We consider the Fourier multiplier operator $T_0$ defined by the explicit expression $$(T_0f)(x)=\int_{\mathbb{R}^n}{e^{ix\cdot \xi}m(\xi)\hat{f}(\xi)d\xi}, \ f\in S(\mathbb{R}^n),$$ where $S(\mathbb{R}^n)$ is the Schwartz function space. Here we assume that the multiplier $m(\xi)\in L^\infty(\mathbb{R}^n)$ satisfies the conditions in the H\"ormander's multiplier theorem, which implies that $T_0$ can be extended to a bounded operator $T$ from $L^p(\mathbb{R}^n)$ to $L^p(\mathbb{R}^n)$, $1<p<\infty$. Then it is natural to ask the following question. Do we have \begin{equation}(1)\quad\quad\quad (Tf)(x)=\int_{\mathbb{R}^n}{e^{ix\cdot \xi}m(\xi)\hat{f}(\xi)d\xi},\ a.e., \ f\in L^p(\mathbb{R}^n)\cap L^1(\mathbb{R}^n),\end{equation} whenever $|m(\xi)\hat{f}(\xi)|\in L^1(\mathbb{R}^n)$?

  • $\begingroup$ Can you be more specific about how the words "makes sense" are defined? $\endgroup$ – fedja Sep 16 '16 at 6:17
  • $\begingroup$ @fedja It means the integral is finite a.e.. $\endgroup$ – Mr.right Sep 16 '16 at 13:36
  • $\begingroup$ Just for others reading: @fedja's comment was in response to an earlier version of this qyestion $\endgroup$ – Yemon Choi Sep 16 '16 at 15:28
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    $\begingroup$ @ChristianRemling We define $Tf$ by the $L^p$-limit of $T_0f_n$, but we don't know whether this limit equals to the right hand side of (1), which is just what we want to prove. $\endgroup$ – Mr.right Sep 17 '16 at 4:02
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    $\begingroup$ @Mr.right: Why don't you try to flesh out the sketch I provided, I think that'll answer all your questions. (As for your last concern, if the RHS has a pointwise limit [as I showed it has], then this is the $L^p$ limit that we also know exists because we can pass to an a.e. convergent subsequence.) $\endgroup$ – Christian Remling Sep 18 '16 at 21:32

We construct a sequence $f_k\in S(\mathbb{R}^n)$ s.t. $||f_k-f||_p\to 0$ and $$T_0f_k-\int_{\mathbb{R}^n}e^{ix\cdot\xi}m(\xi)\hat{f}(\xi)d\xi\to 0,\ a.e..$$ By the bounded extension of the multiplier operator, there is a function $g\in L^p$ s.t. $||T_0f_k- g||_p\to 0$. Then there is a subsequence $T_0f_{k_j}-g\to 0,\ a.e.$. Hence, $g=\int_{\mathbb{R}^n}e^{ix\cdot\xi}m(\xi)\hat{f}(\xi)d\xi$, which gives the desired equality.

The construction of the sequence: Let $\phi\in C_0^\infty$ and $\varphi=\hat{\phi}$. $\phi(0)=1$. $\varphi_\epsilon(x)=\epsilon^{-n}\varphi(x/\epsilon)$, $\phi_\delta=\phi(\delta x)$. Let $f_{\epsilon,\delta}=(\varphi_\epsilon\ast f)\phi_\delta$. Since $$||f_{\epsilon,\delta}-f||_p\le||(\varphi_\epsilon\ast f)\phi_\delta-\varphi_\epsilon\ast f||_p+||\varphi_\epsilon\ast f-f||_p, $$ we can choose some $\epsilon_k$ and $\delta_k$ to make $||f_{\epsilon_k,\delta_k}-f||_p$ smaller than $1/k$.

We also need to show that $$|\int e^{ix\cdot \xi}m(\xi)(\hat{f_{\epsilon,\delta}}-\hat{f})d\xi|$$ can be small.

We see that $$\int|m||\hat{f_{\epsilon,\delta}}-\hat{f}|\le \int|m||(1-\hat{\varphi_\epsilon})\hat{f}|+\int|m||h_\epsilon-h_\epsilon\ast\hat{\phi_\delta}|=I_1+I_2,$$ where $h_\epsilon=\hat{\varphi_\epsilon}\hat{f}$. Since $$|m||(1-\hat{\varphi_\epsilon})\hat{f}|\le |m\hat{f}|\in L^1,$$by dominated convergence, we can choose a subsequence $\epsilon_{k_j}$ s.t. $I_1\le 1/k$.

Fix $\epsilon=\epsilon_{k_j}$. We claim that $$|h_{\epsilon}\ast\hat{\phi_\delta})(\xi)|=|\int h_\epsilon(\xi-\delta y)\hat{\phi}(y)dy|\le C_{\epsilon,N}(1+|\xi|)^{-N},$$ where $C_{\epsilon,N}$ is independent of $\delta$. To see this, we need to use the facts that $||h_\epsilon||_\infty\le ||f||_1$ and ${\rm supp}\ h_\epsilon\subset B(0,\epsilon^{-1})$. Indeed, when $|\xi|\le 10/\epsilon$, $$\int |h_\epsilon(\xi-\delta y)\hat{\phi}(y)|dy\le C\int|\hat{\phi}(y)|dy\le C,$$ and when $|\xi|\ge 10/\epsilon$, $$\int |h_\epsilon(\xi-\delta y)\hat{\phi}(y)|dy\le C\int_{|\xi-\delta y|\le \epsilon^{-1}}|\hat{\phi}(y)|dy\le C_N\int_{|y|\ge (|\xi|-\epsilon^{-1})/\delta}|y|^{-N}dy \le C_N|\xi|^{n-N}.$$ This proves the claim.

So $$|m||h_\epsilon-h_\epsilon\ast\hat{\phi_\delta}|\le |m\hat{f}|+C_{\epsilon,N}|m|(1+|\xi|)^{-N}\in L^1.$$ Since $h_\epsilon$ is bounded and continuous and $\epsilon=\epsilon_{k_j}$ is fixed, we have $${\rm limsup}_{\delta\to 0}|h_\epsilon-h_\epsilon\ast\phi_\delta|(\xi)\le {\rm limsup}_{\delta \to 0}\int |h_\epsilon(\xi-\delta y)-h_\epsilon(\xi)||\hat{\phi}(y)|dy=0,$$ then by dominated convergence, we can choose a subsequence $\delta_{k_{j_l}}$ s.t.$I_2\le 1/k$. Now, the sequence $f_{\epsilon_{k_{j}}, \delta_{k_{j_l}}}$ is what we need.

Any comments are welcome:)

  • $\begingroup$ I haven't checked all the small details, but overall this looks good I think. $\endgroup$ – Christian Remling Sep 19 '16 at 16:57

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