Let $b \in l^1(\mathbb{Z})$. Define $T:l^2(\mathbb{Z}) \rightarrow l^2(\mathbb{Z})$ by $$ (Tx)_n = \sum_{m=-\infty}^{\infty}b_{n-m}x_m $$.
I need to prove that $\|T\| \leq \|b\|_{l^1}$.
I know that this is a direct result of Young's Inequality. Is there a direct way to show this for the discrete convolution above, without using Young's Inequality?
On
I think you can argue that $\|T\|=1$ for the special case that $T$ is a shift operator (that is, $b_i=1$ for one particular $i$ value, otherwise $b_i=0$). Then, since $T\mapsto \|T\|$ is convex, you know your inequality for all finitely supported $b$. If there exists a unit vector $x\in l^2(\mathbb Z)$ for which $\|Tx\|_2>\|b\|_1$, you should be able to derive a contradiction.
Young's inequality holds for any unimodular group. $\mathbb{Z}$ is unimodular. Hence the proof via Young's inequality works for discrete convolution as well.