Let $\Omega$ be a bounded domain of $R^d$ with Lipschitz boundary. If $m>\frac{d}{2}$, such that $H^m(\Omega)$ is continuously embedded in $L^\infty(\Omega)$. Is $L^1(\Omega)$ continuously embedded in the dual space of $H^m(\Omega)$? Thank you very much.



Is $L^1(\Omega)$ continuous embedded in the dual of $H^m(\Omega)$ $(m>\frac{d}{2})$?
Yes. This is a special case of the following result.


**Proposition.** Let $X$ and $Y$ be normed spaces and let $i: X \to Y'$ be a continuous embedding. Then the mapping $j: Y \to X'$, given by $\langle j(y), x\rangle = \langle y, i(x)\rangle$, is continuous.
If, moreover, $i(X)$ is weak${}^\*$-dense in $Y'$, then $j$ is injective, i.e. $j$ is a continuous embedding.


The proof is elementary.


Choose $X = H^m(\Omega)$ and $Y = L^1(\Omega)$ to obtain the situation you are asking for.


