Let $A = [A_1, \ldots, A_m] \in \mathbb{R}^{n \times md}$, where for all $i=1,\ldots,m$, $A_i \in \mathbb{R}^{n \times d}$, $d>1$. Let $x = [x_1,\ldots,x_m]^\top \in \mathbb{R}^m$ with $\|x\|_2 \leq \varepsilon$, then what is a tight upper bound of $\big\| \sum_{i=1}^m A_i x_i \big\|_2$, i.e., the spectral norm of a weighted sum of blocks matrices $A_i$ ($A_i$ is not a vector here), in terms of the largest singular value $\sigma_{\max}(A)$ of $A$ and $\varepsilon$?

A direct calculation gives $\big\| \sum_{i=1}^m A_i x_i \big\|_2 \leq \sum_{i=1}^m \|A_i\|_2 |x_i| \leq \sigma_{\max}(A) \|x\|_1 \leq \sqrt{m} \sigma_{\max}(A) \varepsilon$. But is it possible to get $$\big\| \sum_{i=1}^m A_i x_i \big\|_2 \leq \sigma_{\max}(A) \varepsilon$$

why or why not?

When $d=1$, it simply holds from the Cauchy-Schwarz inequality. When $m=1$, this also holds trivially. But what if $m>1$ and $d>1$?