Hölder's Inequality for Sums/Formulation 1

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

Let $\GF \in \set {\R, \C}$, that is, $\GF$ represents the set of either the real numbers or the complex numbers.

Let $\mathbf x$ and $\mathbf y$ denote the vectors consisting of the sequences:
 * $\mathbf x = \sequence {x_n} \in {\ell^p}_\GF$
 * $\mathbf y = \sequence {y_n} \in {\ell^q}_\GF$

where ${\ell^p}_\GF$ denotes the $p$-sequence space in $\GF$.

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

Then:
 * $\mathbf x \cdot \mathbf y \in {\ell^1}_\GF$

and:
 * $\norm {\mathbf x \cdot \mathbf y}_1 \le \norm {\mathbf x}_p \norm {\mathbf y}_q$

where:
 * $\mathbf x \cdot \mathbf y$ denotes the dot product of $\mathbf x$ and $\mathbf y$
 * $\norm {\mathbf x \cdot \mathbf y}_1$ is the $1$-norm, also known as the taxicab norm.

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:

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

Then:

By the Comparison Test, it follows that:

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