# 数学代写|傅里叶分析代写Fourier analysis代考|Lax–Milgram Theorem

## 数学代写|傅里叶分析代写Fourier analysis代考|Lax–Milgram Theorem

An inner product $\langle\cdot, \cdot\rangle$ on a Hilbert space is clearly a skew-symmetric sesquilinear functional. ${ }^1$

Remark 7.1 $\Phi(x, x)$ is not necessarily real for a sesquilinear functional $\Phi$. However, if $\Phi$ is skew-symmetric, in addition, $\Phi(x, x)$ must be real.

This concept permits us to generalize the Riesz theorem (Theorem 1.1) to represent duals of Hilbert spaces. ${ }^2$

Theorem 7.1 (Lax-Milgram) Let $\mathfrak{5}$ be a Hilbert space. Suppose that a function $B: \mathfrak{F} \times \mathfrak{S} \rightarrow \mathbb{C}$ is a sesquilinear functional which satisfies the following two conditions:
(i) There exists a constant $\alpha>0$ such that
$$|B(x, y)| \leqq \alpha|x| \cdot|y| \quad \text { for all } \quad x, y \in \mathfrak{S} .$$
(ii) There exists a constant $\beta>0$ such that
$$|B(y, y)| \geqq \beta|y|^2 \quad \text { for all } \quad y \in \mathfrak{S}$$
Then there exists a unique $y_{\Lambda} \in \mathfrak{S}$ for each $\Lambda \in \mathfrak{S}^{\prime}$ which satisfies
$$\Lambda(x)=B\left(x, y_{\Lambda}\right) \text { for all } \quad x \in \mathfrak{5} .$$
Proof For a fixed $y \in \mathfrak{5}$, the function $x \mapsto B(x, y)$ is a bounded linear functional on $\mathfrak{5}$. By Riesz’s theorem, there exists some unique $z \in \mathfrak{S}$ such that
$$B(x, y)=\langle x, z\rangle \text { for all } x \in \mathfrak{5} .$$
If we write
$$A y=z$$
(7.1) can be rewritten as
$$B(x, y)=\langle x, A y\rangle, \quad x, y \in \mathfrak{S} .$$

## 数学代写|傅里叶分析代写Fourier analysis代考|Conjugate Operators and Projections

We denote by $\mathcal{L}(\mathfrak{5})$ the space of bounded linear operators $(\mathfrak{5} \rightarrow \mathfrak{5})$ on a complex Hilbert space $\mathfrak{5}$.
Fixing $T \in \mathcal{L}(\mathfrak{G})$, we define a function $\Phi: \mathfrak{S} \times \mathfrak{S} \rightarrow \mathbb{C}$ by
$$\Phi(x, y)=\langle T x, y\rangle$$
Then $\Phi$ is clearly a sesquilinear functional on $\mathfrak{5}$ and it satisfies
$$|\Phi(x, y)| \leqq|T| \cdot|x| \cdot|y|$$
and so $|\Phi| \leqq|T|$. On the other hand, since
$$|T x|^2=\langle T x, T x\rangle=\Phi(x, T x) \leqq|\Phi| \cdot|x| \cdot|T x|,$$
$|T x| \leqq|\Phi| \cdot|x|$ if $T x \neq 0$. Hence $|T| \leqq|\Phi|$. Thus we prove that
$$|T|=|\Phi|$$
It should be observed that any bounded sesquilinear functional on $\mathfrak{5}$ can be represented in the form (7.6). It is easily shown by appealing to the Riesz theorem.

# 傅里叶分析代写

## 数学代写|傅里叶分析代写Fourier analysis代考|Lax–Milgram Theorem

(i) 存在常数 $\alpha>0$ 这样
$$|B(x, y)| \leqq \alpha|x| \cdot|y| \quad \text { for all } \quad x, y \in \mathfrak{S}$$
(ii) 存在常数 $\beta>0$ 这样
$$|B(y, y)| \geqq \beta|y|^2 \quad \text { for all } \quad y \in \mathfrak{S}$$

$$\Lambda(x)=B\left(x, y_{\Lambda}\right) \text { for all } \quad x \in 5$$

$$B(x, y)=\langle x, z\rangle \text { for all } x \in 5 .$$

$$A y=z$$
(7.1) 可以改写为
$$B(x, y)=\langle x, A y\rangle, \quad x, y \in \mathfrak{S}$$

## 数学代写|傅里叶分析代写Fourier analysis代考|Conjugate Operators and Projections

$$\Phi(x, y)=\langle T x, y\rangle$$

$$|\Phi(x, y)| \leqq|T| \cdot|x| \cdot|y|$$

$$|T x|^2=\langle T x, T x\rangle=\Phi(x, T x) \leqq|\Phi| \cdot|x| \cdot|T x|,$$
$|T x| \leqq|\Phi| \cdot|x|$ 如果 $T x \neq 0$. 因此 $|T| \leqq|\Phi|$. 因此我们证明
$$|T|=|\Phi|$$

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