I am studying for my qualifying exams and was asked to prove or disprove that the following norm is equivalent to the $\ell_1$ norm:
$$\lVert x \rVert' = 2\left\lvert \sum_{n=1}^{\infty}x_n \right\rvert + \sum_{n=2}^{\infty}\left(1+\frac{1}{n}\right) \lvert x_n\rvert$$
It was easy to show that $\lVert x\rVert' \leq 4\lVert x\rVert_1$ but I have been trying for quite a while to show there is a constant $C$ such that $C\lVert x\rVert_1 \leq \lVert x\rVert'$.
It has been so difficult I am beginning to believe they are not equivalent, when originally I thought they were.
The $\sum_{n=2}^\infty |x_n|$ part is common to both norms, so we just need to control $x_1$. This can be done as follows- $$|x_1| = \left| \sum_{n=1}^\infty x_n - \sum_{n=2}^\infty x_n\right| \le \left| \sum_{n=1}^\infty x_n\right| + \sum_{n=2}^\infty |x_n|.$$ Hence, $$ \|x\|_{\ell_1} = |x_1| + \sum_{n=2}^\infty |x_n| \le \left| \sum_{n=1}^\infty x_n\right| + 2\sum_{n=2}^\infty |x_n| \le 2 \|x\|'.$$