# A family of Borel sets induces a projection operator on $L^p(\mathbb R^n)$

Let $E_1 ,..., E_n$ be a sequence of disjoint Borel sets in $\mathbb{R} ^n$ of positive finite measure and let $\chi_1,...,\chi_n$ be their characteristic functions.

Given $1 \leq p < \infty$ , prove that the operator $P$ on $L ^p (\mathbb{R}^n )$ defined by: $Pf: = \sum_{r=1}^n |E_r|^{-1} \langle f, \chi_r \rangle \chi_r$ is a projection of finite rank, and find its norm and range.

I was wondering what I need to prove here... What should I prove in order to say this is a projection? Afterwards, how can I prove that the dimension of the range of such an operator is finite?

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projection: linear and $P^2 = P$. The other question is a question about linear alebra: guess a base, take a function in the image and show it's a finite linear combination of the base vectors. (You basically already have them). – user20266 Jun 1 '12 at 17:00
Thanks a lot !!! – joshua Jun 1 '12 at 18:06

The Transormation $P$ very convolution-like: it assigns every point $x$ to the average Value of $f$ over the Borel set that contains $x$. It is not difficult to verify that $P$ is a bounded linear operator and that it's norm does not exceed $1$ by using integration by parts. The range of $P$ is clearly finite, since it is the space of all functions taking constant value on each of the sets Of the partition formed by the sets $E$.

In Order to prove that $P$ is a projection just verify that $P(P(f)) = P(f)$ for a function $f$. This is obvious since an averaged function does remains the same when taking the average again.

Projections like $P$ is very common in probability theory: $P$ is then the conditional expectation operator with respect to the (sigma algebra) spanned by the sets $E_i$.

Hope this helps. Ulrich

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Thanks a lot ! I'll try to complete the details myself tomorrow... If I won't understand something, I'll reply here again Thanks a lot ! – joshua Jun 1 '12 at 18:07

This is not an answer, just a few observations:

A useful inequality for proving that the norm is $\leq 1$ is:

$$\int |f\sum \chi_r|^p = \sum \int |f\chi_r|^p = \sum ||f \chi_r||^p \leq ||f||^p.$$

Also, just from the formula, you have $Pf \in \mathbb{sp}\{ \chi_r \}$. Try computing $Pf$ for a function of the form $f = \sum \alpha_r \chi_r$. These facts will establish the range and its dimension (and that the norm of $P$ is indeed $1$).

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