Problem statement: $$\underset{D}{\text{argmin} }\lVert D\Sigma - G \rVert_F^2 \text{ with constraint } D>O$$
$D$ and $\Sigma$ are positive diagonal, therefore the constraint interprets $\text{diag}(D)>0$ vector.
I have seen $\ge$ and $\le$ in examples, but my concern is having $>$ in this constraint since $D$ is positive.
Please guide. Thanks in advance.
It's a free minimum; then we do the calculation without any conditions.
Let $f(D)=tr((D\Sigma-G)(\Sigma D-G^T)),D=diag(d_{i}),\Sigma=diag(\sigma_i),G=[g_{i,j}]$. Let $\Delta$ be the set of diagonal matrices. The derivative is
$Df_D:H\in\Delta\mapsto 2tr(H^T(D\Sigma -G)\Sigma)$. If $f$ admits a local extremum in $D$, then, for every $H\in \Delta$, $Df_D(H)=0$; that is equivalent to:
for every $i$, $((D\Sigma -G)\Sigma)_{i,i}=0$ or (since $\sigma_{i}>0$),
for every $i$, $(D\Sigma-G)_{i,i}=0$ or, for every $i$, $d_i\sigma_i=g_{i,i}$, that is,
for every $i$, $d_i=g_{i,i}/\sigma_{i,i}$. In particular, if $f$ reaches its minimum in $D$, then, necessarily, $signum(d_i)=signum(g_{i,i})>0$.
$\textbf{Conclusion}$. If there is $i$, s.t. $g_{i,i}\leq 0$, then $\min(f)$ does not exist on $D>0$.
If, for every $i$, $g_{i,i}>0$, then $\min(f)$ is reached for $D=diag(g_{i,i}/\sigma_{i})$ and $f(D)=\sum_{i\not= j}{g_{i,j}}^2$.
EDIT. Of course, we can do the job directly, canceling the entries of the diagonal of $D\Sigma-G$ (because the other entries are always $-g_{i,j}$).