Interpolation Technique in Analysis

Question: Let f belong to both L^{p_1} and L^{p_2}, with 1\leq p_1<p_2<\infty. 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<p<p_2, 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.

Undergraduate Math Books



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One Response to Interpolation Technique in Analysis

  1. Pingback: Lp Interpolation | Singapore Maths Tuition

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