Norm $|\!|\!|x|\!|\!|=\sum_{i=1}^{\infty} 2^{-i}|x_i|$ in $l_2$ is not equivalent to the norm $\|\cdot \|_2$

Question

Show that the norm $|\!|\!|x|\!|\!|=\sum_{i=1}^{\infty} 2^{-i}|x_i|$ in $l_2$ is not equivalent to the norm $\|\cdot\|_2$

We must show that we cannot do

$$a\sum_{i=1}^{\infty} 2^{-i}|x_i|\le \|\cdot\|_2\le b\sum_{i=1}^{\infty} 2^{-i}|x_i|$$

for $a$ and $b$ positive.

I'm having problems understanding what I should do. The norm $|\!|\!|x|\!|\!|$ is only applicable to points in $x$ that converge. For example, $(1,1,1,\dotsc)$ would work for the first norm, but not for $\|x\|_2$. So what should be done here at all?

Answer 1

If $x$ has $1$ at the n-th position and $0$ elsewhere then the right hand inqaulity gives $1 \leq b\frac 1 {2{n}}$. This must hold for every $n$. Do yo see a contr4adiction?

Answer 2

Here is a hint: you could, as you've already noted, use the fact that the $|||\cdot |||$ norm has a built-in damping effect caused by the $2^{-i}$ factors. So perhaps you could assume that the norms are indeed equivalent, and then use a strategic choice of vectors to contradict the existence of the constants $a$ and $b$ which you mention in your post. Your choice of $(1,1,1,\ldots)$ won't work since it is not in $\ell_2$, as you mentioned, but perhaps some vectors which look similar and are in $\ell_2$ could be of use to you.