Given some function $\phi:D\to\mathbb{R}$, with $D\subseteq \mathbb{R}^d$. I was wondering whether we can find an estimate of the form
$$ \|\phi\otimes\phi\|_{\dot{H}^1(D\times D)} \lesssim \|\phi\|_{\dot{H}^{1/2}(D)}^2. $$
Many thanks.
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2$\begingroup$ Let $D = [0,2\pi] \subset \mathbb{R}$. Let $\phi = \sin(k x)$ with $k \gg 1$, then any reasonable definition of $\|\phi\|_{\dot{H}^{1/2}}$ would be $\approx \sqrt{k}$. On the other hand $$ \|\phi \otimes \phi\|_{\dot{H^1}} \approx \|\phi\|_{L^2} \|\phi\|_{\dot{H}^1} \approx k $$ so what you want is impossible. $\endgroup$– Willie WongCommented Jul 31, 2023 at 13:47
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$\begingroup$ Did you perhaps want the RHS to be squared? $\endgroup$– Willie WongCommented Jul 31, 2023 at 13:49
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1$\begingroup$ In fact, if you just double $\phi$, then the left hand side quadruples while the right hand side doubles, based on what you wrote. So I shall assume that you meant for the RHS to be squared. $\endgroup$– Willie WongCommented Jul 31, 2023 at 14:01
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$\begingroup$ @WillieWong You're totally right, I want the RHS squared, thank you very much for pointing that out! $\endgroup$– VíctorCommented Jul 31, 2023 at 15:13
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Assuming the inequality you hoped for is $\| \phi \otimes \phi \|_{\dot{H}^1} \lesssim \|\phi\|_{\dot{H}^{1/2}}^2$, I claim that this is still impossible.
Let $D = [0,2\pi]$ again. Set $$ \phi_N(x) = \sum_{k = 1}^N k^{-1/2} \sin(kx) $$ Then $$ \|\phi_N\|_{\dot{H}^{1/2}}^2 \approx \sum_{k = 1}^N 1 = N $$ But $$ \|\phi_N\otimes \phi_N\|_{\dot{H}^1} \approx \left( \sum_{k = 1}^N k^{-1} \right)^{\frac12} \left( \sum_{k = 1}^N k \right)^{\frac12} \approx \sqrt{\ln(1+N)} N $$ showing that the desired inequality is impossible.
answered Jul 31, 2023 at 14:02
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$\begingroup$ Thank you very much. Could you please elaborate a bit on the last estimte for the $\dot{H}^1$ norm of the tensor product? If I try to compute that myself with your example function I still obtain something of order $N$. If it's related to your first comment under my question, how do you know that $\|\phi\otimes\phi\|_{\dot{H}^1} \sim \|\phi\|_{L^2}\|\phi\|_{\dot{H}^1}$? I try to derive an estimate like that I always need to use some higher $L^p$ norms for $\phi$ and/or $\nabla\phi$. $\endgroup$– VíctorCommented Jul 31, 2023 at 15:36
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1$\begingroup$ The tensor product $\phi\otimes \phi: [0,2\pi]^2\ni(x,y) \mapsto \phi(x) \phi(y)$. So $\partial_x (\phi\otimes \phi)(x,y) = \phi'(x)\phi(y)$, whose $L^2$ norm is just the product of the (one dimensional) $L^2$ norms. $\endgroup$ Commented Jul 31, 2023 at 15:42