∞-范数等于各分量的最大值

hbghlyj

LA_4_VectorLpNorm.pdf
使用 $p$-范数的极限定义∞-范数，$$\|x\|_{\infty}=\lim _{p \to \infty}\left(\sum_{i=1}^{n}\left|x_{i}\right|^{p}\right)^{\frac{1}{p}}$$有趣的是，∞-范数实际上等于向量各分量的最大值！$$\|x\|_{\infty}=\max _{1<j<n}\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$$
证明
Part 1. 提取最大元素
设$x_j$为向量的最大的分量，由定义
\[
\|x\|_{\infty}=\lim _{p \to \infty}\left(\left|x_1\right|^p+\left|x_2\right|^p+\ldots+\left|x_j\right|^p+\ldots+\left|x_n\right|^p\right)^{\frac{1}{p}}
\]
提取因式 $x_j$
\begin{aligned}
&\|x\|_{\infty}=\lim _{p \to \infty}\left(x_j^p\left[\left|\frac{x_1}{x_j}\right|^p+\ldots+1+\ldots+\left|\frac{x_n}{x_j}\right|^p\right]\right)^{\frac{1}{p}} \\
&\|x\|_{\infty}=x_j\left(\lim _{p \to \infty}\left[\left|\frac{x_1}{x_j}\right|^p+\ldots+1+\ldots+\left|\frac{x_n}{x_j}\right|^p\right]^{\frac{1}{p}}\right)
\end{aligned}剩下的工作是处理 ( ) 项。
Part 2. 证明极限是 $0^0$ 不定式
Since $x_j$ is the largest element in the vector, thus$$\frac{x_{i}}{x_{j}}<1$$
And thus
\[
\left|\frac{x_i}{x_j}\right|^p \longrightarrow 0 \quad \text { for } p \rightarrow \infty
\]
Also
\[
\left[\left|\frac{x_1}{x_j}\right|^p+\ldots+1+\ldots+\left|\frac{x_n}{x_j}\right|^p\right]^{\frac{1}{p}} \longrightarrow 0^0 \quad \text { for } p \longrightarrow \infty
\]
Which is an indetermined form.

hbghlyj

Also
\[
\left[\left|\frac{x_1}{x_j}\right|^p+\ldots+\color{#f00}1+\ldots+\left|\frac{x_n}{x_j}\right|^p\right]^{\frac{1}{p}} \longrightarrow 0^0 \quad \text { for } p \longrightarrow \infty
\]
Which is an indetermined form.

我对这步有疑问: [ ]里面有1, 怎么说是$0^0$呢

hbghlyj

Part 3. 用L'Hospital法则计算极限(The following is a old trick in HK A-Level pure2 mathematics to tackle $0^0$)
To handle this indetermined form, consider
\[
z=\frac{1}{y}^{\frac{1}{y}}
\]
Thus
\[
\lim _{y \to \infty} z=0^0
\]
In stead of evaluate the limt directly, consider the limit of $\ln z$, first, take $\ln$
\[
\ln z=\frac{1}{y} \ln \frac{1}{y}=\frac{-1}{y} \ln y
\]
Thus
\[
\lim _{y \to \infty} \ln z=-\lim _{y \to \infty} \frac{\ln y}{y} \quad \text { (in the form } \frac{\infty}{\infty}, \text { still indetermined form) }
\]
But now the indetermined form is in the form $\frac{f(y)}{g(y)}$, we can now apply L'Hospital Rule
\[
-\lim _{y \to \infty} \frac{\ln y}{y}=-\lim _{y \to \infty} \frac{1 / y}{1}=0
\]
Thus
\[
\lim _{y \to \infty} \ln z=0
\]
And therefore
\[
\lim _{y \to \infty} z=e^{\lim _{y \to \infty} \ln z}=1
\]
Thus
\[
\lim _{y \rightarrow \infty} \frac{1}{y}^{\frac{1}{y}}=1
\]
Part 4. 完成证明
And thus
\[
\lim _{p \to \infty}\left[\left|\frac{x_1}{x_j}\right|^p+\ldots+1+\ldots+\left|\frac{x_n}{x_j}\right|^p\right]^{\frac{1}{p}}=1
\]
And therefore
\begin{gathered}
\|x\|_{\infty}=\lim _{p \to \infty}\left(\sum_{i=1}^n\left|x_i\right|^p\right)^{\frac{1}{p}}=x_j\left(\lim _{p \to \infty}\left[\left|\frac{x_1}{x_j}\right|^p+\ldots+1+\ldots+\left|\frac{x_n}{x_j}\right|^p\right]^{\frac{1}{p}}\right) \\
\|x\|_{\infty}=x_j \cdot 1=\max _{1 \leq j \leq n}\left\{x_1, x_2, \ldots, x_n\right\}
\end{gathered}

hbghlyj

\[
\|x\|_{\infty} \leq\|x\|_2 \leq \sqrt{n}\|x\|_{\infty}
\]
Pf.
The left hand side is easy
\[
\|x\|_{\infty}=\max _{1 \leq j \leq n}\left\{x_1, x_2, \ldots, x_n\right\}=\sqrt{\left(\max _{1 \leq j \leq n}\left\{x_1, x_2, \ldots, x_n\right\}\right)^2} \leq \sqrt{\sum_{i=1}^n\left|x_i\right|^2}
\]
For the right hand side
\[
n\|x\|_{\infty}^2=n\left(\max _{1 \leq j \leq n}\left\{x_1, x_2, \ldots, x_n\right\}\right)^2=\sum_{i=1}^n\left(\max _{1 \leq j \leq n}\left\{x_1, x_2, \ldots, x_n\right\}\right)^2 \geq \sum_{i=1}^n\left|x_i\right|^2=\|x\|_2^2 \geq 0
\]
Thus, taking the square-root
\[
\sqrt{n}\|x\|_{\infty} \geq\|x\|_2
\]

hbghlyj

hbghlyj 发表于 2022-10-15 10:33
Part 3. 用L'Hospital法则计算极限(The following is a old trick in HK A-Level pure2 mathematics to tac ...

Part 4似乎没有用到Part 3证明的$\displaystyle\lim _{y \rightarrow \infty} \frac{1}{y}^{\frac{1}{y}}=1$啊? 莫名其妙

hbghlyj

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hbghlyj

Wilson A. Sutherland - Introduction to Metric and Topological Spaces-Oxford University Press (2009)
Exercise 4.11. Given a set of $r$ non-negative real numbers $\left\{a_1, a_2, \ldots, a_r\right\}, a=\max \left\{a_1, a_2, \ldots, a_r\right\}$. For any positive integer $n$,
\[
a^n \leqslant a_1^n+a_2^n+\ldots+a_r^n \leqslant r a^n .
\]
By taking $n$th roots throughout,
\[
a \leqslant\left(a_1^n+a_2^n+\ldots+a_r^n\right)^{1 / n} \leqslant r^{1 / n} a,
\]
$\lim _{n \rightarrow \infty}r^{1/n}=1⇒\lim _{n \rightarrow \infty}\left(a_1^n+a_2^n+\ldots+a_r^n\right)^{1 / n}=a$.

hbghlyj

suhogrozdje的回答
Lemma: Let $\lVert x\rVert_p:=(|x_1|^p+\dots+|x_n|^p)^{\frac 1 p}$ be the $p$-norm on $\mathbb R^n$. For all $x\in\mathbb R^n$, we have
$$
\lim_{p\rightarrow\infty}\lVert x\rVert_p=\max_{i\in\mathbb N_n} |x_i|^p=:\lVert x \rVert_\infty
$$
Proof.
$$\lVert x \rVert_{\lower.7ex\infty}^p=\big(\max_{i\in\mathbb N_n}|x_i|\big)^p=\max_{i\in\mathbb N_n}|x_i|^p\leq\underbrace{|x_1|^p}_{\leq \lVert x \rVert_{\lower.7ex\infty}^p}+\dots+\underbrace{|x_n|^p}_{\leq \lVert x \rVert_{\lower.7ex\infty}^p}=(\lVert x\rVert_p)^p\leq n\lVert x \rVert_{\lower.7ex\infty}^p,
$$
and if we take the $p$-th root of this expression, we get
\begin{align*}
\lVert x \rVert_\infty \leq \lVert x \rVert_p \leq n^{\frac 1 p}\lVert x \rVert_\infty,
\end{align*}
Taking the limit as $p\rightarrow \infty$ of the last equation, we get the desired result.