Decompression is the reverse: take these 8x8 quantized DCT coefficients, perform an inverse 8x8 DCT top get pixel values.
The 11-bit dynamic range part you claim is merely from a color profile, which takes the resulting 8 bits per channel (i.e., 256 possible values) and spreads them over a gamma curve to an 11-bit range. But there are still only 256 possible levels per channel, too few for quality image editing.
Think of it as squaring: taking [0,255] as your input and squaring every value gives you a range of 0 to 255^2 = 65025, but that does not allow you to store any value in that range. It only allows you the 256 values that are squares.
So the dynamic range is 11 stops, but the number of representable levels per channel is still 8 bit: 256 levels. This makes gradients band no matter how you do them in JPEG.
It's why photo processing software wants RAW, not JPEG. JPEG, besides being lossy, does not allow enough steps.
One example: given a so-so RAW, at 11 bits, you can pull out dark things or darken bright things smoothly. This is not possible once you go to jpeg, for any implementation of jpeg.
The cosine values are transcendental numbers, they can have an infinite number of decimal places, yes? So adding up 64 products of (cosine values * 8 bit integer) to get 1 pixel value can obviously have more than 8 bits.
Or simply read the libjpeg source.
Don't like that, read this [1]: "JPEG images are always recorded with 8-bit depth. This means the files can record 256 (28) levels of red, green and blue."
Don't like that, here [2] is the JPEG ISO standard, section 4.11, baseline jpeg, "Source image: 8-bit samples within each component".
A DCT takes an 8x8 pixel input, 8 bits per channel, and transforms them into an 8x8 output. It matters not what these are - the information theory content is nothing more than what was put into it. There is not suddenly magically more information content.
More simply, appending zeroes to a number does not mean you can represent more numbers. You simply can represent the exact same numbers, just wasting more space.
None of what you wrote adds more resolution at the output. It simply isn't there.
If I give you 5 possible inputs to a function, then you have 5 possible outputs, not matter how many digits you finagle into representing the output.
Jpeg has 8 bits of resolution per channel. End of story. That is why professional photos are taken and edited in raw - you get more bits of resolution per channel.
I'm not sure why you're still arguing this. It's a longstanding, well known issue, and I explained it all again very simply.
If you think it isn't true, encode one of your magic jepgs with more than 256 levels of gray and post it here. Good luck :)
If you cannot do that, then maybe you should consider that you're wrong.
It is very precise terminology, used correctly. It's also covered in your links; you can read it there.
Now, if you encode gray levels in RGB, at 8 bits per channel, you do indeed end up with only 256 gray levels in the image, because for each pixel, R=G=B.
The most common JPEG decoders though (in particular libjpeg-turbo) are using a cheap but not super precise iDCT that has 8-bit YCbCr as output, which then gets chroma-upsampled if needed and converted to 8-bit RGB. That causes the effective precision in reds and blues to be only 7-bit. But in principle you could have about 10 bits of effective RGB precision, it just requires a sufficiently precise JPEG decoder.
The JPEG spec does specify both 8 and 12 bit sample precision for lossy, but I don't think anyone ever implemented 12-bit since libjpeg never cared about it.
To save others from Googling, the Wikipedia page: https://en.wikipedia.org/wiki/JPEG