# Problem with in defining a function to be even or odd

1. Jun 6, 2010

### matteo86bo

If I have
$$\int_0^{+\infty}h(x)g(x)dx=0$$
where $$h(x)>0$$ in $$[0,+\infty]$$. Can I conclude that $$g(x)$$ might be zero or an odd function in such interval?

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

$$g(x)=(f(x)-q)$$
where $$f(x)$$ is positive function in the interval. Can I say that $$g(x)$$ must be zero anywhere?

2. Jun 6, 2010

### mnb96

In my opinion your conditions do not force g(x) to be odd.
In fact:

$$\int_0^{+\infty}h(x)g(x)dx = \int_{-\infty}^{+\infty}\overline{h(x)}g(x)dx = 0$$

where:
$$\overline{h(x)}=h(x) \\ ; \mathrm{x\geq 0}$$
$$\overline{h(x)}=0 \\ ; \mathrm{x < 0}$$

Assuming g(x) is odd, you have g(x)=-g(-x); this doesn´t introduce anything useful since $$f(x)\overline{h(x)}=0$$ 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.

Last edited: Jun 6, 2010
3. Jun 6, 2010

### matteo86bo

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 $$\int_0^{+\infty}h(x)g(x)dx$$, and $$h(x)>0$$ in $$[0,+\infty]$$ and not defined for x<0
what can you tell about g(x)?

Furthermore, if the second condition is valid, I mean $$g(x)=f(x)-q$$ with $$f(x)>0$$ in $$[0,+\infty]$$ and not defined elsewhere, is there something interesting to say abot the g(x)??

4. Jun 6, 2010

### mnb96

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. Jun 6, 2010

### matteo86bo

Re: Problem in defining a function to be even or odd

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. Jun 6, 2010

### mnb96

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 $\int_{\mathbbl{R}}f(x)dx > 0[/tex], which contradicts your hypothesis. You can at least say that g cannot be of the form [itex]g:\mathbb{R}^+ \rightarrow \mathbb{R}^+$

7. Jun 7, 2010

### Klockan3

Re: Problem in defining a function to be even or odd

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. Jun 7, 2010

### Staff: Mentor

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.