# 数学代写|泛函分析作业代写Functional Analysis代考|MAT4450

#### Doug I. Jones

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## 数学代写|泛函分析作业代写Functional Analysis代考|Inner Product and Hilbert Spaces

Much of functional analysis involves abstracting and making precise ideas that have been developed and used over many decades, even centuries, in physics and classical mathematics. In this regard, functional analysis makes use of a great deal of “mathematical hindsight” in that it seeks to identify the most primitive features of elementary analysis, geometry, calculus, and the theory of equations in order to generalize them, to give them order and structure, and to define their interdependencies. In doing this, however, it simultaneously unifies this entire collection of ideas and extends them to new areas that could never have been completely explored within the framework of classical mathematics or physics.

The final abstraction we investigate in this book is of geometry: We add to the idea of vector spaces enough structure to include abstractions of the geometrical terms direction, orthogonality, angle between vectors, and length of a vector. Once these ideas are established, we have the framework for not only a geometry of function spaces but also a theory of linear equations, variational methods, approximation theory, and numerous other areas of mathematics.
We begin by recalling the definition of scalar product (comp. Section 2.14).
Scalar (Inner) Product. Let $V$ be a vector space defined over the complex number field $\boldsymbol{C}$. A scalarvalued function $p: V \times V \longrightarrow \mathbb{C}$ that associates with each pair $\boldsymbol{u}, \boldsymbol{v}$ of vectors in $V$ a scalar, denoted $p(\boldsymbol{u}, \boldsymbol{v})=(\boldsymbol{u}, \boldsymbol{v})$, is called a scalar (inner) product on $V$ iff
(i) $(\boldsymbol{u}, \boldsymbol{v})$ is linear with respect to the first argument
$$\left(\alpha_1 \boldsymbol{u}_1+\alpha_2 \boldsymbol{u}_2, \boldsymbol{v}\right)=\alpha_1\left(\boldsymbol{u}_1, \boldsymbol{v}\right)+\alpha_2\left(\boldsymbol{u}_2, \boldsymbol{v}\right) \quad \forall \alpha_1, \alpha_2 \in \boldsymbol{C}, \quad \boldsymbol{u}_1, \boldsymbol{u}_2, \boldsymbol{v} \in V$$
(ii) $(\boldsymbol{u}, \boldsymbol{v})$ is symmetric (in the complex sense)
$$(\boldsymbol{u}, \boldsymbol{v})=\overline{(\boldsymbol{v}, \boldsymbol{u})}, \quad \forall \boldsymbol{u}, \boldsymbol{v} \in V$$
(iii) $(\boldsymbol{u}, \boldsymbol{v})$ is positive definite, i.e.,
$$(\boldsymbol{u}, \boldsymbol{u})>0 \quad \forall \boldsymbol{u} \neq \mathbf{0}, \boldsymbol{u} \in V$$
Note that the first two conditions imply that $(\boldsymbol{u}, \boldsymbol{v})$ is antilinear with respect to the second argument
\begin{aligned} \left(\boldsymbol{u}, \beta_1 \boldsymbol{v}_1+\beta_2 \boldsymbol{v}_2\right) & =\overline{\left(\beta_1 \boldsymbol{v}_1+\beta_2 \boldsymbol{v}_2, \boldsymbol{u}\right)} \ & =\bar{\beta}_1 \overline{\left(\boldsymbol{v}_1, \boldsymbol{u}\right)}+\bar{\beta}_2 \overline{\left(\boldsymbol{v}_2, \boldsymbol{u}\right)} \ & =\bar{\beta}_1\left(\boldsymbol{u}, \boldsymbol{v}_1\right)+\bar{\beta}_2\left(\boldsymbol{u}, \boldsymbol{v}_2\right) \end{aligned}
for every $\beta_1, \beta_2 \in \boldsymbol{C}, \boldsymbol{v}_1, \boldsymbol{v}_2 \in V$.
In the case of a real vector space $V$, condition (ii) becomes one of symmetry
$$(\boldsymbol{u}, \boldsymbol{v})=(\boldsymbol{v}, \boldsymbol{u}) \quad \forall \boldsymbol{u}, \boldsymbol{v} \in V$$
and then $(\boldsymbol{u}, \boldsymbol{v})$ is linear with respect to both arguments $\boldsymbol{u}$ and $\boldsymbol{v}$. Note also that, according to the second condition,
$$(\boldsymbol{u}, \boldsymbol{u})=\overline{(\boldsymbol{u}, \boldsymbol{u})}$$
is a real number and therefore condition (iii) makes sense.

## 数学代写|泛函分析作业代写Functional Analysis代考|Orthogonality and Orthogonal Projections

Orthogonal Complements. Let $V$ be an inner product space and let $V^{\prime}$ be its topological dual. If $M$ is any subspace of $V$, recall that (see Section 5.16) we have defined the space
$$M^{\perp} \stackrel{\text { def }}{=}\left{f \in V^{\prime}:\langle f, \boldsymbol{u}\rangle=0 \quad \forall \boldsymbol{u} \in M\right}$$
as the orthogonal complement of $M$ with respect to the duality pairing $\langle\cdot, \cdot\rangle$.
Since $V$ is an inner product space, the inner product can be used to construct orthogonal subspaces of $V$ rather than its dual. In fact, we also refer to the space
$$M_V^{\perp} \stackrel{\text { def }}{=}{\boldsymbol{v} \in V:(\boldsymbol{u}, \boldsymbol{v})=0 \quad \forall \boldsymbol{u} \in M}$$
as the orthogonal complement of $M$ with respect to the inner product $(\cdot, \cdot)$.
The situation is really not as complicated as it may seem, because the two orthogonal complements $M^{\perp}$ and $M_V^{\perp}$ are algebraically and topologically equivalent. We shall take up this equivalence in some detail in the next section. In this section we shall investigate some fundamental properties of orthogonal complements with respect to the inner product $(\cdot, \cdot)$. Taking for a moment the equivalence of two notions for the orthogonal complements for granted, we shall denote the orthogonal complements $M_V^{\perp}$ simply as $M^{\perp}$.

# 泛函分析代写

## 数学代写|泛函分析作业代写Functional Analysis代考|Inner Product and Hilbert Spaces

(i) $(\boldsymbol{u}, \boldsymbol{v})$相对于第一个参数是线性的
$$\left(\alpha_1 \boldsymbol{u}_1+\alpha_2 \boldsymbol{u}_2, \boldsymbol{v}\right)=\alpha_1\left(\boldsymbol{u}_1, \boldsymbol{v}\right)+\alpha_2\left(\boldsymbol{u}_2, \boldsymbol{v}\right) \quad \forall \alpha_1, \alpha_2 \in \boldsymbol{C}, \quad \boldsymbol{u}_1, \boldsymbol{u}_2, \boldsymbol{v} \in V$$
(ii) $(\boldsymbol{u}, \boldsymbol{v})$是对称的(在复杂意义上)
$$(\boldsymbol{u}, \boldsymbol{v})=\overline{(\boldsymbol{v}, \boldsymbol{u})}, \quad \forall \boldsymbol{u}, \boldsymbol{v} \in V$$
(iii) $(\boldsymbol{u}, \boldsymbol{v})$是肯定的，即
$$(\boldsymbol{u}, \boldsymbol{u})>0 \quad \forall \boldsymbol{u} \neq \mathbf{0}, \boldsymbol{u} \in V$$

\begin{aligned} \left(\boldsymbol{u}, \beta_1 \boldsymbol{v}_1+\beta_2 \boldsymbol{v}_2\right) & =\overline{\left(\beta_1 \boldsymbol{v}_1+\beta_2 \boldsymbol{v}_2, \boldsymbol{u}\right)} \ & =\bar{\beta}_1 \overline{\left(\boldsymbol{v}_1, \boldsymbol{u}\right)}+\bar{\beta}_2 \overline{\left(\boldsymbol{v}_2, \boldsymbol{u}\right)} \ & =\bar{\beta}_1\left(\boldsymbol{u}, \boldsymbol{v}_1\right)+\bar{\beta}_2\left(\boldsymbol{u}, \boldsymbol{v}_2\right) \end{aligned}

$$(\boldsymbol{u}, \boldsymbol{v})=(\boldsymbol{v}, \boldsymbol{u}) \quad \forall \boldsymbol{u}, \boldsymbol{v} \in V$$

$$(\boldsymbol{u}, \boldsymbol{u})=\overline{(\boldsymbol{u}, \boldsymbol{u})}$$

## 数学代写|泛函分析作业代写Functional Analysis代考|Orthogonality and Orthogonal Projections

$$M^{\perp} \stackrel{\text { def }}{=}\left{f \in V^{\prime}:\langle f, \boldsymbol{u}\rangle=0 \quad \forall \boldsymbol{u} \in M\right}$$

$$M_V^{\perp} \stackrel{\text { def }}{=}{\boldsymbol{v} \in V:(\boldsymbol{u}, \boldsymbol{v})=0 \quad \forall \boldsymbol{u} \in M}$$

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