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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.

flag	asked Mar 17 2010 at 14:46 Nirman 19●1

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This is explained in most good real analysis books. For example, Knapp's Basic real analysis. – Mariano Suárez-Alvarez Mar 17 2010 at 15:01

I would agree to this comment in the case p=1, but otherwise? – anton Mar 17 2010 at 15:09

(In particular, this follows from Minkowski's inequality, en.wikipedia.org/wiki/…) – Mariano Suárez-Alvarez Mar 17 2010 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 2010 at 3:27

Keenan Kidwell · Answer 1 · 2010-03-17 15:49:06Z UTC

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.

anton · Answer 2 · 2010-03-17 15:08:34Z UTC

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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.

answered Mar 17 2010 at 15:08

anton
2,598●7●12

You can take any positive $C$ here; when $C=1$ the conclusion is particularly obvious. – Robin Chapman Mar 17 2010 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 2010 at 15:42

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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 2010 at 18:56

Defined Almost Everywhere

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