Let $f\in C^r(A\subset \mathbb R^n,\mathbb R^m)$,

$Df:A\subset\mathbb R^n\to L(\mathbb R^n,\mathbb R^m)$ so that $Df(x):\mathbb R^n\to\mathbb R^m$ is $f$'s total derivative,

(abusing notation) $D^2f(x): \mathbb R^n \times \mathbb R^n\to\mathbb R^m$ be the bilinear map defined by $$[D^2f(x)](x_1,x_2)=[[D(Df)(x)] (x_2)](x_1),$$ (still abusing notation) $D^3f(x): \mathbb R^n \times \mathbb R^n\times \mathbb R^n\to\mathbb R^m$ be the trilinear map defined by $$[D^3f(x)](x_1,x_2,x_3)=[[D(D^2f)(x)] (x_3)](x_1,x_2),$$ and so on.

For $ r\geq 3:$

How do I show that $D^rf(x)$ is symmetric (i.e. returns the same value for every permutation of inputs)?

And why is $D^rf(x)$ being symmetric equivalent to the fact that for each component, $f$'s $r$th partials can be taken in any order?

P.S. I have read the proof for the case of second-order derivatives.

Reference: Marsden's Elementary Classical Analysis

  • $\begingroup$ Can you clarify exactly what your question is? Do you want to know why the symmetry of $D^2 f$ is equivalent to the symmetry of the mixed partials? $\endgroup$ – Anthony Carapetis Oct 31 '13 at 6:33
  • $\begingroup$ What book is this from? $\endgroup$ – littleO Oct 31 '13 at 7:03
  • $\begingroup$ Could someone tell me what definition they give for $D^2 f$? $\endgroup$ – littleO Oct 31 '13 at 7:04
  • $\begingroup$ This can't be the definition of $D^2 f$ - $Df$ is already linear, if you take the derivative of that it's exactly the same. Rather, you want to say that $D^2 f$ is the unique bilinear form such that f(a) + Df(a) (x-a) + D^2 f(a) (x-a) approximates $f(x)$ well enough, if $D^2 f$ represents the associate quadratic form as well. $\endgroup$ – user27126 Oct 31 '13 at 17:16
  • 1
    $\begingroup$ @Sanchez: that's only true when considering $D$ as an operator at one point. Here we consider it as operator on functions -- to a function you assign a position dependent linear (at a point!) operator. The total operator can surely be non-linear. It's really a generalization of standard 1D derivative $f, f', f'', \dots$. $\endgroup$ – Marek Oct 31 '13 at 17:20

The tensor components of the $D^r$ are higher partial derivatives of the $f_i$. A simple induction proof starting with $g_{.12}=g_{.21}$ shows that these higher partial derivatives have the claimed symmetries, i.e., that $f_{.\bf k}$, where ${\bf k}\in[n]^r$, depends only on how often each $\ell\in[n]$ appears as an entry in ${\bf k}$.

  • $\begingroup$ Thanks, and by considering the definitions of $D^2f(x)$ and of partial derivatives, I eventually was able to answer the first question in the pink box too. $\endgroup$ – Ryan G Nov 3 '13 at 11:57

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