User:Addem/Holder/Summation

Theorem
Let $p, q \in \R_{>0}$ be strictly positive real numbers such that:
 * $\dfrac 1 p + \dfrac 1 q = 1$

Let:
 * $\mathbf x = \sequence {x_n} \in \ell^p$
 * $\mathbf y = \sequence {y_n} \in \ell^q$

where $\ell^p$ denotes the $p$-sequence space.

Let $\norm {\mathbf x}_p$ denote the $p$-norm of $\mathbf x$.

Then $\mathbf x \mathbf y = \sequence {x_n y_n} \in \ell^1$, and:
 * $\norm {\mathbf x \mathbf y}_1 \le \norm {\mathbf x}_p \norm {\mathbf y}_q$

Proof
, assume that $\mathbf x$ and $\mathbf y$ are non-zero.

Define:
 * $\mathbf u = \sequence {u_n} = \dfrac {\mathbf x} {\norm {\mathbf x}_p}$
 * $\mathbf v = \sequence {v_n} = \dfrac {\mathbf y} {\norm {\mathbf y}_q}$

Then:
 * $\ds \norm {\mathbf u}_p = \dfrac 1 {\norm {\mathbf x}_p} \paren {\sum_{n \mathop = 0}^\infty \size {x_n}^p}^{1/p} = 1$

Similarly:
 * $\norm {\mathbf v}_q = 1$

By Young's Inequality for Products:
 * $(1): \quad \forall n \in \N: \size {u_n v_n} \le \dfrac 1 p \size {u_n}^p + \dfrac 1 q \size {v_n}^q$

By the comparison test, it follows that:
 * $\mathbf u \mathbf v = \sequence {u_n v_n} \in \ell^1$
 * $\mathbf x \mathbf y = \norm {\mathbf x}_p \norm {\mathbf y}_q \mathbf u \mathbf v \in \ell^1$

From $(1)$, it follows that:
 * $\norm {\mathbf u \mathbf v}_1 \le \dfrac 1 p \norm {\mathbf u}_p + \dfrac 1 q \norm {\mathbf v}_q = 1$

Therefore:
 * $\norm {\mathbf x \mathbf y}_1 = \norm {\mathbf x}_p \norm {\mathbf y}_q \norm {\mathbf u \mathbf v}_1 \le \norm {\mathbf x}_p \norm {\mathbf y}_q$

as desired.

Also see

 * Minkowski's Inequality for Sums