Limit of a discrete convolution is equal to zero The problem:
Let $ (u_n)_{n \in \mathbb{N} } $ and $ (v_n)_{n \in \mathbb{N} } $ be two sequences of real numbers convergents to zero. Suppose that there exists a
number $M>0$ such that
\begin{equation}
\label{q06c01}
 \forall n \in \mathbb{N} ,  \mid v_0 \mid +\mid v_1\mid+\ldots +\mid v_n\mid \leq M \quad .
\end{equation}
Prove that
\begin{equation}
\label{q06c02}
 \lim_{n\to \infty}u_0v_n+u_1v_{n-1}+\ldots +u_nv_0=0 \quad .
\end{equation}
My comments: I  have been reading the Wikipedia article for Cauchy product and it looks like the equation to be proved is a discrete convolution. That same article mentions Mertens's theorem (by Franz Mertens), but one of the conditions is that the series converges. The problem statement only says that the sequence converges, but that does not imply that the series will converge, $\frac{1}{n}$ being the most common counterexample to show that.
I also used ApproachZero to find relevant questions, but the closest thing I found was some tangential talk about the Cauchy product.
What mathematical tools should I use to solve the problem?
 A: This is a standard two-step solution when you have a strong condition $\sum |v_n| \le M < \infty$ and a weaker condition ($u_n \to 0$).
Let $x_n=u_0v_n+u_1v_{n-1}+\ldots +u_nv_0$
Let $C>0, |u_m| <C, m \ge 0$ (since $u_m$ is convergent, hence bounded)
Let $\epsilon >0$ arbitrary and using the strong condition, we pick $N(\epsilon), \sum_{n \ge N(\epsilon)} |v_n| < \epsilon /(2C)$ which means that the part of $x_Q$ that has $v$'s from $N(\epsilon)$ to $Q$ is at most $\epsilon /2$ for any $Q > N(\epsilon)$
But the rest of the terms in $x_Q$ are only $N(\epsilon)$ in number and that is now fixed for given $\epsilon$; also the $u$'s indices are high for each of them (they go from $Q-N(\epsilon)+1$ to $Q$).
So if we pick $Q_1(\epsilon), |u_Q| \le \epsilon/(2N(\epsilon)M), Q >Q_1(\epsilon)$ and then $Q(\epsilon)$ a bit higher to ensure that $Q>Q(\epsilon)$ implies $Q-N(\epsilon)+1 > Q_1(\epsilon)$, so for example $Q(\epsilon)=Q_1(\epsilon)+N(\epsilon)$ will do, the remaining terms in $x_Q$ are at most $\epsilon/(2N(\epsilon))$ so they sum to at most $\epsilon /2$ making the full $|x_Q| \le \epsilon, Q \ge Q(\epsilon)$
Since $\epsilon >0$ arbitrary, we get $x_Q \to 0, Q \to \infty$ so we are done!
