Partial chain rule

[sonni]

Hi,

this question is about something in the book Spivak: "Calculus on Manifolds" (1995) which I don't quite understand. It is theorem 2-9 on page 32:

Let \(\displaystyle g_1, \dots, g_m : \mathbb{R}^n\rightarrow\mathbb{R}\) be continuously differentiable at \(\displaystyle a\), and let \(\displaystyle f: \mathbb{R}^m\rightarrow\mathbb{R}\) be differentiable at \(\displaystyle (g_1(a), \dots, g_m(a))\). Define the function \(\displaystyle F: \mathbb{R}^n\rightarrow \mathbb{R}\) by \(\displaystyle F(x)=f(g_1(x), \dots, g_m(x))\). Then
\(\displaystyle D_i F(a) = \sum_{j=1}^m D_j f(g_1(a), \dots, g_m(a))\cdot D_i g_j(a)\)


Why is it required that the functions \(\displaystyle g_1, \dots, g_m\) are continuously differentiable?
He already proved the general multidimensional chain rule \(\displaystyle D (g\circ f)(a) = Dg(f(a))\circ Df(a)\) only requiring \(\displaystyle f\) and \(\displaystyle g\) to be differentiable functions.

In the proof of theorem 2-9 he seems to need it to show that the function \(\displaystyle g = (g_1, \dots, g_m)\) is differentiable... but isn't "component functions differentiable \(\displaystyle \Rightarrow\) function differentiable" already a basic result that follows from the general chain rule, so that \(\displaystyle f, \,g\) don't have to be continuously differentiable, just differentiable?

Thanks in advance for any help! (maybe I'm lucky and somebody worked with this book...)

 

[daon2]

In the proof of theorem 2-9 he seems to need it to show that the function \(\displaystyle g = (g_1, \dots, g_m)\) is differentiable... but isn't "component functions differentiable \(\displaystyle \Rightarrow\) function differentiable" already a basic result

Section 6 of Munkres' "Analysis on Manifolds" may be of interest to you. He explains that "existence of partial derivatives does not imply differentiability." And we need the partial derivatives to be continuous.

An counter-example is:

\(\displaystyle \displaystyle f(x,y) = \dfrac{x^2y}{x^4+y^2}\) if (x,y) is not (0,0), and 0 if (x,y)=(0,0).

The partial derivatives exist at 0, but f is not continuous there.

 

[sonni]

Thanks for your answer. I still don't understand it, the \(\displaystyle g_1, \dots, g_m\) are components of the function \(\displaystyle g\), not the partials. On the other hand, it follows from differentiability of \(\displaystyle g_1, \dots, g_m\) that the partials \(\displaystyle D_i g_j\) must exist, it's just, as you said, that they don't need to be continuous. But why exactly do we need them to be continuous? Because if all the components of a function are differentiable, the function itself is differentiable, too.

Here is the full proof:

 

[daon2]

If you look at the proof of theorem 2.8, the hypothesis that the partial derivatives are continuous is needed at the end.

The proof of the chain rule assumes that Df(a) exists. Theorem 2.8 gives a condition for it to exist, and Theorem 2.9 uses this condition.