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I'm working on a problem involving the Fourier transform and have encountered an inequality that I am unsure how to prove. I would greatly appreciate any help or guidance you can provide.

Let $\gamma\in L^{\infty}(\mathbb{R}^{2n})$ and $v\in L^{1}(\mathbb{R}^{n})\cap L^{2}(\mathbb{R}^{n})$. Also $\text{supp}\gamma(\cdot,\xi)\subset U$ and $|U|<+\infty$ for all $\xi\in \mathbb{R}^{n}$. Prove that there exist a constant $C=C(U)>0$ such that \begin{equation*} \left(\int_{\mathbb{R}^{n}}|\mathcal{F}(\gamma(x,\cdot)v(\cdot))(x)|^{2}dx\right)^{\frac{1}{2}}\leq C\Vert{\gamma}\Vert_{\infty}\left (\int_{\mathbb{R}^{n}}|v(y)|dy\right), \end{equation*} where $\mathcal{F}$ denotes the Fourier transform on $L^2$.

Initially, I tried to use Plancherel's theorem, this is

\begin{equation} \Vert \mathcal{F}(\gamma(x,\cdot)v(\cdot)) \Vert_{L^2}=\Vert{\gamma (x,\cdot)v(\cdot)}\Vert_{L^{2}} \end{equation} and then use the hypothesis that $\gamma\in L^{\infty}(\mathbb{R}^{2n})$. However, this idea is totally incorrect, since it is not possible to use plancherel in that case.

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  • $\begingroup$ I have very hard time understanding your question, not least because the number of parentheses in your formula is odd... $\endgroup$
    – Aleksei Kulikov
    Commented Jul 14 at 16:03
  • $\begingroup$ @AlekseiKulikov, I edited it $\endgroup$
    – Julian Bejarano
    Commented Jul 14 at 16:06
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    $\begingroup$ There is a quadrillion reasons why this can't hold. To begin with, the scaling: left-hand side is quadratic in $\gamma$ and $v$ while the right-hand side is linear (should there be a square root on the left-hand side?) Second of all, if you do use plancherel, as you planned, then you essentially want to bound the $L^2$-norm of the compactly supported function by its $L^1$-norm, which is also not possible. Please, carefully examine whether this is really what you want or not, and if it is I can post an answer. $\endgroup$
    – Aleksei Kulikov
    Commented Jul 14 at 16:33
  • $\begingroup$ @AlekseiKulikov, You're right, I had a typing error on the left side. $\endgroup$
    – Julian Bejarano
    Commented Jul 14 at 16:42
  • $\begingroup$ The idea of ​​the problem is that for this case it is not possible to apply Plancherel, since it depends on two variables, however it is possible to find an estimate $\endgroup$
    – Julian Bejarano
    Commented Jul 14 at 16:47

1 Answer 1

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I'm sorry if I misunderstood the question, but isn't this just $$ |\mathcal{F}(\gamma(x,\cdot)v(\cdot))(x)| = |\mathcal{F}(\gamma(x,\cdot)v(\cdot))(x)| \mathbb{1}_{U}(x) \leq \Vert \gamma(x, \cdot) v(\cdot) \Vert_{L^1(\mathbb{R}^n)} \mathbb{1}_{U}(x)\\ \leq \Vert \gamma\Vert_{L^\infty(\mathbb{R}^{2n})} \Vert v \Vert_{L^1(\mathbb{R}^n)} \mathbb{1}_{U}(x) $$ and therefore $$ \left(\int_{\mathbb{R}^{n}}|\mathcal{F}(\gamma(x,\cdot)v(\cdot))(x)|^{2}dx\right)^{\frac{1}{2}} \leq \vert U\vert^{1/2} \Vert \gamma\Vert_{L^\infty(\mathbb{R}^{2n})} \Vert v \Vert_{L^1(\mathbb{R}^n)} $$?

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  • $\begingroup$ How did you get the very first equality? When you take the Fourier transform, the support does not stay the same... $\endgroup$
    – Aleksei Kulikov
    Commented Jul 14 at 17:25
  • $\begingroup$ @AlekseiKulikov Isn't $\gamma(x, \cdot) = 0$ for $x \notin U$ by assumption of OP? $\endgroup$
    – unwissen
    Commented Jul 14 at 17:28
  • $\begingroup$ Oh, I interpreted this line differently. Yes, if you look at it this way you are correct. $\endgroup$
    – Aleksei Kulikov
    Commented Jul 14 at 17:29
  • $\begingroup$ @AlekseiKulikov Yeah I think the assumption is written down in a confusing way. $\endgroup$
    – unwissen
    Commented Jul 14 at 17:31
  • $\begingroup$ @unwissen,Can you explain the first inequality to me? $\endgroup$
    – Julian Bejarano
    Commented Jul 15 at 13:09

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