# 数学代写|泛函分析作业代写Functional Analysis代考|MATH597

## 数学代写|泛函分析作业代写Functional Analysis代考|Connected and Compact Subsets

Recall that connected sets may be complicated objects in general metric spaces. This is still true in normed spaces, but at least for open subsets, connectedness reduces to path-connectedness, which is more intuitive and usually easier to prove.

Proof Let $C$ be a non-empty open connected set in $X$. Recall that “path-connected” means that any two points in $C$ can be joined by a continuous path $r:[0,1] \rightarrow C$ starting at one point and ending at the other. Fix any $x \in C$, and let $P$ be the subset of $C$ consisting of those points that are path-connected to $x$. We wish to show that $P=C$.
$P$ has no boundary in $C$ : Given any boundary point $z$ of $P$, there is a ball $B_\epsilon(z) \subseteq C$ since $C$ is open, and thus a point $y \in P$ in the ball. This means that there is a path $r$ from $x$ to $y$. In normed spaces, it is obvious that balls, like all convex sets, are path-connected (by straight paths). So we can extend the path $r$ to one that starts from $x$ and ends at any other $w \in B_\epsilon(z)$, simply by adjoining the straight line at the end. More rigorously, the function $\tilde{r}:[0,1] \rightarrow C$ defined by
$$\tilde{r}(t):= \begin{cases}r(2 t) & t \in\left[0, \frac{1}{2}\right] \ y+(2 t-1)(w-y) & \left.t \in] \frac{1}{2}, 1\right]\end{cases}$$
is continuous. So $z$ is surrounded by points of $P$, a contradiction.
But a connected set such as $C$, cannot contain a subset, such as $P$, without a boundary (Proposition 5.3), unless $P=\varnothing$ (which is not the case here) or $P=C$.

There is quite a bit to say about bounded and totally bounded sets. As we will see later on, they are the same in finite dimensional normed spaces, but in infinite dimensional ones, no open set can be totally bounded, although balls are bounded sets. For now, let us show that translations and scalings of bounded and totally bounded sets remain so.

## 数学代写|泛函分析作业代写Functional Analysis代考|Every normed space can be completed to a Banach space

Proof Let $\tilde{X}$ be the completion of the normed space $X$ (Theorem 4.6). We need to prove that vector addition, scalar multiplication and the norm on $X$ can be extended to $\widetilde{X}$. Using the notation of Theorem 4.6, let $\boldsymbol{x}=\left[x_n\right], \boldsymbol{y}=\left[y_n\right]$ be elements of $\widetilde{X}$, with $\left(x_n\right),\left(y_n\right)$ Cauchy sequences in $X$. Since
\begin{aligned} \left|x_n+y_n-x_m-y_m\right| & \leqslant\left|x_n-x_m\right|+\left|y_n-y_m\right| \rightarrow 0 \ \left|\lambda x_n-\lambda x_m\right| &=|\lambda|\left|x_n-x_m\right| \rightarrow 0 \ \left|\left|x_n\right|-\left|x_m\right|\right| & \leqslant\left|x_n-x_m\right| \rightarrow 0, \end{aligned}
as $n, m \rightarrow \infty$, we find that $\left(x_n+y_n\right),\left(\lambda x_n\right)$ and $\left(\left|x_n\right|\right)$ are all Cauchy sequences. For the same reasons, if $\left(x_n^{\prime}\right)$ is asymptotic to $\left(x_n\right)$, and $\left(y_n^{\prime}\right)$ to $\left(y_n\right)$, then $\left(x_n^{\prime}+y_n^{\prime}\right)$ and $\left(x_n+y_n\right),\left(\lambda x_n^{\prime}\right)$ and $\left(\lambda x_n\right)$, and $\left|x_n^{\prime}\right|$ and $\left|x_n\right|$, are asymptotic to each other, respectively. So we can define
$$\boldsymbol{x}+\boldsymbol{y}:=\left[x_n+y_n\right], \quad \lambda \boldsymbol{x}:=\left[\lambda x_n\right], \quad|\boldsymbol{x}|:=\lim {n \rightarrow \infty}\left|x_n\right| .$$ Note that $\tilde{d}(\boldsymbol{x}, \boldsymbol{y})=|\boldsymbol{x}-\boldsymbol{y}|$. It is easy to check that they give a legitimate vector addition, scalar multiplication and a norm; the required axioms follow from the same properties in $X$ and the continuity of these operations, e.g. \begin{aligned} |\boldsymbol{x}+\boldsymbol{y}| &=\lim {n \rightarrow \infty}\left|x_n+y_n\right| \leqslant \lim _{n \rightarrow \infty}\left(\left|x_n\right|+\left|y_n\right|\right)=|\boldsymbol{x}|+|\boldsymbol{y}|, \ |\boldsymbol{x}| &=0 \Rightarrow\left|x_n\right| \rightarrow 0 \Rightarrow \boldsymbol{x}=\left[x_n\right]=[0]=\mathbf{0} . \end{aligned}
Note that the zero can be represented by the Cauchy sequence (0), and $-\boldsymbol{x}$ by $\left(-x_n\right)$. Furthermore, recall that there is a copy of $X$ in $\tilde{X}$ (as constant sequences); the operations just defined on $\tilde{X}$ reduce to the given operations on $X$, when restricted to it.

# 泛函分析代考

## 数学代写|泛函分析作业代写Functional Analysis代考|Connected and Compact Subsets

$P$ 没有边界 $C$ : 给定任何边界点 $z$ 的 $P$ ，有一个球 $B_\epsilon(z) \subseteq C$ 自从 $C$ 是开放的，因此有一点 $y \in P$ 在球中。 这意味着有一条路径 $r$ 从 $x$ 至 $y$. 在范数空间中，很明显球，像所有凸集一样，是路径连接的（通过直线路 径)。所以我们可以扩展路径 $r$ 开始于 $x$ 并在任何其他结束 $w \in B_\epsilon(z)$ ，只需在末端连接直线即可。更严格 地说，函数 $\tilde{r}:[0,1] \rightarrow C$ 被定义为
$$\left.\tilde{r}(t):=\left{r(2 t) \quad t \in\left[0, \frac{1}{2}\right] y+(2 t-1)(w-y) \quad t \in\right] \frac{1}{2}, 1\right]$$

## 数学代写|泛函分析作业代写Functional Analysis代考|Every normed space can be completed to a Banach space

$$\left|x_n+y_n-x_m-y_m\right| \leqslant\left|x_n-x_m\right|+\left|y_n-y_m\right| \rightarrow 0\left|\lambda x_n-\lambda x_m\right| \quad=|\lambda|\left|x_n-x_m\right| \rightarrow 0|| x_n|-| x_m|| \leqslant$$

$$\boldsymbol{x}+\boldsymbol{y}:=\left[x_n+y_n\right], \quad \lambda \boldsymbol{x}:=\left[\lambda x_n\right], \quad|\boldsymbol{x}|:=\lim n \rightarrow \infty\left|x_n\right| .$$

$$|\boldsymbol{x}+\boldsymbol{y}|=\lim n \rightarrow \infty\left|x_n+y_n\right| \leqslant \lim _{n \rightarrow \infty}\left(\left|x_n\right|+\left|y_n\right|\right)=|\boldsymbol{x}|+|\boldsymbol{y}|,|\boldsymbol{x}| \quad=0 \Rightarrow\left|x_n\right| \rightarrow 0 \Rightarrow \boldsymbol{x}=\left[x_n\right]=[0]$$

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