Function to satisfy this integral for all parameters?

[Gerenuk]

I wonder:

Is it possible to find a function f(x) which satisfies this integral for *arbitrary* parameters a,b (but maybe of known domain)?

$$\int_{-\infty}^\infty e^{-x^2}f(ax+b)\mathrm{d}x=0$$

I want $$f(\pm\infty)=0$$ and I can deduce
$$\int_{-\infty}^\infty x^n e^{-x^2}f^{(m)}(ax+b)\mathrm{d}x=0$$

It would have to be oscillary I guess?!

 

[Manchot]

Yes, and no. Obviously, f(x)=0 satisfies the criterion, but this is probably not what you were looking for. If you do a change of variable in the expression, you'll see that the integrand is exp(-((x-b)/a)^2)*f(x), i.e., a Gaussian times some function. However, if you let a and b be arbitrary, you're allowing the Gaussian's position and width to vary as well, and by making a arbitrarily large, you can make the Gaussian as thin as you want. The net effect is that the integral would "pick out" the value of f(b), so unless f is 0 everywhere, the integral cannot be 0 everywhere.

This, of course, assumes that f is well-behaved. Any function which did satisfy the criterion would be nowhere continuous, as it would have to oscillate infinitely at every point.

 

[Gerenuk]

And I an estimate for b and a is known? Say it's in a range around some a0 and b0.

 

[epenguin]

Since coming here I have found questions I could answer, but from the other answers it looked like I hadn't understood the question. With that reservation:

Firstly, why f(ax + b) ? Call it g(x). If you can get that you can translate into a function of (ax + b) afterwards.

Now $$e^{-x^2}$$ is symmetrical about x = 0. So if g is any continuous and finite function of x that is antisymmetrical i.e. g(x) = - g(-x) , and defined from - to + infinity, it will satisfy your requirement won't it? Simplest example g(x) = x.

If this shows I have not understood the question (guess no-one would ask one that easy) someone please tell me as other times I was left suspended.

 

[epenguin]

Ps

sorry, g(x) = x doesn't satisfy your $$f(\pm\infty)=0$$ condition, but you can easily find one that does, e.g. any function symmetrical about x = 0 that does, multiplied by x.