Hölder's Inequality for Sums

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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

Without loss of generality, 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.

$\blacksquare$


Also see


Source of Name

This entry was named for Otto Ludwig Hölder.


Historical Note

Hölder's Inequality for Sums was first found by Leonard James Rogers in $1888$, and discovered independently by Otto Ludwig Hölder in $1889$.


Sources

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