# Quod Erat Demonstrandum

## 2009/02/01

### 卡爾松（Carleson）不等式

Filed under: Pure Mathematics — johnmayhk @ 9:25 下午
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Carleson first inequality

Let $a_1, a_2, \dots , a_n \in \mathbb{R}$, then

$(a_1 + a_2 + \dots + a_n)^2 < \frac{\pi^2}{6}(a_1^2 + 2^2a_2^2 + \dots + n^2a_n^2)$

Carleson second inequality

Let $a_1, a_2, \dots , a_n \in \mathbb{R}$, then

$(a_1 + a_2 + \dots + a_n)^4 < \pi^2(a_1^2 + a_2^2 + \dots + a_n^2)(a_1^2 + 2^2a_2^2 + \dots + n^2a_n^2)$

Carleson first inequality

$(a_1 + a_2 + \dots + a_n)^2$
$= (a_1c_1\times \frac{1}{c_1} + a_2c_2\times \frac{1}{c_2} + \dots + a_nc_n\times \frac{1}{c_n})^2$
$\le (a_1^2c_1^2 + a_2^2c_2^2 + \dots + a_n^2c_n^2)(\frac{1}{c_1^2} + \frac{1}{c_2^2} + \dots + \frac{1}{c_n^2})$

$b_n = \frac{1}{c_1^2} + \frac{1}{c_2^2} + \dots + \frac{1}{c_n^2}$，得

$(a_1 + a_2 + \dots + a_n)^2 \le b_n(a_1^2c_1^2 + a_2^2c_2^2 + \dots + a_n^2c_n^2)$ – – – – – – (*)

$c_n = n$，有

$(a_1 + a_2 + \dots + a_n)^2 \le b_n(a_1^2 + 2^2a_2^2 + \dots + n^2a_n^2)$

$b_n \rightarrow \frac{\pi^2}{6}$ as $n \rightarrow \infty$，知 $b_n < \frac{\pi^2}{6}$，即卡爾松第一不等式成立。

Carleson second inequality

$a_1^2c_1^2 + a_2^2c_2^2 + \dots + a_n^2c_n^2 = tp + \frac{q}{t}$

(*) 變為

$(a_1^2 + a_2^2 + \dots + a_n^2)^2 \le b_n(tp + \frac{q}{t})$

$b_n = (t + \frac{1^2}{t})^{-1} + (t + \frac{2^2}{t})^{-1} + \dots + (t + \frac{n^2}{t})^{-1} = \frac{t}{t^2 + 1^2} + \frac{t}{t^2 + 2^2} + \dots + \frac{t}{t^2 + n^2}$

$\frac{t}{2}$
$= \frac{1}{2}|OM_{k-1}||OM_{k}|\sin\alpha_k$
$= \frac{1}{2}\sqrt{t^2 + (k - 1)^2}\sqrt{t^2 + k^2}\sin\alpha_k$

$\sin\alpha_k = \frac{t}{\sqrt{t^2 + (k - 1)^2}\sqrt{t^2 + k^2}} > \frac{t}{t^2 + k^2}$

$b_n = \frac{t}{t^2 + 1^2} + \frac{t}{t^2 + 2^2} + \dots + \frac{t}{t^2 + n^2} < \alpha_1 + \alpha_2 + \dots + \alpha_n < \frac{\pi}{2}$，故

$(a_1 + a_2 + \dots + a_n)^2 < \frac{\pi}{2}(tp + \frac{q}{t})$

$t = \sqrt{\frac{q}{p}}$，得 $tp + \frac{q}{t} = 2\sqrt{pq}$，即

$(a_1 + a_2 + \dots + a_n)^2 < \pi\sqrt{pq}$，亦即卡爾松第二不等式。

Let $a_n \ge 0$, then $(\sum a_n)^2 \le \pi[(\sum a_n^2\sum(n - 0.5)^2a_n^2)^2 - 0.25B_3^2]^{0.25}$