# Existence of convolution

1. Jan 6, 2008

### quasar987

1. The problem statement, all variables and given/known data
It is a theorem in my book that if f and g are two Lebesgue integrable complex valued functions on R, then the integral

$$\int_{-\infty}^{+\infty}|f(x-y)g(y)|dy$$

is finite for almost all x in R.

Why not all? f is integrable, hence bounded, say, by M. Therefor, whatever x, we have |f(x-y)g(y)|<M|g(y)|, which is integrable, so |f(x-y)g(y)| is too, independantly of x!

Last edited: Jan 6, 2008
2. Jan 6, 2008

### HallsofIvy

I'm a bit puzzled by your question. If f and g are "Lebesque" integrable that only means the are "nice" everywhere except on a set of measure 0. In particular your statemnt "f is integrable, hence bounded" is not true. If f is integrable, it is bounded everywhere except on a set of measure 0. For example, since the rational numbers, between 0 and 1, are countable, the set of all such rational numbers has measure 0. Also, since they are countable, the can be put in a "list", r1[/sup], r2, etc. Define f(x), between 0 and 1, by f(x)= 1 if x is irrational, f(x]) = i for r= ri. That function is integrable on [0, 1] (in fact, its integral is 1) and bounded on the set of irrational numbers between 0 and 1, , but not on [0, 1].

3. Jan 6, 2008

4. Jan 6, 2008

### quasar987

replace "bounded" with "bounded almost everywhere".

5. Jan 6, 2008

### olgranpappy

consider f(x)=g(x)=1/Sqrt[x]

6. Jan 6, 2008

### quasar987

I think 1/sqrt{x} is not integrable on ]0,a], because 1/x isn't and 1/sqrt{x} is bigger than 1/x.

So what is your counter-example for oldgranpappy? (I assume this is what you'Re hinting at)

7. Jan 6, 2008

### olgranpappy

1/Sqrt[x] is integrable on [a,b] for any 'a' and 'b', pos, neg, whatever.

not near zero.

Oh, I was just saying that if f(x-y)=1/Sqrt[x-y] and g(y)=1/Sqrt[y]
then for x=0
f(x-y)g(y) is 1/y which blows up too fast at zero to integrate... but maybe that's a dumb example.