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Let

  • $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ be open
  • $H_0^k(\Lambda)$ denote the topological closure of $C_c^\infty(\Lambda)$ with respect to $\left\|\;\cdot\;\right\|_{H^k(\Lambda)}$ for $k\in\mathbb N$

I've frequently read $H_0^1(\Lambda)\cap H^2(\Lambda)$. Isn't this space the same as $H_0^2(\Lambda)$?

It's not clear from definition, but we can show that $$H_0^k(\Lambda)=\left\{u\in H^k(\Lambda)\mid\exists(\phi_n)_{n\in\mathbb N}:\left\|\phi_n-u\right\|_{H^k(\Lambda)}\xrightarrow{n\to\infty}0\right\}\;.\tag 1$$ With $(1)$ in mind, it should be obvious that $$H_0^1(\Lambda)\cap H^2(\Lambda)=H_0^2(\Lambda)\;.\tag 2$$

So, am I missing something? Isn't $(1)$ true for any open $\Lambda$?

  • 1
    How do you conclude (2)? If $f$ is in the intersection then you know that $f\in H^2$ and that there exists a sequence $\phi_n$ s.t. $\phi_n \to f$ in $H^1$ but not that $\phi_n \to f$ in $H^2$ what would be nec. to have $f\in H^2_0$. – 2017-01-05

1 Answers 1

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No, these are different spaces. To see the difference, focus on the gradient $\nabla u$: $$u\in H^1_0\cap H^2 \iff u\in H^1_0 \text{ and } \nabla u\in H^1$$ $$u\in H^2_0 \iff u\in H^1_0 \text{ and } \nabla u\in H^1_0$$ So, an example demonstrating the difference of two spaces is given by a function where $\nabla u\in H^1\setminus H_0^1$. Perhaps the simplest one is $u(x) = 1-\|x\|^2$ on the unit ball. This function is in $H^1_0$ but its gradient is not in $H^1_0$, so $u$ is not in $H^2_0$.