Upper and lower bound of Bayesian Error in 2 category case In Chapter 2 of [Duda,Hart,Stork]'s Pattern Classification book, it was discussed that full error of Bayes' decision rule in 2 category case is $P(error)=\int P(error\mid x) P(x) dx$. In part (a) of question 1 of the same chapter, the author asked us to replace $P(error \mid x)=2P(\omega_1 \mid x)P(\omega_2 \mid x)$ to get an upper bound on the full error.
Can somebody help me to see how the hint / replacement leads to the upper bound? I have tried to search around and suspect that it might have to do with the Chernoff bound but I can't establish the link. 
Thank in in advance.
 A: Unfortunately as you have stated the question above, it is difficult to follow without referring directly to the original reference.
In the above you miss a key fact which is that in the two category case (that a variable is $\omega_1$ or $\omega_2$), if we use the (Bayes) decision rule
$$
R(x) = \begin{cases}
\omega_1 & \text{if $\mathbf P(\omega_1 \, | \, x) > \mathbf P(\omega_2 \, | \, x)$,} \\
\omega_2 & \text{else.}
\end{cases}$$
Then the probability of an error (i.e. $\omega_1$ when the decision rule $R(x) = \omega_2$, and visa versa) conditioned on $x$ is given by
\begin{align*}
\mathbf{P}(\text{error}\, | \, x) & = \min \big(\mathbf P(\omega_1 \, | \,x), \mathbf P (\omega_2 \, | \, x)  \big) \\
& = \min \big(\mathbf P(\omega_1 \, | \,x), 1 - \mathbf P(\omega_1 \, | \,x) \big)
\end{align*}
This formula is essential to solving the problem, since now if we denote $p = \mathbf P(\omega_1 \, | \,x) \in [0,1]$ then the above formula says $\mathbf P(\text{error} | x) = \min(p,1-p)$.
If we denote $f_\alpha(p) = \alpha p(1-p)$, the question (parts a-d, not just part a) is asking you to prove that for all $p \in [0,1]$
$$ f_1(p)  \leq \min(p,1-p) \leq f_2(p),$$
so that $f_1$ is a lower bound and $f_2$ an upper bound on the error. Further that for any $\alpha \in (0,2)$, $f_\alpha$ is not a bound. That is, there exist points $q,r \in [0,1]$ such that 
\begin{align*}
f_\alpha(q) & < \min\big(q,(1-q)\big) \\
f_\alpha(r) & > \min\big(r,(1-r)\big).
\end{align*} 
The graph below demonstrates this. Hopefully this is enough of a hint for you to now solve the problem.

