quot; for $|\alpha| \leq k-1$. The sense in which this property understood is made precise via [[the trace theorem|trace theory]]. > [!note] Remark. > Some definitions use an [[norm|equivalent norm]]. Most commonly this is ($k=1$ for simplicity) $\|u\|_{k,p}:=\|u\|_{p}+\|Du\|_{p}, \, (\text{alternative definition})$ where [[Lp-norm|we recall]] $\|Du\|_{L^{p}}^{p} = \int |Du(x)|_{2}^{p} \, dx$. (Have equality when $p=2$.) ^note **Completeness.** Observe that $(u_{m}) \subset W^{k,p}$ Cauchy implies $(\partial^{\alpha}u_{m}) \subset L^{p}$ [[Cauchy sequence|Cauchy]] for all $|\alpha| \leq k$, and since $L^{p}$ is [[complete]], have $(\partial^{\alpha}u_{m}) \to v_{\alpha}$ for some $v_{\alpha} \in L^{p}$. In particular for $\alpha=0$ we get $u_{m} \to v_{0}=:u$ in $L^{p}$. The claim is then that in fact $u \in W^{k,p}$ with weak derivatives equalling precisely the $v_{\alpha}$. For each $m$, $\partial^{\alpha}u_{m}$ is the $\alpha$-weak derivative of $u_{m}$ and so $\int_{\Omega}u_{m} \, \partial^{\alpha} \varphi= (-1)^{|\alpha|} \int_{\Omega} \partial^{\alpha}u_{m}\, \varphi$ for all $\varphi \in C_{c}^{\infty}$. If we can pass $m \to \infty$ on both sides we obtain $\int_{\Omega} u \partial^{\alpha} \varphi = \int_{\Omega} v_{\alpha} \varphi$ for all $\varphi \in C_{c}^{\infty}$ which is precisely the statement that $v_{\alpha}$ is the $\alpha$-weak derivative of $u$. We can indeed do this via [[Hölder's inequality|Holder]], see the relevant property and proof in [[weak derivative]] (short). Hence $u \in W^{k,p}$, and $\|u_{m}- u\|_{W^{k,p}} = \sum_{|\alpha| \leq k}\|\partial^{\alpha} u_{m}- \partial^{\alpha} u\|_{L^{p}} =\sum_{|\alpha| \leq k}\|\partial^{\alpha} u_{m} - v_{\alpha}\|_{L^{p}} \to 0.$ > [!basicproperties] > - For $k \geq 0$ and $1< p < \infty$, $W^{k,p}(\Omega)$ is [[reflexive space|reflexive]], though we will not prove it here. ^properties $C_{c}^{\infty}(\mathbb{R}^{n})$ a subset of $C^{\infty}(\overline{\Omega}) \cap W^{k,p}(\Omega)$ ---- #### [^1]: Here the summation ranges over all multi-indices $\alpha=(\alpha_{1},\dots,\alpha_{k})$ satisfying $|\alpha|=\alpha_{1}+\dots+\alpha_{k} \leq k$. Since this includes $\alpha=(0,\dots,0)$, for which the corresponding term is $\|u\|_{p}^{p}$, the Sobolev norm can be though about quantifying both the size ($\|u\|_{p}^{p}$) and regularity $(\|\partial ^{\alpha}u\|_{p}^{p}, \alpha \neq 0)$ of $u$. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```