# Integral of product of two functions = 0

1. Nov 13, 2011

### resolvent1

If $f \in L^1(0,1)$ and if $\int_0^1 fg = 0$ for any continuous $g \colon (0,1) \rightarrow \mathbb{R}$, then is it the case that $f = 0$ almost everywhere in $(0,1)$?

Thanks.

2. Nov 13, 2011

### awkward

Consider the case g = f.

3. Nov 13, 2011

### resolvent1

I've got it, thanks.

4. Nov 13, 2011

### TylerH

Can you post your proof? This is an interesting problem, and I've been unable to prove it so far.

5. Nov 14, 2011

### disregardthat

It is not necessarily so that the integral of f*g is 0 if g = f, since f isn't necessarily continuous.

6. Nov 14, 2011

### TylerH

Oh! Would g(x)=0 also be a counter example? With g(x)=0, f(x)g(x)=0 for all x and thus $\int^1_0 f(x)g(x)=0$.

7. Nov 14, 2011

### disregardthat

A counter-example must be an f which is 0 on a set of non-zero measure such that for any continuous function g, the integral of fg from 0 to 1 is 0...

By the way, is L^1(0,1) the space of riemann-integrable functions, or lebesgue integrable functions?

Last edited: Nov 14, 2011
8. Nov 14, 2011

### pwsnafu

Similarly, g is not necessarily L1. The domain is (0,1) not [0,1].

Err...

In order to be a counter example you need to produce a $f \in L^1 (0,1)$ and $g \in C(0,1)$ such that
1. the integral of fg is 0,
2. the set of x such that $f(x) \neq 0$ is not equal to zero.
In measure notation $\mu(f^{-1}(\mathbb{R}\setminus \{0\})) > 0$.

That would be Lebesgue.

Tyler, consider g(x) = 1.

9. Nov 14, 2011

### resolvent1

Let \eps>0. Then since f is L-integrable, f^2 also is, so there is a \delt>0 such that if m(E)<\delt, then

$\int_E f^2 < \eps$

Also, since f is L-integrable, it's measurable, so Lusin implies that there exists a measurable F contained in (0,1) such that M((0,1)-F) < \delt and for which the restriction of f to F is continuous. Define the following:

$u(x) = f|_F(x)$ if $x \in F$, $0$ if $x \in F^C$
$v(x) = f|_{F^C}(x)$ if $x \in F^C$, $0$ if $x \in F$

Then:

$\int_0^1 f^2 = \int_0^1 f \cdot (u+v) = \int_0^1 f \cdot u + \int_0^1 f \cdot v = \int_F f \cdot u + \int_{F^C} f^2 = \int_{F^C} f^2 < \eps$

So $\int_0^1 f^2 = 0$. Since f^2 is nonnegative on (0,1), Tschebyshev's inequality implies f^2 = 0 ae on (0,1). So f = 0 ae on (0,1).

I guess u would actually be piecewise continuous, but I think it can be patched up fairly easily.

Last edited: Nov 14, 2011
10. Nov 14, 2011

### disregardthat

Nice proof resolvent.

I don't think so

The premise is that f is integrable, and that fg has integral = 0 for any continuous function g. A counter-example must satisfy this (you can't just bring about a single continuous g), but also not satisfy that the set of zeroes has measure 0.

Last edited: Nov 14, 2011
11. Nov 14, 2011

### pwsnafu

You're right.

Edit: No, you are right about g. But not about zeroes of f. The theorem concludes that "f = 0 a.e." That means the set of x where f is not 0 has zero measure.

So for a counter example you need "not f = 0 a.e." which is saying "the set of f=/=0 has non-zero measure". You can have "f=0" have measure 1/2 and "f=/=0" also have measure half, which isn't zero a.e.

Last edited: Nov 14, 2011