INEQUALITY'S BLOG

December 11, 2010

A GENERALIZATION OF HOLDER’S INEQUALITY

Filed under: Holder's inequality — KKKVVV @ 3:59 pm

A GENERALIZATION OF HOLDER’S INEQUALITY

BY VAN KHEA

CAMBODIAN BUILDING ENGINEER

December 12 , 2010

Email: van_khea@yahoo.com

Let $\displaystyle p, q>0 ; \frac{1}{p}+\frac{1}{q}\leq 1$ and let $\displaystyle a_i\geq 0 ; \sum_{i=1}^{n}\alpha_i=1$ then $\displaystyle \sum_{i=1}^{n}\alpha_i |a_ib_i|\leq \biggl(\sum_{i=1}^{n}\alpha_i |a_i|^p\biggl)^{\frac{1}{p}}\biggl(\sum_{i=1}^{n}\alpha_i |b_i|^q\biggl)^{\frac{1}{q}} ; \forall{(a_1, a_2, ..., a_n) ; (b_1, b_2, ..., b_n)\in \mathbb{R}^n}$ or $\mathbb{C}^n$

Proof

Theorem: (VAN KHEA): Let $a_i\geq 0$ and $x_i, y_i, z_i>0 ; (i=1, 2, ..., n)$ and let $k\geq 1 ; \alpha , \beta , \gamma\geq 0$ such that $k-\alpha-\beta-\gamma=1$ then $\displaystyle \frac{a_1^k}{x_1^{\alpha}y_1^{\beta}z_1^{\gamma}}+...+\frac{a_n^k}{x_n^{\alpha}y_n^{\beta}z_n^{\gamma}}\geq \frac{(a_1+...+a_n)^k}{(x_1+...+x_n)^{\alpha}(y_1+...+y_n)^{\beta}(z_1+...+z_n)^{\gamma}}$

From above theorem  we have $\displaystyle \sum_{i=1}^{n}\alpha_i |a_ib_i|=\sum_{i=1}^{n}\frac{(\alpha_i |a_ib_i|)^2}{\alpha_i^{1-\frac{1}{p}-\frac{1}{q}}(\alpha_i|a_i|^p)^{\frac{1}{p}}(\alpha_i|b_i|^q)^{\frac{1}{q}}}$ $\displaystyle \geq \frac{(\sum_{i=1}^{n}\alpha_i |a_ib_i|)^2}{(\sum_{i=1}^{n}\alpha_i)^{1-\frac{1}{p}-\frac{1}{q}}(\sum_{i=1}^{n}\alpha_i |a_i|^p)^{\frac{1}{p}}(\sum_{i=1}^{n}\alpha_i |b_i|^q)^{\frac{1}{q}}}$ $\displaystyle \Leftrightarrow 1\geq \frac{\sum_{i=1}^{n}\alpha_i |a_ib_i|}{(\sum_{i=1}^{n}\alpha_i |a_i|^p)^{\frac{1}{p}}(\sum_{i=1}^{n}\alpha_i |b_i|^q)^{\frac{1}{q}}}$ $\Rightarrow$ $\displaystyle \sum_{i=1}^{n}\alpha_i |a_ib_i|\leq \biggl(\sum_{i=1}^{n}\alpha_i |a_i|^p\biggl)^{\frac{1}{p}}\biggl(\sum_{i=1}^{n}\alpha_i |b_i|^q\biggl)^{\frac{1}{q}} ; \forall{(a_1, a_2, ..., a_n) ; (b_1, b_2, ..., b_n)\in \mathbb{R}^n}$ or $\mathbb{C}^n$

Special cases

• If $\displaystyle \frac{1}{p}+\frac{1}{q}\leq 1$ and $\displaystyle \alpha_1=\alpha_2=...=\alpha_n=\frac{1}{n}$ then: $\displaystyle \frac{1}{n}\sum_{i=1}^{n}|a_ib_i|\leq \biggl(\frac{1}{n}\sum_{i=1}^{n}|a_i|^p\biggl)^{\frac{1}{p}}\biggl(\frac{1}{n}\sum_{i=1}^{n}|b_i|^q\biggl)^{\frac{1}{q}}$

• If $\displaystyle \frac{1}{p}+\frac{1}{q}=1$ and $\displaystyle \alpha_1=\alpha_2=...=\alpha_n$ then: $\displaystyle \sum_{i=1}^{n}|a_ib_i|\leq \biggl(\sum_{i=1}^{n}|a_i|^p\biggl)^{\frac{1}{p}}\biggl(\sum_{i=1}^{n}|b_i|^q\biggl)^{\frac{1}{q}}$