I was reading the famous paper of Kenig, Ponce and Vega about Oscillatory integrals and in page 8 they say that a direct application of Sobolev's embedding provide the following inequality: for $u(t,x):\mathbb{R}^2\to\mathbb{R}$ we have $$ \qquad \qquad \Vert u(t,x)\Vert_{L^8(\mathbb{R}^2)}\leq c\big\Vert\vert D_x^\gamma\vert\,\vert D_t^\gamma\vert u(t,x)\big\Vert_{L^6(\mathbb{R}^2)}, \qquad \qquad (*) $$ where $\gamma=\tfrac{1}{24}$. Here, the operator $D_x$ and $D_t$ are given in terms of their Fourier symbol $\vert \xi\vert$ and $\vert\tau\vert$. Now, in order to explain my doubt let me recall Sobolev embedding: Let $k>\ell$ and $1\leq p<q<\infty$ satisfying: $$ \dfrac{1}{p}-\dfrac{k}{n}=\dfrac{1}{q}-\dfrac{\ell}{n}. $$ Then, $W^{k,p}(\mathbb{R}^n)\subset W^{\ell,q}(\mathbb{R}^n)$. Hence, I tried to apply the latter embedding with $n=2$, $p=6$, $q=8$, $\ell=0$ from where I got $k=\tfrac{1}{12}$, what (if I am not wrong) means $$ \Vert u(t,x)\Vert_{L^8(\mathbb{R}^2)}\leq c\Vert u(t,x)\Vert_{W^{1/12,6}(\mathbb{R}^2)}. $$ Now, my problem is that I don't know how to use the latter inequality in order to obtain $(*)$. For me it seems that somehow the factor $k=\tfrac{1}{12}$ is distributed into two terms of size $\tfrac{k}{2}=\gamma=\tfrac{1}{24}$ (related to these two factors $\vert D_x^\gamma\vert$ and $\vert D_t^\gamma\vert$ in $(*)$), but it is not clear to me how to rigorously do that.

  • $\begingroup$ Your second inequality is weaker than $(*)$, I don't think you can use it to prove that. $\endgroup$
    – LL 3.14
    Aug 25, 2020 at 7:37
  • $\begingroup$ @LL3.14 I have the same feeling, however, that is argument Kenig-Ponce-Vega give. I am not sure how they use Sobolev's embedding, but they explicitly wrote inequality $(*)$ for general functions (has nothing to do with a solution of any PDE) (of course, whenever both sides of the ineq make sense). $\endgroup$
    – W2S
    Aug 25, 2020 at 9:37


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