# Fourier transform in 3d

1. Feb 25, 2012

### dikmikkel

1. The problem statement, all variables and given/known data
Show that $|\tilde{I}(\vec{k})|^2\leq CP^2$
Where C denotes a constant,
Using this inequality: $\int f(\vec{r})^*g(\vec{r})\,\text{d}\vec{r}\leq \int f^*f\text{d}\vec{r}\,\int gg^*\text{d}\vec{r}$
Where k denotes the fourier transform from r->k(in 3d)
R is assumed positive definite, real, symmetric and normalized like $\int R(r)dr = 1$.

2. Relevant equations
I already know that $P = \int\limits_{-\infty}^{\infty} I\text{d}r$
And that $\tilde{N}(I) = \tilde{R}(\vec{k})\tilde{I}(\vec{k})$
And that $\int N(I)I\,\text{d}\vec{r} = (2\pi)^{3/2}\int \tilde(R(\vec{k})|\tilde{I}(\vec{k})|^2\text{d}\vec{k}$
The Nonlocal Gross-Pitaveski equation:
$i\dfrac{\partial u(x,t)}{\partial t} +\nabla^2 u(x,t) -V(r)u + N(I)u = 0$
The convolution theorem and Parsevals theorem is also relevant.

3. The attempt at a solution
I tried many attempts, if anyone just could give a hint, for example this was one of the wrongs:
$|\int \dfrac{d\tilde{I}}{d\vec{k}}\,\text{d}\vec{k}|^2 = |\tilde{I}(\vec{k})|^2\leq \int 1\times 1^*\text{d}\vec{k} \int \dfrac{d\tilde{I}^*}{d\vec{k}}\dfrac{d\tilde{I}}{d\vec{k}}$
Or maybe with a delta function on it:
$|\int \tilde{I}\delta^3(\vec{y}-\vec{k})\text{d}\vec{k}|^2 = |\tilde{I}(\vec{k})|^2 \leq \int |I(k)|^2\text{d}\vec{k}$
But i really can't see what i should start with under the absolute squared sign.
Any ideas or better hints will be like gold for me.

Last edited: Feb 25, 2012