When I read the book Linear and Nonlinear Functional Analysis with Applications, I came across J.L. Lion's Lemma (the book doesn't give a proof), which states

Let $\Omega \subset \mathbb R^n$ be a Lipschitz domain. If $f \in H^{-1}(\Omega)$ and $\nabla f\in H^{-1}(\Omega)$. Then $f \in L^2(\Omega)$.

I have looked for the proof of this lemma on the Internet and found several links, such as this, this and this. It seems all proofs of the lemma have something to do with Fourier transform. To be more precise, we first prove the special case where $\Omega = \mathbb R^n$ using Fourier transform, then we try to prove the general case using the special case. The full proof is quite tedious, and I think there is a lack of intuition.

I think the lemma itself is quite concise and promising. Maybe we can prove it in a more direct way, which is, not invoking Fourier transform? I think at least it should be possible when $n = 1$, though I don't know how to prove it, either.

Thank you very much if you would like to help.

When I read the book Linear and Nonlinear Functional Analysis with Applications, I came across J.L. Lions's Lemma (the book doesn't give a proof), which states

Let $\Omega \subset \mathbb R^n$ be a Lipschitz domain. If $f \in H^{-1}(\Omega)$ and $\nabla f\in H^{-1}(\Omega)$. Then $f \in L^2(\Omega)$.

I have looked for the proof of this lemma on the Internet and found several links, such as this, this and this. It seems all proofs of the lemma have something to do with Fourier transform. To be more precise, we first prove the special case where $\Omega = \mathbb R^n$ using Fourier transform, then we try to prove the general case using the special case. The full proof is quite tedious, and I think there is a lack of intuition.

I think the lemma itself is quite concise and promising. Maybe we can prove it in a more direct way, which is, not invoking Fourier transform? I think at least it should be possible when $n = 1$, though I don't know how to prove it, either.

Thank you very much if you would like to help.

When I read the book Linear and Nonlinear Functional Analysis with Applications, I came across J.L. Lion's Lemma (the book doesn't give a proof), witch states

Let $\Omega \subset \mathbb R^n$ be a Lipschitz domain. If $f \in H^{-1}(\Omega)$ and $\nabla f\in H^{-1}(\Omega)$. Then $f \in L^2(\Omega)$.

I have looked for the proof of this lemma on the Internet and found several links, such as this, this and this. It seems all proofs of the lemma have something to do with Fourier transform. To be more precise, we first prove the special case where $\Omega = \mathbb R^n$ using Fourier transform, then we try to prove the general case using the special case. The full proof is quite tedious, and I think there is a lack of intuition.

I think the lemma itself is quite concise and promising. Maybe we can prove it in a more direct way, which is, not invoking Fourier transform? I think at least it should be possible when $n = 1$, though I don't know how to prove it, either.

Thank you very much if you would like to help.

When I read the book Linear and Nonlinear Functional Analysis with Applications, I came across J.L. Lion's Lemma (the book doesn't give a proof), which states

Let $\Omega \subset \mathbb R^n$ be a Lipschitz domain. If $f \in H^{-1}(\Omega)$ and $\nabla f\in H^{-1}(\Omega)$. Then $f \in L^2(\Omega)$.

I have looked for the proof of this lemma on the Internet and found several links, such as this, this and this. It seems all proofs of the lemma have something to do with Fourier transform. To be more precise, we first prove the special case where $\Omega = \mathbb R^n$ using Fourier transform, then we try to prove the general case using the special case. The full proof is quite tedious, and I think there is a lack of intuition.

I think the lemma itself is quite concise and promising. Maybe we can prove it in a more direct way, which is, not invoking Fourier transform? I think at least it should be possible when $n = 1$, though I don't know how to prove it, either.

Thank you very much if you would like to help.