$H-\Lambda H \Lambda > 0$ if $H>0$ and $\mathbf{0}_n<\Lambda<\mathbf{1}_n$? Let $H>0$ be a $(n\times n)$-positive definite Hermitian matrix. Let $\mathbf{0}_n<\Lambda=\{\lambda_1,\ldots,\lambda_n\}<\mathbf{1}_n$ be a diagonal matrix with entries $0<\lambda_j<1\,(1\leq j \leq n)$. Then can we say that $H-\Lambda H \Lambda > 0$?
 A: The answer in general is no and you can see this by tackling an easier problem and wielding continuity.
First I assume $H$ is not diagonal, and then fix some selection of $\mathbf{0}_n<\Lambda=\{\lambda_1,\lambda_2,\ldots,\lambda_n\}<\mathbf{1}_n$  now define
$\Lambda' :=\{1,\lambda_2,\ldots,\lambda_n\}$

technical nit:
if column 1 of $H$ is propotional to the 1st std basis vector, then instead 'attack' column $j$ which is not proportional to $\mathbf e_j$ and do this by instead selecting
$\Lambda' :=\{\lambda_1,\ldots,\lambda_{j-1},1,\lambda_{j+1},\ldots,\lambda_n\}$. Since $H$ is not diagonal, it is easiest to just assume WLOG that column 1 of $H$ has at least 2 non-zero components-- and the argument here tacitly does this.

Then
$\big(H-\Lambda' H \Lambda'\big)$
is a Hermitian matrix with a zero in the top left corner but that first column is non-zero, hence the matrix is not positive semi-definite.  (i.e. it necessarily has mixed signature).   Finally for $\tau \in [0,1]$ define
$\Lambda(\tau):= \tau \cdot \Lambda + (1-\tau)\cdot \Lambda '$
By topological continuity of eigenvalues
$\Big(H-\Lambda(\tau) H \Lambda(\tau)\Big)$ is not positive semi-definite for $\tau \in [0,\delta)$ for some $\delta \gt 0$. And setting e.g. $\tau:= \frac{\delta}{2}$ gives a counter-example.
