 Solution

Let $\alpha$ be a multi-index with $|\alpha |\le k$. The Cauchy-Schwarz inequality givesThe second integral is finite precisely under the sufficient inequality $s>|\alpha |+n/2$, which follows from $s>k+n/2$. Hence $(i\xi {)}^{\alpha }\hat{u}$ is Lebesgue integrable for every $|\alpha |\le k$.

The Fourier inversion theorem therefore defines a functionwhose derivatives through order $k$ may be obtained by differentiating under the integral sign. Those derivatives are continuous, so $u^{\ast }\in C^k({\mathbb{R}}^n)$. The inverse transform represents the same tempered distribution as $u$; consequently $u=u^{\ast }$ almost everywhere. This is the Sobolev embedding theorem $H^s({\mathbb{R}}^n)\hookrightarrow C^k({\mathbb{R}}^n)$ in this range.