Higher Order Mixed Partial Derivatives

• Oct 20th 2009, 06:22 PM
6DOM
Higher Order Mixed Partial Derivatives
If f∈C³(S) for open S⊆R^n and f: S -> R^m, prove that all the third order mixed partial derivatives are equal.

My textbook has the proof for the second order mixed partial derivatives, but the third order seems harder.

Or can I somehow reduce the problem to the second partial derivative problem and use the fact that all the second order mixed partial derivatives are equal to prove this?

Thank you
• Oct 20th 2009, 06:28 PM
Jose27
Quote:

Originally Posted by 6DOM

Or can I somehow reduce the problem to the second partial derivative problem and use the fact that all the second order mixed partial derivatives are equal to prove this?

Thank you

Exactly, just use the fact that the third partial derivatives are just the second partial derivatives of the first p.d. and the fact that these last ones are $C^2$
• Oct 20th 2009, 06:37 PM
6DOM
Quote:

Originally Posted by Jose27
and the fact that these last ones are $C^2$

Could you please elaborate?
I am still confused as to how to satisfy all conditions for the "new" second p.d to follow the fact.
• Oct 20th 2009, 06:58 PM
Jose27
Take for instance $\frac{ \partial ^3}{ \partial x_i \partial x_j \partial x_k }f(x_1,...,x_n)$ and $\frac{ \partial ^3}{ \partial x_j \partial x_k \partial x_i }f(x_1,...,x_n)$ $= \frac{ \partial }{ \partial x_j } (\frac{ \partial ^2 }{\partial x_k \partial x_i} f(x_1,...,x_n))$ $= \frac{ \partial }{ \partial x_j } (\frac{ \partial ^2 }{\partial x_i \partial x_k} f(x_1,...,x_n))$ $=\frac{ \partial ^2}{ \partial x_j \partial x_i } (\frac{ \partial }{\partial x_k } f(x_1,...,x_n))$ $=\frac{ \partial ^2}{ \partial x_i \partial x_j } (\frac{ \partial }{\partial x_k } f(x_1,...,x_n))$ $= \frac{ \partial ^3}{ \partial x_i \partial x_j \partial x_k }f(x_1,...,x_n)$

Notice that the only thing you do is switching two adjacent derivatives at a time (we can since the first p.d. are $C^2$)