If $\sum_{n=0}^{\infty} |x_n| = \infty$ can convolution with $x$ map bounded sequences to bounded sequences?

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Suppose we have a sequence of real numbers $x_0, x_1, x_2, \ldots $ such that $\sum_n |x_n|$ is divergent.

Must there exist a bounded sequence $(y_n)_{n=0}^{\infty}$ such that the convolution of $(x_n)$ and $(y_n)$ is unbounded? That is, is there a bounded sequence $(y_n)$ such that the sequence $(z_n)$, where $$ z_n = \sum_{i=0}^{n} x_i y_{n-i},$$ is unbounded?

This is certainly true if all the $x_n$ are positive for example, but I can't see how to prove it in the general case.

The question is inspired by some readings in signal processing, but it felt more like a math question than a signal processing question.

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If $|x_j|$ is unbounded then simply take $y=1,0,0,\ldots$

Otherwise let $w_j = sign(x_j)$ and construct $y$ as

$$y=w_{e_1},w_{e_1-1},\ldots,w_0,w_{e_2},w_{e_2-1},\ldots,w_0, w_{e_3},w_{e_3-1},\ldots,w_0,w_{e_4},\ldots$$

Where $e_k$ is constructed inductively such that $\sum_{j=0}^{e_k} |x_j| > 2^j+\|x\|_\infty \sum_{l=1}^{k-1} (e_l+1)$

so that $|z_{\sum_{l=1}^k (e_l+1)}| > 2^j$.