Quod Erat Demonstrandum

2009/02/01

卡爾松(Carleson)不等式

Filed under: Pure Mathematics — johnmayhk @ 9:25 下午
Tags: , , ,

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)

沒有額外添加的人工化前提,得出不平凡的結果。

證明

Carleson first inequality

由 Cauchy-Schwarz,得

(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)

其中,b_n = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{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

在 (*) 中取 c_n^2 = t + \frac{n^2}{t},得

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

其中 p = a_1^2 + a_2^2 + \dots + a_n^2q = a_1^2 + 2^2a_2^2 + \dots + n^2a_n^2

(*) 變為

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

其中

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}

現在計算 b_n 的一個上限,考慮下面直角三角形,

carleson-inequality

考慮某個 \Delta OM_{k-1}M_{k} 的面積,有

\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}

又因 \sin\alpha_k < \alpha_k,得

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。

不少有關卡爾松不等式的推廣,諸如

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}

其中 B_3 = \sum a_n^2\sum (n^2 - 0.25)a_na_{n+1} - \sum(n - 0.25)^2a_n^2\sum a_na_{n+1}

注:卡爾松,瑞典數學家,生於 1888-07-23

參考書目:數學奧林匹克與數學文化(劉培傑 主編)P.363~364 (書中有些 typo…)

Also read

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19087

利用平行四邊形證明 Cauchy-Schwarz 不等式

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