📂Lebesgue Spaces

Generalized Hölder's Inequality, Corollaries of Hölder's Inequality

Description

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

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

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

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

The inequality in the above theorem is called the Hölder’s inequality. From Hölder’s inequality, the following two corollaries can easily be shown to hold.

Theorem¹

Theorem 1

Suppose three constants $p>0, q>0, r>0$ satisfy $\dfrac{1}{p} + \dfrac{1}{q} = \dfrac{1}{r}$ and if $u \in {L}^{p}(\Omega), v \in {L}^{q}(\Omega)$, then $uv \in L^{r}(\Omega)$ and the inequality below holds.

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

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In the case of $r=1$, it equals Hölder’s inequality.

Proof

By assumption,

$$ \dfrac{1}{p}+\dfrac{1}{q}=\dfrac{1}{r} \implies \dfrac{1}{p/r}+\dfrac{1}{q/r}=1 $$

And since we assumed that $u \in L^p(\Omega)$, $\left( \int_{\Omega}|u|^p dx \right)^{1/p} < \infty$, therefore,

$$ \left( \int_{\Omega}|u^r|^{\frac{p}{r}} dx \right)^{1/p} < \infty \implies \left( \int_{\Omega}|u^r|^{\frac{p}{r}} dx\right)^{r/p} < \infty $$

Thus, $u^r \in {L}^{p/r}(\Omega)$ and $v^r \in{L}^{q/r}(\Omega)$ can be confirmed in the same way. So by Hölder’s inequality,

$$ \int_{\Omega} |u(x)v(x)|^{r} dx = \int_{\Omega} |u^{r}(x)v^{r}(x) | dx \le \| u^r \|_{p/r} \|v^r\|_{q/r} $$

Rewriting the right side in integral form,

$$ \begin{align*} \int_{\Omega} |u(x)v(x)|^{r} dx \le& \left(\int_{\Omega} |u(x)^{r}|^{p/r} dx \right)^{q/p} \left(\int_{\Omega} |v(x)^r|^{q/r} dx \right)^{r/q} \\ =&\ \left(\int_{\Omega}|u(x)|^{p} dx \right)^{r/p} \left(\int_{\Omega} |v(x)|^{q} dx \right)^{r/q} \end{align*} $$

Taking the power of $\dfrac{1}{r}$ to both sides,

$$ \left( \int_{\Omega} |u(x)v(x)|^rdx \right)^{1/r} \le \left(\int_{\Omega}|u(x)|^{p} dx \right)^{1/p} \left(\int_{\Omega} |v(x)|^{q} dx \right)^{1/q} $$

Therefore,

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

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Theorem 2

Let’s say for $1\le j \le N$, $p_{j}>0$ and $\sum\limits_{j=1}^N\dfrac{1}{{p}_{j}}=\dfrac{1}{{p}_{1}}+\dfrac{1}{{p}_2}+\cdots+\dfrac{1}{{p}_{N}}=\dfrac{1}{r}$. And suppose $u=\prod _{j=1}^N u_{j}=u_{1}u_2\dots u_{N}$ and $u_{j}\in L^{{p}_{j}}(\Omega)$. Then $u\in {L}^r (\Omega)$ and the inequality below holds.

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

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The above Theorem 1 can be seen to hold not only for two functions but also for any $N$ number of functions.

Proof

We use mathematical induction. First, when $N=2$, it holds by Theorem 1. Then, assuming it holds for $N=k$, showing it also holds for $N=k+1$ completes the proof.

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Let’s say $\sum\limits_{j=1}^k \dfrac{1}{{p}_{j}}=\dfrac{1}{r}$ and it holds for $N=k$. Then, the following holds.

$$ \left\| \prod_{j=1}^N u_{j} \right\|_{r} \le \| u_{1} \|_{p_{1}} \| u_{2} \|_{p_{2}} \cdots \| u_{k} \|_{p_{k}} $$

Now, let’s say $\sum_{j=1}^{k+1}\dfrac{1}{{p}_{j}}=\dfrac{1}{r}+\dfrac{1}{{p}_{k+1}}=\dfrac{1}{r^{\prime}}$. Then,

$$ \begin{align*} \| u \|_{r^{\prime}} =&\ \left\| \left( \prod_{j=1}^k u_{j} \right) u_{k+1} \right\|_{r^{\prime}} \\ \le& \left\| \prod \limits_{j=1}^{k+1}u_{j} \right\|_{r} \| u_{k+1} \|_{p_{k+1}} \\ \le& \| u_{1} \|_{p_{1}} \| u_{2} \|_{p_{2}} \cdots \| u_{k} \|_{p_{k}} \| u_{k+1} \| _{p_{k+1}} \\ =&\ \prod \limits_{j=1}^{k+1} \| u_{j}\|_{p_{j}} \end{align*} $$

The second line holds by Theorem 1. The third line holds by assumption. Therefore, assuming it holds for $N=k$ means it also holds for $N=k+1$. Thus, the proof is completed by mathematical induction.

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See also

• Hölder’s inequality in Euclidean space
• Hölder’s inequality