Hi. I am trying to prove an observation in Stein, singular integrals.
Observation. $\mathcal{M}_{1}$ (class of Fourier multiplier in $L^1$) is the class of Fourier trasnforms of elements of $\mathcal{B}(\mathbb{R}^n)$ (finite borel measures), and the norm of $\mathcal{M}_{1}$ is identical with the norm of $\mathcal{B}(\mathbb{R}^n)$.
I have this:
Let $m\in \mathcal{M}_{1}$, i.e., the operator $T_{m}=\mathcal{F}^{-1}\left( m(x)\mathcal{F}(f)(x)\right)\in L^p(\mathbb{R}^n)$ for all $f\in L^2(\mathbb{R}^n)\cap L^1(\mathbb{R}^n)$ and is bounded. Since $T_m$ is traslation invariant, by the proposition 1 (image), exists $\mu\in\mathcal{B}(\mathbb{R}^n)$ such that $T_{m}(f)=f\ast \mu$ for all $f\in L^2(\mathbb{R}^n)\cap L^1(\mathbb{R}^n)$, then, $m(x)\mathcal{F}(f)(x)=\mathcal{F}(f\ast \mu)$ , $\mathcal{F}(f\ast \mu)=\mathcal{F}(f)\mathcal{F}(\mu)$. Now, $m(x)\mathcal{F}(f)(x)=\mathcal{F}(\mu)\mathcal{F}(f)(x)$ for all $f\in L^2(\mathbb{R}^n)\cap L^1(\mathbb{R}^n)$, this implies $m(x)=\mathcal{F}(\mu)(x)$.
conversely, let $\mu\in\mathcal{B}(\mathbb{R}^n)$, then the Fourier transform of $\mu$ is in $\mathcal{M}_{1}$.
Indeed,
\begin{align*}&T_{\mu} f(x)\\
&=\mathcal{F}^{-1}\left( \mathcal{F}(\mu)(x)\mathcal{F}(f)(x)\right)\\
&=\mathcal{F}^{-1}\left( \mathcal{F}((\mu\ast f)(x))\right)\\
&=(\mu\ast f)(x)
\end{align*}
Therefore, $T_{\mu}(f)=(\mu\ast f)$.
Now, $\left\|T_{\mu}(f)\right\|_{L^1(\mathbb{R}^n)}\leq \ldots$
The observation about the norm follows of the proposition 1, i.e. $\left\|T\right\|=\left\|\mu\right\|$.
I don't know how to prove that the operator $T_{\mu}$ is in $L^1$ and bounded. I would like to have something like $|\mu\ast f|_{L^1}\leq |\mu|_{L^1}|f|_{L^1}$, but I don't know if the inequality is applicable when you have a measure …
Can finite Borel measures be considered as a subspace of $L^1$?
Best Answer
You can define the convolution of two bounded variation Borel measures as follows: $\int h(x) d(\mu \ast \nu) (x) = \int \int h(y+z) d\mu(y) d\nu(z)$ for each bounded measurable $h$. In particular, it is true that $\|\mu\ast\nu\|\leq\|\mu\|\|\nu\|$. In your exercise, you can let $\nu$ be defined by $d\nu = fdx$, where $dx$ is the Lebesgue measure. It then follows that $\| \mu \ast f \|_1 \leq \| fdx \| \| \mu \| = \| \mu \| \|f\|_1$