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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?

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  • $\begingroup$ It is a basic theorem of mathematics that $\frac{\partial^2f}{\partial x\partial y}= \frac{\partial^2f}{\partial y\partial x}$ $\endgroup$ Jul 14, 2022 at 19:11

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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 $v_\alpha$.

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  • $\begingroup$ So writing $D^{\alpha}$ means a weak derivative, not the "ordinary" derivative? $\endgroup$
    – mathslover
    Jul 15, 2022 at 15:02
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    $\begingroup$ The notation is certainly used for both. In this answer, it is used twice: First, at the beginning as $D^\alpha f$. As $f$ is not defined, it could be either. Later, the notation is used in the integral: $D^\alpha \phi$. Here, $\phi$ is a test function (from the definition of weak derivatives). Therefore, the ordinary derivative is meant in this case (recall that the weak derivative is defined by “shifting” the differentiation to a smooth function under an integral sign, so we can differentiate normally). $\endgroup$ Jul 15, 2022 at 15:16
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It is a basic threm of Calculus that $\frac{\partial^2f}{\partial x\partial y}= \frac{\partial^2f}{\partial y\partial x}$.

The order DOES NOT MATTER!

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    $\begingroup$ Of course, you still need to think for a moment how the basic theorem implies the version for weak derivatives. $\endgroup$ Jul 15, 2022 at 6:01
  • $\begingroup$ Is this generally true? I thaught that we need to assume, for example, that $f$ is twice continuously differentiable? $\endgroup$
    – mathslover
    Jul 15, 2022 at 7:07
  • $\begingroup$ @mathslover You’re Right, there are conditions (like continuity of the derivatives) for the classic theorem. For the weak derivative, you will need the classic theorem in the proof that the order does not matter, but the conditions are rather obviously satisfied in this application. (See the other answer.) $\endgroup$ Jul 15, 2022 at 8:02
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Usually the order of differentiation does not matter due to the Clairaut's Theorem. But there are some conditions that should be satisfied -- for instance function should have continuous partial derivatives on a neighbourhood of the point at which we take the derivatives.

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