% file is s25.tex
THEOREM\\
The bound $B_1 =  B_2$.

In fact:
\[
\begin{array}{ccl}
 f_2(u)&=& q_u(y_u) \\
  &\geq&  \max_{(-1\leq y \leq 1)} q_{(u+\lambda e)}(y)\\
 & = & f_1(u+ \lambda e) \\
\end{array}
\]
and\\
$B_2=f_2(u)$, with Lagrange multiplier $\lambda$ for
$(RP^2_u)$ if and only if  $B_2=f_2(u)=f_1(u+\lambda e)=B_1$.