The maximum obtainable angular resolution of an optical system with some given aperture is well known, but it seems to me that this isn't a real theoretical limit. The assumption is that you are going to take a picture using the system and no further processing will take place. However, given the known point spread function of the optical system, you could perform a deconvolution calculation. The final resolution of the processed image should be limited by the noise. So, what then seems to matter is the observation time (the longer you integrate the signal the better the signal/noise ratio will be).
So, what is the correct theoretical limit of the maximum resolution in terms of brightness of the two sources to be resolved, the aperture and the observation time (assuming that the only noise comes from the fluctuations in the finite numbers of photons from the sources)?
Answer
I'm going to focus on the information contained in the light field itself. This excludes from the discussion many if not all "superresolution" techniques, which directly or indirectly make use of information further to that in the imaging light field[footnote 1].
It is true that you can do a deconvolution to get somewhat below the traditional diffraction "limit" if the signal to noise ratio is very high. But there is a fundamental limit even to this, and it leads to tradeoff between distance from the object and resolution. Near field microscopy can image arbitrarily small features with light, but herein lies the catch - you need to be near to the object, and, for a given noise level, no matter how small, the imageable feature size decreases exponentially with distance from the object. The notion of what leads to the notion a hard limit arises from:
Only nonevanescent waves (corresponding to truly free photons) can convey Fourier component information to an imaging system that is arbitrarily far from the object
The phenomenon is indeed best understood through evanescent waves. If you want to encode Fourier component of a transverse feature into the light field and that component's spatial angular frequency $k_f>k$ (here $k$ is the light's wavenumber), then as the plane wave encoding this component propagates away from the object (call this the $z$ direction), its amplitude varies as $\exp(-\sqrt{k_f^2-k^2}\,z)$, i.e. the wavevector component becomes imaginary and the amplitude swiftly drops off with distance. As $z\to\infty$, only the nonevanescent waves are left, so the system transfer function looks more and more like a hard limitting lowpass filter with cutoff spatial frequency $k$ as $z$ increases. If you want to image features of characteristic length $d<\lambda$, then the loss in signal to noise ratio is:
$$\begin{array}{lcl}SNR &=& SNR_0-40\,\pi\,z\,\sqrt{\frac{1}{d^2}-\frac{1}{\lambda^2}}\,\log_{10}e\quad\text{(decibel)}\\&\approx& SNR_0-40\,\pi\,\frac{z}{d}\,\log_{10}e\quad (d\ll\lambda)\end{array}$$
where $SNR_0$ is the signal to noise ratio you would get if you held the SNOM right up on the imaged object and $z$ is the distance of the SNOM tip from the object. This is a horrifically fast dropoff. If you probe scans $1{\rm \mu m}$ from the imaged object and we wish to see $50{\rm nm}$ sized objects, the signal to noise lost by the mere $1{\rm \mu m}$ standoff is 1000 decibels (a power factor of $10^{100}$!). Practically speaking, your probe must be within a distance $d$ or less of the imaged object, where $d$ is the subwavelength feature length you wish to see; the above formula then gives an SNR dropoff of about $54{\rm dB}$ when $z=d$.
Footnotes
[1]. For example STED depletes fluorophores out of focus before taking the final light reading, thereby disabling anything more than a few tens of nanometres from the focus from registering)
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