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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?

Mandelbrot
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  • Hello and welcome to math.stackexchange. This estimate may be shown as follows: $|Df(u) w| \le |Df(u)|\infty |w|$ for all $w,u$ where $|\cdot|\infty$ is the maximum norm for matrices. Now show that $u \mapsto |Df(u)|_\infty$ is a continuous function by using the assumption $f \in C^1$ and apply the extreme value theorem. – Hans Engler Jun 18 '20 at 19:03
  • Hello Hans,

    thank you for your answer. $u\mapsto Df(u)$ is continuous, since the entrys of $Df(u)$ are the partial derivatives of $f$, which are continuous since $f$ is $C^1$. Replacing every entry with its absolute value is also continuous. The function $max(x_1,x_2)$ is also continuous, since $$max(x_1,x_2)=\frac{x_1+x_2+\mid x_1-x_2,\mid}{2}$$ For more than 2 variables, one can write $max(x_1,...,x_n)=max(x_1,max(x_2,...,x_n))$ and as a composition of continuous functions, this is also continuous. Is this correct?

     – Mandelbrot Jun 18 '20 at 19:25
  • What would be well-known is that $f \in C^1$ means that $u \mapsto Df(u)$ is continuous with respect to the supremum norm. – RRL Jun 18 '20 at 21:01

2 Answers2

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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,

$$| \, \|Df(u)\|_{mn} - \|Df(v)\|_{mn} \, | \leqslant \|Df(u) - Df(v)\|_{mn} \leqslant \left(\sum_{i=1}^n\sum_{j=1}^m |D_jf_i(u)- D_jf_i(v)|^2 \right)^{1/2}$$

RRL
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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 $$ |Df(p)\cdot w| = \bigg|Df(p)\cdot \frac{w}{|w|}\bigg||w| \leq L|w|. $$

Neal
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