# Lp Interpolation

It turns out that there are two types of Lp interpolation: One is called “Lyapunov’s inequality” which is addressed in this previous blog post.

The other one is called Littlewood’s inequality: If $f\in L^p\cap L^q$, then $f\in L^r$ for any intermediate $p.

The proof is here:

Case 1) $q<\infty$. We have $\frac{1}{q}<\frac{1}{r}<\frac{1}{p}$. There exists $0<\lambda<1$ such that $\frac{1}{r}=\frac{\lambda}{p}+\frac{1-\lambda}{q}$.

\begin{aligned} \|f\|_r^r&=\int |f|^r\\ &=\int |f|^{\lambda r}|f|^{(1-\lambda)r}\\ &\leq\left(\int |f|^{\lambda r(\frac{p}{\lambda r})}\right)^{\lambda r/p}\left(\int |f|^{(1-\lambda)r(\frac{q}{(1-\lambda)r})}\right)^{r(1-\lambda)/q}\\ &=\left(\int |f|^p\right)^{\lambda r/p}\left(\int |f|^q\right)^{r(1-\lambda)/q}\\ &=\|f\|_p^{\lambda r}\|f\|_q^{r(1-\lambda)}. \end{aligned}

Thus $\|f\|_r\leq\|f\|_p^\lambda\|f\|_q^{1-\lambda}$.

Case 2) $q=\infty$. We have $0<\frac{1}{r}<\frac{1}{p}$. There exists $0<\lambda<1$ with $\frac{1}{r}=\frac{\lambda}{p}$. Then
\begin{aligned} \|f\|_r^r&=\||f|^{\lambda r}|f|^{(1-\lambda)r}\|_1\\ &\leq\||f|^{\lambda r}\|_{\frac{p}{\lambda r}}\||f|^{(1-\lambda)r}\|_\infty\\ &=\left(\int |f|^p\right)^{\lambda r/p}\|f\|_\infty^{(1-\lambda)r}\\ &=\|f\|_p^{\lambda r}\|f\|_\infty^{(1-\lambda)r}. \end{aligned}

Hence $\|f\|_r\leq\|f\|_p^\lambda\|f\|_\infty^{1-\lambda}$.

## Author: mathtuition88

https://mathtuition88.com/

This site uses Akismet to reduce spam. Learn how your comment data is processed.