According to the sub-multiplicative property of (some) matrix norms, we know that
\begin{equation} \| \boldsymbol I \| \leq \|\boldsymbol A\| \|\boldsymbol A^{-1}\| \ , \end{equation}
for some invertible matrix $\boldsymbol A$.
Let $\boldsymbol A(t) = \frac{1}{t}\sum\limits_{s=1}^t \boldsymbol H(s)$ be a positive-definite matrix such that $\boldsymbol H(t)$ is bounded for $t\geq 1$. Given that, I would like to know if
\begin{equation} \lim_{t \rightarrow \infty} \|\boldsymbol A(t)\| \|\boldsymbol A^{-1}(t)\| < \infty \ . \end{equation}
And out of curiosity, I would also like to know if the term $\|\boldsymbol A\| \|\boldsymbol A^{-1}\|$ is bounded for all eligible matrices $\boldsymbol A$.
According to the bounded inverse theorem, https://en.wikipedia.org/wiki/Bounded_inverse_theorem, the term $\|A\|\|A^-1\|$ is bounded, whenever $A$ is a linear bounded operator that maps from one banach space to another.
Since $A(t) = \sum_{s=1}^tH(s)$ is a finite sum of bounded linear operators, $A(t)$ is also a bounded linear operator. This means that, given $H(s)$ maps from a banach space to another, $\|A\|\|A^-1\|$ is bounded.
In the limit of $t\to\infty$ this probably breaks, since $A(t)$ is no longer guaranteed to be a bounded linear operator, which implies that the bounded inverse theorem might not hold.