Is g(x) Equal to g(a) If Their Integrals Are Equivalent?

[Steve Zissou]

TL;DR

wondering if integrand terms can be cancelled

Howdy all,

Let's say we have, in general an expression:

$$ \int f(x) g(x) dx $$

But in through some machinations, we have, for parameter $a$,

$$ \int f(x) g(x) dx = \int f(x) g(a) dx $$

...can we conclude that $g(x) = g(a)$ ????

Thanks

 

[BvU]

No.

 

[Steve Zissou]

No.

Thank you. Is there any chance you can help me understand?

 

[Frabjous]

let f(x)=1
let g(x)=x
integrate from 0 to 1
then a=½ satisfies your relation

 

[Steve Zissou]

let f(x)=1
let g(x)=x
integrate from 0 to 1
then a=½ satisfies your relation

Thank you Frabjous.
How about this? Let's say we have
$$ \int f(x) g(a) dx = \int f(x) g(b) dx $$
...where a and b are parameters. Can we say that $g(a)=g(b)$??
...or even that $a = b$?
Thanks again

 

[Frabjous]

Thank you Frabjous.
How about this? Let's say we have
$$ \int f(x) g(a) dx = \int f(x) g(b) dx $$
...where a and b are parameters. Can we say that $g(a)=g(b)$??

Assuming that ∫f(x) is well behaved, yes. g(a) and g(b) are constants that can be pulled out of the integral.

Edit: By well behaved, I mean not equal to ±∞ or 0.

 

[PeroK]

TL;DR Summary: wondering if integrand terms can be cancelled

Howdy all,

Let's say we have, in general an expression:

$$ \int f(x) g(x) dx $$

But in through some machinations, we have, for parameter $a$,

$$ \int f(x) g(x) dx = \int f(x) g(a) dx $$

...can we conclude that $g(x) = g(a)$ ????

Thanks

Are these indefinite integrals?

 

[BvU]

On the left, $g$ is a function of $x$. On the right $g(a)$ is a number. $g(x) = g(a)$ would mean $g(x)$ is a constant.

How about this? Let's say we have
$$ \int f(x) g(a) dx = \int f(x) g(b) dx $$
...where a and b are parameters. Can we say that $g(a)=g(b)$??
...or even that $a = b$?
Thanks again

No. If $\int f(x) = 0$ nothing can be said about $a$ and $b$

$\$

 

[Steve Zissou]

Are these indefinite integrals?

I was hoping to be able to make definite integrals here.

 

[Steve Zissou]

I'm very sorry guys, allow me to make my examples firmer. I don't want to waste your time. I'll be back with a "better question."

 

[PeroK]

I was hoping to be able to make definite integrals here.

There's a general mathematical idea that $A = B$ if and only if $A - B = 0$. You can apply that to your questions. Also, $g(a)$ and $g(b)$ are just numbers. For example:
$$\int_0^1f(x)g(x) \ dx = \int_0^1kf(x) \ \Rightarrow \ \int_0^1f(x)(g(x)-k) \ dx = 0$$But, lots of definite integrals are zero. So, for any function $f$, there will be lots of functions $g$ where that integral is zero. In fact, if $h$ is some function where $\int_0^1 h(x) \ dx = 0$, then $g(x) = k + \frac{h(x)}{f(x)}$ will be a counteraxample to your claim.

 

[Steve Zissou]

Thank you all for looking at this. I wish I had a more concrete example but I'm just trying to understand a certain concept here. So please allow me to rephrase my original question.
How about this? Let's say $f = f(x,y)$ and also $g = g(x,y)$.
$$\int_{0}^{\infty}f(x,y)g(x,y)dx =\int_{0}^{\infty}f(x,y)g(x,a)dx$$
where $a$ is some number.
Can we conclude that $g(x,y)=g(x,a)$?

 

[PeroK]

Thank you all for looking at this. I wish I had a more concrete example but I'm just trying to understand a certain concept here. So please allow me to rephrase my original question.
How about this? Let's say $f = f(x,y)$ and also $g = g(x,y)$.
$$\int_{0}^{\infty}f(x,y)g(x,y)dx =\int_{0}^{\infty}f(x,y)g(x,a)dx$$
where $a$ is some number.
Can we conclude that $g(x,y)=g(x,a)$?

Are you just making this stuff up? What is the point of all, one wonders!

 

[Frabjous]

g(x,y)=g(x,a)

Since a is some number, you do realize that this means g(x,y)=g(x)? What concept are you actually trying to understand?

 

[renormalize]

Let's say $f = f(x,y)$ and also $g = g(x,y)$.
$$\int_{0}^{\infty}f(x,y)g(x,y)dx =\int_{0}^{\infty}f(x,y)g(x,a)dx$$
where $a$ is some number.
Can we conclude that $g(x,y)=g(x,a)$?

Nope! A trivial counterexample: $f\left(x,y\right)=e^{-x},\;g\left(x,y\right)=\left(x-1\right)y$. Both of your integrals equal zero, but $g\left(x,y\right)\neq g\left(x,a\right)$.

 

[Steve Zissou]

renormalize, thanks for your reply. In your example, can we not conclude that $(x-1)y=(x-1)a$?

 

[Steve Zissou]

I'm imagining that the two integrals are Riemann sums in a limit. So I'm seeing the $f(x,y)$'s all cancelling out as additive terms.

 

[renormalize]

renormalize, thanks for your reply. In your example, can we not conclude that $(x-1)y=(x-1)a$?

No, the left integral vanishes for any arbitrary $y$. Nothing forces $y=a$.

 

[Steve Zissou]

Thank you to those who have shared their insights here.
Let me rephrase one more time.
$$\int_{0}^{\infty}f(x)g(x)dx=\int_{0}^{\infty}f(x)h(x)dx$$
1) Can we "cancel" the $f(x)$ terms on each side, thereby now having
$$\int_{0}^{\infty}g(x)dx=\int_{0}^{\infty}h(x)dx$$
2) Can we also now say
$$g(x)=h(x)$$
Thanks

 

[Hill]

Thank you to those who have shared their insights here.
Let me rephrase one more time.
$$\int_{0}^{\infty}f(x)g(x)dx=\int_{0}^{\infty}f(x)h(x)dx$$
1) Can we "cancel" the $f(x)$ terms on each side, thereby now having
$$\int_{0}^{\infty}g(x)dx=\int_{0}^{\infty}h(x)dx$$
2) Can we also now say
$$g(x)=h(x)$$
Thanks

No to both questions. Plenty of counterexamples have been given above.

 

[Steve Zissou]

No to both questions. Plenty of counterexamples have been given above.

Thank you Hill.