# 数学代写|泛函分析作业代写Functional Analysis代考|MATH3909

## 数学代写|泛函分析作业代写Functional Analysis代考|Orthogonal Complements

Throughout this section we fix a Hilbert space $H$.
Definition 3.10 (Orthogonality). The elements $x, x^{\prime} \in H$ are said to be orthogonal, notation
$$x \perp x^{\prime},$$
if $\left(x \mid x^{\prime}\right)=0$. Two subsets $A$ and $B$ of $H$ are called orthogonal if $a \perp b$ for all $a \in A$ and $b \in B$
Orthogonal elements $x \perp x^{\prime}$ satisfy the Pythagorean identity
$$\left|x+x^{\prime}\right|^2=|x|^2+\left|x^{\prime}\right|^2,$$

as is seen by expanding the square norms in terms of inner products.
Definition 3.11 (Orthogonal complement). The orthogonal complement of a subset $A$ of $H$ is the set
$$A^{\perp}:={x \in H: x \perp a \text { for all } a \in A} .$$
The orthogonal complement $A^{\perp}$ of a subset $A$ is a closed subspace of $H$. Indeed, it is trivially checked that $A^{\perp}$ is a vector space. To prove its closedness, let $x_n \rightarrow x$ in $H$ with $x_n \in A^{\perp}$. Then, by the continuity of the inner product, for all $a \in A$ we obtain $(x \mid a)=\lim _{n \rightarrow \infty}\left(x_n \mid a\right)=0$.

The most important result on orthogonality is certainly the fact that every closed subspace $Y$ of a Hilbert space is orthogonally complemented by $Y^{\perp}$. This is the content of Theorem $3.13$ below. For its proof we need the following approximation theorem for convex closed sets in Hilbert space. Recall that a subset $C$ of a vector space is called convex if for all $x_0, x_1 \in C$ we have $(1-\lambda) x_0+\lambda x_1 \in C$ for all $0 \leqslant \lambda \leqslant 1$.

Theorem $3.12$ (Best approximation). Let $C$ be a nonempty convex closed subset of $H$. Then for all $x \in H$ there exists a unique $c \in C$ that minimises the distance from $x$ to the points of C.:
$$|x-c|=\min {y \in C}|x-y| .$$ Proof Let $\left(y_n\right){n \geqslant 1}$ be a sequence in $C$ such that
$$\lim {n \rightarrow \infty}\left|x-y_n\right|=\inf {y \in C}|x-y|=: D$$

## 数学代写|泛函分析作业代写Functional Analysis代考|The Trigonometric System

In this example T denotes the unit circle in the complex plane, parametrised by the interval $[\pi, \pi]$ and equipped with the normalised Lebesgue measure $\mathrm{d} \theta / 2 \pi$. We shall prove that the functions
$$e_n(\theta):=\exp (i n \theta), \quad \theta \in[-\pi, \pi], n \in \mathbb{Z},$$
form an orthonormal basis for $L^2(\mathrm{~T})$.
That $\left(e_n\right){n \in \mathbb{Z}}$ is an orthonormal sequence in $L^2(\mathbb{T})$ is evident from $$\left(e_j \mid e_k\right)=\frac{1}{2 \pi} \int{-\pi}^\pi \exp (i j \theta) \overline{\exp (i k \theta)} \mathrm{d} \theta=\frac{1}{2 \pi} \int_{-\pi}^\pi \exp (i(j-k) \theta) \mathrm{d} \theta=\delta_{j k} .$$
To prove that $\left(e_n\right){n \in \mathbb{Z}}$ is an orthonormal basis, by Theorem $3.21$ it remains to be proved that the trigonometric polynomials, i.e., the functions of the form $\sum{n=-N}^N c_n e_n$, are dense in $L^2(T)$. This can be deduced from the Stone-Weierstrass theorem (see Problem 3.11), but we prefer the following argument from Fourier Analysis which gives explicit approximants and some error bounds.

Definition 3.26 (Fourier coefficients). The Fourier coefficients of a function $f \in L^1(\mathbb{T})$ are defined as
$$\widehat{f}(n):=\left(f \mid e_n\right)=\frac{1}{2 \pi} \int_{-\pi}^\pi f(\theta) \exp (-i n \theta) \mathrm{d} \theta, \quad n \in \mathbb{Z} .$$
Theorem 3.27. For all $f \in C(\mathbb{T})$ we have
$$\lim {N \rightarrow \infty}\left|f-\frac{1}{N} \sum{n=0}^{N-1} \sum_{k=-n}^n \widehat{f}(k) e_k\right|_{\infty}=0 .$$
Proof Fix $f \in C(\mathbb{T})$ with $|f|_{\infty}=1$. We have
$$\widehat{f}(n) \exp (i n \theta)=\frac{1}{2 \pi} \int_{-\pi}^\pi \exp (i n(\theta-\sigma)) f(\sigma) \mathrm{d} \sigma=\frac{1}{2 \pi} \int_{-\pi}^\pi \exp (i n \sigma) f(\theta-\sigma) \mathrm{d} \sigma$$
and therefore
$$f_N(\theta):=\frac{1}{N} \sum_{n=0}^{N-1} \sum_{k=-n}^n \widehat{f}(k) \exp (i k \theta)=\frac{1}{2 \pi} \int_{-\pi}^\pi K_N(\sigma) f(\theta-\sigma) \mathrm{d} \sigma$$
where the Fejer kernel $K_N$ is defined by
$$K_N(\theta):=\frac{1}{N} \sum_{n=0}^{N-1} \sum_{k=-n}^n \exp (i k \theta)=\frac{1}{N} \frac{\sin ^2\left(\frac{1}{2} N \theta\right)}{\sin ^2\left(\frac{1}{2} \theta\right)}$$

# 泛函分析代考

## 数学代写|泛函分析作业代写Functional Analysis代考|Orthogonal Complements

$$x \perp x^{\prime},$$

$$\left|x+x^{\prime}\right|^2=|x|^2+\left|x^{\prime}\right|^2,$$

$$A^{\perp}:=x \in H: x \perp a \text { for all } a \in A .$$

$$|x-c|=\min y \in C|x-y| .$$

$$\lim n \rightarrow \infty\left|x-y_n\right|=\inf y \in C|x-y|=: D$$

## 数学代写|泛函分析作业代写Functional Analysis代考|The Trigonometric System

$$e_n(\theta):=\exp (i n \theta), \quad \theta \in[-\pi, \pi], n \in \mathbb{Z},$$

$$\left(e_j \mid e_k\right)=\frac{1}{2 \pi} \int-\pi^\pi \exp (i j \theta) \overline{\exp (i k \theta)} \mathrm{d} \theta=\frac{1}{2 \pi} \int_{-\pi}^\pi \exp (i(j-k) \theta) \mathrm{d} \theta=\delta_{j k} .$$

$\sum n=-N^N c_n e_n$ ，密集于 $L^2(T)$. 这可以从 Stone-Weierstrass 定理推导出来（参见问题 3.11），但我 们更喜欢傅里叶分析中的以下论点，它给出了明确的近似值和一些误差界限。

$$\widehat{f}(n):=\left(f \mid e_n\right)=\frac{1}{2 \pi} \int_{-\pi}^\pi f(\theta) \exp (-i n \theta) \mathrm{d} \theta, \quad n \in \mathbb{Z} .$$

$$\lim N \rightarrow \infty\left|f-\frac{1}{N} \sum n=0^{N-1} \sum_{k=-n}^n \widehat{f}(k) e_k\right|{\infty}=0 .$$ 证明修复 $f \in C(\mathbb{T})$ 和 $|f|{\infty}=1$. 我们有
$$\widehat{f}(n) \exp (i n \theta)=\frac{1}{2 \pi} \int_{-\pi}^\pi \exp (i n(\theta-\sigma)) f(\sigma) \mathrm{d} \sigma=\frac{1}{2 \pi} \int_{-\pi}^\pi \exp (i n \sigma) f(\theta-\sigma) \mathrm{d} \sigma$$

$$f_N(\theta):=\frac{1}{N} \sum_{n=0}^{N-1} \sum_{k=-n}^n \widehat{f}(k) \exp (i k \theta)=\frac{1}{2 \pi} \int_{-\pi}^\pi K_N(\sigma) f(\theta-\sigma) \mathrm{d} \sigma$$
Fejer内核在哪里 $K_N$ 定义为
$$K_N(\theta):=\frac{1}{N} \sum_{n=0}^{N-1} \sum_{k=-n}^n \exp (i k \theta)=\frac{1}{N} \frac{\sin ^2\left(\frac{1}{2} N \theta\right)}{\sin ^2\left(\frac{1}{2} \theta\right)}$$

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