## Interpolation Technique in Analysis

Question: Let $f$ belong to both $L^{p_1}$ and $L^{p_2}$, with $1\leq p_1. Show that $f\in L^p$ for all $p_1\leq p\leq p_2$.

There is a pretty neat trick to do this question, known as the “interpolation technique”. The proof is as follows.

For $p_1, there exists $0<\alpha<1$ such that $\displaystyle\boxed{p=\alpha p_1+(1-\alpha)p_2}$. This is the key “interpolation step”. Once we have this, everything flows smoothly with the help of Holder’s inequality. \displaystyle\begin{aligned} \int |f|^p\ d\mu&=\int (|f|^{\alpha p_1}\cdot |f|^{(1-\alpha)p_2})\ d\mu\\ &\leq\||f|^{\alpha p_1}\|_\frac{1}{\alpha}\||f|^{(1-\alpha)p_2}\|_\frac{1}{1-\alpha}\\ &=(\int |f|^{p_1}\ d\mu)^\alpha\cdot (\int |f|^{p_2}\ d\mu)^{1-\alpha}\\ &<\infty \end{aligned}

Thus $f\in L^p$.

Note that the magical thing about the interpolation technique is that $p=\frac{1}{\alpha}$ and $q=\frac{1}{1-\alpha}$ are Holder conjugates, since $\frac{1}{p}+\frac{1}{q}=1$ is easily verified. 