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

Proof of Hölder's Inequality in Lebesgue Spaces

Theorem¹

Let $\Omega \subset \mathbb{R}^{n}$ be an open set. Suppose two constants $1 \lt p \lt \infty, 1 \lt p^{\prime} \lt \infty$ satisfying the following equation are given.

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

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

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

The above inequality is called Hölder’s inequality.

Explanation

$p^{\prime}$ is called the Hölder conjugate or the conjugate exponent of $p$. It is also often denoted by $q$.

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

It is essentially the same as Hölder’s inequality in Euclidean space, and likewise it becomes the Cauchy-Schwarz inequality when $p=p^{\prime}=2$. The proof itself is nothing more than the proof of the Cauchy-Schwarz inequality with Young’s inequality added.

A generalization of the following form is also possible.

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

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

Proof

Young’s inequality
For two constants $p, p^{\prime}$ greater than 1 satisfying $\dfrac{1}{p} + \dfrac{1}{p^{\prime}} = 1$ 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 $u(x) = 0$ almost everywhere in $\Omega$ or $v(x) = 0$ almost everywhere in $\Omega$, we have $u(x)v(x) = 0$ almost everywhere in $\Omega$. Therefore
$$ \left| \int_{\Omega} u(x) v(x) dx \right| = \| uv \|_{1} = 0 $$
and
$$ \| u \|_{p} \| v \|_{p^{\prime}} = 0 $$
so the inequality is satisfied.
Case 2. Otherwise
Substitute $a = \dfrac{\left| u(x) \right|}{\| u \|_{p}}$ and $b = \dfrac{\left| v(x) \right|}{\| v \|_{p^{\prime}}}$ into Young’s inequality. Then
$$ \dfrac{\left| u(x) \right|}{\| u \|_{p}} \dfrac{\left| v(x) \right|}{\| v \|_{p^{\prime}}} \le \dfrac{ \left| u(x) \right|^{p}}{ p \| u \|_{p}^{p}} + \dfrac{\left| v(x) \right|^{p^{\prime}}}{ p^{\prime} \| v \|_{p^{\prime}}^{p^{\prime}}} $$
Integrating both sides gives the following.
$$ \begin{align*} \dfrac{1}{\| u \|_{p} \| v \|_{p^{\prime}}} \int_{\Omega}\left| u(x)v(x) \right| dx \le & \dfrac{1}{p \| u \|_{p}^{p}} \int_{\Omega} \left| u(x) \right|^{p} dx + \dfrac{1}{ p^{\prime} \| v \|_{p^{\prime}}^{p^{\prime}}} \int_{\Omega} \left| v(x) \right|^{p^{\prime}} dx \\ \le & \dfrac{1}{p \| u \|_{p}^{p}} \| u \|_{p}^{p} + \dfrac{1}{ p^{\prime} \| v \|_{p^{\prime}}^{p^{\prime}}} \| v \|_{p^{\prime}}^{p^{\prime}} \\ \le & \dfrac{1}{p} + \dfrac{1}{ p^{\prime} } \\ =& 1 \end{align*} $$
Moving the constant on the left-hand side to the other side yields
$$ \| uv \|_{1} = \int_{\Omega} |u(x)v(x)| dx \le \| u \|_{p} \| v \|_{p^{\prime}} $$
Therefore $uv \in L^{1}(\Omega)$, and the inequality holds.

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