7
$\begingroup$

I had a little difficulty understanding Theorem 12.13 of Apostol's Mathematical Analysis, and was wondering if someone who had read this book could give me some pointers. Specifically, on page 360, the author writes

"As a consequence of Theorem 12.11 and 12.12 we have

Theorem 12.13 If both partial derivatives $D_{r}f$ and $D_{k}f$ exist in an n-ball $B(c)$ and if both $D_{k,r}f$ and $D_{r,k}f$ are continuous at c, then $D_{k,r}f(c) = D_{r,k}f(c)$."

My understanding is that we're using Theorem 12.11 to establish the differentiability of $D_{r}f$ and $D_{k}f$ so that we can use Theorem 12.12 to assert the equality of $D_{k,r}f(c) = D_{r,k}f(c)$. If so, my question is: why shouldn't we require $D_{r,r}f$ and $D_{k,k}f$ also exist at c?

For convenience, I typed Theorem 12.11 & 12.12 below:

"Theorem 12.11 Assume that one of the partial derivatives $D_{1}f,...,D_{n}f$ exists at c and that the remaining n-1 partial derivatives exist in some n-ball $B(c)$ and are continuous at c. Then $f$ is differentiable at c.

Theorem 12.12 If both partial derivatives $D_{r}f$ and $D_{k}f$ exist in an n-ball $B(c)$ and if both are differentiable at c, then $D_{k,r}f(c) = D_{r,k}f(c)$."

$\endgroup$
6
  • 1
    $\begingroup$ He does require the mixed partials to exist at $c$! He assumes they are continuous at $c$; to be continuous at $c$ requires existence in a neighborhood of $c$, which includes $c$. $\endgroup$ Commented Aug 3, 2015 at 15:33
  • $\begingroup$ @DavidC.Ullrich My question is about $D_{r,r}f$ & $D_{k,k}f$. $\endgroup$
    – syeh_106
    Commented Aug 3, 2015 at 15:40
  • $\begingroup$ Oh. Sorry, hard to read the subscripts... $\endgroup$ Commented Aug 3, 2015 at 16:02
  • $\begingroup$ Looking at it more carefully, I see your point. Note that 12.13 is certainly true, and not that hard to prove. But offhand I don't see how it's immediate from 12.11 and 12.12, for exactly the reason you give, There's no further explanation? $\endgroup$ Commented Aug 3, 2015 at 16:10
  • $\begingroup$ @DavidC.Ullrich Thanks, David, for your comments. The author didn't provide proof for Thm 12.13, but simply stated "As a consequence of Thm 12.11 & 12.12 we have Thm 12.13." That's why I got stuck. I couldn't infer it from Thm 12.11 & 12.12, and was wondering where I missed. $\endgroup$
    – syeh_106
    Commented Aug 3, 2015 at 16:43

0

You must log in to answer this question.

Browse other questions tagged .