Derivation of KL divergence by Bregman divergence

This confuses me a little because if we take a system with state $x_t=(p,q)$ following the Hamiltonian which yields KL divergence as its Bregman divergence we would need the sum of $p$ and $q$ to be equal to 1, otherwise, we won't actually get KL divergence but an additional term. (See the proof in the above link, assume $p_1$ and $p_2$ there is our $p$ and $q$).

The Hamiltonian for this case is:

$$ H(p,q) = p\log p + q \log q $$

The loss would be the KL divergence

$$ c(x_1,x_2) = p_1\log (p_1/p_2) + q_1\log (q_1/q_2) $$

The state transitions would be such,

$$ x_{t+1} = J\nabla H(x_{t})+x_{t} +u_{t} + w_{t} $$

Let's assume our control to also be non-linear and a function of $\nabla H(x_t)$

$$ u_t = M\nabla H(x_t) $$

Furthermore, for convergence, if the cost function is Lipschitz continuous we would need

$$ c(x_{t+1}) - c(x_t) < L|x_{t+1} - x_t|< 1

$$

$\implies$

$$ \left|J\nabla H(x_{t}) + M\nabla H(x_t) + w_{t}\right| < 1/L $$

If $w_t$ is bounded by $|w|$, and the $\nabla H(x_t)$ by $G$, then

$$ |(J+M)\nabla H(x_t)| + |w| < 1/L $$