Sobolev space $H_0^m$ and sobolev norm and seminorm

Question

I have problems understanding the definitions of $H_0^m(\Omega)$-spaces. What does the $0$ stand for? Does it mean that the functions are zero at $\partial\Omega$? Or does it mean that it is non-zero on all of $\Omega$? Also the sum notation of sobolev seminorm and norm confuses me. The seminorm is defined as $ |v|_m = \sum_{\alpha=m} <D^\alpha v,D^\alpha v>_0 $ where as the norm $\|v\|_m = \sum_{\alpha\leq m} <D^\alpha v,D^\alpha v>_0$

Answer 1

Yes, the subscript zero usually denotes that the functions are zero on the boundary. For the semi-norm, the summation is taken over all multi-indices $\alpha$ of order $m$ (also I think you are missing some square roots). This means you take all derivatives of order $m$, individually take their $L^2$ norms and then sum those. For example, if $m = 2$, and $v = v(x,y)$, then \begin{align*} \lvert v \rvert_{H^2}^2 &= \| v_{xx}\|_{2}^2 + \| v_{xy}\|_2^2 + \| v_{yx}\|_2^2 + \| v_{yy}\|_2^2 \\ &= \langle v_{xx}, v_{xx}\rangle + \langle v_{xy}, v_{xy}\rangle + \langle v_{yx}, v_{yx}\rangle + \langle v_{yy}, v_{yy}\rangle \end{align*} where $\langle \cdot, \cdot \rangle$ denotes the $L^2$ inner product.

For the norm, you do the same thing, but do it with all derivatives up to the given order. In my example, we would have $$\lvert \lvert v \rvert \rvert_{H^2}^2 = \| v\|_{2}^2 + \| v_{x}\|_{2}^2+ \| v_{y}\|_{2}^2 + \| v_{xx}\|_{2}^2 + \| v_{xy}\|_2^2 + \| v_{yx}\|_2^2 + \| v_{yy}\|_2^2.$$ Does this help at all? If not, please explain what about the notation confuses you.