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We've learned today that if $f$ has second order partial derivatives that are continuous at some point $a$, then they're all equal to each other at that point.

Then there's a short remark that says this holds for higher order as well, i.e. $$D_1D_2D_3f= D_3D_2D_1f$$

What's the proof of this? I mean generally, and not just in above example?

Also, what kind of function wouldn't satisfy these conditions? I've been trying to think of a simple function that wasn't $C^2$, even if just in some point, and it's proving to be more difficult than I'd thought. Does such a simple function exist, without entering the realm of complicated piecewise functions?

Jani Tyen
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1 Answers1

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For the proof of why it holds for higher order derivatives, remember that the derivatives are themselves functions so $$\begin{align}D_1D_2D_3f &= D_1D_2(D_3f) = D_2D_1(D_3 f) \\&= D_1(D_2 D_3 f) = D_1 (D_2 D_3 f)\end{align}$$ by the fact we can interchange the order for the second derivative of a function. In general applying the above gives us the permutations $D_iD_jD_k = D_jD_iD_k$ and $D_iD_jD_k = D_iD_kD_j$. Using these two permutations repeatedly we can arrive at any order of $D_1$, $D_2$ and $D_3$.

For an example of a function which has a non-symmetric partial derviative at one point you can refer to this Wikipedia page.

Dan
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