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Notation:

  • $B^n_R$ stands for the ball of radius $R$ in $\mathbb{R}^n$.
  • $\hat{f}$ stands for the Fourier transform of $f$.

Question. The following inequality holds true for all $f\in H^s(\mathbb{R}^n)$ and all measurable $t=t(x)$ provided that $s>1/2$ ($^1$): $$\tag{1}\left\lVert \int_{\mathbb{R}^n}\, \hat{f}(\xi)e^{i(x\cdot\xi-t(x)\lvert\xi\rvert^2)}\, d\xi\right\rVert_{L^2_x(B^n_R)}\le C_s R^{\frac{1}{2}}\lVert f \rVert_{H^s(\mathbb{R}^n)}.$$ Can we use a scaling argument (or a variation of it) to show that the exponent of $R$ in the right hand side of (1) cannot be different from $1/2$?

It is mostly the presence of the inhomogeneous norm $$\lVert f\rVert_{H^s}^2=\int_{\mathbb{R}^n}\lvert \hat{f}(\xi)\rvert^2\left( 1+ \lvert \xi\rvert^2\right)^s\, d\xi$$ which confuses me. However, I think that there should be a scaling explanation for that term $R^{1/2}$. Any hint is welcome.


($^1$) Cfr. Rogers-Vargas-Vega, Pointwise convergence of solutions of the non-elliptic Schrödinger equation, Theorem 3.

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  • $\begingroup$ To who's intersted: I asked one of the authors, and she replied that she's inclined to think that indeed a scaling argument is not applicable here. $\endgroup$ – Giuseppe Negro Feb 26 '13 at 0:29

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