By imagining that the surface is NOT Lambertian but is a special surface with constant intensity (the first line of my first post)

But such a surface CANNOT EXIST. IT IS NOT PHYSICAL. Any physical surface will have a falloff with view angle eventually at least as large as the Lambertian reflectance cosine term falloff and radiance shall never tend to infinity. A "special surface with constant intensity" has by definition infinite density and hence zero area, and is thus a point light source which cannot exist in real life and for which radiance is not defined anyway (as it has solid angle zero). If the surface had any area, the power would have to be infinite!

For instance, if you try to evaluate irradiance with your "constant intensity" surfaces, the integral will not converge. Period. For instance, just try to measure the radiant exitance from any point on your theoretical surface! The radiance is proportional to 1 / cos(theta), so let's calculate the radiant exitance by summing up radiance in every direction:

And as you can see, this integral does not converge. Yet the result (radiant exitance) has units W/m^2, and hence should be a physically valid amount of intensity.

Your thought experiment is meaningless. It's like saying "consider this rectangle with negative surface area" or "imagine if this ball of infinite mass collides with the ground". It doesn't mean anything. The theory does not even apply to such surfaces which are outside the realm of physics. I am sure you can see that now.

it is the BRDF that results in constant radiance ...

No, it is NOT. The BRDF is exactly the opposite and is a parameter to help you configure your reflectance distribution to depend on view angle (within physically plausible limits) in order to approximate materials better. Check out the rendering equation. There is a cosine term outside the BRDF. Where do you think it comes from? It's Lambert's cosine law, applied backwards (for irradiance). It must be there.

On the topic of non-lambertian surfaces:
"Lamberts cosine term" is different to "lambertian surfaces". Lambertian surfaces use BRDF(...) = 1 (not BRDF(...) = cos(theta), where theta is the incident angle). The incident cosine term is not a property of the surface at all, but a fact of geometry.

That is to say, that Lambert proved the cosine term was necessary, and then a Lambertian BRDF is one that does nothing (multiplies by 1), relying on just this mandatory cosine term to give a somewhat correct result.

[edit] this is not true... [/edit]
Using BRDF(...) = 1 ("lambertian") is very common, but it violates helmholtz reciprocity, so it isn't physically plausible. The fix isn't to remove the cosine term, but add another one!

To make it obey helmholtz reciprocity, you actually need to use BRDF(N,V) = dot(N,V) = cos(theta) (where theta is the exitance angle), so that your rendering equation has two cos(theta) multiplications -- one for the path where the light enters the lambertian reflector to begin with (the irradiance at the reflector) and one for the path where the light exits the reflector towards the viewer (the radiance in some direction).

i.e. Seeing as the incident cosine term is mandatory, then helmoltz reciprocity makes an exitance cosine term also mandatory (you can also reverse that, depending on which one you've proven to be required first ), so you can't just remove this term and be physically plausible.

Using BRDF(...) = 1 (lambertian) is very common, but it violates helmholtz reciprocity, so it isn't physically plausible.
To make it obey helmholtz reciprocity, you actually need to use BRDF(N,V) = dot(N,V) = cos(theta) (where theta is the exitance angle), so that your rendering equation has two cos(theta) multiplications -- one for the path where the light enters the lambertian reflector to begin with (the irradiance at the reflector) and one for the path where the light exits the reflector towards the viewer (the radiance in some direction).

Are you sure that's right? BRDF is ratio of radiance to irradiance, so the exitance cosine term is already handled in the incidence cosine term of the next "bounce" of the light ray. And BRDF(..) = 1 clearly obeys Hemlholtz reciprocity, since BRDF(incident, exitant) = BRDF(exitant, incident) = 1. No?

That said for a constant BRDF we need the constant (albedo) to be between 0 and 1 / pi otherwise energy conservation is not achieved, I believe. In theory, anyway. This is usually baked into shaders.

I agree about the cosine term, though. Taking it out makes the definitions meaningless as they no longer stand for "radiant flux", "radiance", but "radiant flux with the cosine term missing", "radiance with the cosine term missing", etc.. it doesn't make sense. The theory doesn't make any sense anymore.

Are you sure that's right?

Ah yeah I've gotten mixed up.

Basic Lambertian is reciprocal, yes; you only run into issues when you try to make it energy conserving by incorporating both reflection and refraction according to Fresnel's law.

The common approach is to use N•L (incident theta) to calculate the reflection/refraction ratio, with only the refracted part using the lambertian (constant) BRDF. However, this doesn't obey helmholtz reciprocity. You also need to calculate the above ratios using N•V (exitant theta) and account for the fraction of the "lambertian reflectance" (which is incident light that is refracted and diffused into exit light) that actually reflects off the inside of the surface.

If you only perform the former calculations, then when swapping L and V, you get different results from the BRDF, which is non-physical.