📂Lebesgue Spaces

Holder's Inequality

Theorem¹

Let $\Omega \subset \mathbb{R}^{n}$ be an open set. Assume $0 < p < 1$ and $p^{\prime} = \dfrac{p}{p-1} < 0$. If $u \in$ $L^{p}(\Omega)$, $uv\in$ $L^{1}(\Omega)$, then

$$ \begin{equation} 0 \lt \int_{\Omega} |v(x)|^{p^{\prime}}dx \lt \infty \end{equation} $$

the following inequality is established:

$$ \int_{\Omega} |u(x)v(x)|dx \ge \left( \int_{\Omega} |u(x)|^{p} dx \right)^{1/p} \left( \int_{\Omega} |v(x)|^{p^{\prime}} dx \right) ^{1/p^{\prime}} $$

Explanation

This is called the reverse Höelder’s inequality. It’s not a converse of the Höelder’s inequality, but a case where the direction of the inequality is reversed. Compared to the Höelder’s inequality, the direction of the inequality is exactly opposite.

• Höelder’s inequality: When $1 \le p \le \infty$, the right side is greater,
• Reverse Höelder’s inequality: When $0 < p < 1$, the left side is greater.

It’s important to note that $\| u \|_{p}$ is defined as follows, which becomes the norm of space $L^{p}$ only when $1 \le p < \infty$:

$$ \| u \|_{p} :=\left( \int |u(x)|^p dx \right)^{1/p} $$

In other cases, $\| u \|_{p}$ is not the norm of the $L^{p}$ space. Therefore, when $0 < p < 1$, be aware that the integral on the right side of the inequality is not the norm $\left\| \cdot \right\|_{p}$, $\left\| \cdot \right\|_{p^{\prime}}$ respectively.

Also, it’s a natural assumption that $uv \in L^{1}$ is required for the inequality to have meaning.

Proof

Assume $\phi = | v |^{-p}$ and $\psi = | uv |^{p}$. Then, $\phi\psi=| u |^{p}$. If $q = \dfrac{1}{p}$ is considered, since $0 < p < 1$, $1 < q < \infty$. Moreover, due to the assumption of $uv \in L^1$, it can be confirmed that $\psi \in L^q$.

$$ \int |\psi|^q dx=\int |uv|^{pq}dx=\int |uv| dx <\infty $$

And if $q^{\prime} = q/(q-1)$ is considered, since $1 < q < \infty$, then $1 < q^{\prime} < \infty$ and $p^{\prime} = -pq^{\prime}$.

$$ p^{\prime}=\frac{p}{p-1}=\frac{1}{1-\frac{1}{p}}=\frac{1}{1-q}=-\frac{1}{p}\frac{p}{q-1}=-q\frac{p}{q-1}=-pq^{\prime} $$

Then, according to the assumption $(1)$, it can be shown that $\phi \in L^{q^{\prime}}$.

$$ \int |\phi |^{q^{\prime}} dx = \int |v|^{-pq^{\prime}} dx =\int |v|^{p^{\prime}} dx<\infty $$

Therefore, $1 < q, q^{\prime} < \infty$ and $\psi \in L^{p}, \phi \in L^{q^{\prime}}$, and by applying the Höelder’s inequality,

$$ \begin{align*} \int |u(x)|^pdx =&\ \int \left| \phi (x)\psi (x) \right| dx \\ \le& \| \psi \|_{q} \| \phi \|_{q^{\prime}} \\ =&\ \left( \int |u(x)v(x)|^{pq} dx\right)^{1/q} \left( \int | v(x) |^{-pq^{\prime}}dx \right)^{1/q^{\prime}} \\ =&\ \left( \int |u(x)v(x)| dx\right)^{p} \left( \int | v(x) |^{p^{\prime}}dx \right)^{-p / p^{\prime}} \end{align*} $$

Multiplying both sides by $\displaystyle \left(\int |v|^{p^{\prime}}dx \right)^{p/p^{\prime}}$ yields

$$ \left( \int |u(x)|^{p} dx \right) \left( \int | v(x) |^{p^{\prime}}dx \right)^{p / p^{\prime}} \le \left( \int |u(x)v(x)| dx\right)^{p} $$

Lastly, multiplying the exponents of both sides by $\dfrac{1}{p}$ results in

$$ \left( \int |u(x)|^pdx \right)^{1/p} \left( \int | v(x) |^{p^{\prime}}dx \right)^{ 1 / p^{\prime}} \le \int |u(x)v(x)| dx $$

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