# Is there any counterexample to this ?

is there any counterexample to this ??

let be the Fourier transform

$$G(s) = \int_{-\infty}^{\infty}dxf(x)exp(isx)$$

with the properties

$$f(x)$$ and $$D^{2}f(x)$$ are EVEN funnctions of 'x'

$$f(x) > 0$$ and $$D^{2}f(x) > 0$$ on the whole interval (-oo,oo)

then G(s) has only REAL roots

is there any counterexample to this ?? thanks

let be the Fourier transform

$$G(s) = \int_{-\infty}^{\infty}dxf(x)exp(isx)$$

with the properties

$$f(x)$$ and $$D^{2}f(x)$$ are EVEN funnctions of 'x'

$$f(x) > 0$$ and $$D^{2}f(x) > 0$$ on the whole interval (-oo,oo)

then G(s) has only REAL roots

is there any counterexample to this ?? thanks
Are you sure you mean that you want f''(x)>0? No function satisfying this can decay to zero for both small and large x (try drawing a graph), and hence no such function will have a Fourier Transform (except in a distributional sense).

It means a different thing to have f even: it means f(x)=f(-x) for all x. In that case, if f is twice differentiable, then f'' will automatically also be even (just use the chain rule).

The Fourier transform is only usually defined for real s. It maps complex functions of real variables to complex functions of real variables. If your function f is very well behaved, so it decays faster than any exponential, you could extend the transform to complex s, but I'm not sure what benefit you get from this.