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Thanks.

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- Thread starter resolvent1
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Thanks.

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Consider the case g = f.

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I've got it, thanks.

- #4

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I've got it, thanks.

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

- #5

disregardthat

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disregardthat

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A counter-example must be an f which is 0 on a set of non-zero measure such that for

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

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- #8

pwsnafu

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Similarly, g is not necessarily L

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

Err...

In order to be a counter example you need to produce a [itex]f \in L^1 (0,1)[/itex] and [itex]g \in C(0,1)[/itex] such that

- the integral of fg is 0,
- the set of x such that [itex]f(x) \neq 0[/itex] is not equal to zero.

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

That would be Lebesgue.

Tyler, consider g(x) = 1.

- #9

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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

[itex] \int_E f^2 < \eps [/itex]

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:

[itex] u(x) = f|_F(x) [/itex] if [itex] x \in F [/itex], [itex] 0 [/itex] if [itex] x \in F^C [/itex]

[itex] v(x) = f|_{F^C}(x) [/itex] if [itex] x \in F^C [/itex], [itex] 0 [/itex] if [itex] x \in F [/itex]

Then:

[itex]

\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

[/itex]

So [itex] \int_0^1 f^2 = 0 [/itex]. 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.

[itex] \int_E f^2 < \eps [/itex]

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:

[itex] u(x) = f|_F(x) [/itex] if [itex] x \in F [/itex], [itex] 0 [/itex] if [itex] x \in F^C [/itex]

[itex] v(x) = f|_{F^C}(x) [/itex] if [itex] x \in F^C [/itex], [itex] 0 [/itex] if [itex] x \in F [/itex]

Then:

[itex]

\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

[/itex]

So [itex] \int_0^1 f^2 = 0 [/itex]. 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.

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- #10

disregardthat

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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.

Err...

In order to be a counter example you need to produce a [itex]f \in L^1 (0,1)[/itex] and [itex]g \in C(0,1)[/itex] such that

In measure notation [itex]\mu(f^{-1}(\mathbb{R}\setminus \{0\})) > 0[/itex].

- the integral of fg is 0,
- the set of x such that [itex]f(x) \neq 0[/itex] is not equal to zero.

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.

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- #11

pwsnafu

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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.

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.

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