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Let $\phi\in C^{\infty}_0(\mathbb{R}^n)$ be such that $$\{\lvert \xi\rvert \le 1\} \prec \phi \prec \{\lvert \xi \rvert < 2\}^{[1]} $$ and define the Littlewood-Paley projectors as \begin{equation} (P_{2^j}f)^\wedge=\left[\phi\left( \frac{\xi}{2^j} \right)- \phi\left( \frac{\xi}{2^{j-1}}\right)\right] \widehat{f}(\xi) \end{equation} where $j\in \mathbb{Z}$ and $f$ is Schwartz. ($\,^\wedge$ denotes Fourier transform.)

Question. Suppose that $f$ is a Schwartz function such that $\widehat{f}$ is supported in $\{\lvert \xi\rvert >2\}$, and let $s>0$ be fixed. Does there exist an absolute constant $C$ such that the following inequality is true? $$\sum_{j=1}^\infty \left( 2^j\right)^s \lVert P_{2^j} f\rVert_2\le C \lVert f \rVert_{H^s}.$$ EDIT I now believe that the answer is negative, see comments.

Some motivation(you can safely omit reading this):

I want to prove a maximal estimate for the Schrödinger group $S_t=e^{it\Delta}$, namely $$\tag{1}\lVert S^\star f\rVert_{L^p(B_1)}\le C \lVert f\rVert_{H^s},$$ where $S^\star f=\sup_{t\in (0, 1)}\lvert S_tf\rvert$ and $B_1$ is the unit ball. All the articles I am consulting claim without further explanation that it is enough to prove this apparently weaker fact: $$\tag{2}\lVert S^\star f\rVert_{L^p(B_1)}\le C \lambda^s\lVert f\rVert_2,\qquad \operatorname{Spt}\widehat{f}\subset \{\lvert \xi\rvert \sim \lambda\}$$ (where $\lvert \xi\rvert \sim \lambda$ means that $\lambda\le \lvert \xi \rvert \le 2\lambda$). I would like to prove the implication $(2)\Rightarrow (1)$.

Now if $f$ is a Schwartz function that is Fourier supported in $\{\lvert \xi \rvert> 2\}$, then $$f=\sum_{j=1}^\infty P_{2^j}f.$$ Applying the sublinearity of $S^\star$ and (2) we arrive at \begin{equation} \begin{split} \lVert S^\star f\rVert_{L^p(B_1)} &\le\sum_{j=1}^\infty \lVert S^\star P_{2^j} f\rVert_{L^p(B_1)} \\ &\le C \sum_{j=1}^\infty \left(2^j\right)^s\lVert P_{2^j}f\rVert_2, \end{split} \end{equation} and this is where the Question comes in.


$\,^{[1]}$ Meaning that $\phi\ge 0$ everywhere, that $\phi(\xi)=1$ for all $\xi \in \{\lvert \xi\rvert \le 1\}$ and that the support of $\phi$ is contained in $\{\lvert \xi \rvert < 2\}$.

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  • $\begingroup$ I think you are dealing with inhomogeneous case, there are should be two functions used to decompose, $\chi$, which is supported in a ball, and $\phi$, which is supported in an annulus. By dyadic relation($\phi_i = \phi(2^{-i}\cdot)-\phi(2^{-(i-1)}\cdot$), you can extend $\chi + \sum_i \phi = 1, \forall \xi\in\mathbb{R}^d$. $\endgroup$
    – newbie
    Jul 6, 2013 at 13:36
  • $\begingroup$ You could consult the section 2.2, in This book. If you have access or a 'walkaround'. $\endgroup$
    – newbie
    Jul 6, 2013 at 13:39
  • $\begingroup$ You are absolutely right. I have overlooked the parameter. So I deleted the answer. Since we have $B^s_{p,1}\subset H^s_p$, the estimation is the other way around. On the otherhand, $B^s_{2,2} = H^s_2$ (in this book, it is somehow inconvenient that you need more than one book to study Besov spaces). You may try to play with that norm a bit. $\endgroup$
    – newbie
    Jul 8, 2013 at 22:58
  • $\begingroup$ I'm starting to think that the inequality does not hold. Assuming that it does, and putting $$\widehat{f}_N(\xi)=\sum_{k=1}^N \chi_{\{\lvert \xi \rvert \sim 2^{2k}\}}, $$ I think that we reach a contradiction. $\endgroup$ Jul 8, 2013 at 23:16
  • $\begingroup$ @newbie: The book you suggested has proven to be very useful. Thank you. $\endgroup$ Jul 22, 2013 at 12:11

1 Answer 1

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After a talk with my advisor I believe that I have a clear view of the matter.

Suppose that $f$ is a Schwartz function whose Fourier transform is supported away from the origin${}^{[1]}$, so that $$ f=\sum_{j=0}^\infty P_{2^j}f.$$ The series converges in the sense that the sum is locally finite at Fourier side. By a simple version of the Littlewood-Paley's inequality we have ${}^{[2]}$: $$\tag{LP}\lVert f\rVert_{H^s}^2\sim \sum_{j=0}^\infty (2^j)^{2s} \lVert P_{2^j} f\rVert_2^2$$ that is, the $H^s$-norm of $f$ is comparable with the $\ell^2$-norm of the sequence $$\mathbf{F}=\left\{ (2^j)^s\lVert P_{2^j}f\rVert_2\ :\ j=0,1,2\ldots\right\}.$$

With this newly introduced notation we can rephrase the Question above:

Question (rephrased): Is it true that $\lVert \mathbf{F}\rVert_{\ell^1}\le C \lVert f\rVert_{H^s}?$

That this needs not be true can be easily seen now by inserting relation (LP) into the question: $$\lVert \mathbf{F}\rVert_{\ell^1}\le C \lVert \mathbf{F}\rVert_{\ell^2} \tag{?!}$$ and this is clearly false as inequalities between $\ell^p$ spaces go the opposite way. For a concrete example just take a function $f$ such that $\mathbf{F}=\{\frac{1}{j}\}$.

However, not everything is lost. If we are ready to lose an $\varepsilon$ in the exponent $s$ we can apply Cauchy-Schwarz's inequality as follows: $$ \begin{split} \sum_{j=0}^\infty (2^j)^s\lVert P_{2^j}f\rVert_2 & = \sum_{j=0}^\infty (2^j)^{s+\varepsilon}\lVert P_{2^j}f\rVert_2 (2^j)^{-\varepsilon} \\ &\le \left( \sum_{j=0}^\infty (2^j)^{2(s+\varepsilon)}\lVert P_{2^j}f\rVert_2^2\right)^{\frac{1}{2}}\left(\sum_{j=0}^\infty (2^{-\varepsilon})^j\right)^{\frac{1}{2}} \\ &=C_\varepsilon \left( \sum_{j=0}^\infty (2^j)^{2(s+\varepsilon)}\lVert P_{2^j}f\rVert_2^2\right)^{\frac{1}{2}} \le C_\varepsilon \lVert f\rVert_{H^{s+\varepsilon}}. \end{split} $$ This way we get an estimate not into the desired space $H^s$ but into a slightly worse space $H^{s+\varepsilon}$.


${}^{[1]}$ This is just a simplifying assumption and can be removed at the cost of some minor technicality.

${}^{[2]}$ Here $A\sim B$ means that $cB\le A\le CB$ where $c, C$ are absolute constants. For this particular inequality the constants can be taken to be $c=\frac{1}{2}$ and $C=1$.

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