Let $f,g\in\ell^1(\mathbb{Z})$, where $$\ell^1(\mathbb{Z}):=\lbrace f:\mathbb{Z}\rightarrow\mathbb{C}:\sum_{n=-\infty}^\infty|f(n)|<\infty\rbrace$$ and convolution is defined as: $$(f*g)(n):=\sum_{m=-\infty}^\infty f(m)g(n-m).$$ Let $n\in\mathbb{Z}$. I have proved that $f*g:\mathbb{Z}\rightarrow\mathbb{C}$ is well-defined. So I need to prove that $$\sum_{n=-\infty}^\infty|(f*g)(n)|\leq\Big{(}\sum_{n=-\infty}^\infty|f(n)|\Big{)}\Big{(}\sum_{n=-\infty}^\infty|g(n)|\Big{)}.$$
I have that $$\sum_{n=-\infty}^\infty|(f*g)(n)|\leq\sum_{n=-\infty}^\infty\sum_{m=-\infty}^\infty|f(m)|\,|g(n-m)|,\:\:\:\:\:\:\:\:(*)$$ provided that the RHS converges. Since $$\sum_{j=-n}^n\sum_{i=-m}^m|f(i)|\,|g(j-i)|\leq\sum_{i=-m-n}^{m+n}\sum_{j=-m-n}^{m+n}|f(i)|\,|g(j)|=(\sum_{i=-m-n}^{m+n}|f(i)|)(\sum_{j=-m-n}^{m+n}|g(j)|)$$ Is it possibly to simply take limits as $m,n\rightarrow\infty$ to get the result? Something doesn't seem quite right...
We can simply reorder summands because they are all non-negative. $$\Bigg(\sum_{-\infty}^\infty |f(n)|\Bigg)\cdot\Bigg(\sum_{-\infty}^\infty |g(n)|\Bigg)=\sum_{(m,n)\in\Bbb Z\times\Bbb Z}|f(m)||f(n)|=\sum_{-\infty}^\infty\sum_{-\infty}^\infty |f(m)||g(n-m)|$$