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Let $(\varphi_k, \lambda_k)$ be the eigenelements of the Neumann Laplacian. It's possible to define a space $$H(\Omega) = \{ u \in L^2(\Omega) \mid \sum_{k \geq 1}\lambda_k^{\frac 12}|(u,\varphi_k)_{L^2(\Omega)}|^2 < \infty \}$$ with the norm $$\lVert u \rVert_{H(\Omega)}^2 = \lVert u \rVert_{L^2(\Omega)}^2 + \sum_{k \geq 1}\lambda_k^{\frac 12}|(u,\varphi_k)_{L^2(\Omega)}|^2,$$ and also define the space $$H^{\frac 12}(\Omega)$$ defined by the double integral over $\Omega$ (the "usual definition") endowed with the obvious norm.

One can show that these spaces are equal, i.e., there is a map $I:H(\Omega) \to H^{\frac 12}(\Omega)$ such that $$c_0\lVert I(u) \rVert_{H(\Omega)} \leq \lVert u \rVert_{H^{\frac 12}(\Omega)} \leq c_1\lVert I(u) \rVert_{H(\Omega)}.$$

The question is, is $I$ just the identity map? Because both spaces are by definition subspaces of $L^2(\Omega)$, so I am led to believe that this is true (that $I$ literally does nothing to $u$). Or is $I$ some sort of remapping in $L^2$?

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  • $\begingroup$ I don't think these spaces are equal: $H^{1/2}$ functions have $1/2$ derivative, while being in the domain of $(-\Delta)^{1/2}$ gives you one derivative. $\endgroup$ Apr 24, 2015 at 16:02
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    $\begingroup$ @ChristianRemling But it's weak derivative, no? So like $H^1$ functions have 1 derivative and the domain of the weak Laplacian $-\Delta$ is $H^1$. $\endgroup$
    – C_Al
    Apr 24, 2015 at 16:23

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