# Is this norm equivalent to the $\ell_1$ norm?

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.

## 1 Answer

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, $$\|x\|_{\ell_1} = |x_1| + \sum_{n=2}^\infty |x_n| \le \left| \sum_{n=1}^\infty x_n\right| + 2\sum_{n=2}^\infty |x_n| \le 2 \|x\|'.$$

• Thanks! I thought it had something to do with the size of $x_1$. Jun 29, 2020 at 3:57
• @Aphyd you're welcome! Jun 29, 2020 at 3:58