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Is the Sobolev space $H^s(\mathbb{R}^3)$ dense in $L^2(\mathbb{R}^3)$ for $s>0$ ?

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1 Answer 1

We denote by $\mathcal D(\Bbb R^d)$ the set of smooth functions with compact support.

  • As the Fourier transform of a function in $\mathcal D(\Bbb R^d)$ has an arbitrary decay, we have the inclusion $\mathcal D(\Bbb R^d)\subset H^s(\Bbb R^d)$ for any $s>0$.
  • By Plancherel's formula, we have $H^s(\Bbb R^d)\subset L^2(\Bbb R^d)$.
  • By a truncation and regularization argument, it can be shown that $\mathcal D(\Bbb R^d)$ in dense in $L^2(\Bbb R^d)$ with its natural norm.
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