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Thanks.

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In summary, the conversation discusses the question of whether a function f is equal to 0 almost everywhere in the interval (0,1) if the integral of fg is 0 for any continuous function g. It is proven that this is not necessarily the case, as f may not be continuous and g may not be integrable. A counter-example is given and it is clarified that the space L^1(0,1) refers to Lebesgue integrable functions.

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Thanks.

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Consider the case g = f.

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I've got it, thanks.

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resolvent1 said: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.

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TylerH said:

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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disregardthat said:

Similarly, g is not necessarily L

disregardthat said: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.

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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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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.

pwsnafu said: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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disregardthat said: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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The integral of the product of two functions is a mathematical operation that calculates the area under the curve of the product of the two functions. It is denoted as ∫(f(x)g(x))dx, where f(x) and g(x) are the two functions being multiplied and dx is the differential of x.

The integral of the product of two functions equals 0 when the area under the curve of the product is equal to 0. This can occur when either one of the functions is equal to 0 over the interval of integration, or when the two functions have opposite signs and cancel each other out.

To solve for the integral of the product of two functions, you can use integration techniques such as integration by parts, substitution, or partial fractions. It is important to first identify the type of function being multiplied and then choose the appropriate integration method.

Yes, the integral of the product of two functions can be negative if the area under the curve of the product is below the x-axis. This occurs when both functions have the same sign, but one of the functions has a greater magnitude than the other.

The integral of the product of two functions is commonly used in physics, engineering, and economics to calculate quantities such as work, force, and profit. It can also be used to find the area under a curve in probability and statistics, or to calculate the volume of irregular shapes in geometry.

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