# 数学代写|泛函分析作业代写Functional Analysis代考|МАTH3402

## 数学代写|泛函分析作业代写Functional Analysis代考|Continuous Linear Maps

In every branch of mathematics which concerns itself with sets having some particular structure, the functions which preserve that structure, called morphisms, feature prominently. Such maps allow us to transfer equations from one space to another, to compare them with each other and state when two spaces are essentially the same, or if not, whether one can be embedded in the other, etc. Even in applications, it is often the case that certain aspects of a process are conserved. For example, a rotation of geometric space yields essentially the same space. The morphisms on normed spaces are formalized by the following definition.

Proof The definition of a Lipschitz map reads, when applied for normed spaces, $|f(x)-f(y)| \leqslant c|x-y|$ for some $c>0$. When $f$ is in fact a linear map $T$, it becomes $|T(x-y)| \leqslant c|x-y|$, or equivalently, $|T a| \leqslant c|a|$ for all $a \in X$. That Lipschitz maps are (uniformly) continuous is true in every metric space (Examples $4.15(3)$ ), but can easily be seen in this context. If $x_n \rightarrow x$, then $T x_n \rightarrow T x$, since
$$\left|T x_n-T x\right|=\left|T\left(x_n-x\right)\right| \leqslant c\left|x_n-x\right| \rightarrow 0 .$$
Conversely, suppose the ratios $|T x| /|x|$ are unbounded. Since scaling $x$ does not affect this ratio (because $T$ is linear), there must be vectors $x_n$ such that $\left|T x_n\right|=1$ but $\left|x_n\right| \leqslant 1 / n$. So $x_n \rightarrow 0$ yet $T x_n \nrightarrow 0$, and $T$ is not continuous.

## 数学代写|泛函分析作业代写Functional Analysis代考|Riesz’s lemma

For any non-trivial closed linear subspace $M$, and $0 \leqslant c<1$, there is a unit vector $x$ such that $|x+M|=c$.

Proof Let $y \notin M$ so that $|y+M|>0$; by re-scaling $y$ if necessary, one can assume $|y+M|=c$. The map $f: M \rightarrow \mathbb{R}$, defined by $f(a):=|y+a|$, takes values close to $c$, as well as arbitrarily large values $(|y+\lambda a| \geqslant|\lambda||a|-|y| \rightarrow \infty$ as $\lambda \rightarrow \infty$, for $M \neq 0$ ). Since $M$ is connected, and $f$ is continuous, its image must include $] c, \infty[$ by the intermediate value theorem (Proposition 5.6). In particular there is an $a \in M$ such that $|y+a|=1$, so letting $x:=y+a$ gives $|x+M|=|y+M|=c$.
Exercises 8.21

1. The mapping $x \mapsto x+M, X \rightarrow X / M$, is linear and continuous.
2. Let $M:={f \in C[0,1]: f(0)=0}$, then $2+M={f \in C[0,1]: f(0)=2}$, and $C[0,1] / M \cong \mathbb{C}$.
3. (a) $X / X \equiv 0, X / 0 \equiv X$.
4. (b) If $X, Y$ are normed spaces, then $\frac{X \times Y}{X \times 0} \equiv Y$.
5. Let $X$ be a finite-dimensional space generated by a set of unit vectors $E:=$ $\left{e_i: i=1, \ldots, n\right}$, and let $M_i:=\mathbb{E} \backslash\left{e_i\right} \mathbb{1}$. Then the coefficient $\left|\alpha_i\right|$ in $x=\sum_{i=1}^n \alpha_i e_i$ is at most $|x| /\left|e_i+M_i\right|$. Thus, in finding a basis for $X$, it is best to select unit vectors that are as ‘far’ from each other as possible.
6. Let $M$ be a closed subspace of $X$. If both $M$ and $X / M$ are separable, then so is $X$.

# 泛函分析代考

## 数学代写|泛函分析作业代写Functional Analysis代考|Continuous Linear Maps

$$\left|T x_n-T x\right|=\left|T\left(x_n-x\right)\right| \leqslant c\left|x_n-x\right| \rightarrow 0 .$$

## 数学代写|泛函分析作业代写Functional Analysis代考|Riesz’s lemma

1. 映射 $x \mapsto x+M, X \rightarrow X / M$ ，是线性且连续的。
2. 让 $M:=f \in C[0,1]: f(0)=0$ ， 然后 $2+M=f \in C[0,1]: f(0)=2$ ，和 $C[0,1] / M \cong \mathbb{C}$.
3. (一个) $X / X \equiv 0, X / 0 \equiv X$.
4. (b) 如果 $X, Y$ 是范数空间，那么 $\frac{X \times Y}{X \times 0} \equiv Y$. $|x| /\left|e_i+M_i\right|$. 因此，在寻找基础 $X$ ，最好选择彼此屈可能”远”的单位向量。
5. 让 $M$ 是一个闭子空间 $X$. 如果两者 $M$ 和 $X / M$ 是可分离的，那么也是 $X$.

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