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数学代写|几何测度论代写geometric measure theory代考|H1 and the classical notion of length

A set $\Gamma \subset \mathbb{R}^n$ is a curve if there exist $a>0$ and a continuous, injective function $\gamma:[0, a] \rightarrow \mathbb{R}^n$ such that $\Gamma=\gamma([0, a])$. The function $\gamma$ is called a parametrization of $\Gamma$. Given a parametrization $\gamma:[0, a] \rightarrow \mathbb{R}^n$ and a sub-interval $[b, c]$ of $[0, a]$ we define the length of $\gamma$ over $[b, c]$ as
$$\ell(\gamma ;[b, c])=\sup \left{\sum_{h=1}^N\left|\gamma\left(t_h\right)-\gamma\left(t_{h-1}\right)\right|: b=t_0<t_{h-1}<t_h<t_N=c, N \in \mathbb{N}\right} .$$
It is easily seen that $\ell(\gamma ;[0, a])$ is independent of the parametrization $\gamma$ of $\Gamma$. Therefore, the length of $\Gamma$ is defined as
$$\text { length }(\Gamma)=\ell(\gamma ;[0, a]) .$$
Whether length $(\Gamma)$ is finite or not, the following theorem holds true.
Theorem $3.8$ If $\Gamma$ is a curve, then $\mathcal{H}^1(\Gamma)=$ length $(\Gamma)$.
Proof The theorem is proved by Remark $3.7$ if $\Gamma$ is a segment. We now consider a parametrization $\gamma:[0, a] \rightarrow \mathbb{R}^n$ of $\Gamma$ and set $\ell=\ell(\gamma ;[0, a])=\operatorname{length}(\Gamma)$. We divide the proof into three steps, and notice that
(i) $\ell(\gamma ;[b, c]) \geq|\gamma(b)-\gamma(c)|$, whenever $0 \leq b \leq c \leq a$;
(ii) $\ell(\gamma ;[b, c])=\ell(\gamma ;[b, d])+\ell(\gamma ;[d, c])$ whenever $0 \leq b \leq d \leq c \leq a$.
Step one: We show that $\mathcal{H}^1(\Gamma) \geq|\gamma(a)-\gamma(0)|$. Since the projection $\mathbf{p}: \mathbb{R}^n \rightarrow$ $\mathbb{R}^n$ of $\mathbb{R}^n$ onto the line defined by $\gamma(0)$ and $\gamma(a)$ satisfies $\operatorname{Lip}(\mathbf{p}) \leq 1$, by Proposition $3.5$ we have $\mathcal{H} \mathcal{H}^1\left(\mathbf{p}\left(I^{\prime}\right)\right) \leq \mathcal{H}^1(1)$. At the same time, $\mathbf{p}(1)$ must contain the segment $[\gamma(0) \gamma(a)]$ : otherwise, $\Gamma=\gamma([0, a])$ would be disconnected, against the continuity of $\gamma$. Thus $\mathcal{H}^1(\mathbf{p}(\Gamma)) \geq \mathcal{H}^1([\gamma(0) \gamma(a)])=|\gamma(a)-\gamma(0)|$.
Step two: If $\left{t_h\right}_{h=0}^N$ is a competitor in the definition of $\ell$, then, setting $\Gamma_h=$ $\gamma\left(\left[t_{h-1}, t_h\right]\right)$, we have $\Gamma=\bigcup_{h=1}^N \Gamma_h$ and, by the injectivity of $\gamma, \mathcal{H}^1\left(\Gamma_h \cap \Gamma_{h+1}\right)=$ $\mathcal{H}^1\left(\left{\gamma\left(t_h\right)\right}\right)=0$. We thus find $\mathcal{H}^1(\Gamma) \geq \ell$ as, by step one,
$$\mathcal{H}^1(\Gamma)=\sum_{h=1}^N \mathcal{H}^1\left(\Gamma_h\right) \geq \sum_{h=1}^N\left|\gamma\left(t_h\right)-\gamma\left(t_{h-1}\right)\right| .$$

数学代写|几何测度论代写geometric measure theory代考|Hn = Ln and the isodiametric inequality

We show here equivalence of the Lebesgue measure and the $n$-dimensional Hausdorff measure $\mathcal{H}^n$ on $\mathbb{R}^n$.
Theorem 3.10 If $E \subset \mathbb{R}^n$, and $\delta \in(0, \infty]$, then $|E|=\mathcal{H}^n(E)=\mathcal{H}\delta^n(E)$. A first tool used in proving Theorem $3.10$ is Vitali’s property of Lebesgue measure: if $A \subset \mathbb{R}^n$ is open and $\delta>0$, then a countable disjoint family $\mathcal{F}$ of closed balls contained in $A$ with diameter less than $\delta$ exists such that $$|A \backslash \bigcup{\bar{B}: \bar{B} \in \mathcal{F}}|=0 .$$ Postponing until Section $5.1$ the proof of this result, we now introduce the second tool used in proving Theorem $3.10$, namely, the isodiametric inequality. Theorem $3.11$ (Isodiametric inequality) Among all sets of fixed diameter, balls have maximum volume. In other words. $$|E| \leq \omega_n\left(\frac{\operatorname{diam}(E)}{2}\right)^n, \quad \forall E \subset \mathbb{R}^n .$$ Proof of Theorem $3.10$ Step one: We first notice that $$\omega_n\left(\frac{\sqrt{n}}{2}\right)^n|E| \geq \mathcal{H}{\infty}^n(E) .$$

If the covering $\mathcal{F}$ is a competitor in the definition of $|E|$, and $r(F)$ denotes the side length of the cube $F \in \mathcal{F}$, then $\operatorname{diam}(F)=\sqrt{n} r(F)$, so that, in particular,
$$\mathcal{H}{\infty}^n(E) \leq \omega_n \sum{F \in \mathcal{F}}\left(\frac{\operatorname{diam}(F)}{2}\right)^n=\omega_n\left(\frac{\sqrt{n}}{2}\right)^n \sum_{F \in \mathcal{F}} r(F)^n .$$
By the arbitrariness of $\mathcal{F}$, we find (3.6).

几何测度论代考

数学代写|几何测度论代写geometric measure theory代考|H1 and the classical notion of length

$\backslash 1 e f t$ 的分隔符缺失或无法识别

$$\text { length }(\Gamma)=\ell(\gamma ;[0, a]) \text {. }$$

(二) $\ell(\gamma ;[b, c])=\ell(\gamma ;[b, d])+\ell(\gamma ;[d, c])$ 每当 $0 \leq b \leq d \leq c \leq a$. 时， $\mathbf{p}(1)$ 必须包含段 $[\gamma(0) \gamma(a)]$ : 否则， $\Gamma=\gamma([0, a])$ 将被断开，反对的连续性 $\gamma$. 因此 $\mathcal{H}^1(\mathbf{p}(\Gamma)) \geq \mathcal{H}^1([\gamma(0) \gamma(a)])=|\gamma(a)-\gamma(0)|$.

$$\mathcal{H}^1(\Gamma)=\sum_{h=1}^N \mathcal{H}^1\left(\Gamma_h\right) \geq \sum_{h=1}^N\left|\gamma\left(t_h\right)-\gamma\left(t_{h-1}\right)\right| .$$

数学代写|几何测度论代写geometric measure theory代考|Hn = Ln and the isodiametric inequality

$$|A \backslash \bigcup \bar{B}: \bar{B} \in \mathcal{F}|=0 .$$

$$|E| \leq \omega_n\left(\frac{\operatorname{diam}(E)}{2}\right)^n, \quad \forall E \subset \mathbb{R}^n$$

$$\omega_n\left(\frac{\sqrt{n}}{2}\right)^n|E| \geq \mathcal{H} \infty^n(E) \text {. }$$

$$\mathcal{H} \infty^n(E) \leq \omega_n \sum F \in \mathcal{F}\left(\frac{\operatorname{diam}(F)}{2}\right)^n=\omega_n\left(\frac{\sqrt{n}}{2}\right)^n \sum_{F \in \mathcal{F}} r(F)^n$$

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assignmentutor™您的专属作业导师
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