The discounting condition is as follow:
There exists some $\beta \in (0, 1)$ such that $[T(f + a)](x) ≤ (T f)(x) + βa$, for all $f ∈ B(X), a ≥ 0, x ∈ X$.
While the monotonicity condition makes sense, I can't give a nice meaning to this property.
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2 Answers
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Without discounting, you cannot show either
$$ T(g + || f - g||) \leq Tg + \beta || f - g|| $$
or
$$ T(f + || g - f||) \leq Tf + \beta || g - f|| $$
and thus you cannot demonstrate that $T$ is a contraction mapping in the traditional proof.
For intuition, note that Blackwell's sufficiency conditions for a contraction mapping is a fixed-point theorem used to establish the convergence of value function iterations. In the simplest case possible of a collapsed state space with the linear operator $T(W) = \sigma + \beta W$, it is clear that $\beta \in (0,1)$ will always lead to an intersection of $T(W)$ (from above) with the 45-degree line $T(W) = W$ and thus converge at a fixed point independent of the initial guess. The discounting condition that pins down such a $\beta$ assures this.
In systems with fully specified state spaces, however, discounting is no longer sufficient to establish uniqueness. Monotonicity is additionally required in these cases to establish the sufficiency of Blackwell's conditions because of additional concerns over dynamics.
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In SLP, $T(v)=\sup \{f(x,y)+ \beta v(y)\}$, where $v(y)$ is the value function in the next period and $\beta\in(0,1)$ is the discounting factor. Check Blackwell's discounting condition in order to get the intuition.
\begin{align} T(v+a) = Tv + \beta a \end{align} That is, if there is an increase by $a$ in the next-period value function, the increase should be discounted to get the present value.
But the above intuition may be specific. Generally, the discounting condition should be termed as the shortening condition. That is, $a$ should be shortened after mapping.
