Does this embedding holds?

Question

Could you please tell me whether the following continuous embedding holds?

$L^1(R^d)\hookrightarrow H^{-s}(R^d)$ $(s>\frac{d}{2})$

Answer 1

Because $u\in L^1$, then $\hat{u}\in L^\infty$. This implies that there exist a constant $C>0$, such that

\begin{eqnarray} \int_{\mathbb{R}^d}\frac{\hat{u}(\xi)^2}{(1+|\xi|^2)^s}d\xi &\leq& C \int_{\mathbb{R}^d}\frac{|\hat{u}(\xi)|}{(1+|\xi|^2)^s}d\xi \nonumber \\ &\leq& C_1\|u\|_1\int_{\mathbb{R}^d}\frac{1}{(1+|\xi|^2)^s}d\xi \nonumber \end{eqnarray}

where $C_1>0$ is another constant. Because $s>\frac{d}{2}$, you can conclude that the last integral is finite and hance you have proved that $L^1$ is continuously embedded in $H^{-s}$.