Reading from this Wikipedia article about Sobolev spaces.
Let $k \in \mathbb{N}$ and $1 \le p \le \infty$. The Sobbolev space $W^{k,p}(\Omega)$ is defined to be the set of all functions $f$ on $\Omega $ such that for every multi-index $\alpha = (\alpha_1, \dots, \alpha_n)$ with $|\alpha| \le k$,the mixed partial derivative $$ f^{(\alpha)} = \frac{\partial^{\alpha_1+\dots+\alpha_n} f}{\partial x_1^{\alpha_1} \dots \partial x_n^{\alpha_n}} $$ exists in the weak sense and is in $L^p (\Omega)$.
When it says for every multi-index, does it mean that for every partial derivative of $f$ of order at most $k$; or for every partial derivative of $f$ of order at most $k$, where we differentiate in this particular order: first with respect to $x_n$, then wrt $x_{n-1}$, $\dots$, and finally wrt $x_1$?
In addition, how to denote partial derivatives in this way when the order of differentiation is different?
Best Answer
So you are right about the order: The notation $f^{(\alpha)}$ or more commonly $D^\alpha f$ implies that "you are allowed" to take every possible combination of partial derivative such that $\vert \alpha \vert := \sum_{i=1}^N \alpha_i \leq k$ where $N = \text{dim}(\boldsymbol x)$ is the dimension of the "argument" (not function) space.
Concerning the question regarding the order: For sufficiently often continuously differentiable functions the order of derivatives does indeed not matter, this is stated by Generalized Young's Theorem.
To make a statement on weak derivatives, we should take a look at the definition:
We say $v_\alpha$ is the weak derivative of $u$ for a particular $\alpha : \vert \alpha \vert \leq k$ if $$ \int_\Omega u D^\alpha \phi \text{d} \boldsymbol x = (-1)^{\vert \alpha \vert} \int_\Omega v_\alpha \phi \text{d} \boldsymbol x \quad \forall \: \phi \in C_c^\infty(\Omega). $$
Since $\phi$ is by definition a smooth = infinitely often continuously differentiable function, the order of the derivatives in the differential operator $D^\alpha$ is indeed arbitrary and leads to the same weak derivative $v_\alpha$.