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<r<q.

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


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

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