Let $A_i$ be matrices such that
$$\left\Vert \sum_i A_i \right\Vert_\infty \leq \varepsilon,$$
where $\Vert\cdot\Vert_\infty$ is the operator norm and is equal to the largest singular value of its argument and $\dagger$ denotes the conjugate transpose. Can one say anything about an upper bound for
$$\left\Vert \sum_i A^\dagger_iA_i \right\Vert_\infty$$
in terms of $\varepsilon$? Moreover, is there a converse statement too i.e. given
$$\left\Vert \sum_i A^\dagger_iA_i \right\Vert_\infty\leq \delta$$
is there an upper bound for $\left\Vert \sum_i A_i \right\Vert_\infty$ in terms of $\delta$?
Suppose that we have finitely many matrix $A_1,\dots,A_n$ to consider. Let $\tilde A$ denote the matrix $$ \tilde A = \pmatrix{A_1\\ \vdots\\ A_n}. $$ We see that $\sum_{i=1}^n A_i^\dagger A_i = \tilde A ^\dagger \tilde A$, so that $$ \left\| \sum_{i=1}^n A_iA_i^\dagger\right\|_{\infty} = \left\|\tilde A^\dagger \tilde A \right\|_\infty = \left\| \tilde A\right\|_\infty^2 \leq \delta. $$ On the other hand, we can write $\sum_{i=1}^n A_i$ as the product $M \tilde A$, where $$ M = \pmatrix{I & \cdots & I}. $$ We find that $\|M\|_\infty = \sqrt{n}$, so that $$ \left\|\sum_{i=1}^n A_i \right\|_\infty = \| M \tilde A\|_\infty \leq \|M\|_\infty \cdot \|\tilde A\|_\infty = \sqrt{n} \cdot \|\tilde A\|_\infty \leq \sqrt{n\delta}. $$