A version of Hörmander multiplier theorem

Question

Let $m>n/2$ be an integer. Let $h\in H^m_{loc}(\mathbb{R}^n)$ satisfy that

$\displaystyle \exists M>0,\forall R>0,\sum_{|\alpha|\le m}\int_{\frac R2\le|w|\le2R}R^{2|\alpha|}|\partial^\alpha h(w)|^2dw\le MR^n$

Prove that the linear operator $T: u(x)\mapsto \mathscr{F}^{-1}[h(w)e^{-|w|}\hat{u}(w)](x)$ is a bounded operator in $L^p(\mathbb{R}^n)$ and in $[\dot{C}^\mu\cap\mathscr{E}](\mathbb{R}^n)\to\dot{C}^\mu(\mathbb{R}^n)$

where $\dot{C}^\mu$ means homogeneous Hölder space, $\mathscr{E}$ means compact support.

Is the above statement true? I read some books and find that some of the version require $\hat{h}\in L^1_{loc}(\mathbb{R}^n)$, and some require that $h\in L^\infty(\mathbb{R}^n)$. It seems that my statement lack some condition. Can it still be true or is there any counter-example?

Answer 1

Let us start by recalling the statement of Hörmander's multiplier theorem as it appears in Hörmander's The Analysis of Linear Partial Differential Operators I – Distribution Theory and Fourier Analysis (2nd edition, Springer-Verlag, 1990), Theorems 7.9.5, pages 243-245 and 7.9.6, pages 245-246.

Let $\mathbb{N}\ni m>\frac{n}{2}$ and $k\in\mathscr{S}'(\mathbb{R}^n)$. Assume that $\hat{k}\in L^1_{\text{loc}}$ and that there is $C>0$ such that $$\sum_{|\alpha|\leq m}\int_{\frac{R}{2}<\|\xi\|<2R}\left|R^{|\alpha|}D^\alpha\hat{k}(\xi)\right|^2\mathrm{d}\xi\leq CR^n<\infty\text{ for all } R>0,\text{ }D^\alpha=i^{-|\alpha|}\partial^\alpha.$$ Then for all $1<p<\infty$, $0<\mu<1$, we have $C_p$, $C_\mu>0$ such that that $$\|k*u\|_p\leq C_p\|u\|_p \text{ for all } u\in L^p_\text{c}=L^p\cap\mathscr{E}',$$ $$\|k*u\|_{\dot{C}^\mu}\leq C_\mu\|u\|_{\dot{C}^\mu} \text{ for all }u\in\dot{C}^\mu_\text{c}=\dot{C}^\mu\cap\mathscr{E}'.$$ Moreover, if $|\cdot|$ is the $n$-dimensional Lebesgue measure, then for some $C_1>0$,$$\left|\{x\in\mathbb{R}^n : |k*u(x)|>\tau\}\right|\leq\frac{C_1}{\tau} \text{ for all } u\in L^1_\text{c}=L^1\cap\mathscr{E}',\text{ }\tau>0.$$

If we make the stronger assumption $\hat{k}\in L^\infty\subset L^1_{loc}\cap\mathscr{S}'$, we no longer need to take compactly supported $u$ above, and we recover our conclusion for $1<p<\infty$ and $0<\mu<1$. See e.g. Theorem 0.2.6, pages 15-19 of Sogge's Fourier Integrals in Classical Analysis (Cambridge University Press, 1993). In fact, $\hat{k}\in L^\infty$ is also a necessary condition for $T$ to be bounded in $L^p$ with $1<p<\infty$, and $T$ is bounded in $L^1$ if and only if $k$ is a complex valued Borel measure, see e.g. Section 2.5 of Grafakos's Classical Fourier Analysis (2nd edition, Springer-Verlag, 2008), which by Bochner's theorem is the same as requiring that $\hat{k}$ has the form$$\hat{k}=\sum^3_{l=0}i^l\hat{k}_l,$$where $\hat{k}_0$, $\hat{k}_1$, $\hat{k}_2$, $\hat{k}_3$ are bounded and continuous functions which are positive definite: for all $\xi_1$, $\ldots$, $\xi_m\in\mathbb{R}^n$, $\zeta_1$, $\ldots$, $\zeta_m\in\mathbb{C}^n$, $m\in\mathbb{N}$, $l=0$, $1$, $2$, $3$, we must have $$\sum^m_{j,\,k=1}\overline{\zeta}_j\zeta_k\hat{k}_l(\xi_j-\xi_k)\geq 0.$$

In our instance of Hörmander's multiplier theorem, we take $k(w)=\mathscr{F}^{-1}[e^{-\|w\|}h(w)]$ with $h\in H^m_{\text{loc}}$, which entails that $\hat{k}\in C^0\subset L^1_{\text{loc}}$ by Sobolev's embedding theorem. That being said, we then see that $h$ indeed has a missing hypothesis, and the conclusion is not as strong as claimed. We state this more precisely.

1. We need that $\hat{k}\in\mathscr{S}'$ in order for it to be a valid multiplier. Just requiring $h\in H^m_{\text{loc}}$ might still make $\hat{k}$ grow faster than polynomially, even with the exponential damping factor, hence its inverse Fourier transform may no longer be defined as a distribution.
2. Even when $\hat{k}\in L^1_{\text{loc}}\cap\mathscr{S}'$, the operator $T$ is only guaranteed to be a bounded operator from the subspace $L^p_\text{c}$ into $L^p$ for $1<p<\infty$, just as you correctly did for Hölder spaces. Moreover, for $p=1$, we can only guarantee $T$ to be a bounded operator from $L^1_\text{c}$ into $L^1_{\text{weak}}$.

Again, requiring in addition that $\hat{k}(w)=h(w)e^{-\|w\|}\in L^\infty$ takes care of both problems in one stroke for $1<p<\infty$ and $0<\mu<1$, and it is clear that such a condition is necessary for addressing the second point above. in those cases. For the remaining sharp case $p=1$, it suffices to find a continuous function $h$ such that $\hat{k}$ is bounded, say, real valued and not the difference of two positive definite functions to provide a counterexample to boundedness of $T$ in $L^1$. Unfortunately, I am not aware of such an example at the moment.