Are the norms $\|\cdot\|_1$ and $\|\cdot\|$ equivalent?

Question

Let $\|x\|_1=\sum_{n=1}^{\infty}|x_n|$ is a norm for $\ell^1$, suppose $x=\{x_n\}\in\ell^1$ and $\|x\|=\sup|\sum_{k=1}^{n}x_k|$.

Are the norms $\|\cdot\|_1$ and $\|\cdot\|$ equivalent?

Answer 1

As the sequence $\sum_{n=1}^N |x_n|$ is monotone increasing (since $|x_n|\geq 0$) they are in fact equal.

EDIT: For the modified question: This new norm is not equivalent to $||-||_1$. Take for example the sequence $x^l=(\frac{1}{2^l},-\frac{1}{2^l},\frac{1}{2^l},\cdots,-\frac{1}{2^l},0,0,0,\cdots)$ (there are $2^l$ non zero terms). Then $||x^l||_1=1$ and $||x^l||=\frac{1}{2^l}$. So $x^l$ converges to $0$ for $||-||$ but does not converge to $0$ for $||-||_1$.

Answer 2

It's not clear to me what you mean by $\sup \sum_{n=1}^n \|x_n\|$.

If you mean $\sup_n |x_n|$, then consider the following:

Let $x(N)$ be the sequence in $\ell_1$ which starts with $N$ 1's and is 0 thereafter. So $x(N)_n=1$ for $n\leq N$ and $x(N)_n=0$ for $n>N$.

Then $\|x(N)\|_1=N$ but $\|x(N)\|_\infty =1$.

So the two norms are not equivalent.

If you mean $\sup_N \sum_{n=1}^N |x_n|$, then (as Quimey says in his answer) this is $\sum_{n=1}^\infty |x_n|$ and so the two norms are equal.