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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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.) $\endgroup$ 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. $\endgroup$ Sep 19 '16 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.