COMMENT ⊗   VALID 00002 PAGES
C REC  PAGE   DESCRIPTION
C00001 00001
C00002 00002	\input basic
C00005 ENDMK
C⊗;
\input basic
\magnify{1300}
 
$$\eqalign{I↓{ij}⊗= -\int↓{R↑p} {\partial \phi (\underline{x})\over \partial x↓i}
f(\underline{x}) \sigma↓{ij} {\partial g(\lambda ,\underline{x})\over 
\partial x↓j} d\underline{x}
-\int↓{R↑p} \phi(\underline{x}) {\partial f(\underline{x}) \over\partial x↓i} 
\sigma↓{ij} {\partial g(\lambda ,\underline{x}) \over\partial x↓j} d
\underline{x}\cr 
⊗\eqv -J↓{ij} -
K↓{ij}.\cr}\eqno(9)$$
(We used the formula $[\int\phi fG↑\prime = \phi fG]↑\infty↓-∞ - \int
\phi↑\prime fG - \int\phi f↑\prime G$  with the first term in the right
hand side vanishing).  Now  ${\partial\phi(\underline{x})\over\partial x↓i}
= -\sum↓k c↓{ki} x↓k \phi(\underline{x})$  and thus
$$\eqalign{
\sum↓i J{ij}⊗= -\int↓{R↑p} \sum↓k \sum↓i c↓{ki} \sigma↓{ij} x↓k
\phi(\underline{x})f(\underline{x}) {\partial g\over\partial x↓j} 
d\underline{x}\cr⊗=-\int↓{R↑p}x↓j\phi (\underline{x})f(\underline{x})
{\partial g(\lambda ,\underline{x})\over \partial x↓j}d\underline{x}
\quad  (\hbox{since}\quad 
C \Sigma = I).\cr}\eqno(10)$$
Summing in (9) over $i$ and $j$ we see by (10) and (8) that
$$\eqalign{F↑\prime(\lambda)⊗= -{1\over\lambda} \sum↓{ij} (I↓{ij} + J↓{ij}) =
{1\over\lambda} \sum↓{ij} K↓{ij}\cr
⊗= {1\over\lambda} \sum↓{ij} \sigma↓{ij}
\int↓{R↑p} \phi(\underline{x}) {\partial f(\underline{x}) \over
 \partial x↓i} 
{\partial g(\lambda,\underline{x})\over\partial x↓j} d\underline{x},\cr}
\eqno(11)$$
which is nonnegative by (4a) and since  ${\partial f\over\partial x↓i} ≥ 0$
by the monotonicity assumption on $f$  and the proof is complete.
 
\par\vfill\end