# Capacity region of 'symbol asynchronous' Gaussian multiple access channel

I was interested to learn what theoretical limits have been determined so far for the 'multiple access channel' with additive Gaussian noise. This brought me to the 1989 paper by Sergio Verdu, named "The Capacity Region of the Symbol-Asynchronous Gaussian Multiple-Access Channel". Found it here.

If anyone here is familiar with this area, can you help clear up a confusion I am facing regarding what Figure 7 and Figure 5 in this paper seem to be saying?

Say there are two transmitters 1 & 2, and the noise is $N$. The 'capacity region' to be the set of all possible combinations of $R_1$ and $R_2$, and can be drawn in a graph of $R_1$ against $R_2$. In such a graph, the 'Cover-Wyner pentagon' is defined as all points satisfying the following: $$R_1 \le {1 \over 2} \log {\left( 1 + \frac {P_1} {N}\right)} \\ R_2 \le {1 \over 2} \log {\left( 1 + \frac {P_2} {N}\right)} \\ R_1+R_2 \le {1 \over 2} \log {\left(1 + \frac{P_1+P_2}{N}\right)} \\ R_1 \ge 0 \\ R_2 \ge 0$$

Now my understanding of the conclusions of the paper is that the capacity region coincides with the Cover-Wyner pentagon when one of the following is true:

1. The two users are exactly symbol synchronous
2. The two users are symbol asynchronous, but both transmit the same waveform.

The Cover-Wyner pentagon makes intuitive sense to me because, as long as the rate sum $R_1+R_2$ is within that pentagon, no single user can 'pretend to be' two users and obtain a total rate higher than the Shannon capacity ${1 \over 2} \log {\left(1 + \frac{P_1+P_2}{N}\right)}$.

However, then the paper goes on to illustrate two cases where the rate sum seems to be higher than ${1 \over 2} \log {\left(1 + \frac{P_1+P_2}{N}\right)}$ as follows:

1. In Figure 5, where the symbol timing of the two users are offset by exactly half symbol duration.
2. In Figure 7, where the users are totally asynchronous, the waveforms transmitted by the two users are different, and the waveforms would be orthogonal if they were to be synchronous.

My question is, this apparently seems to create a 'loophole' by which a single user can divide its transmit power between two 'virtual' users, and end up with a rate that is higher than the upper bound of the channel capacity ${1 \over 2} \log {\left(1 + \frac{P_1+P_2}{N}\right)}$. Since that can't be possible, what am I missing here?

Or am I mixing up my understanding of the 'discrete time Gaussian channel' with the 'continuous time Gaussian channel' in some stupid way?

Thanks...