$\mid Df(u)\cdot w\mid \leq L \mid w\,\mid$ Suppose $f:C\to \mathbb{R}^n$ is a $C^1$-function, where $C$ is a compact subset of $\mathbb{R}^m$. I want to show that there exists an $L\in\mathbb{R}_{>0}$ such that
$$\mid Df(u)\cdot w\mid\leq L\mid w\,\mid$$
for all $u\in C$ and $w\in \mathbb{R}^m$, where $D$ denotes the Jacobian of $f$ at $u$. I have already shown this using the extreme value theorem and the Heine-Borel-theorem, but my proof is too long for my purposes. I am looking for a proof, which is as short as possible. Could that be a well known fact?
 A: In terms of partial derivatives, we have  for all $w = (w_1,\ldots,w_m)\in \mathbb{R}^m$
$$Df(u) \cdot w = \left(\sum_{j=1}^mD_jf_1(u) w_j, \ldots, \sum_{j=1}^mD_jf_n(u) w_j\right)$$
Whence,
$$|Df(u) \cdot w |^2 = \sum_{i=1}^n\left(\sum_{j=1}^mD_jf_i(u)w_j\right)^2\leqslant\sum_{i=1}^n\left(\sum_{j=1}^m |D_jf_i(u)|^2\sum_{j=1}^m w_j^2\right)\\ = \sum_{i=1}^n\sum_{j=1}^m |D_jf_i(u)|^2 \, |w|^2 \\ \implies |Df(u) \cdot w | \leqslant \left(\sum_{i=1}^n\sum_{j=1}^m |D_jf_i(u)|^2 \right)^{1/2}|w|$$
Since, all partial derivatives are continuous on the compact set $C$, they are bounded and so too is $ \left(\sum_{i=1}^n\sum_{j=1}^m |D_jf_i(u)|^2 \right)^{1/2}$. Hence, there exists $L$ such that for all $u \in C$
$$|Df(u) \cdot w | \leqslant \left(\sum_{i=1}^n\sum_{j=1}^m |D_jf_i(u)|^2 \right)^{1/2}|w| \leqslant L |w|$$

The proof is more concise using well-known properties of the normed, linear space of operators from $\mathbb{R}^m$ to $\mathbb{R}^n$.
We have $Df(u) \in \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n) )$ and $|Df(u) \cdot w| \leqslant \|Df(u)\|_{mn}\, |w|$.
The map $u \mapsto \|Df(u)\|_{mn}$ is continuous since, by the reverse triangle inequality,
$$| \, \|Df(u)\|_{mn} - \|Df(v)\|_{mn} \, | \leqslant \|Df(u) - Df(v)\|_{mn} \leqslant \left(\sum_{i=1}^n\sum_{j=1}^m |D_jf_i(u)- D_jf_i(v)|^2 \right)^{1/2}$$
A: Here's a compact proof. Define $F: C\times S^m\to \mathbb{R}$ by $(p,v)\mapsto |Df(p)\cdot v|$. As a composition/product/sum of continuous functions, $F$ is manifestly continuous. Because $C\times S^m$ is the product of compact spaces, it itself is compact, hence $F$ attains a maximum value $L$ on $C\times S^m$. Now we are done, for letting $w\in\mathbb{R}^m$ be arbitrary we have
$$ |Df(p)\cdot w| = \bigg|Df(p)\cdot \frac{w}{|w|}\bigg||w| \leq L|w|. $$
