Proof of Hölder's Inequality in Lebesgue Spaces 📂Lebesgue Spaces

Proof of Hölder's Inequality in Lebesgue Spaces

Theorem¹

Let us consider $\Omega \subset \mathbb{R}^{n}$ as an open set. Suppose we’re given two constants $1 \lt p \lt \infty, 1 \lt p^{\prime} \lt \infty$ satisfying the following equation.

$$ \dfrac{1}{p}+\dfrac{1}{p^{\prime}} = 1 \left(\text{or } p^{\prime} = \frac{p}{p-1} \right) $$

If $u \in L^p(\Omega)$ and $v\in L^{p^{\prime}}(\Omega)$, then $uv \in L^1(\Omega)$, and the inequality below holds.

$$ \| uv \|_{1} = \int_{\Omega} |u(x)v(x)| dx \le \| u \|_{p} \| v \|_{p^{\prime}} $$

This inequality is known as Hölder’s inequality.

Explanation

$p^{\prime}$ is known as the Hölder conjugate or conjugate exponent of $p$. It is often denoted as $q$.

If $| u(x) |^{p}$ and $| v(x) |^{p^{\prime}}$ are proportional almost everywhere in $\Omega$, equality holds.

It is essentially the same as Hölder’s inequality in Euclidean spaces, and it also becomes the Cauchy-Schwarz inequality when $p=p^{\prime}=2$. The proof is no different than that of the Cauchy-Schwarz inequality, with the addition of Young’s inequality.

It can also be generalized in the following form.

$$ \| uv \|_{r} = \left( \int_{\Omega} |u(x)v(x)|^{r} dx \right)^{1/r} \le \| u \|_{p} \| v\|_{p^{\prime}} $$

$$ \| u \|_{r} = \left( \int_{\Omega} |u(x)|^{r} dx \right)^{1/r} \le \prod_{j=1}^{N} \| u_{j} \|_{{p}_j} = \| u_{1} \|_{{p}_1} \cdots \| u_{N} \|_{p_{N}} $$

Proof

Young’s Inequality
Given $\dfrac{1}{p} + \dfrac{1}{p^{\prime}} = 1$ satisfies and for two constants greater than 1, $p, p^{\prime}$, and two positive numbers $a,b$,
$$ ab \le { {a^{p}} \over {p} } + {{b^{p^{\prime}}} \over {p^{\prime}}} $$

Case 1. $\| u \|_{p} = 0$ or $\| v \|_{p^{\prime}} = 0$
Since almost everywhere in $\Omega$ is $u(x) = 0$ or almost everywhere in $\Omega$ is $v(x) = 0$, almost everywhere in $\Omega$ is $u(x)v(x) = 0$. Therefore,
$$ \left| \int_{\Omega} u(x) v(x) dx \right| = \| uv \|_{1} = 0 $$
and
$$ \| u \|_{p} \| v \|_{p^{\prime}} = 0 $$
hence the inequality is satisfied.
Case 2. Other cases
By substituting $a = \dfrac{\left| u(x) \right|}{\| u \|_{p}}$ and $b = \dfrac{\left| v(x) \right|}{\| v \|_{p^{\prime}}}$ in Young’s inequality, we have
$$ \dfrac{\left| u(x) \right|}{\| u \|_{p}} \dfrac{\left| v(x) \right|}{\| v \|_{p^{\prime}}} \le \dfrac{ \left| u(x) \right|^{p}}{ p \| u \|_{p}^{p}} + \dfrac{\left| v(x) \right|^{p^{\prime}}}{ p^{\prime} \| v \|_{p^{\prime}}^{p^{\prime}}} $$
Integrating both sides yields
$$ \begin{align*} \dfrac{1}{\| u \|_{p} \| v \|_{p^{\prime}}} \int_{\Omega}\left| u(x)v(x) \right| dx \le & \dfrac{1}{p \| u \|_{p}^{p}} \int_{\Omega} \left| u(x) \right|^{p} dx + \dfrac{1}{ p^{\prime} \| v \|_{p^{\prime}}^{p^{\prime}}} \int_{\Omega} \left| v(x) \right|^{p^{\prime}} dx \\ \le & \dfrac{1}{p \| u \|_{p}^{p}} \| u \|_{p}^{p} + \dfrac{1}{ p^{\prime} \| v \|_{p^{\prime}}^{p^{\prime}}} \| v \|_{p^{\prime}}^{p^{\prime}} \\ \le & \dfrac{1}{p} + \dfrac{1}{ p^{\prime} } \\ =& 1 \end{align*} $$
Moving the constant to the left side gives
$$ \| uv \|_{1} = \int_{\Omega} |u(x)v(x)| dx \le \| u \|_{p} \| v \|_{p^{\prime}} $$
Thus, $uv \in L^{1}(\Omega)$ holds, and the inequality is satisfied.

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