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I am trying to understand the proof below: screenshot, transcribed below

8.4.2 Continuously differentiable functions

Definition 8.4.5. Let $U \subset \mathbb{R}^{n}$ be open. We say $f: U \rightarrow \mathbb{R}^{m}$ is continuously differentiable or $C^{1}(U)$, if $f$ is differentiable and $f^{\prime}: U \rightarrow L\left(\mathbb{R}^{n}, \mathbb{R}^{m}\right)$ is continuous.

Proposition 8.4.6. Let $U \subset \mathbb{R}^{n}$ be open and $f: U \rightarrow \mathbb{R}^{m} .$ The function $f$ is continuously differentiable if and only if the partial derivatives $\frac{\partial f_{j}}{\partial x_{\ell}}$ exist for all $j$ and $\ell$ and are continuous.

Without continuity the theorem does not hold. Just because partial derivatives exist does not mean that $f$ is differentiable, in fact, $f$ may not even be continuous. See the exercises for the last section and also for this section.

Proof. We proved that if $f$ is differentiable, then the partial derivatives exist. The partial derivatives are the entries of the matrix of $f^{\prime}(x) .$ If $f^{\prime}: U \rightarrow L\left(\mathbb{R}^{n}, \mathbb{R}^{m}\right)$ is continuous, then the entries are continuous, and hence the partial derivatives are continuous.

In single variable calculus, the functions are from R to R and the derivative is also a function from R to R, but in the multi variable case, the derivative is a linear map, and if it exist, it is the Jacobian matrix. Now according to the definition, to say that the derivative is continuous in a region means that the mapping from U to $L(R^n,R^m)$ is continuous. So we are mapping points of $R^n$ to the space of linear transformations and we are saying that if this mapping is continuous, then the partial derivatives are continuous in the region. First, I'm not even sure what it means for a mapping from $R^n$ to $L(R^n,R^m)$ to be continuous. Don't we need to define a norm on the space of linear transformation first? Second, why would the continuity of that mapping imply that the partial derivatives, which are mapping from $R^n$ to R, are continuous?

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    $\begingroup$ please don't use screenshots of plain text; i have transcribed the image $\endgroup$ Aug 4 at 14:24
  • $\begingroup$ When you post excerpts like this, you should always cite the source. (And this will also help to provide context to your question.) $\endgroup$ Aug 4 at 19:24
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The space $\mathbb{R}^{n \times m}$ of real-valued $n \times m$ matrices is a normed vector space, and is of (finite!) dimension $nm$, so in fact every norm on it is equivalent. There are several commonly used matrix norms, most of which can be seen as extensions of the vector norms we know and love.

For your second question, recall that a function into a product is continuous if and only if it is continuous in each of its components. Considering the derivative $f'$ to be a function into $nm$ components, if $f'$ is continuous, all of its components must be as well.

However, we know that if $f'$ exists, its components are just the partials!

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You claim that “the derivative is a linear map, and if it exist, it is the Jacobian matrix”. There is some confusion here, since linear maps and matrices are distinct things. What happens is that the Jacobian matrix of $f'(p)$ is the matrix of $f'(p)$ with respect to the standard bases.

Now, let us talk about your main questions. Yes, it is natural to define first a norm on $L(\Bbb R^n,\Bbb R^m)$. It doesn't really matter which norm, since $L(\Bbb R^n,\Bbb R^m)$ is finite-dimensional and, on a finite-dimensional real vector space all norms are equivalent. And, no matter which norm you will take, each map$$\begin{array}{ccc}L(\Bbb R^n,\Bbb R^m)&\longrightarrow&\Bbb R\\f&\mapsto &a_{ij}(f),\end{array}$$will be continuous, where $a_{ij}(f)$ is the entry of the matrix of $f$ with respect to the standard bases which belongs to the $i$th row and to the $j$th column. But if $f=f'(p)$, then $a_{ij}(f)=\frac{\partial f_j}{\partial x_i}$.

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  • $\begingroup$ thanks! By the way, if we do not use the standard basis, does the Jacobian matrix still represent the derivative or does it get a lot more complicated? $\endgroup$
    – Bill
    Aug 5 at 1:09
  • $\begingroup$ By definition, the Jacobian matrix is the matrix of $f'(p)$ with respect to the standard basis. So, if we use some other basis, all that we'll get is some matrix which is similar to the Jacobian matrix. $\endgroup$ Aug 5 at 1:26

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