How Can the Convolution Theorem Help Prove a Fourier Transform Inequality in 3D?

[dikmikkel]

Homework Statement

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 $\int R(r)dr = 1$.

Homework 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.

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.

 

[mark726]

One possible approach to solving this problem is to start by using the convolution theorem to rewrite the Fourier transform of the intensity, \tilde{I}(\vec{k}), as a convolution of the Fourier transforms of the intensity and the real-space function R(\vec{r}):

\tilde{I}(\vec{k}) = \int \tilde{R}(\vec{k}-\vec{k}')\tilde{I}(\vec{k}')\,\text{d}\vec{k}'

Then, use the given inequality for the convolution of two functions to write:

|\tilde{I}(\vec{k})|^2 = \left|\int \tilde{R}(\vec{k}-\vec{k}')\tilde{I}(\vec{k}')\,\text{d}\vec{k}'\right|^2 \leq \int \tilde{R}^*(\vec{k}-\vec{k}')\tilde{R}(\vec{k}-\vec{k}')\,\text{d}\vec{k}' \int \tilde{I}^*(\vec{k}')\tilde{I}(\vec{k}')\,\text{d}\vec{k}'

Next, use Parseval's theorem to rewrite the integrals in terms of the real-space functions:

\int \tilde{R}^*(\vec{k}-\vec{k}')\tilde{R}(\vec{k}-\vec{k}')\,\text{d}\vec{k}' = \int R^*(\vec{r}-\vec{r}')R(\vec{r}-\vec{r}')\,\text{d}\vec{r}' = \int R^*(\vec{r}')R(\vec{r}')\,\text{d}\vec{r}' = 1

\int \tilde{I}^*(\vec{k}')\tilde{I}(\vec{k}')\,\text{d}\vec{k}' = \int I^*(\vec{r}')I(\vec{r}')\,\text{d}\vec{r}' = P^2

Substituting these results back into the inequality, we get:

|\tilde{I}(\vec{k})|^2 \leq \int \tilde{R}^*(\vec{k}-\vec{k}')\tilde{R}(\vec{k