The Sobolev norm descends to equivalence classes

The set ${\mathcal{A}}_k$ is finite because each coordinate of every $\alpha$ lies in $\{0,\dots ,k\}$, and is nonempty because it contains $0$. By [F1], every derivative class in the formula has finite $L^p$ norm. If $k=0$, then ${\mathcal{A}}_0=\{0\}$ and $D^{0}u=u$, so the expression is exactly $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p}$. If $\mathrm{\Omega }=\mathrm{\varnothing }$, [F1] says the only class is zero and the expression is zero.

Under Countable Choice [F8], changing representatives of $u$ does not change the weak-derivative property or its value classes by [F2]. If two locally integrable values represent the weak derivative of the same input, [F3] makes them equal almost everywhere. Thus each $D^{\alpha }u$ is intrinsic as an $L^p$ class, and the real or complex quotient norm in [F5] or [F6] depends only on that class. Hence the formula is independent of all representative choices.

For $u,v\in W^{k,p}(\mathrm{\Omega })$ and scalars $a,b$ in the chosen field, [F4] gives $D^{\alpha }(au+bv)=a{D}^{\alpha }u+b{D}^{\alpha }v$ as weak-derivative classes for every $\alpha \in {\mathcal{A}}_k$. The right side is an $L^p$ class by the vector-space properties in [F5] or [F6], so $au+bv\in W^{k,p}(\mathrm{\Omega })$. This covers $a=0$ or $b=0$ directly. Thus the usual operations make $W^{k,p}(\mathrm{\Omega })$ a real or complex vector space.

For a scalar $c$ and finite $p$, class homogeneity in [F5] or [F6] gives $\mathrm{\parallel }cu{\mathrm{\parallel }}_{W^{k,p}}^p={\sum }_{\alpha \in {\mathcal{A}}_k}\mathrm{\parallel }c{D}^{\alpha }u{\mathrm{\parallel }}_{L^p}^p=\mathrm{\mid }c{\mathrm{\mid }}^p{\sum }_{\alpha \in {\mathcal{A}}_k}\mathrm{\parallel }{D}^{\alpha }u{\mathrm{\parallel }}_{L^p}^p=\mathrm{\mid }c{\mathrm{\mid }}^p\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{k,p}}^p$; taking nonnegative $p$th roots proves absolute homogeneity. At $p=\mathrm{\infty }$, the maximum formula and $\mathrm{\parallel }cf{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\mid }c\mathrm{\mid }\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ give the same conclusion. For $c=0$ both formulas give zero directly.

Suppose $1\le p<\mathrm{\infty }$ and put $a_{\alpha }=\mathrm{\parallel }{D}^{\alpha }u{\mathrm{\parallel }}_{L^p}$, $b_{\alpha }=\mathrm{\parallel }{D}^{\alpha }v{\mathrm{\parallel }}_{L^p}$. By [F4] and the componentwise $L^p$ triangle inequality in [F5] or [F6], $\mathrm{\parallel }{D}^{\alpha }(u+v){\mathrm{\parallel }}_{L^p}\le a_{\alpha }+b_{\alpha }$ for each $\alpha$. Raising to $p$, summing, and taking the $p$th root bounds $\mathrm{\parallel }u+v{\mathrm{\parallel }}_{W^{k,p}}$ by ${({\sum }_{\alpha \in {\mathcal{A}}_k}(a_{\alpha }+b_{\alpha }{)}^p)}^{1\mathrm{/}p}$. Give the finite set ${\mathcal{A}}_k$ its counting measure; the real sequences $(a_{\alpha })$ and $(b_{\alpha })$ belong to this $L^p$ space, and their norms are the finite sums in [F1]. Applying [F7] yields ${({\sum }_{\alpha \in {\mathcal{A}}_k}(a_{\alpha }+b_{\alpha }{)}^p)}^{1\mathrm{/}p}\le {\left({\sum }_{\alpha \in {\mathcal{A}}_k}{a}_{\alpha }^p\right)}^{1\mathrm{/}p}+{\left({\sum }_{\alpha \in {\mathcal{A}}_k}{b}_{\alpha }^p\right)}^{1\mathrm{/}p}=\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{k,p}}+\mathrm{\parallel }v{\mathrm{\parallel }}_{W^{k,p}}$. This proves subadditivity also at $p=1$.

At $p=\mathrm{\infty }$, put $a_{\alpha }=\mathrm{\parallel }{D}^{\alpha }u{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$ and $b_{\alpha }=\mathrm{\parallel }{D}^{\alpha }v{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$. By [F4] and [F5] or [F6], $\mathrm{\parallel }{D}^{\alpha }(u+v){\mathrm{\parallel }}_{L^{\mathrm{\infty }}}\le a_{\alpha }+b_{\alpha }\le {\max }_{\beta \in {\mathcal{A}}_k}{a}_{\beta }+{\max }_{\beta \in {\mathcal{A}}_k}{b}_{\beta }$. Taking the maximum over $\alpha$ and using [F1] gives $\mathrm{\parallel }u+v{\mathrm{\parallel }}_{W^{k,\mathrm{\infty }}}\le \mathrm{\parallel }u{\mathrm{\parallel }}_{W^{k,\mathrm{\infty }}}+\mathrm{\parallel }v{\mathrm{\parallel }}_{W^{k,\mathrm{\infty }}}$.

If the Sobolev expression is zero, then at finite $p$ each nonnegative summand is zero, and at $p=\mathrm{\infty }$ every component norm is zero. Since $0\in {\mathcal{A}}_k$ and $D^{0}u=u$ by [F1], in either case $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p}=0$. The real or complex $L^p$ norm separates its quotient classes by [F5] or [F6], so $u$ is the zero class. Conversely, if $u=0$ as an $L^p$ class, zero is a weak derivative of zero at every order; uniqueness under [F8] and [F3] makes every $D^{\alpha }u$ the zero class, and [F1] gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{k,p}}=0$. Together with the homogeneity and triangle inequalities already proved, this establishes the norm and both directions of the zero equivalence for real and complex scalars and all endpoint exponents.