A proof on smooth function that I don't know what to proof. Here's the question:
Suppose $f: U \rightarrow V$ is a smooth map, for $U \subset R^k$ and $V \subset R^\ell$ open sets. That is, all partial derivatives (of all orders) of $f$ exist and are continuous. Show that the map $Df: U \rightarrow \mathrm{Mat}_{\ell \times k} \simeq R^{k\ell}$ given by $Df(x) = df_x \in \mathrm{End}(R^k,R^\ell) = \mathrm{Mat}_{\ell \times k}$, is smooth.
And here's my thought:
Give all partial derivatives (of all orders) of $f$ exist and are continuous, that is $d^n f_x$ exists and continuous. And by definition, Df is so.
So my doubt is - I didn't really do anything here, right? Thank you very much for helping me clear this out!
 A: I think you could be more explicit in relating the partial derivatives of $f$ with the partial derivatives of $Df$. We want $Df$ to be smooth; this requires each one of its coordinate functions, $(Df)_{i,j}:U\to\mathbb R$ for indices $i\leq k$ and $j\leq\ell$, to be smooth. In other words, we want each partial derivative $\partial_\alpha(Df)_{i,j}$ to exist and be continuous for each multi-index $\alpha$.
Now, $(Df)_{i,j}=\partial_i f_j$, where $\partial_i$ is the partial derivative with respect to the $i$-th coordinate, and $f_j:U\to\mathbb R$ is the $j$-th coordinate of $f$. By hypothesis, $f$ is smooth, so $f_j$ is smooth. That means that every partial derivative of $f_j$ exists and is continuous. In particular, $\partial_\alpha \partial_i f_j$ exists and is continuous for all $\alpha$, which is exactly what we want!
(Note that I haven't been terribly careful about using subscripts versus superscripts. You might want to clean up the notation, and you might also want to specify just what kind of partial derivative $\partial_\alpha \partial_i$ is.)
