Minkowski's Inequality for Sums/Index Greater than 1

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be non-negative real numbers.

Let $p \in \R$ be a real number such that $p > 1$.


Then:

$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / p} \le \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p}$


Proof 1

Define:

$q = \dfrac p {p - 1}$

Then:

$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$


It follows that:

\(\ds \sum_{k \mathop = 1}^n \paren {a_k + b_k}^p\) \(=\) \(\ds \sum_{k \mathop = 1}^n a_k \paren {a_k + b_k}^{p - 1} + \sum_{k \mathop = 1}^n b_k \paren {a_k + b_k}^{p - 1}\)
\(\ds \) \(\le\) \(\ds \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \paren {\paren {a_k + b_k}^{p - 1} }^q}^{1 / q} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \paren {\paren {a_k + b_k}^{p - 1} }^q}^{1 / q}\) Hölder's Inequality for Sums (twice)
\(\ds \) \(=\) \(\ds \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q}\) Power of Power, and by hypothesis: $\paren {p - 1} q = p$
\(\ds \) \(=\) \(\ds \paren {\paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p} } \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q}\)
\(\ds \leadsto \ \ \) \(\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 - 1 / q}\) \(\le\) \(\ds \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p}\) dividing by $\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q}$
\(\ds \leadsto \ \ \) \(\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / p}\) \(\le\) \(\ds \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p}\) as $1 - \dfrac 1 q = p$


Proof 2

Let $\mathbf a$ and $\mathbf b$ be real finite sequences:

\(\ds \mathbf a\) \(=\) \(\ds \sequence {a_k}_{1 \mathop \le k \mathop \le n}\)
\(\ds \mathbf b\) \(=\) \(\ds \sequence {b_k}_{1 \mathop \le k \mathop \le n}\)


Let $\norm {\mathbf a}_p$ denote the $p$-norm of $\mathbf a$:

$\norm {\mathbf a}_p := \paren {\ds \sum_{k \mathop = 1}^n {a_k}^p}^{1 / p}$


Define:

$q = \dfrac p {p - 1}$

Then:

$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$


Then:

\(\ds \paren {\norm {\mathbf a + \mathbf b}_p}^p\) \(=\) \(\ds \norm {\paren {\mathbf a + \mathbf b}^p}_1\)
\(\ds \) \(=\) \(\ds \norm {\mathbf a \paren {\mathbf a + \mathbf b}^{p - 1} + \mathbf b \paren {\mathbf a + \mathbf b}^{p - 1} }_1\)
\(\ds \) \(=\) \(\ds \norm {\mathbf a \paren {\mathbf a + \mathbf b}^{p - 1} }_1 + \norm {\mathbf b \paren {\mathbf a + \mathbf b}^{p - 1} }_1\)
\(\ds \) \(\le\) \(\ds \norm {\mathbf a}_p \norm {\paren {\mathbf a + \mathbf b}^{p - 1} }_q + \norm {\mathbf b}_p \norm {\paren {\mathbf a + \mathbf b}^{p - 1} }_q\) Hölder's Inequality for Sums (twice)
\(\ds \) \(=\) \(\ds \paren {\norm {\mathbf a}_p + \norm {\mathbf b}_p} \norm {\paren {\mathbf a + \mathbf b}^{p - 1} }_q\) simplifying
\(\ds \) \(=\) \(\ds \paren {\norm {\mathbf a}_p + \norm {\mathbf b}_p} \norm {\paren {\mathbf a + \mathbf b}^{p - 1} }_{p / \paren {p - 1} }\) by hypothesis: $\dfrac p {p - 1} = q$
\(\ds \) \(=\) \(\ds \paren {\norm {\mathbf a}_p + \norm {\mathbf b}_p} \paren {\norm {\mathbf a + \mathbf b}_p}^{p - 1}\) Transformation of P-Norm
\(\ds \leadsto \ \ \) \(\ds \norm {\mathbf a + \mathbf b}_p\) \(\le\) \(\ds \norm {\mathbf a}_p + \norm {\mathbf b}_p\) dividing both sides by $\paren {\norm {\mathbf a + \mathbf b}_p}^{p - 1}$

$\blacksquare$


Source of Name

This entry was named for Hermann Minkowski.