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# 数学代写|傅里叶分析代写Fourier Analysis代考|Equivalent Formulations of Convergence in Norm

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## 数学代写|傅里叶分析代写Fourier Analysis代考|Equivalent Formulations of Convergence in Norm

The question we pose is for which $1 \leq p<\infty$ we have
$$|D(n, N) * f-f|_{L^p\left(\mathbf{T}^n\right)} \rightarrow 0 \quad \text { as } N \rightarrow \infty$$
and similarly for the circular Dirichlet kernel $\widetilde{D}(n, N)$. We tackle this question by looking at an equivalent formulation of it.
Theorem 3.5.1. Fix $1 \leq p<\infty$ and $\left{a_m\right}$ in $\ell^{\infty}\left(\mathbf{Z}^n\right)$. For each $R \geq 0$, let $\left{a_m(R)\right}_{m \in \mathbf{Z}^n}$ be a compactly supported sequence (whose support depends on $R$ ) that satisfies $\lim {R \rightarrow \infty} a_m(R)=a_m$. For $f \in L^p\left(\mathbf{T}^n\right)$ define $$S_R(f)(x)=\sum{m \in \mathbf{Z}^n} a_m(R) \widehat{f}(m) e^{2 \pi i m \cdot x}$$
and for $h \in \mathscr{C}^{\infty}\left(\mathbf{T}^n\right)$ define
$$A(h)(x)=\sum_{m \in \mathbf{Z}^n} a_m \widehat{h}(m) e^{2 \pi i m \cdot x}$$

## 数学代写|傅里叶分析代写Fourier Analysis代考|Functions with Absolutely Summable Fourier Coefficients

Decay for the Fourier coefficients can also be indirectly deduced from a certain knowledge about the summability of these coefficients. The simplest such kind of summability is in the sense of $\ell^1$. It is therefore natural to consider the class of functions on the torus whose Fourier coefficients form an absolutely summable series.
Definition 3.2.15. An integrable function $f$ on the torus is said to have an absolutely convergent Fourier series if
$$\sum_{m \in \mathbf{Z}^n}|\widehat{f}(m)|<+\infty$$
We denote by $A\left(\mathbf{T}^n\right)$ the space of all integrable functions on the torus $\mathbf{T}^n$ whose Fourier series are absolutely convergent. We then introduce a norm on $A\left(\mathbf{T}^n\right)$ by setting
$$|f|_{A\left(\mathbf{T}^n\right)}=\sum_{m \in \mathbf{Z}^n}|\widehat{f}(m)|$$
It is straightforward that every function in $A\left(\mathbf{T}^n\right)$ must be bounded. The following theorem gives us a sufficient condition for a function to be in $A\left(\mathbf{T}^n\right)$.

Theorem 3.2.16. Let $s$ be a nonnegative integer and let $0 \leq \alpha<1$. Assume that $f$ is a function defined on $\mathbf{T}^n$ all of whose partial derivatives of order $s$ lie in the space $\dot{\Lambda}\alpha$. Suppose that $s+\alpha>n / 2$. Then $f \in A\left(\mathbf{T}^n\right)$ and $$|f|{A\left(\mathbf{T}^n\right)} \leq C \sup {|\beta|=s}\left|\partial^\beta f\right|{\dot{\Lambda}_\alpha}$$
where $C$ depends on $n, \alpha$, and $s$.
Proof. For $1 \leq j \leq n$, let $e_j$ be the element of $\mathbf{R}^n$ with zero entries except for the $j$ th coordinate, which is 1 . Let $l$ be a positive integer and let $h_j=2^{-l-2} e_j$.

Then for a multi-index $m=\left(m_1, \ldots, m_n\right)$ satisfying $2^l \leq|m| \leq 2^{l+1}$ and for $j$ in ${1, \ldots, n}$ chosen such that $\left|m_j\right|=\sup _k\left|m_k\right|$ we have
$$\frac{\left|m_j\right|}{2^l} \geq \frac{|m|}{2^l \sqrt{n}} \geq \frac{1}{\sqrt{n}}$$

## 数学代写|傅里叶分析代写Fourier Analysis代考|Decay of Fourier Coefficients of Smooth Functions

3.2.5.定义对于$0 \leq \gamma<1$
$$|f|{\dot{\Lambda}\gamma}=\sup {x, h \in \mathbf{T}^n} \frac{|f(x+h)-f(x)|}{|h|^\gamma}$$ 和 $$\dot{\Lambda}\gamma\left(\mathbf{T}^n\right)=\left{f: \mathbf{T}^n \rightarrow \mathbf{C} \text { with }|f|{\dot{\lambda}\gamma}<\infty\right}$$我们称$\dot{\Lambda}\gamma\left(\mathbf{T}^n\right)$为环面上阶为$\gamma$的齐次Lipschitz空间。在$\mathbf{T}^n$上与$|f|{\dot{\Lambda}\gamma}<\infty$的函数$f$称为$\gamma$阶的齐次Lipschitz函数。有些话是适当的。3.2.6.备注$\dot{\Lambda}\gamma\left(\mathbf{T}^n\right)$在$\mathbf{T}^n$上被称为$\gamma$阶的齐次Lipschitz空间，而$\Lambda_\gamma\left(\mathbf{T}^n\right)$则被称为$\gamma$阶的Lipschitz空间。后一个空间定义为
$$\Lambda_\gamma\left(\mathbf{T}^n\right)=\left{f: \mathbf{T}^n \rightarrow \mathbf{C} \text { with }|f|{\Lambda\gamma}<\infty\right},$$

$$|f|{\Lambda\gamma}=|f|{L^{\infty}}+|f|{\dot{\Lambda}_\gamma} .$$

## 数学代写|傅里叶分析代写Fourier Analysis代考|Functions with Absolutely Summable Fourier Coefficients

3.2.15.定义环面上的可积函数$f$有绝对收敛的傅立叶级数，如果
$$\sum_{m \in \mathbf{Z}^n}|\widehat{f}(m)|<+\infty$$

$$|f|{A\left(\mathbf{T}^n\right)}=\sum{m \in \mathbf{Z}^n}|\widehat{f}(m)|$$

$$\frac{\left|m_j\right|}{2^l} \geq \frac{|m|}{2^l \sqrt{n}} \geq \frac{1}{\sqrt{n}}$$

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