Given a sequence of matrices $F_k\in \mathbb{R}^{n\times n}$, $H_k\in \mathbb{R}^{m\times n}$, $m< n$. If $\underline{f}^2 I_{n\times n}\leq F_kF_k^T\leq {\overline{f}}^2 I_{n\times n}$, and $\underline{h}^2 I_{m\times m} \leq H_k H_k^T\leq \overline{h}^2 I_{m\times m}$. Can the following inequality hold?
$$F_k H_k^T H_k F_k^T \geq \underline{f}^2 H_k^T H_k$$
Remark: From the condition, it indicates $H_k^TH_k$ is positive semi-definite.
No, it can't, I found a counterexample.
Let $F_k=\begin{bmatrix}1 &2\\ 0 &2 \end{bmatrix}$, $H_k=[0 \quad 1]$, Accordingly, we can find $\underline{f}^2=0.2$.
$H_k^T H_k= \begin{bmatrix} 0 &0 \\ 0 &1 \end{bmatrix}$, $F_kH_k^T H_k F_k^T= \begin{bmatrix} 4 &4 \\ 4 &4 \end{bmatrix}$
while $\underline{f}^2 H_k^T H_k=\begin{bmatrix} 0 &0 \\ 0 &0.2 \end{bmatrix}$.
To this end, we have $F_kH_k^T H_k F_k^T-\underline{f}^2 H_k^T H_k= \begin{bmatrix} 4 &4 \\ 4 &3.8 \end{bmatrix}$, which doesn't $\geq0$.
Remark: the eigenvalues of $F_k F_k^T$ are $\frac{9}{2}-\frac{\sqrt{65}}{2}$ and $\frac{9}{2}+\frac{\sqrt{65}}{2}$. Choosing $\underline{f}=\frac{9}{2}-\frac{\sqrt{65}}{2}$, such that $\underline{f}^2=0.2$