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I came across this inequality without any proof, saying it is straightforward. But I found difficulties to see why it is true. Namely, for a nice function $w(x,y)$ (Schwartz. etc.) on $\mathbb{R}^2,$ and a symbol $m(\xi,\mu)>0$ on $\mathbb{R}^2,$ we have
$$\|w\|_{L^{\infty}_{xy}}\leq C\left\|\frac{1}{m}\right\|_{L^{4}_{\xi\mu}}\left\|\check{(m\hat{w})}\right\|_{L^{4}_{xy}}$$ where $\hat{f}$. respectively $\check{f}$ are the Fourier transform, respectively the inverse Fourier transform of $f.$ It seems the choice of $m$ does not matter here. I tried to apply Hausdorff-Young inequality here, but it does not work. Thank you for any help.

Daniele Tampieri
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Frank Zermelo
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  • I believe this follows from the Hausdorff-Young inequality, convolution identities with the Fourier transform, and viewing the map $f\mapsto m^{-1}f$ as a linear operator. I will write an answer later. – Eli Seamans Jan 21 '25 at 03:06
  • It seems easy enough to get to $||w||{L^{\infty}} \le C||m^{-1}||{L^{4}}||m\hat{w}||_{L^{4/3}}$ which means that the inequality scales correctly, but it is hard for me to see how you could get all the way to what you want without more information about m and/or w. – LVFToENlCx Jan 27 '25 at 02:24
  • Since the case of $m$ a polynomial is essentially Sobolev embedding for fractional sobolev spaces, this answer should be useful. https://math.stackexchange.com/questions/654811/fractional-sobolev-embedding-into-l-infty – kieransquared Jan 27 '25 at 13:19

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