Problem with in defining a function to be even or odd

In summary, the conversation discusses the conditions needed for g(x) to be zero or an odd function in the interval [0, +inf) when given the equation \int_0^{+\infty}h(x)g(x)dx=0 where h(x)>0 in [0,+\infty]. It is concluded that the behavior of g(x) in the interval (-inf,0) is not relevant and that g(x) cannot be of the form g:\mathbb{R}^+ \rightarrow \mathbb{R}^+ for the integral to hold. The impact of g(x) being odd or not is also discussed.
  • #1
matteo86bo
60
0
If I have
[tex]\int_0^{+\infty}h(x)g(x)dx=0

[/tex]
where [tex]h(x)>0[/tex] in [tex][0,+\infty][/tex]. Can I conclude that [tex]g(x)[/tex] might be zero or an odd function in such interval?

If the condition above is still valid and I add this request:

[tex]
g(x)=(f(x)-q)
[/tex]
where [tex]f(x)[/tex] is positive function in the interval. Can I say that [tex]g(x)[/tex] must be zero anywhere?
 
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  • #2
In my opinion your conditions do not force g(x) to be odd.
In fact:

[tex]
\int_0^{+\infty}h(x)g(x)dx = \int_{-\infty}^{+\infty}\overline{h(x)}g(x)dx = 0
[/tex]

where:
[tex]\overline{h(x)}=h(x) \\ ; \mathrm{x\geq 0}[/tex]
[tex]\overline{h(x)}=0 \\ ; \mathrm{x < 0}[/tex]

Assuming g(x) is odd, you have g(x)=-g(-x); this doesn´t introduce anything useful since [tex]f(x)\overline{h(x)}=0[/tex] for any function f which is defined in the interval (0, -inf).

In other words, if you are interested in integrating your function in [0,+inf) the behavior of your function in (-inf,0) does not make any difference.

I hope someone else will answer the rest of your question.
 
Last edited:
  • #3
mnb96 said:
In my opinion your conditions do not force g(x) to be odd.
... since [itex]f(x)\overline{h(x)}=0[/itex] for any function f which is defined in the interval (0, -inf).

sorry, I don't understand how you can say that since I haven't told nothing about f(x) or h(x) in the interval (0,-inf).

Let's pose the problem in another way ...
If [tex]\int_0^{+\infty}h(x)g(x)dx[/tex], and [tex]h(x)>0[/tex] in [tex][0,+\infty][/tex] and not defined for x<0
what can you tell about g(x)?

Furthermore, if the second condition is valid, I mean [tex]g(x)=f(x)-q[/tex] with [tex]f(x)>0[/tex] in [tex][0,+\infty][/tex] and not defined elsewhere, is there something interesting to say abot the g(x)??
 
  • #4
My point was:
Whatever your function looks like in the interval (-inf,0) is not relevant. Simply because you compute the integral in [0,+inf).
Also, the fact that g is (or is not) odd is irrelevant, for the same reason: when you compute the integral, you discard the "mirrored" part of the function g(-|x|).
 
  • #5


ok, I got this

but if h(x)>0 and the integral of h(x)*g(x) is null g(x) must be negative somewhere between 0 and +infinite or null anywhere, mustn't it?
 
  • #6
I would say yes.
Let's suppose that g(x)>0 and consider f(x)=h(x)*g(x).
Clearly we have f(x)>0.
Assuming f(x) is integrable you can deduce, using the definition of integral, that [itex] \int_{\mathbbl{R}}f(x)dx > 0[/tex], which contradicts your hypothesis.

You can at least say that g cannot be of the form [itex]g:\mathbb{R}^+ \rightarrow \mathbb{R}^+ [/itex]
 
  • #7


matteo86bo said:
ok, I got this

but if h(x)>0 and the integral of h(x)*g(x) is null g(x) must be negative somewhere between 0 and +infinite or null anywhere, mustn't it?
No, g(x) can be zero almost everywhere and positive everywhere else and it would still hold. If you are talking about continuous functions, then yes.
http://en.wikipedia.org/wiki/Almost_everywhere
 
  • #8
Going back to the original post, what does g(x) being odd or not have to do with anything? Since you are integrating over [0, inf), I don't see how it matters that g(-x) = -g(x). If the interval were (-inf, inf) then the oddness might be pertinent.
 

1. What is the definition of an even function?

An even function is a mathematical function where the output value remains the same when the input value is replaced by its negative equivalent. In other words, if f(x) is an even function, then f(-x) = f(x).

2. How do you determine if a function is even or odd?

To determine if a function is even or odd, you can use the symmetry test. If the function remains unchanged when its graph is reflected over the y-axis, then it is an even function. If the function remains unchanged when its graph is rotated 180 degrees around the origin, then it is an odd function.

3. Can a function be both even and odd?

No, a function cannot be both even and odd. An even function has symmetry about the y-axis, while an odd function has rotational symmetry about the origin. These two types of symmetry are not possible to have at the same time.

4. Why is it important to define a function as even or odd?

Defining a function as even or odd can help in simplifying and solving mathematical equations. It also allows us to understand the behavior of the function and make predictions about its graph.

5. Are there any real-life applications of even and odd functions?

Yes, even and odd functions have various real-life applications in fields such as physics, engineering, and economics. For example, even functions can describe the motion of a pendulum, while odd functions can be used to model the relationship between supply and demand in economics.

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