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Glorfindel
Glorfindel
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This is a cross post in continuation to this questionthis question on MathstackexchangeMathematics Stack Exchange. I wanted to know if this inequality holds true in two or three dimensions,

$\|\nabla\phi\|_{L^{\infty}(\Omega)}\leq C\|\phi\|_{H^2(\Omega)}.$

Where $\Omega$ is an open-bounded domain and $\phi$ is a test function, so we can assume $H^2(\Omega)$ regularity. We also have some leeway in putting extra conditions on the domain(I think convexity) or the boundaries.

Thank you!

This is a cross post in continuation to this question on Mathstackexchange. I wanted to know if this inequality holds true in two or three dimensions,

$\|\nabla\phi\|_{L^{\infty}(\Omega)}\leq C\|\phi\|_{H^2(\Omega)}.$

Where $\Omega$ is an open-bounded domain and $\phi$ is a test function, so we can assume $H^2(\Omega)$ regularity. We also have some leeway in putting extra conditions on the domain(I think convexity) or the boundaries.

Thank you!

This is a cross post in continuation to this question on Mathematics Stack Exchange. I wanted to know if this inequality holds true in two or three dimensions,

$\|\nabla\phi\|_{L^{\infty}(\Omega)}\leq C\|\phi\|_{H^2(\Omega)}.$

Where $\Omega$ is an open-bounded domain and $\phi$ is a test function, so we can assume $H^2(\Omega)$ regularity. We also have some leeway in putting extra conditions on the domain(I think convexity) or the boundaries.

Thank you!

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Mainak
Mainak
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Sobolev estimates $\|\nabla\phi\|_{\infty}\leq C\|\phi\|_{H^2}$

This is a cross post in continuation to this question on Mathstackexchange. I wanted to know if this inequality holds true in two or three dimensions,

$\|\nabla\phi\|_{L^{\infty}(\Omega)}\leq C\|\phi\|_{H^2(\Omega)}.$

Where $\Omega$ is an open-bounded domain and $\phi$ is a test function, so we can assume $H^2(\Omega)$ regularity. We also have some leeway in putting extra conditions on the domain(I think convexity) or the boundaries.

Thank you!