Demeter's book Fourier Restriction, Decoupling, and Applications give a principle that one cannot decouple in a direction where the manifold is flat. Which is the below proposition:

Proposition 9.5 Let $T$ be a tube in $\mathbb{R}^n$. Consider a partition $\mathcal{T}_N$ of $T$ into $N$ shorter, essentially congruent tubes. Then $${\rm{Dec}}(\mathcal{T}_N,p)\sim N^{1/2-1/p},$$ with implicit similarity constants independent of $N$.

The proof claim that considering the function $f$ whose fourier transform is a smooth approximation of $1_T$ can get the lower bound.

I have tried to assume that $T=\left\{(x_1,x^*)\in\mathbb{R}^n: x_1\in[0,1],x^*\in\mathbb{R}^n\right\}$. So $\mathcal{T}_N=\left\{T_1,T_2,\cdots,T_N\right\}$, where $T_k=\left\{(x_1,x^*)\in\mathbb{R}^n: x_1\in\left[\frac{k-1}{N},\frac{k}{N}\right],x^*\in\mathbb{R}^n\right\}$, $k=1,2,\cdots,N$. And $\left\{f_\varepsilon\right\}_{\varepsilon>0}\subset \mathcal{S}(\mathbb{R}^n)$ such that $1_T\leq \hat{f_\varepsilon}\leq 1_{(1+\varepsilon)T}$.

But for $k=2,3,\cdots,N-1$, $\mathcal{P}_{T_k}f_\varepsilon(x)=\int_{T_K}\hat{f_\varepsilon}(\xi) e^{2\pi i x\cdot\xi}{\operatorname{d}}\xi=\int_{\frac{k-1}{N}}^{\frac{k}{N}}e^{2\pi i x_1\xi_1}{\operatorname{d}}\xi_1\cdot\int_{\mathbb{R}^{n-1}}e^{2\pi i x^*\cdot\xi^*}{\operatorname{d}}\xi^*$ is disconverge.

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    $\begingroup$ Doesn't a tube have compact cross section? I thought in this business a tube is something like $[a,b]\times B_r(x^*)$. $\endgroup$ Dec 1, 2023 at 15:31
  • $\begingroup$ Willie is correct - tubes should in particular be bounded sets. After fixing some typos ($n-1$ in the set-builder notation instead of $n$), you have instead defined a decomposition of thickened hyperplanes. Adding on to this, your calculation of the Fourier projections is incorrect; you have forgotten to pair with the Schwartz function $\hat{f}$, so the integrals will of course not classically converge. $\endgroup$ Dec 1, 2023 at 18:38

1 Answer 1


Thanks to Willie Wong and Ben Johnsrude.

Now I set $T=[0,1]\times B_{n-1}(0,1)=\left\{(x_1,x^*)\in\mathbb{R}^n: x_1\in[0,1],x^*\in B_{n-1}(0,1)\right\}$, here $B_{n-1}(0,1)$ is the unit ball in $\mathbb{R}^{n-1}$. And $T_k=\left\{(x_1,x^*)\in T: x_1\in\left[\frac{k-1}{N},\frac{k}{N}\right]\right\}$, $\left\{f_\varepsilon\right\}_{\varepsilon>0}\subset \mathcal{S}(\mathbb{R}^n)$ such that $1_{(1-\varepsilon)T}\leq \hat{f_\varepsilon}\leq 1_{T}$.

For $\left\lVert f_\varepsilon\right\rVert_{L^p(\mathbb{R}^n)}$, since it is independent to $N$ hence $\left\lVert f_\varepsilon\right\rVert_{L^p(\mathbb{R}^n)}\sim 1$.

On the other hand, for each $k$:

\begin{align*} \mathcal{P}_{T_k}f_\varepsilon(x)&=\int_{\frac{k-1}{N}}^{\frac{k}{N}}\int_{B_{n-1}(0,1)}\hat{f}(\xi)e^{2\pi i x\cdot \xi}{\operatorname{d}}\xi_1 {\operatorname{d}}\xi^*\\ &\to \int_{\frac{k-1}{N}}^{\frac{k}{N}}e^{2\pi i x_1\xi}{\operatorname{d}}\xi_1\cdot\int_{B_{n-1}(0,1)}e^{2\pi i x^*\cdot \xi^*}{\operatorname{d}}\xi^*\\ &\sim \frac{e^{2\pi i (k-1)\frac{x_1}{N}}-e^{2\pi i k \frac{x_1}{N}}}{x_1}\cdot \prod_{k=2}^n \frac{\sin(2\pi x_k)}{x_k}. \end{align*}

So \begin{align*} \left\lVert\mathcal{P}_{T_k}f_{\varepsilon}\right\rVert^2_{L^p(\mathbb{R}^n)}&\sim \left(\int_\mathbb{R}\left(\frac{e^{2\pi i (k-1)\frac{x_1}{N}}-e^{2\pi i k \frac{x_1}{N}}}{x_1}\right)^p{\operatorname{d}}x_1\right)^{\frac{2}{p}}\\ &= \left(\int_\mathbb{R}\left(\frac{e^{2\pi i (k-1)t}-e^{2\pi i k t}}{N t}\right)^p N {\operatorname{d}}t\right)^{\frac{2}{p}}, \quad t=\frac{x_1}{N}\\ &\sim N^{2/p-2} \end{align*}

Then $\left(\sum_{k=1}^N \left\lVert\mathcal{P}_{T_k}f_{\varepsilon}\right\rVert^2_{L^p(\mathbb{R}^n)} \right)^{\frac{1}{2}} \sim N^{1/p-1/2}$ which is inverse.


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