Consider the discounting condition in the Blackwell's sufficient conditions:
(Reference: A related question).
There exists some $\beta \in (0, 1)$ such that $[T(f + a)](x) ≤ (T f)(x) + βa$, for all $f ∈ B(X), a ≥ 0, x ∈ X$.
Here $f$ is a function from a space of bounded functions and $a$ is a positive constant. Without the positive constant, I obtain:
$$(Tf)(x)=b+\delta f(x)$$
With the constant, I obtain:
$$T(f+a)(x)=b+\delta[f(x)+a]=b+\delta f(x)+\delta a=(Tf)(x)+\delta a$$
Is my second line correct?
$f+a$ is a function that maps $x$ to $f(x)+a$. The operator $T$ is defined as a map between functions $$T:g\mapsto Tg \\ Tg(x)=b+\delta ~ g(x)$$ So taking $g=f+a$, $$T(f+a)(x)=b+\delta(f(x)+a)\\ =b+\delta ~f(x)+a\delta $$ Since $Tf(x)=b+\delta ~f(x)$, $$T(f+a)(x)=Tf(x)+a\delta$$