I'm having doubts on the definition of the norm in the space $H_0^2$ defined over the (rectangular) domain $\Omega \subset \mathbb{R}^2$ as follows $$ H_0^2(\Omega) = \left\{ v \in L^2(\Omega): v = \frac{\partial v}{\partial x_1} = \frac{\partial v}{\partial x_2} = 0 \text{ on } \partial \Omega \right\} $$

The definition of the norm on this space is given as $$ \left\|v\right\|_{H_0^2(\Omega)} = \left(\sum_{i = 0}^2 \left\|D^i v\right\|^2\right)^{1/2} $$

Now, according to my logic, in expanded form said norm would be $$ \left\|v\right\|_{H_0^2(\Omega)} = \left( \left\|v\right\|^2 + \left\|\frac{\partial v}{\partial x_1} + \frac{\partial v}{\partial x_2}\right\|^2 + \left\| \frac{\partial^2 v}{\partial x_1^2} + \frac{\partial^2 v}{\partial x_2^2} + 2\frac{\partial^2 v}{\partial x_1 \partial x_2}\right\|^2 \right)^{1/2} $$

If it's not too much trouble, I'd appreciate someone confirming that this is correct, or knock me over the head if it's not :)

Thank you in advance! Cheers!

  • $\begingroup$ Why "homogeneous" Sobolev space? In my limited experience, homogeneous Sobolev spaces are a different thing. Can you explain? $\endgroup$ – Giuseppe Negro Apr 2 '16 at 22:00

This is not correct, the norm of $Du$, $D^2u$, etc, takes the norm of all individual entries (partial derivatives). So $$ \|v\|_{H^2}^2 = \|v\|_{L^2}^2 + \|\frac{\partial v}{\partial x_1}\|_{L^2}^2 + \|\frac{\partial v}{\partial x_1}\|_{L^2}^2 + \|\frac{\partial^2 v}{\partial x_1^2}\|_{L^2}^2 +\|\frac{\partial^2 v}{\partial x_1\partial x_2}\|_{L^2}^2 + \|\frac{\partial^2 v}{\partial x_2^2}\|_{L^2}^2 , $$ which is usually written with multi-indices as $$ \|v\|_{H^2}^2 = \sum_{|\alpha|\le 2} \|D^\alpha v\|_{L^2}^2. $$

  • $\begingroup$ Thank you very much! I was thinking that perhaps this is the correct form but was unsure how the notation goes. $\endgroup$ – Iordan Iordanov Jun 11 '15 at 12:24

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