Multivariable Differentiation - Component Functions ....

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 2: Differentiation ... ...

I need help with the proof of Proposition 2.2.9 ... ...

Duistermaat and Kolk's Proposition 2.2.9 read as follows:

In the above text D&K state that Lemma 1.1.7 (iv) implies Proposition 2.2.9 ...

Can someone please indicate how/why ths is the case ...

Peter
===========================================================================================The above post mentions Lemma 1.1.7 ... so I am providing the text of the same ... as follows:

 

[fresh_42]

Differentiability of a function $g(x)$ means: $g(a+v)-g(a)-((D_a)(g))(v) = \varepsilon_a(v)$ with $\lim_{v \to 0}\dfrac{\varepsilon_a(v)}{||\varepsilon_a(v)||}=0\,.$
Now $D_a(f_i)$ are all linear iff $D_a(f)$ is. With the lemma we get
$$
|f_i(a+v)-f_i(a)-((D_a)(f_i))(v)| \leq \sqrt{n}\cdot ||f(a+v)-f(a)-((D_a)(f))(v)|| = \sqrt{n} \cdot ||\varepsilon_a(v)|| =: || \psi_a(v)||
$$
with $\psi_a = \sqrt{n}\cdot \varepsilon_a$ and $\lim_{v\to 0}\dfrac{\psi_a(v)}{||\psi_a(v)||} = 0$, i.e. the differentiability of $f$ gives the differentiability of the component functions $f_i\,.$ The other inequality $||y|| \leq |\sum_{i=1}^n\,y_i|$ gives the other estimation $||f(a+v)-f(a)-((D_a)(f))(v)|| \leq \sum \ldots$ with $\varepsilon_a(v)= \sum (\varepsilon_i)_a (v_i)\,.$

As an important note here: Proposition 2.2.9 is about the component functions of $f$, not the partial derivatives, i.e. not about the components of $x$. The situation with partial derivatives is less strong:

• Total differentiability implies continuity.
• Total differentiability implies partial differentiability in all coordinates.
• Partial differentiability does not imply continuity and thus not total differentiability.
• Continuous partial differentiability, i.e. functions with continuous partial derivatives are also continuous total differentiable.

 

[Math Amateur]

Differentiability of a function $g(x)$ means: $g(a+v)-g(a)-((D_a)(g))(v) = \varepsilon_a(v)$ with $\lim_{v \to 0}\dfrac{\varepsilon_a(v)}{||\varepsilon_a(v)||}=0\,.$
Now $D_a(f_i)$ are all linear iff $D_a(f)$ is. With the lemma we get
$$
|f_i(a+v)-f_i(a)-((D_a)(f_i))(v)| \leq \sqrt{n}\cdot ||f(a+v)-f(a)-((D_a)(f))(v)|| = \sqrt{n} \cdot ||\varepsilon_a(v)|| =: || \psi_a(v)||
$$
with $\psi_a = \sqrt{n}\cdot \varepsilon_a$ and $\lim_{v\to 0}\dfrac{\psi_a(v)}{||\psi_a(v)||} = 0$, i.e. the differentiability of $f$ gives the differentiability of the component functions $f_i\,.$ The other inequality $||y|| \leq |\sum_{i=1}^n\,y_i|$ gives the other estimation $||f(a+v)-f(a)-((D_a)(f))(v)|| \leq \sum \ldots$ with $\varepsilon_a(v)= \sum (\varepsilon_i)_a (v_i)\,.$

As an important note here: Proposition 2.2.9 is about the component functions of $f$, not the partial derivatives, i.e. not about the components of $x$. The situation with partial derivatives is less strong:

• Total differentiability implies continuity.
• Total differentiability implies partial differentiability in all coordinates.
• Partial differentiability does not imply continuity and thus not total differentiability.
• Continuous partial differentiability, i.e. functions with continuous partial derivatives are also continuous total differentiable.

Thanks fresh_42 ... for such a clear explanation ...

Just a point of clarification ... ...

You write:

" ... ... Now $D_a(f_i)$ are all linear iff $D_a(f)$ is. ... ... "Can you explain how we know this is true ...

Peter

 

[fresh_42]

If we have a linear map $D_a(f)$ then this is in coordinates a matrix. And each row $D_a(f_i)$ defines a linear map $v \mapsto D_a(f_i) \cdot v^\tau = \langle D_a(f_i),v\rangle\,.$ And this goes back in the other direction the same way: If we have $p$ linear maps $D_a(f_i) \, : \, \mathbb{R}^n \longrightarrow \mathbb{R}^1$ then they can always be written as $D_a(f_i)(v)=v_i\cdot v^\tau = \langle v_i,v \rangle$ and the $v_i$ give us the rows for $D_a(f)$. (Here vectors $w$ are rows and $w^\tau$ columns.)

 

[Math Amateur]

Thanks fresh_42 ... appreciate the help ...

Reflecting on what you have written ...

Thanks again ...

Peter