Let $x \in \ell^1$ and $z \in \ell^2$ taking values in $\mathbb{R}$ and define a linear map $T_z: \ell^1 \rightarrow \ell^2$ as follows:
$y_1=0$ and $y_n=\sum_{k=1}^{n-1}z_{n-k}x_k$ for $n\geq 2$. Then, let $(T_z(x))_n=y_n$ for all $n$. I want to show that $\Vert T_z(x) \Vert_2 \leq \Vert z \Vert_2 \cdot \Vert x \Vert_1$. My present work is as follows:
$\Vert T_z(x) \Vert_2 = \left( \sum_{n} \vert (T_z(x))_n \vert^{2} \right)^{1/2} = \left( \sum_{n} \vert y_n \vert^{2} \right)^{1/2} = \left( \sum_{n} \lvert \sum_{k=1}^{n-1}z_{n-k}x_k \rvert^{2} \right)^{1/2} \leq \left( \sum_{n} ( \sum_{k=1}^{n-1} \lvert z_{n-k}x_k \rvert)^{2} \right)^{1/2}$.
But I can't seem to get further with this, since I want to use the Cauchy-Schwarz inequality here but it gets really messy quickly. Am I going in the right direction or have I made some flaw somewhere?
Well, $T_zx$ is just the convolution of $x$ and $z$, so the inequality you want is just a special case of Young's inequality. In a discrete case like this Young's inequality should follow just from the triangle inequality in $\ell^2$ if you look at it right. Let's see: $$\begin{aligned}T_zx&=(0,x_1z_1,x_1z_2+x_2z_1,\dots) \\&=x_1(0,z_1,z_2,\dots) \\&+x_2(0,0,z_1,z_2,\dots) \\&+x_3(0,0,0,z_1,z_2,\dots) \\&+\dots \end{aligned}$$
Sure enough; apply the triangle inequality in $\ell^2$ to that sum and you're done.