📂Lebesgue Spaces

Interpolation Inequalities in Lebesgue Spaces

Theorem¹

Let’s say $\Omega \subset \mathbb{R}^{n}$ is an open set. Suppose that for some $\theta$, $1 \le p \lt q\lt r \le \infty$ satisfies the following equation being $0 \lt \theta \lt 1$.

$$ \dfrac{1}{q} = \frac{\theta}{p} + \frac{1-\theta}{r} $$

Assume that $u \in L^p(\Omega) \cap L^r(\Omega)$. Then, $u\in L^{q}(\Omega)$ holds, and the following inequality is established.

$$ \left\| u \right\|_{q} \le \left\| u \right\|_{p}^{\theta} \left\| u \right\|_{r}^{1-\theta} $$

This is called the interpolation inequality.

Explanation

Translating interpolation means interpolating, which implies the meaning of filling in the gaps.

For any $p, r$ that is greater than or equal to 1, if $u\in L^{p}$ and $u\in L^{r}$, then it is guaranteed that for all $q$ between $p$ and $r$, $u\in L^q$ holds.

Proof

First, by multiplying both sides of the given assumption by $q$, we have

$$ \begin{align*} && 1 =&\ \dfrac{ \theta q}{p}+\dfrac{(1-\theta) q}{r} \\ \implies && 1 =&\ \dfrac{1}{\frac{p}{\theta q}}+\dfrac{1}{\frac{r}{(1-\theta)q}} \end{align*} $$

Thus, $\dfrac{1}{\frac{p}{\theta q}}$ and $\dfrac{1}{\frac{r}{(1-\theta)q}}$ are conjugate exponents satisfying the Hölder’s inequality. Now, let each be as follows.

$$ s=\dfrac{p}{q\theta} \quad \text{and} \quad s^{\prime}=\dfrac{r}{(1-\theta)q} $$

Hölder’s inequality

For any $\dfrac{1}{p} + \dfrac{1}{p^{\prime}} = 1$ being $1 \le p, p^{\prime} < \infty$, if $u \in L^p(\Omega)$ and $v\in L^{p^{\prime}}(\Omega)$, then

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

• Case 1. $r \lt \infty$

  It can be verified that $|u|^{\theta q} \in L^{s}$ and $|u|^{q(1-\theta)} \in L^{s^{\prime}}$.

  $$ \left\| u^{\theta q} \right\|_{s}^{s} = \int_{\Omega} \left( |u(x)|^{q\theta} \right)^{s} dx = \int_{\Omega} |u(x)|^p dx \lt \infty $$

  $$ \left\| u^{q(1-\theta)} \right\|_{s^{\prime}}^{s^{\prime}}=\int_{\Omega} \left( |u(x)|^{q(1-\theta)} \right)^{s^{\prime}} dx=\int_{\Omega} |u(x)|^r dx \lt\infty $$

  Hence, we can use Hölder’s inequality. Calculating $\left\| u \right\|_{q}^{q}$ yields

  $$ \begin{align*} \left\| u \right\|_{q}^{q} =&\ \int_{\Omega} |u(x)|^{q} dx \\ =&\ \int_{\Omega} |u(x)|^{\theta q} |u(x)|^{(1-\theta)q} dx \\ \le& \left\| u^{\theta q} \right\|_{s} \left\| u^{(1-\theta)q} \right\|_{s^{\prime}} \quad \text{by Hoelder’s inequality} \\ =&\ \left( \int_{\Omega} |u(x)|^{\theta q s } dx \right)^{1/s} \left( \int_{\Omega} |u(x)|^{(1-\theta) q s^{\prime} } dx \right)^{1/s^{\prime}} \\ =&\ \left( \int_{\Omega} |u(x)|^{p} dx \right)^{\frac{1}{p}q\theta} \left( \int_{\Omega} |u(x)|^{r} dx \right)^{\frac{1}{r}(1-\theta)q} \\ =&\ \| u \|_{p}^{\theta q} \left\| u \right\|_{r}^{(1-\theta)q} \end{align*} $$

  By multiplying the exponent of both sides by $\dfrac{1}{q}$, we get

  $$ \left\| u \right\|_{q} \le \left\| u \right\|_{p}^{\theta} \left\| u \right\|_{r}^{1-\theta} $$

• Case 2. $r = \infty$

  The condition of the assumption is such that $\dfrac{1}{q} = \dfrac{\theta}{p}$. Similarly, if we let $s=\frac{p}{\theta q}=1$ having $s^{\prime}=\infty$ and satisfying $1=\frac{1}{s}+\frac{1}{s^{\prime}}$ with $|u|^{\theta q} \in L^s=L^1$ and $|u|^{(1-\theta)q} \in L^{s^{\prime}}=L^{\infty}$.

  $$ \left\| u^{\theta q} \right\|_{1}=\int_{\Omega} |u(x)|^{q\theta} dx=\int_{\Omega} |u(x)|^p dx \lt\infty $$

  If $u\in L^{\infty}$, it is easily understood by the definition of $| \cdot|_\infty$ that for any positive number $k$, $|u|^k \in L^{\infty}$ holds. Since $(1-\theta)\gt 0$, thus $|u|^{(1-\theta)q} \in L^{\infty}$. Therefore, using Hölder’s inequality,

  $$ \begin{align*} \left\| u \right\|_{q}^{q} =&\ \int_{\Omega} |u|^q dx \\ =&\ \int_{\Omega} |u|^{\theta q} |u|^{(1-\theta)q}dx \\ \le& \left\| u^{\theta q} \right\|_{1} \left\| u^{(1-\theta)q} \right\|_{\infty} \\ =&\ \left( \int_{\Omega} |u|^{\theta q} dx \right)^{1} \left\| u^{(1-\theta)q} \right\|_{\infty} \\ =&\ \left( \int_{\Omega} |u|^{p} dx \right)^{\frac{1}{p}q\theta} \left\| u^{(1-\theta)q} \right\|_{\infty} \\ =&\ \| u \|_{p}^{\theta q} \left\| u^{(1-\theta)q} \right\|_{\infty} \\ =&\ \| u \|_{p}^{\theta q}\ \left\| u \right\|_{\infty}^{(1-\theta)q} \end{align*} $$

  The third line is established using Hölder’s inequality. The last line is established by the property of $\left\| \cdot \right\| _{\infty}$. Hence,

  $$ \left\| u \right\|_{q} \le \| u \|_{p}^{\theta } \left\| u \right\|_{\infty}^{(1-\theta)}= \| u \|_{p}^{\theta }\ |u|_{r}^{(1-\theta)} $$

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