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.


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.

  • 1
    $\begingroup$ it might be instructive to just prove it directly using the imbedding of the Sobolev space into $L^\ifty$. $\endgroup$ – Math604 Dec 5 '17 at 3:00
  • $\begingroup$ @ Jochen Glueck. Thank you so much for your answer. $\endgroup$ – Jane Dec 5 '17 at 9:08
  • $\begingroup$ @ Math604. Thank you so much. Could you please show a proof by using the imbedding of Sobolev space into $L^\infty$? $\endgroup$ – Jane Dec 5 '17 at 9:11
  • $\begingroup$ fix $f \in L^1(\Omega)$ and consider the linear functional $ T_f( \phi):= \int_\Omega f(x) \phi(x) dx$. Now try and show $ T_f \in (H^m)^*$; so do Holder on the integral with $ 1 $ and $ \infty$ and use imbedding $\endgroup$ – Math604 Dec 5 '17 at 16:36
  • $\begingroup$ In fact, I have ever tried to prove in the same way. But I met a problem on one step. More precisely, for $f\in L^1$, define $T_f(\phi):=\int_\Omega f(x)\phi(x)dx$ $\forall \phi\in H^m$. It follows from $H^m\hookrightarrow L^\infty$ that $T_f\in (H^m)^*$ and $\|T_f\|\leq \|f\|_{L^1}$. Moreover, the mappling $T:f\longrightarrow T_f$ is a bijection from $L^1$ onto $T(L^1)$ which is a subset of $(H^m)^*$. We are left to prove that $\|T_f\|=\|f\|_{L^1}$ such that $T$ is a Isometric isomorphism. My question is how to prove that $\|f\|_{L^1}\leq \|T_f\|$? Thank you very much. $\endgroup$ – Jane Dec 5 '17 at 21:23

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