TITLE: Proof that Bellman update is a contraction
QUESTION [5 upvotes]: Following the proof that Bellman update is a contraction from Instructor's resources to Artificial Intelligence: A Modern Approach I fail to understand 1 step. Some definitions and proofs that were used in the final step:

definition of max norm: $||U||=\max_s|U(s)|$
definition of Bellman update: $U(s) = R(s) + \gamma\max_{a\in A(s)}\sum_{s'}P(s'|s,a)U(s')$
$U_i$ - vector of utilities for all states at the $i$th iteration.
statement: $|max_a f(a) - max_a g(a)| \leq max_a|f(a)-g(a)|$

$|(BU_i-BU'_i)(s)| = |R(s) + \gamma\underset{a\in A(s)}{\max}|\sum_{s'}P(s'|s,a)U_i(s') - R(s) \\- \gamma\max_{a\in A(s)}\sum_{s'}P(s'|s,a)U'_i(s')| \\= \gamma|\max_{a\in A(s)}\sum_{s'}P(s'|s,a)U_i(s')-\max_{a\in A(s)}\sum_{s'}P(s'|s,a)U'_i(s')| \\
\leq \gamma \max_{a\in A(s)}|\sum_{s'}P(s'|s,a)U_i(s')-\sum_{s'}P(s'|s,a)U'_i(s')| \\
= \gamma|\sum_{s'}P(s'|s,a^*(s))U_i(s')-\sum_{s'}P(s'|s,a^*(s))U'_i(s')|\\
= \gamma|\sum_{s'}P(s'|s,a^*(s))(U_i(s')-U'_i(s')|$

then the final part of proof where we use the above:
$||BU_i-BU'_i|| = \max_s|(BU_i-BU'_i)(s)| \\
\leq \gamma\max_s|\sum_{s'}P(s'|s,a^*(s))(U_i(s')-U'_i(s'))| \\
\leq \gamma\max_s|U_i(s)-U'_i(s)|\\
=\gamma||U_i-U'_i||$
How do I get the inequality $\gamma\max_s|\sum_{s'}P(s'|s,a^*(s))(U_i(s')-U'_i(s'))|
\leq \gamma\max_s|U_i(s)-U'_i(s)|$?

REPLY [5 votes]: It's a bit of an old question, but thought I'd answer it, anyways.
Note that (a)
$$
P(s'|s,a^*(s)) \ge 0, \sum_{s'} P(s'|s,a^*(s)) = 1
$$
and (b), that
$$
U_i(s') - U_i'(s') \le \max_s | U_i(s) - U_i'(s)|
$$
using both of these statements yields
$$
\begin{align}
\sum_{s'} \left(P(s'|s,a^*(s))(U_i(s') - U_i'(s'))\right) &\le \sum_{s'}\left( P(s'|s,a^*(s))\max_s | U_i(s) - U_i'(s)|\right)\\
&=\left( \sum_{s'}P(s'|s,a^*(s))\right)\max_s | U_i(s) - U_i'(s)|\\
&=\max_s | U_i(s) - U_i'(s)|
\end{align}
$$
for all $s$, from which the final result follows.