1
$\begingroup$

Let $U \subseteq {\rm I\!R}^{n}$ be an open subset and let $f: U \longrightarrow {\rm I\!R}^{m}$ be a function, where $m \geqslant 1$. Assuming the Jacobian matrix $\textbf{J}$ of $f$ exists, when is it continuous and how would one best show continuity? I know the Jacobian can be thought of as the following mapping:

$\textbf{J}: U \longrightarrow Hom({\rm I\!R}^{n}, {\rm I\!R}^{m}), \ x_{0} \mapsto(v \mapsto \textbf{J}_{x_{0}}(v))$

and $\textbf{J}_{x_{0}}: {\rm I\!R}^{n} \longrightarrow{\rm I\!R}^{m}$ is continuous, since it is a linear map on a finite-dimensional vector space. But when is $\textbf{J}$ continuous in $U$ and how does one show this? I already know that if all partial derivatives exist and are continuous, then $f$ is continuously differentiable, i.e. $\textbf{J}$ is continuous. So does that mean one could check continuity by verfying that $\partial_{j}f_{k}$ is continuous for all $j \in \{1,...,n \}$ and $k \in \{1,...,m \}$, i.e. if every component of $\textbf{J}$ is continuous, then $\textbf{J}$ is continuous? Surely not?

EDIT: Given that vector-valued functions are continuous if and only if their component functions are continuous and the partial derivative $\partial_{k} f: U \longrightarrow {\rm I\!R}^{m}$, $x \mapsto \partial_{k} f(x) = (\partial_{k} f_{1}(x), ..., \partial_{k} f_{m}(x))$ is a vector-valued function for all $k \in \{1,...n \}$, perhaps it isn't such a bad idea after all to deduce continuity of the Jacobian by checking every component, since the columns of the Jacobian are exactly the vector-valued partial derivatives. Any objections?

$\endgroup$
4
  • 1
    $\begingroup$ math.stackexchange.com/questions/3003126/… $\endgroup$ – Calvin Khor Jun 20 '19 at 17:58
  • 1
    $\begingroup$ Even though they translate nicely into one another, it should be noted that the Jacobian matrix and the total differential are not the same object and the Jacobian matrix is not an element of $\mathrm{Hom}(\mathbb{R}^n,\mathbb{R}^m)$. The continuity of the Jacobian matrix is by the component-wise characterization exactly what you describe. That this is equivalent to the continuity of the map $x\mapsto\mathrm{d}f(x)$ is (in my opinion) not immediate. However it remains true that this map is continuous if and only if all the partial derivatives are continuous. $\endgroup$ – Thorgott Jun 20 '19 at 23:45
  • $\begingroup$ @Thorgott Could you please explain how the Jacobian matrix and the total differential differ? $\endgroup$ – playdis Jun 21 '19 at 8:22
  • 1
    $\begingroup$ @playdis The Jacobian is a matrix, the total differential is a linear map. I'm not trying to imply they aren't closely related (because they are), but one should not forget they are different objects. The continuity of these maps translates into one another because mapping a linear operator to it's transformation matrix is a homeomorphism. $\endgroup$ – Thorgott Jun 21 '19 at 10:46
1
$\begingroup$

I already know that if all partial derivatives exist and are continuous, then $f$ is continuously differentiable, i.e. $\textbf{J}$ is continuous. So does that mean one could check continuity by verfying that $\partial_{j}f_{k}$ is continuous for all $j \in \{1,...,n \}$ and $k \in \{1,...,m \}$, i.e. if every component of $\textbf{J}$ is continuous, then $\textbf{J}$ is continuous?

This is correct.

Perhaps you were just slightly unconvinced that checking continuity of a "simple" object like $\partial_jf_k: U \to \Bbb{R}$ (for all $j,k$) implies the continuity of a more "complicated" object like $\boldsymbol{J}: U \to \text{Hom}(\Bbb{R}^n, \Bbb{R}^m)$.

In fact, using induction, you can replace "continuously differentable" with $\mathcal{C}^k$, for any $k \in \Bbb{N}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.