# Showing L^p=L^1+L^2 for 1<p<2

1. Mar 26, 2013

### jpriori

Analysis professor gave the following problem as a thought exercise:

Show that an Lp function for 1<p<2 can be written as the sum of an L1 and and L2 function.

Last edited: Mar 26, 2013
2. Mar 26, 2013

### jbunniii

If $|f| \leq 1$ then $|f|^2 \leq |f|^p$, whereas if $|f| > 1$ then $|f| < |f|^p$. So consider the restriction of $f$ to the sets where $|f| \leq 1$ and $|f| > 1$.

3. Mar 27, 2013

### micromass

Your title is not very accurate. It should be $L^p\subseteq L^1 + L^2$ and not $=$.

This can be generalized to the following: if $p<r<q$, then $L^r \subseteq L^p + L^q$. And curiously enough, a dual statement holds as well: $L^p\cap L^q\subseteq L^r$.

Many of the properties of the $L^p$ spaces can be seen when you draw a small diagram.
Consider a square in $\mathbb{R}^2$ with vertices $(1,0), (0,1), (-1,0), (0,-1)$.
The $L^p$ space is then given by the rectangle with vertices $(1/p,1-1/p), (-1/p, 1-1/p), (1/p, -1+1/p), (-1/p, -1+1/p)$.

We see that $L^2$ is a square, which indicates self-duality. We see that $L^p$ and $L^q$ are dual spaces for $\frac{1}{p}+\frac{1}{q}=1$. And we see that $L^p+L^q\subseteq L^r$ for $p<r<q$.