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From the Sobolev embedding theorem we know that for $\alpha = \frac{d}{2}$, $W^{\alpha, 2}(\mathbb{R}^d)$ is continuously embedded in $C^0(\mathbb{R}^d)$. Especially the point evaluations are in the dual of $W^{\alpha, 2}(\mathbb{R}^d)$ and thus well defined.

But is it also necessary to be in $W^{\alpha, 2}(\mathbb{R}^d)$ to have well-defined point evaluations? Or is there maybe some $\beta < \alpha = \frac{d}{2}$ s.t. the point evaluations are well defined functions on $W^{\beta, 2}(\mathbb{R}^d)$ even if they aren't continuous anymore.

Edit: By well-definedness I mean that if there is a $x \in \mathbb{R}^d$ s.t. $f(x) \not= g(x)$ for two representants of functions $f, g \in W^{\beta, 2}(\mathbb{R}^d)$ then $||f-g||_\beta \not= 0$ (they do not have the same equivalence class)

K.Doe
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  • Denoting by $C^0_0(R^d)$ the continuous functions vanishing at infinity, then point evaluation are a continuous functional on this space, with the uniform norm. However, not all functions in $C^0_0(R^d)$ are in L2. So Sobolev regularity is not necessary. – pcp Jun 04 '18 at 06:33

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