de Bruijn 的Cauchy型不等式

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(c) [de Bruijn] 设 $\underline w$ 为实数 $n$-元组, $\underline z$ 为复数 $n$-元组, 则$$\left|\sum_{k=1}^{n} w_{k} z_{k}\right| \leq \frac{1}{\sqrt{2}}\left(\sum_{k=1}^{n} w_{k}^{2}\right)^{1 / 2}\left(\sum_{k=1}^{n}\left|z_{k}\right|^{2}+\left|\sum_{k=1}^{n} z_{k}^{2}\right|\right)^{1 / 2}$$取等当且仅当存在 $λ\in\Bbb C$, 使 $w_k = ℜ(λz_k), 1 ≤ k ≤ n$, 且 $∑^n_{k=1} λ^2z^2_k$是非负实数.

原帖子

Peter S. Bullen - Dictionary of Inequalities-Chapman & Hall_CRC (2015) page 54
(a) If $\underline z$ is complex $n$-tuple$$\left|\sum_{k=1}^{n} z_{k}\right| \leq \sum_{k=1}^n\left|z_{k}\right|$$with equality if and only if for all non-zero terms $z_j/z_k>0$
(b) For all $z_1,z_2,z_3\in\Bbb C$, $1 ≤ |1 + z_1| + |z_1 + z_2| + |z_2 + z_3| + |z_3|$
(c) [de Bruijn] If $\underline w$ is a real $n$-tuple and $\underline z$ a complex $n$-tuple then$$\left|\sum_{k=1}^{n} w_{k} z_{k}\right| \leq \frac{1}{\sqrt{2}}\left(\sum_{k=1}^{n} w_{k}^{2}\right)^{1 / 2}\left(\sum_{k=1}^{n}\left|z_{k}\right|^{2}+\left|\sum_{k=1}^{n} z_{k}^{2}\right|\right)^{1 / 2}$$with equality if and only if for some complex $λ$, $w_k = ℜλz_k, 1 ≤ k ≤ n$, and $∑^n_{k=1} λ^2z^2_k ≥ 0$.
(d) If $z, w ∈\Bbb C, r ≥ 0$ then$$|z+w|^{r} \leq\left\{\begin{array}{ll}|z|^{r}+|w|^{r}, & \text { if } r \leq 1 \\ 2^{r-1}\left(|z|^{r}+|w|^{r}\right), & \text { if } r>1\end{array}\right.$$
对$\underline z=(-1,z_1,-z_2,z_3)$应用(a)就证明了(b).
如何证明(c)呢?
(d)见Minkowski inequality
当$0\le r\le1$时 见范数不等式
由$\left(\abs{z}\over \abs{z}+\abs{w}\right)^r+\left(\abs{w}\over \abs{z}+\abs{w}\right)^r\ge{\abs{z}\over \abs{z}+\abs{w}}+{\abs{w}\over \abs{z}+\abs{w}}=1$与三角不等式得$\abs{z}^r+\abs{w}^r\ge(\abs{z}+\abs{w})^r\ge\abs{z+w}^r$
当$r>1$时
由三角不等式$\abs{z+w\over2}^r\le\left(\abs z+\abs w\over2\right)^r$
对convex function $f(x)=x^r,x>0,r>1$使用Jensen不等式,$\left(\abs z+\abs w\over2\right)^r\le\frac{\abs{z}^r+\abs{w}^r}2$
所以$\abs{z+w\over2}^r\le\frac{\abs{z}^r+\abs{w}^r}2$
所以$|z+w|^r\le2^{r-1}(|z|^r+|w|^r)$

色k

建议把 (c) 单独抽出来并翻译成中文，(abd) 都显然，不值一提，统统去掉，这样才容易引起大家的兴趣。

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相关: AOPS–An inequality with complex numbers
Some Applications of De Bruijn's Inequality for Power Series Lemma 1 式(2.2)
On some inequalities of Cauchy-Bunyakovsky-Schwarz type and applications
Cauchy–Bunyakovsky–Schwarz type inequalities related to Möbius operations
Reverses of the Cauchy-Bunyakovsky-Schwarz Inequality for n-tuples of Complex Numbers


在Further Inequalities for Sequences and Power Series of Operators in Hilbert Spaces Via Hermitian Forms的式(33) :

Remark 3.2. We notice that (32) is an operator version of de Bruijn inequality obtained in 1960 in [1], which provides the following refinement of the Cauchy-Bunyakovsky-Schwarz inequality...

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Dragoslav S. Mitrinovic, J. Pecaric, A.M Fink - Classical and New Inequalities in Analysis-Kluwer Academic (1993) page 89
Theorem 2. If $a_1, \ldots, a_n$ are real and $z_1, \ldots, z_n$ complex numbers, then
\[\tag{2.8}
\left|\sum_{k=1}^n a_k z_k\right|^2 \leq \frac{1}{2}\left(\sum_{k=1}^n\left|z_k\right|^2+\left|\sum_{k=1}^n z_k^2\right|\right)\left(\sum_{k=1}^n a_k^2\right) .
\]
Equality holds if and only if, for $k=1, \ldots, n, a_k=\operatorname{Re}\left(\lambda z_k\right)$, where $\lambda$ is a complex number and $\sum_{k=1}^n \lambda^2 z_k^2$ is real and nonnegative.

Proof. By a simultaneous rotation of all the $z_k$ 's about the origin, we get $\sum_{k=1}^n a_k z_k \geq 0$. This rotation does not affect the moduli
\[
\left|\sum_{k=1}^n a_k z_k\right|, \quad\left|\sum_{k=1}^n z_k^2\right|, \quad\left|z_k\right| \quad(k=1, \ldots, n) .
\]
Hence, it is sufficient to prove inequality (2.8) for the case $\sum_{k=1}^n a_k z_k \geq 0$.
If we put $z_k=x_k+i y_k(k=1, \ldots, n)$, then
\[
\left|\sum_{k=1}^n a_k z_k\right|^2=\left(\sum_{k=1}^n a_k x_k\right)^2 \leq\left(\sum_{k=1}^n a_k^2\right)\left(\sum_{k=1}^n x_k^2\right),
\]
where we make use of Cauchy's inequality for real numbers. Since
\[
2 x_k^2=\left|z_k\right|^2+\operatorname{Re} z_k^2,
\]
we obtain
\[
\left|\sum_{k=1}^n a_k z_k\right|^2 \leq \frac{1}{2}\left(\sum_{k=1}^n a_k^2\right)\left(\sum_{k=1}^n\left|z_k\right|^2+\sum_{k=1}^n \operatorname{Re} z_k^2\right) .
\]
From this inequality we get (2.8), taking into account that
\[
\sum_{k=1}^n \operatorname{Re} z_k^2=\operatorname{Re} \sum_{k=1}^n z_k^2 \leq\left|\sum_{k=1}^n z_k^2\right| \text {. }
\]