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# 数学代写|傅里叶分析代写Fourier Analysis代考|The Spaces $\mathscr{M}^{p, q}\left(\mathbf{R}^n\right)$

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## 数学代写|傅里叶分析代写Fourier Analysis代考|The Spaces $\mathscr{M}^{p, q}\left(\mathbf{R}^n\right)$

Definition 2.5.5. Given $1 \leq p, q \leq \infty$, we denote by $\mathscr{M}^{p, q}\left(\mathbf{R}^n\right)$ the set of all bounded linear operators from $L^p\left(\mathbf{R}^n\right)$ to $L^q\left(\mathbf{R}^n\right)$ that commute with translations.

By Theorem 2.5.2 we have that every $T$ in $\mathscr{M}^{p, q}$ is given by convolution with a tempered distribution. We introduce a norm $|\cdot|$ on $\mathscr{M}^{p, q}$ by setting
$$|T|_{\mathscr{M}^{p, q}}=|T|_{L^p \rightarrow L^q},$$
that is, the norm of $T$ in $\mathscr{M}^{p, q}$ is the operator norm of $T$ as an operator from $L^p$ to $L^q$. It is a known fact that under this norm, $\mathscr{M}^{p, q}$ is a complete normed space (i.e., a Banach space).

Next we show that when $p>q$ the set $\mathscr{M}^{p, q}$ consists of only one element, namely the zero operator $T=0$. This means that the only interesting classes of operators arise when $p \leq q$
Theorem 2.5.6. $\mathscr{M}^{p, q}={0}$ whenever $1 \leq q<p<\infty$.
Proof. Let $f$ be a nonzero $\mathscr{C}0^{\infty}$ function and let $h \in \mathbf{R}^n$. We have $$\left|\tau^h(T(f))+T(f)\right|{L^q}=\left|T\left(\tau^h(f)+f\right)\right|_{L^q} \leq|T|_{L^p \rightarrow L^q}\left|\tau^h(f)+f\right|_{L^p} .$$
Now let $|h| \rightarrow \infty$ and use Exercise 2.5.1. We conclude that
$$2^{\frac{1}{q}}|T(f)|_{L^q} \leq|T|_{L^p \rightarrow L^q} 2^{\frac{1}{p}}|f|_{L^p},$$
which is impossible if $q<p$ unless $T$ is the zero operator.
Next we have a theorem concerning the duals of the spaces $\mathscr{M}^{p, q}\left(\mathbf{R}^n\right)$.

## 数学代写|傅里叶分析代写Fourier Analysis代考|Characterizations of $\mathscr{M}^{1,1}\left(\mathbf{R}^n\right)$ and $\mathscr{M}^{2,2}\left(\mathbf{R}^n\right)$

It would be desirable to have a characterization of the spaces $\mathscr{M}^{p, p}$ in terms of properties of the convolving distribution. Unfortunately, this is unknown at present (it is not clear whether it is possible) except for certain cases.

Theorem 2.5.8. An operator $T$ is in $\mathscr{M}^{1,1}\left(\mathbf{R}^n\right)$ if and only if it is given by convolution with a finite Borel (complex-valued) measure. In this case, the norm of the operator is equal to the total variation of the measure.

Proof. If $T$ is given with convolution with a finite Borel measure $\mu$, then clearly $T$ maps $L^1$ to itself and $|T|_{L^1 \rightarrow L^1} \leq|\mu|_{\mathscr{M}}$, where $|\mu|_{\mathscr{M}}$ is the total variation of $\mu$.
Conversely, let $T$ be an operator bounded from $L^1$ to $L^1$. By Theorem 2.5.2, $T$ is given by convolution with a tempered distribution $u$. Let
$$f_{\varepsilon}(x)=\varepsilon^{-n} e^{-\pi|x / \varepsilon|^2}$$
Since the functions $f_{\varepsilon}$ are uniformly bounded in $L^1$, it follows from the boundedness of $T$ that $f_{\varepsilon} * u$ are also uniformly bounded in $L^1$. Since $L^1$ is naturally embedded in the space of finite Borel measures, which is the dual of the space $C_{00}$ of continuous functions that tend to zero at infinity, we obtain that the family $f_{\varepsilon} * u$ lies in a fixed multiple of the unit ball of $C_{00}^$. By the Banach-Alaoglu theorem, this is a weak ${ }^$ compact set. Therefore, some subsequence of $f_{\varepsilon} * u$ converges in the weak ${ }^*$ topology to a measure $\mu$. That is, for some $\varepsilon_k \rightarrow 0$ and all $g \in C_{00}\left(\mathbf{R}^n\right)$ we have
$$\lim {k \rightarrow \infty} \int{\mathbf{R}^n} g(x)\left(f_{\varepsilon_k} * u\right)(x) d x=\int_{\mathbf{R}^n} g(x) d \mu(x)$$

## 数学代写|傅里叶分析代写Fourier Analysis代考|The Spaces $\mathscr{M}^{p, q}\left(\mathbf{R}^n\right)$

2.5.5.定义给定$1 \leq p, q \leq \infty$，我们用$\mathscr{M}^{p, q}\left(\mathbf{R}^n\right)$表示从$L^p\left(\mathbf{R}^n\right)$到$L^q\left(\mathbf{R}^n\right)$的所有有界线性算子的集合。

$$|T|{\mathscr{M}^{p, q}}=|T|{L^p \rightarrow L^q},$$

## 数学代写|傅里叶分析代写Fourier Analysis代考|Characterizations of $\mathscr{M}^{1,1}\left(\mathbf{R}^n\right)$ and $\mathscr{M}^{2,2}\left(\mathbf{R}^n\right)$

$$\lim {k \rightarrow \infty} \int{\mathbf{R}^n} g(x)\left(f_{\varepsilon_k} * u\right)(x) d x=\int_{\mathbf{R}^n} g(x) d \mu(x)$$

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