Distributional Limits of Fourier Transforms

Question

This is a problem for a practice applied analysis qualifying exam:

Let $f(x,y)$ be the indicator function of the rectangle $R = [-A,A]\times [-B,B]$ in $\mathbb{R}^2$. Compute $\hat f$, the Fourier transform of $f$. Evaluate the limit of $\hat f$ in the sense of distributions as $A \to \infty$ for fixed $B$.

After some fun manipulation, I calculated $\hat f (x_1, x_2) = \frac{2(\sin A x_1)(\sin B x_2)}{\pi x_1 x_2}$. I'm having trouble with the limit of $\hat f$ in the sense of distributions. It seems counterproductive in this case to use $\hat f \phi = f \hat \phi$ for $\phi \in \mathcal{D}(\mathbb{R}^2)$, as we just did all the work to find $\hat f$. However, when I've tried to calculate the quantity directly, I haven't made any meaningful progress.

$$\lim_{A \to \infty} \hat f \phi = \int_{\mathbb{R}^2} \hat f(x) \phi(x) dx = \int_{\mathbb{R}^2} \frac{2(\sin A x_1)(\sin B x_2)}{\pi x_1 x_2} \phi(x_1, x_2) dx_1dx_2$$

Answer 1

Not a full solution, but hopefully a help.

Let's do a variable change: $$\begin{align} \left\langle \frac{\sin Ax}{x}, \phi(x) \right\rangle & = \int \frac{\sin Ax}{x} \phi(x) dx \\ & = \{ \text{ set $\xi=Ax$ } \} \\ & = \int \frac{\sin\xi}{\xi/A} \phi\left(\frac{\xi}{A}\right) \frac{d\xi}{A} \\ & = \int \frac{\sin\xi}{\xi} \phi\left(\frac{\xi}{A}\right) d\xi \end{align}$$

Now, $\phi(\xi/A)\to\phi(0)$ as $A\to\infty$, so if swapping integral and limit can be justified, $$ \left\langle \frac{\sin Ax}{x}, \phi(x) \right\rangle \to \int \frac{\sin\xi}{\xi} \phi(0) d\xi = \to \int \frac{\sin\xi}{\xi} d\xi \, \phi(0) = C \phi(0) = \langle C\delta(x), \phi(x) \rangle $$ where $C = \int \frac{\sin\xi}{\xi} d\xi$.

Thus, $$ \frac{\sin Ax}{x} \to C\delta $$

So, if you can justify swapping integral and limit, then you only have to determine $C$ and take this to your two-dimensional case.