重构

杂谈 | 共 454 字 | 2021/8/20 发表 | 2021/8/20 更新

稍微重构了一下网站，本来想写点东西记录一下，懒得写了。

部署

git config --global core.quotePath false && hugo --minify

音乐

{name: "%title%",artist: "%artist%",url: "https://yy.halu.lu/%date%-%album%/%track%-%title%-%artist%.flac",cover: "/头像/头像.png"},

LaTeX\LaTeX

数学

a

Since exp⁡(x)=ex\exp(x) = e^x is a convex function, by Jensen's inequality from MATH 147, we have for t∈[0,1]t \in [0,1]

exp⁡(ta+(1−t)b)≤texp⁡(a)+(1−t)exp⁡(b)\exp(ta+(1-t)b) \le t\exp(a) + (1-t)\exp(b)

Then since 1p+1q=1\frac1p + \frac1q = 1,

ab=exp⁡(log⁡a+log⁡b)=exp⁡(1pplog⁡a+1qqlog⁡b)≤1pexp⁡(plog⁡a)+1qexp⁡(qlog⁡b)=app+bqq\begin{align*} ab &= \exp(\log a + \log b) \\\\ &= \exp(\frac1pp\log a + \frac1qq\log b) \\\\ &\le \frac1p\exp(p\log a) + \frac1q\exp(q\log b) \\\\ &= \frac{a^p}p + \frac{b^q}q \end{align*}

b

The inequality holds when x=0x = 0 or y=0y = 0, so we assume x≠0x \neq 0 and y≠0y \neq 0, then we have

∣∣xy∣∣1∣∣x∣∣p∣∣y∣∣q=∑n=1∞∣xnyn∣∣∣x∣∣p∣∣y∣∣q=∑n=1∞∣xn∣∣∣x∣∣p∣yn∣∣∣y∣∣q≤by(a)∑n=1∞∣xn∣pp(∣∣x∣∣p)p+∑n=1∞∣yn∣qq(∣∣y∣∣q)q=∑n=1∞∣xn∣pp(∣∣x∣∣p)p+∑n=1∞∣yn∣qq(∣∣y∣∣q)q=(∣∣x∣∣p)pp(∣∣x∣∣p)p+(∣∣y∣∣q)qq(∣∣y∣∣q)q=1p+1q=1\begin{align*} \frac{||xy||_1}{||x||_p ||y||_q} &= \sum_{n=1}^\infty \frac{|x_ny_n|}{||x||_p ||y||_q} \\\\ &= \sum_{n=1}^\infty \frac{|x_n|}{||x||_p}\frac{|y_n|}{||y||_q} \\\\ &\stackrel{by(a)}\le \sum_{n=1}^\infty \frac{|x_n|^p}{p(||x||_p)^p} + \sum_{n=1}^\infty \frac{|y_n|^q}{q(||y||_q)^q} \\\\ &= \frac{\sum_{n=1}^\infty|x_n|^p}{p(||x||_p)^p} + \frac{\sum_{n=1}^\infty|y_n|^q}{q(||y||_q)^q} \\\\ &= \frac{(||x||_p)^p}{p(||x||_p)^p} + \frac{(||y||_q)^q}{q(||y||_q)^q} \\\\ &= \frac1p + \frac1q\\\\ &= 1 \end{align*}

Multiplying both side by ∣∣x∣∣p∣∣y∣∣q||x||_p ||y||_q, we get ∣∣xy∣∣1≤∣∣x∣∣p∣∣y∣∣q||xy||_1 \le ||x||_p ||y||_q.

c

Let (x+y)p−1=((x1+y1)p−1,(x2+y2)p−1,...)(x+y)^{p-1} = ((x_1+y_1)^{p-1},(x_2+y_2)^{p-1},...), then

∑n=1∞∣xn+yn∣p≤∑n=1∞∣xn∣∣xn+yn∣p−1+∑n=1∞∣yn∣∣xn+yn∣p−1=∣∣x(x+y)p−1∣∣1+∣∣y(x+y)p−1∣∣1≤by(b)∣∣x∣∣p∣∣(x+y)p−1∣∣pp−1+∣∣y∣∣p∣∣(x+y)p−1∣∣pp−1=(∣∣x∣∣p+∣∣y∣∣p)∣∣(x+y)p−1∣∣pp−1=(∣∣x∣∣p+∣∣y∣∣p)(∑n=1∞(∣xn+yn∣p−1)pp−1)p−1p=(∣∣x∣∣p+∣∣y∣∣p)(∑n=1∞∣xn+yn∣p)p−1p\begin{align*} \sum_{n=1}^\infty |x_n+y_n|^p &\le \sum_{n=1}^\infty |x_n||x_n+y_n|^{p-1} + \sum_{n=1}^\infty |y_n||x_n+y_n|^{p-1} \\\\ &= ||x(x+y)^{p-1}||_1 + ||y(x+y)^{p-1}||_1 \\\\ &\stackrel{by (b)}\le ||x||_p||(x+y)^{p-1}||_{\frac{p}{p-1}} + ||y||_p||(x+y)^{p-1}||_{\frac{p}{p-1}} \\\\ &= (||x||_p+||y||_p)||(x+y)^{p-1}||_{\frac{p}{p-1}} \\\\ &= (||x||_p+||y||_p) (\sum_{n=1}^\infty (|x_n+y_n|^{p-1})^{\frac{p}{p-1}})^\frac{p-1}p \\\\ &= (||x||_p+||y||_p) (\sum_{n=1}^\infty |x_n+y_n|^{p})^\frac{p-1}p \end{align*}

Multiplying both side by (∑n=1∞∣xn+yn∣p)1−pp(\sum_{n=1}^\infty |x_n+y_n|^{p})^\frac{1-p}p, we get

∣∣x+y∣∣p=(∑n=1∞∣xn+yn∣p)1p=(∑n=1∞∣xn+yn∣p)1+1−pp≤∣∣x∣∣p+∣∣y∣∣p||x+y||_p=(\sum_{n=1}^\infty |x_n+y_n|^{p})^\frac{1}p = (\sum_{n=1}^\infty |x_n+y_n|^{p})^{1+\frac{1-p}p} \le ||x||_p+||y||_p