Why does the point spread function not violate the linearity of the Fourier transform? In radio astronomy the point spread function is the Fourier inverse of the $uv$-sampling function of a telescope. The $uv$-sampling function is a sum of sampling functions (one for each baseline). So the $uv$-sampling function can be written as $s(u,v) = \sum_{pq} s_{pq}(u,v)$. The individual terms $s_{pq}(u,v)$ is the $uv$-track from a specific baseline. The $s_{pq}(u,v)$ function is one on an elliptical track in the $uv$ plane and zero everywhere else (so a very sparse function). The sampling function is then equal to the sum of these individual elliptical tracks and is also very sparse. The problem is that $\mathcal{F}^{-1}\{s(u,v)\} = \frac{1}{N}\sum_{pq}\mathcal{F}^{-1}\{s_{pq}(u,v)\}$, where $N$ is the amount of terms in the sum. If I do not use the average the Fourier inverse is $N$ times too large. Why is this correct? According to me I am violating the linearity of the Fourier transform... If you do not divide by $N$ then if you had 3 baselines your PSF would be 3 at (0,0) instead of 1.      

I think we can see $s_{pq}'(u,v)$ as a delta sequence. We can also write, $\frac{1}{N}\sum_{pq}\mathcal{F}^{-1}\{s_{pq}(u,v)\}=\frac{1}{N\times n_{uv}}\sum_{ij}\mathcal{F}^{-1}\{s_{ij}(u,v)\}$ , $s_{ij}(u,v)=\delta(u,v)$ and $\mathcal{F}^{-1}\{\delta(u,v)\}=1$, The total number of sampling points is $N\times n_{u,v}$. This imply that $\sum_{ij}\mathcal{F}^{-1}\{s_{ij}(u,v)\}=N\times n_{u,v}$. I think this is why for a source at the phase center, the sky is equal to the average of all visibilities which also result as the PSF. $I=\frac{1}{nomber-of-visibilities}\sum_{vis}s_{ij(u,v)\exp(-2\pi j(u\times 0+v\times 0+w\times 0))}$
