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# 数学代写|傅里叶分析代写Fourier Analysis代考|Equivalence of Lp Norms of Lacunary Series

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## 数学代写|傅里叶分析代写Fourier Analysis代考|Equivalence of Lp Norms of Lacunary Series

We now turn to one of the most important properties of lacunary series, equivalence of their norms. It is a remarkable result that lacunary Fourier series have comparable $L^p$ norms for $1 \leq p<\infty$. More precisely, we have the following theorem:

Theorem 3.7.4. Let $0<\lambda_1<\lambda_2<\lambda_3<\cdots$ be a lacunary sequence with constant $A>1$. Set $\Lambda=\left{\lambda_k: k \in \mathbf{Z}^{+}\right}$. Then for all $1<p<\infty$ there exists a constant $C_p(A)$ such that for all $f \in L^1\left(\mathbf{T}^1\right)$ with $\widehat{f}(k)=0$ when $k \in \mathbf{Z} \backslash \Lambda$ we have
$$|f|_{L^p\left(\mathbf{T}^1\right)} \leq C_p(A)|f|_{L^1\left(\mathbf{T}^1\right)}$$
Note that the converse inequality to (3.7.8) is trivial. Therefore, $L^p$ norms of lacunary Fourier series are all equivalent for $1 \leq p<\infty$.

Proof. We suppose first that $f \in L^2\left(\mathbf{T}^1\right)$ and we define
$$f_N(x)=\sum_{j=1}^N \widehat{f}\left(\lambda_j\right) e^{2 \pi i \lambda_j x}$$
Given a $2 \leq p<\infty$, we pick an integer $m$ with $2 m>p$ and we also pick a positive integer $r$ such that $A^r>m$. Then we can write $f_N$ as a sum of $r$ functions $\varphi_s, s=$ $1,2, \ldots, r$, where each $\varphi_s$ has Fourier coefficients that vanish except possibly on the lacunary set
$$\left{\lambda_{k r+s}: k \in \mathbf{Z}^{+} \cup{0}\right}=\left{\mu_1, \mu_2, \mu_3, \ldots\right}$$

## 数学代写|傅里叶分析代写Fourier Analysis代考|Definition and Basic Properties of the Hilbert Transform

There are several equivalent ways to introduce the Hilbert transform; in this exposition we first define it as a convolution operator with a certain principal value distribution, but we later discuss other equivalent definitions.
We begin by defining a distribution $W_0$ in $\mathscr{S}^{\prime}(\mathbf{R})$ as follows:
$$\left\langle W_0, \varphi\right\rangle=\frac{1}{\pi} \lim {\varepsilon \rightarrow 0} \int{\varepsilon \leq|x| \leq 1} \frac{\varphi(x)}{x} d x+\frac{1}{\pi} \int_{|x| \geq 1} \frac{\varphi(x)}{x} d x$$
for $\varphi$ in $\mathscr{S}(\mathbf{R})$. The function $1 / x$ integrated over $[-1,-\varepsilon] \cup[\varepsilon, 1]$ has mean value zero, and we may replace $\varphi(x)$ by $\varphi(x)-\varphi(0)$ in the first integral in (4.1.1). Since $(\varphi(x)-\varphi(0)) x^{-1}$ is controlled by $\left|\varphi^{\prime}\right|_{L^{\infty}}$, it follows that the limit in (4.1.1) exists. To see that $W_0$ is indeed in $\mathscr{S}^{\prime}(\mathbf{R})$, we go an extra step in the previous reasoning and obtain the estimate
$$\left|\left\langle W_0, \varphi\right\rangle\right| \leq \frac{2}{\pi}\left|\varphi^{\prime}\right|_{L^{\infty}}+\frac{2}{\pi} \sup {x \in \mathbf{R}}|x \varphi(x)| .$$ This guarantees that $W_0 \in \mathscr{S}^{\prime}(\mathbf{R})$. Definition 4.1.1. The truncated Hilbert transform of $f \in \mathscr{S}(\mathbf{R})$ (at height $\varepsilon$ ) is defined by $$H^{(\varepsilon)}(f)(x)=\frac{1}{\pi} \int{|y| \geq \varepsilon} \frac{f(x-y)}{y} d y=\frac{1}{\pi} \int_{|x-y| \geq \varepsilon} \frac{f(y)}{x-y} d y .$$
The Hilbert transform of $f \in \mathscr{S}(\mathbf{R})$ is defined by
$$H(f)(x)=\left(W_0 * f\right)(x)=\lim _{\varepsilon \rightarrow 0} H^{(\varepsilon)}(f)(x)$$

## 数学代写|傅里叶分析代写Fourier Analysis代考|Equivalence of Lp Norms of Lacunary Series

$$|f|{L^p\left(\mathbf{T}^1\right)} \leq C_p(A)|f|{L^1\left(\mathbf{T}^1\right)}$$

$$f_N(x)=\sum_{j=1}^N \widehat{f}\left(\lambda_j\right) e^{2 \pi i \lambda_j x}$$

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