Transmission capacity in "duobinary coding"
If you want to transmit a symbol sequence $A_k$ using baseband pulse amplitude modulation (PAM), the transmitted signal is

$$s(t)=\sum_kA_kp(t-kT)$$

where $p(t)$ is the transmit pulse, and $T$ is the symbol interval. If you want to avoid intersymbol interference (ISI), the pulse function $p(t)$ must satisfy the Nyquist criterion, which says that its value must be zero at multiples of the symbol interval:

$$p(0)=1,\quad p(kT)=0,\quad k\neq 0\tag{1}$$

The minimum bandwidth required to satisfy (1) is $1/2T$, but the corresponding pulse $p(t)$ is an ideal low-pass filter, i.e. a sinc function, which is of course impractical. So in practice, Nyquist pulses with a higher bandwidth (excess bandwidth) are used. This means that for a given bandwidth, the possible symbol rate is reduced as compared to the theoretically maximum possible value.

With duobinary signalling practical systems can be built with zero excess bandwidth, because the data symbols are filtered such that a spectral zero is introduced at half the baud rate. Note that this is a very simple filter unlike the ideal low-pass filter required for the minimum bandwidth Nyquist pulse. The price paid is the deliberate introduction of ISI and the introduction of a ternary (instead of binary) signal, which requires a higher SNR to achieve a given error probability.