)v,w\rangle_E$ where I have renamed the original pseudo Riemannian metric to be $G$,
and where $\tau:\pi^{-1}(U)\to U\times E$ is the "trivializing" map associated with $(U,\phi)$. ($\tau_x$ is a linear isomorphism of $(TX)_x=\pi^{-1}(\{x\})$ (the fiber in $TX$ over $x$)
with $E$.) $G(x)$ is a non-singular symmetric bilinear form on $(TX)_x$ when given
the topology derived from $E$ through $\tau_x$. The non-singularity
implies that the associated operator is invertible, so really
$g\colon U\to\text{Laut}(E)$, the linear automorphisms of E onto itself.
With this, we have enough to write down MS 1:
\begin{equation}
  \begin{split}
    \text{MS 1.}\\
    -2\langle(&B_U(\phi x))(v,w),g(\phi x)z\rangle_E\\
    &=\langle g'(\phi x)\cdot v\cdot z,w\rangle_E+\langle g'(\phi x)\cdot w\cdot z,v\rangle_E
    -\langle g'(\phi x)\cdot z\cdot w,v\rangle_E\qquad\text{(for all $v,w,z\in E$)}.
  \end{split}
\end{equation}
I can show that $B_U(\phi)$ "defined" by MS 1 is symmetric bilinear continuous.
What I'm unable to show, so far, is that $B_U$ is smooth. Certainly everything
involved in MS 1 is smooth, but that's not a proof. I have not been able to come up
with an expression for $B_U(\phi x)$. The best I've been able to do is to come up
with an expression for $(B_U(\phi x))(v,w)$, and it's rather convoluted, and doesn't
seem amenable to massaging so that $(v,w)$ can be pulled out. I've tried using various
tricks like using the linear isomorphism associated with a non-singular bilinear form,
and even tried to use the canonical embedding of $E$ in its double dual. I also tried
putting the expression in a form that could possibly be used with the implicit function
theorem to try to prove smoothness that way, but with no luck. These attempts involve
lots of writing, so I'd prefer to not enter them all to begin with. I'd appreciate any
suggestions on how to show $B_U$ is $C^\infty$.
 A: After a lot more studying, I realized that Lang actually gave two proofs
of the theorem. The second proof, tucked into the last paragraph of the entire
proof solves my problem. He starts with an earlier result, that given a
pseudo Riemannian manifold (X,g), there is a canonical spray over X and
the part of the local representation of the spray that is quadratic,
satisfies a certain equation (Chapter VII Theorem 7.2) given in terms of
the quadratic maps. If one uses the formula for recovering the bilinear
map from the quadratic map and calculates the same equation for the
bilinear map, MS 1 pops out. Incidentally, this also proves that
$B_U$ is $C^\infty$ since it's derived from the local representation
of the spray by taking two partial derivatives and evaluating at 0.
Since the spray is smooth, so is $B_U$.
So, I believe I was wrong in saying that I needed to show that
$B_U$ was $C^\infty$, but even worse, my attempted proof (which was trying
to follow Lang's first proof) was not quite on the right track.
