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# 数学代写|实分析代写real analysis代考|Fourier Series

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## 数学代写|实分析代写real analysis代考|The Fourier Transform on L2(R)

Given $f \in L^{2}(\mathbb{R})$, let $\left{f_{n}\right}_{n \in \mathbb{N}}$ be any sequence in $C_{c}^{2}(\mathbb{R})$ such that $f_{n} \rightarrow f$ in $L^{2}$-norm. Then the Fourier transform of $f$ is the function $\widehat{f} \in L^{2}(\mathbb{R})$ such that $\widehat{f_{n}} \rightarrow \widehat{f}$ in $L^{2}$-norm.

This defines the Fourier transform of every square-integrable function. However, we now have two Fourier transforms, one defined on $L^{1}(\mathbb{R})$ and one on $L^{2}(\mathbb{R})$. We show next that these two definitions coincide for any function that belongs to both spaces. Note that if $f \in L^{1}(\mathbb{R})$, then $\widehat{f}$ is a continuous function that is defined by the integral that appears in equation (9.47). In contrast, if $f \in L^{2}(\mathbb{R})$, then $\widehat{f}$ is only implicitly defined as the $L^{2}$-norm limit of $\widehat{f}{n}$ where $f{n} \in C_{c}^{2}(\mathbb{R})$ and $f_{n} \rightarrow f$ in $L^{2}$-norm. Hence, if $f \in L^{2}(\mathbb{R})$, then its Fourier transform $\hat{f}$ is an element of $L^{2}(\mathbb{R})$, and therefore is only defined up to sets of measure zero.

## 数学代写|实分析代写REAL ANALYSIS代考|Decay of Fourier Coefficients

If $f \in L^{1}(\mathbb{R}) \cap L^{2}(\mathbb{R})$, then the function $\widehat{f}$ given by equation $(9.47)$ is equal almost everywhere to the function $\widehat{f}$ given by Definition 9.4.3.
Proof. Fix a function $f \in L^{1}(\mathbb{R}) \cap L^{2}(\mathbb{R})$. Let $\widehat{f}$ be the function defined by equation (9.47), and let $F$ be the $L^{2}$-Fourier transform of $f$ as given by Definition 9.4.3.

The proof of Theorem $9.1 .12$ shows how to explicitly construct functions $f_{N} \in C_{c}^{\infty}(\mathbb{R})$ that converge to $f$ in $L^{1}$-norm. Specifically, if $f_{N}$ is defined as in equation (9.5), then $\left|f-f_{N}\right|_{1} \rightarrow 0$. Replacing the $L^{1}$-norm by the $L^{2}$-norm, exactly the same proof shows that we also have $\left|f-f_{N}\right|_{2} \rightarrow 0$ (compare Problem 9.1.22).

Now, since $\left|f-f_{N}\right|_{1} \rightarrow 0$, Lemma $9.2 .3$ implies that $\widehat{f_{N}} \rightarrow \widehat{f}$ uniformly, and hence pointwise. On the other hand, since $\left|f-f_{N}\right|_{2} \rightarrow 0$, we have by definition that $\widehat{f_{N}} \rightarrow F$ in $L^{2}$-norm. Hence there is a subsequence of the $\widehat{f_{N}} \widehat{\widehat{f}}$ that converges to $F$ pointwise a.e. But this subsequence also converges to $\widehat{f}$ pointwise, so we conclude that $F=\widehat{f}$ a.e.

In summary, we have defined the Fourier transform of every function in $L^{1}(\mathbb{R}) \cup L^{2}(\mathbb{R})$. For functions in $L^{1}(\mathbb{R})$ the Fourier transform is given by equation (9.47), while for functions in $L^{2}(\mathbb{R})$ it is given by Definition 9.4.3.

For functions that belong to both $L^{1}(\mathbb{R})$ and $L^{2}(\mathbb{R})$ these two definitions coincide in the usual almost everywhere sense.
We show next that the Fourier transform is isometric on all of $L^{2}(\mathbb{R})$.

## 离散数学代写

Partial Differential Equations代写可以参考一份偏微分方程midterm答案解析