Let $p\in[1,\infty)$ and $m$ be an $L^p(\mathbb R^n)$ multiplier. Let $\psi\in L^1(\mathbb R^n)$. Prove that $m\hat{\psi}$ and $m*\psi$ are $L^p(\mathbb R^n)$ multipliers, and
$$\left\|T_{m\hat{\psi}}\right\|_{L^p\to L^p}\leq\|\psi\|_{L^1}\left\|T_{m}\right\|_{L^p\to L^p},\tag{1}$$
$$\left\|T_{m*\psi}\right\|_{L^p\to L^p}\leq\|\psi\|_{L^1}\left\|T_{m}\right\|_{L^p\to L^p}.\tag{2}$$
Let me explain the notations first. $m: \mathbb R^n\to\mathbb C$ is called an $L^p(\mathbb R^n)$ multiplier if $m\in L^\infty(\mathbb R^n)$ and $$\|T_m f\|_{L^p} := \|(m\hat{f})^\vee\|_{L^p} \leq C\|f\|_{L^p}$$
for some $C>0$, where $\hat{f}$ is the Fourier transform of $f$, and $f^\vee$ is its inverse Fourier transform:
$$\hat f(\xi)=\int_{\mathbb{R}^n} f(x)e^{-2\pi ix\cdot\xi}\,dx,\qquad f^\vee(x)=\int_{\mathbb{R}^n} f(\xi)e^{2\pi ix\cdot\xi}\,d\xi.$$
In this case, $T_m$ is a bounded linear operator from $L^p(\mathbb R^n)$ to $L^p(\mathbb R^n)$, and we denote its operator norm by $\left\|T_{m}\right\|_{L^p\to L^p}$.
Here is my attempt. I can prove $(1)$. We have
$$(T_{m\hat{\psi}}f)\hat{ }= m\hat\psi\hat f= \hat\psi(T_mf)\hat{ }=(\psi*T_mf)\hat{},\tag{3}$$
hence $T_{m\hat{\psi}}f=\psi*T_mf$. Therefore, $(1)$ follows from Young's inequality. As for $(2)$, I tried to do something like $(3)$, but it just didn't work. So I think there must be something I missed.
Thanks in advance!
Best Answer
For each $f\in C_c^\infty(\mathbb R^n)$, we have $$\left(T_{m*\psi}f\right)\hat{}(\xi)=(m*\psi)(\xi)\hat f(\xi)=\int_{\mathbb R^n}m(\xi-\eta)\psi(\eta)\hat f(\xi)\,d\eta.$$ Hence \begin{align*} \left(T_{m*\psi}f\right)(x)&=\int_{\mathbb R^n}\int_{\mathbb R^n}m(\xi-\eta)\psi(\eta)\hat f(\xi)\,d\eta e^{2\pi ix\cdot \xi}\,d\xi\\ &=\int_{\mathbb R^n}\left(\int_{\mathbb R^n}m(\xi-\eta)\hat f(\xi)e^{2\pi ix\cdot \xi}\,d\xi\right)\psi(\eta)\,d\eta. \end{align*} Now we calculate the integral in the bracket. We define two operators to simplify our notations: $$M_af(x)=e^{2\pi ia\cdot x}f(x),\qquad \tau_af(\xi)=f(\xi-a),\qquad a\in\mathbb R^n.$$ Then clearly we have $(M_af)\hat{ }=\tau_a(\hat f)$. Now, \begin{align*} \int_{\mathbb R^n}m(\xi-\eta)\hat f(\xi)e^{2\pi ix\cdot \xi}\,d\xi&=\int_{\mathbb R^n}m(\xi-\eta)\int_{\mathbb{R}^n} f(y)e^{-2\pi iy\cdot\xi}\,dy e^{2\pi ix\cdot \xi}\,d\xi\\ &=\int_{\mathbb R^n}\left(\int_{\mathbb R^n}m(\xi-\eta)e^{2\pi i(x-y)\cdot(\xi-\eta)}\,d\xi\right)e^{2\pi i(x-y)\cdot\eta}f(y)\,dy\\ &=\int_{\mathbb R^n}\overset{\vee}{m}(x-y)e^{2\pi i(x-y)\cdot\eta}f(y)\,dy\\ &=\left(\left(M_\eta \overset{\vee}{m}\right)*f\right)(x). \end{align*} We claim that $$\left\|\left(M_a \overset{\vee}{m}\right)*f\right\|_{L^p}\leq \left\|T_{m}\right\|_{L^p\to L^p}\|f\|_{L^p}\qquad\forall a\in\mathbb R^n.\tag{$*$}$$ Assuming the claim, by Minkowski's inequality we get $$\left\|T_{m*\psi}f\right\|_{L^p}\leq\int_{\mathbb R^n} |\psi(\eta)|\left\|\left(M_\eta \overset{\vee}{m}\right)*f\right\|_{L^p}\,d\eta\leq \left\|T_{m}\right\|_{L^p\to L^p}\|\psi\|_{L^1}\|f\|_{L^p}.$$ This gives the desired result.
Finally, we prove the claimed identity $(*)$. Since $\left(M_a \overset{\vee}{m}\right)\hat{}=\tau_a m$, we have $\left(M_a \overset{\vee}{m}\right)*f=T_{\tau_am}f$. It is easy to check that $$T_{\tau_am}f= M_a\left(T_m\left(M_{-a}f\right)\right).$$ By definition of $M_a$, we have $|M_af|=|f|$, so $\|M_af\|_{L^p}=\|f\|_{L^p}$. Therefore, $$\left\|T_{\tau_am}f\right\|_{L^p}=\left\|T_m\left(M_{-a}f\right)\right\|_{L^p}\leq\left\|T_{m}\right\|_{L^p\to L^p}\|M_{-a}f\|_{L^p}=\left\|T_{m}\right\|_{L^p\to L^p}\|f\|_{L^p}.$$ And $(*)$ follows.