Show that the function $f$ defined in $l_1$ by $f(x)=\Big(\Big(\sum|x_i|\Big)^2+\sum\frac{1}{2^i}|x_i|^2\Big)^{1/2}$ is a norm equivalent in $l_1$

Question

Show that the function $f$ defined in $l_1$ by $f(x) = \left(\left(\sum|x_i|\right)^2 + \sum\frac{1}{2^i}|x_i|^2\right)^{1/2}$ is a norm equivalent in $l_1$

We know that $l_1$ is the space of sequences $x_n$ such that $\sum |x_i|^1$ converges. I think therefore that the norm being considered to be equivalent to $f(x)$ is the $\sum|x_i|^1$ norm.

We must show that there exists $a>0, b>0$ such that

$$ a\sum|x_i|\le \left(\left(\sum|x_i|\right)^2 + \sum\frac{1}{2^i}|x_i|^2\right)^{1/2} \le b\sum|x_i|$$

for all $x$.

Any hints?

Answer 1

The left hand inequality, involving $a$, is obvious.

The right hand inequality, involving $b$, has the added trickiness introduced by the term $Q=\sum 2^{-i} |x_i|^2 $. You can bound $Q$ by the square of $\sup|x_i|$ which in turn is bounded by the square of $L=\sum |x_i|$. So you are left with the task of finding a $b$ such that $$ \left( L^2 + Q\right)^{1/2} \le \left(L^2 + L^2\right)^{1/2}\le b L.$$