Smoothness of component functions Let $X$ be a real Banach space and $V$ be a real, finite dimensional vector space. Consinder the map
$F\colon X\rightarrow V$, $F(x)=\sum_{i=1}^{n}\lambda_i(x)e_i(x)$, where $\lambda_i\colon X\rightarrow \mathbb{R}$ and $e_i\colon X\rightarrow V$. Moreover, assume that $(e_1(x),\ldots,e_n(x))$ is a basis of $V$ for all $x\in X$.
Claim: If $F$ is smooth and $e_i$ is smooth for every $i=1,\ldots,n$, then $\lambda_i$ is smooth for every $i=1,\ldots,n$.
I thought about it for a while and wasn't able to come up with something useful. Maybe I did enough math for today and the claim is obviously true or false. However, I would appreciate any help.
 A: Consider the map
$$
\Phi:\Bbb{R}^n \times X \to V, ((a_1,\dots, a_n), x) \mapsto \sum_{i=1}^n a_i \cdot e_i (x).
$$
Then $\Phi$ is smooth and the partial derivative
$$D_1 \Phi (a,x) : \Bbb{R}^n \to V, (b_1, \dots, b_n) \mapsto \sum_{i=1}^n b_i e_i(x)
$$
is invertible for every $(a,x) \in \Bbb{R}^n \times X$ (since the $e_i(x)$ form a basis).
Now apply the implicit function theorem.
EDIT: You are right, the argument following the application of the implicit
function theorem is not as straightforward as I thought. So here goes:
Let us fix a basis $\left(v_{1},\dots,v_{n}\right)$ of $V$. Let
$z\in X$ be arbitrary. Since the map $\Phi\left(\cdot,z\right)$
is surjective, the implicit function theorem implies that there is
for each $i\in\left\{ 1,\dots,n\right\} $ a smooth function $g_{i}:U\to\mathbb{R}^{n}$
defined on a neighorhood of $z$ (which we can take to not depend
on $i$) satisfying
$$
\Phi\left(g_{i}\left(x\right),x\right)=v_{i}\qquad\forall x\in U.
$$
In coordinates, this means
$$
v_{i}=\sum_{\ell}\left(g_{i}\left(x\right)\right)_{\ell}e_{\ell}\left(x\right).
$$
Let $\left(\varphi_{1}\left(x\right),\dots,\varphi_{n}\left(x\right)\right)$
be the dual basis to $\left(e_{1}\left(x\right),\dots,e_{n}\left(x\right)\right)$
and let $\left(\psi_{1},\dots,\psi_{n}\right)$ be the dual basis
for $\left(v_{1},\dots,v_{n}\right)$. We then have $\varphi_{j}(x)=\sum_{\ell}\alpha_{j,\ell}\left(x\right)\cdot\psi_{\ell}$
for certain (uniquely determined) functions $\alpha_{j,\ell}:X\to\mathbb{R}$.
Now note
\begin{eqnarray*}
\left\langle \varphi_{j}\left(x\right),v_{i}\right\rangle  & = & \sum_{\ell}\left(g_{i}\left(x\right)\right)_{\ell}\underbrace{\left\langle \varphi_{j}\left(x\right),e_{\ell}\left(x\right)\right\rangle }_{=\delta_{\ell,j}}\\
 & = & \left(g_{i}\left(x\right)\right)_{j}
\end{eqnarray*}
for $x\in U$. On the other hand,
$$
\left\langle \varphi_{j}\left(x\right),v_{i}\right\rangle =\sum_{\ell}\alpha_{j,\ell}\left(x\right)\cdot\underbrace{\left\langle \psi_{\ell},v_{i}\right\rangle }_{=\delta_{\ell,i}}=\alpha_{j,i}\left(x\right),
$$
so that
$$
\alpha_{j,i}\left(x\right)=\left(g_{i}\left(x\right)\right)_{j}
$$
is smooth on $U$. Since $U$ is a neighborhood of $z$ and since
$z\in X$ was arbitrary, we have shown that the functions $X\to\mathbb{R},x\mapsto\alpha_{j,i}\left(x\right)$
are smooth.
Now, by definition of a dual basis and by choice of $F,\lambda$
and $\left(\alpha_{j,\ell}\right)_{j,\ell}$, we have
\begin{eqnarray*}
\lambda_{i}\left(x\right) & = & \left\langle \varphi_{i}\left(x\right),F\left(x\right)\right\rangle \\
 & = & \left\langle \sum_{\ell}\alpha_{i,\ell}\left(x\right)\psi_{\ell},F\left(x\right)\right\rangle \\
 & = & \sum_{\ell}\underbrace{\alpha_{i,\ell}\left(x\right)}_{\text{smooth}}\underbrace{\left\langle \psi_{\ell},F\left(x\right)\right\rangle }_{\text{smooth since }F\text{ is smooth}},
\end{eqnarray*}
so that each of the functions $\lambda_{i}$ is smooth.
