Prove that, in vector space $l_1$, the expressions $\sum_{n=1}^{\infty} |x_n|$ and $\sup_{n \in \mathbb{N}} |x_n|$ define two non-eq. norms.

Question

Prove that, in vector space $l_1$, the expressions $\sum_{n=1}^{\infty} |x_n|$ and $\sup_{n \in \mathbb{N}} |x_n|$ define two non-eq. norms.

Here is my attempt:

I know this result: If $||.||_{1}$ and $||.||_{2}$ are two equivalent norms, then there exists two constans $A,B>0$ such that: A$||.||_{1} \leq ||.||_{2} \leq B||.||_{1}$

Since the vector space is $l_1$, I can guarantee that there is a constant $C$ such that $\sum_{n=1}^{\infty} |x_n|=C$. $C \geq 0$, because $|x_n|\geq 0 \forall n$. Then $C$ is an upper bound to the set $\{ |x_n| ;n=1,...\}$, so $\sup_{n \in \mathbb{N}} |x_n| \leq C$. Then the inequality that must be false is $A.C \leq \sup_{n \in \mathbb{N}} |x_n|$, but I'm failing to verify this.

Thanks.

Answer 1

One way to see it is to think of a vector which is in one space and not the other, and then "approximate" it by a sequence of vectors in both spaces. The "approximations" will then be bounded in one norm and blow up in the other. You've correctly observed that you want to find a vector in $\ell^\infty$ which is not in $\ell^1$ because the reverse inclusion holds. Now try looking at the vector $x$ with $x_n=1$ for all $n$.