Excessive function

Remarks

Under clause 1, $t \mapsto P_t f$ is decreasing, since $P_{t+s} f = P_t (P_s f) \le P_t f$; so the limit in clause 2 exists and is at most $f$. A function with clause 1 alone is called supermedian; for $\alpha > 0$, the $\alpha$-excessive functions are those that are excessive for the semigroup $e^{-\alpha t} P_t$. Example: let $(P_t)$ be the semigroup of Brownian motion in $\R^3$ and $f(x) = 1/\abs{x}$, with $f(0) = \infty$. Then $f$ is a nonnegative superharmonic function, and averaging it against the Gaussian kernel gives $P_t f \le f$ with $P_t f \uparrow f$ as $t \downarrow 0$, so $f$ is excessive. Non-example: for the same semigroup, let $f$ be the indicator of the single point $0$. Then $P_t f = 0 \le f$, so clause 1 holds, but $P_t f(0) = 0$ does not tend to $f(0) = 1$, so clause 2 fails. For Brownian motion in $\R^d$, the excessive functions are exactly the nonnegative superharmonic functions on $\R^d$, together with the constant $+\infty$.