TITLE: Integrals on a closed ball
QUESTION [3 upvotes]: Prove
$$ \int_{\overline{B}(0,r)}c'D^{-1}c\ \exp(-1/2 y'D^{-1}y)\ dy >\int_{\overline{B}(0,r)}(y'D^{-1}c)^2\ \exp(-1/2 y'D^{-1}y)\ dy $$ where $ D\in\mathbb{R}^{n\times n} $ is a diagonal matrix, $ \overline{B}(0,r) $ is a closed $n$-ball with nonzero radius,  and $c$ is an an arbitrary vector. The ' mark denotes the transpose. 
(I'd like ideas for a start at a solution if you're not  willing to provide a full solution. Thanks!)

REPLY [4 votes]: This can only be true if the entries of $D$ are positive. Otherwise if $D_i<0$, (zero is not allowed as it is invertible) take $c=e_i$ the unit vector in direction $i$ and your inequality says
$$
0> D_i\times \mbox{a positive cst} > \mbox{ a non-negative number}
$$
which is false. And $c$ must be a non-zero vector. 
Thus, suppose now all the $D_i$ are positive. 
Next, note that 
$$
(c^TDy)^2=\sum_{i=1}^n D_{i}^2c_i^2y^2_i +\sum_{i}\sum_{j\neq i} D_iD_jc_ic_jy_iy_j
$$
The first term is an even function in $y_i$, whereas the second one is odd.
Since the domain of integration is symmetric, and $y^TDy =\sum y_i^2D_i$ is even in $y_i$,
$$
\sum_{i}\sum_{j\neq i} \int_{B_r}D_iD_jc_ic_jy_iy_j \textrm{exp}\left(-\frac{1}{2}y^TDy \right) dy=0
$$
Thus,
$$
\int_{B_r} (c^TDy)^2 \,\textrm{exp}\left(-\frac{1}{2}y^TDy \right) 
= \int_{B_r} B(y)^TDy \,\textrm{exp}\left(-\frac{1}{2}y^TDy \right)
$$
with $B(y)_i=D_ic_i^2y_i$ Note that div$(B(y))=c^TDc$. As
$$
u^TDy \textrm{exp}\left(-\frac{1}{2}y^TDy \right) = u\cdot \nabla \left( -\textrm{exp}\left(-\frac{1}{2}y^TDy \right)\right)
$$
and Green's Theorem says
$$
\int_{B_r} u \cdot \nabla v = \frac{1}{r}\int_{C_r} v u^T y  -  \int_{B_r} v \textrm{div}{u}  
$$
therefore, applying this to $u=B(y)$ and $v=-\textrm{exp}\left(-\frac{1}{2}y^TDy \right)$ you obtain
$$
\int_{B_r} (c^TDy)^2 \,\textrm{exp}\left(-\frac{1}{2}y^TDy \right)=-\frac{1}{r}\int_{C_r} \textrm{exp}\left(-\frac{1}{2}y^TDy \right)\sum_{i}(D_ic_i^2)y_i ^2  +\int_{B_r} \textrm{exp}\left(-\frac{1}{2}y^TDy \right)(c^T D  c) dy.
$$
The first term is negative, so the bound
$$
\int_{B_r} (c^TDy)^2 \,\textrm{exp}\left(-\frac{1}{2}y^TDy \right) <   
\int_{B_r} c^T D c \,\textrm{exp}\left(-\frac{1}{2}y^TDy \right) dy 
$$
holds.
{"set_name": "stack_exchange", "score": 3, "question_id": 813187}