Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

How can one prove that the convolution of f \in L^1 and g \in L^p is defined almost everywhere? Here f and g are measurable functions in R^n.

In general what techniques are there for showing some function is defined a.e and where should I look for them in analysis literature.

Background

This came up while I was reading some notes to understand harmonic analysis.

share|improve this question
1  
This is explained in most good real analysis books. For example, Knapp's Basic real analysis. –  Mariano Suárez-Alvarez Mar 17 '10 at 15:01
    
I would agree to this comment in the case p=1, but otherwise? –  anton Mar 17 '10 at 15:09
    
(In particular, this follows from Minkowski's inequality, en.wikipedia.org/wiki/…) –  Mariano Suárez-Alvarez Mar 17 '10 at 15:10
    
Correct me if I'm wrong,measure theory was several semesters ago-but isn't the simplest way to prove a function is defined almost everywhere is just to show the complement of it's support is finite? I don't remember if it's allowed to be countable,but since countable sets have measure 0 (VERY easy to prove) ,I'd be willing to bet the common definition allows for countable sets in the range where f(x)=0. Analysts,please feel free to jump in and correct me at any time here. –  Andrew L Mar 18 '10 at 3:27
add comment

2 Answers

Folland's real analysis text spends a lot of time on various basic $L^p$ inequalities including $L^p$ norms of convolutions. The material on convolutions is (I believe) in Chapter 8 which is titled ``Elements of Fourier Analysis," and the basic $L^p$ stuff is in Chapter 6. Incidentally, this is definitely not a research level question, and in fact, I would say it's a standard exercise (using Hölder\Minkowski\Fubini) in a first course in measure theory.

share|improve this answer
add comment

If $g\in L^p$ for $1\le p\le\infty$, then there exists $C>0$ such that the function $g_1(x)=g(x){\bf 1}_A(x)$ is in $L^1$, where $A$ is the set where $|g(x)|>C$.That means that $g$ is the sum of an $L^1$-function plus a bounded function, so the convolution $f*g$ exists.

share|improve this answer
    
You can take any positive $C$ here; when $C=1$ the conclusion is particularly obvious. –  Robin Chapman Mar 17 '10 at 15:28
    
Only if p is finite. If p is infinity, you have to take C to be equal to the infinity-norm of g. –  anton Mar 17 '10 at 15:42
3  
When $p=\infty$, I wouldn't use this method to prove that an $L^\infty$ function is the sum of an $L^1$ and an $L^\infty$ function :-) –  Robin Chapman Mar 17 '10 at 18:56
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.