重构

杂谈 | 共 249 字 | 20210820 发表 | 20210902 更新

施工中,请稍候

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数学

a

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

$$ \exp(ta+(1-t)b) \le t\exp(a) + (1-t)\exp(b) $$

Then since $\frac1p + \frac1q = 1$,

$$ \begin{align*} ab &= \exp(\log a + \log b) \\
&= \exp(\frac1pp\log a + \frac1qq\log b) \\
&\le \frac1p\exp(p\log a) + \frac1q\exp(q\log b) \\
&= \frac{a^p}p + \frac{b^q}q \end{align*} $$

b

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

$$ \begin{align*} \frac{||xy||_1}{||x||_p ||y||_q} &= \sum_{n=1}^\infty \frac{|x_ny_n|}{||x||_p ||y||_q} \\
&= \sum_{n=1}^\infty \frac{|x_n|}{||x||_p}\frac{|y_n|}{||y||_q} \\
&\stackrel{by(a)}\le \sum_{n=1}^\infty \frac{|x_n|^p}{p(||x||_p)^p} + \sum_{n=1}^\infty \frac{|y_n|^q}{q(||y||_q)^q} \\
&= \frac{\sum_{n=1}^\infty|x_n|^p}{p(||x||_p)^p} + \frac{\sum_{n=1}^\infty|y_n|^q}{q(||y||_q)^q} \\
&= \frac{(||x||_p)^p}{p(||x||_p)^p} + \frac{(||y||_q)^q}{q(||y||_q)^q} \\
&= \frac1p + \frac1q\\
&= 1 \end{align*} $$

Multiplying both side by $||x||_p ||y||_q$, we get $||xy||_1 \le ||x||_p ||y||_q$.

c

Let $(x+y)^{p-1} = ((x_1+y_1)^{p-1},(x_2+y_2)^{p-1},…)$, then

$$ \begin{align*} \sum_{n=1}^\infty |x_n+y_n|^p &\le \sum_{n=1}^\infty |x_n||x_n+y_n|^{p-1} + \sum_{n=1}^\infty |y_n||x_n+y_n|^{p-1} \\
&= ||x(x+y)^{p-1}||_1 + ||y(x+y)^{p-1}||_1 \\
&\stackrel{by (b)}\le ||x||_p||(x+y)^{p-1}||_{\frac{p}{p-1}} + ||y||_p||(x+y)^{p-1}||_{\frac{p}{p-1}} \\
&= (||x||_p+||y||_p)||(x+y)^{p-1}||_{\frac{p}{p-1}} \\
&= (||x||_p+||y||_p) (\sum_{n=1}^\infty (|x_n+y_n|^{p-1})^{\frac{p}{p-1}})^\frac{p-1}p \\
&= (||x||_p+||y||_p) (\sum_{n=1}^\infty |x_n+y_n|^{p})^\frac{p-1}p \end{align*} $$

Multiplying both side by $(\sum_{n=1}^\infty |x_n+y_n|^{p})^\frac{1-p}p$, we get

$$ ||x+y||p=(\sum{n=1}^\infty |x_n+y_n|^{p})^\frac{1}p = (\sum_{n=1}^\infty |x_n+y_n|^{p})^{1+\frac{1-p}p} \le ||x||_p+||y||_p $$

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